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Case Study Questions Class 10 Science Chapter 12 Electricity

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Case study Questions Class 10 Science Chapter 12 are very important to solve for your exam. Class 10 Science Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Science Chapter 12 Electricity

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In CBSE Class 10 Science Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electricity Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser, etc. all are based on the heating effect of current.

(i) What are the properties of heating elements? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point.

Answer: (b) Low resistance, high melting point

(ii) What are the properties of an electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(a) doubled	(b) halved
(c) four times	(d) one fourth times

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(a) 4 times	(b) 2 times
(c) 6 times	(d) 8 times

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

(a) 250 J	(b) 5000J
(c) 750J	(d) 1000J

Answer: (c) 750J

Question 2:

The relationship between potential difference and the current was first established by George Simon Ohm. This relationship is known as Ohm’s law. According to this law, the current passed through a conductor is proportional to the potential difference applied between its ends provided the temperature remains constant i.e. I ∝ V or V = IR where R is the constant for the conductor and it is known as the resistance of the conductor. Although Ohm’s law has been found valid over a large class of materials, there are some materials that do not hold Ohm’s law.

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

Case Study 3

3.1) The current passing through an electric kettle has been doubled. The heat produced will become : (a) half (b) double (c) four times (d) one fourth

Answer(c) four times

3.2) The heat produced in a wire of resistance ‘a’ when a current ‘b’ flows through it in time ‘c’ is given by : (a) a 2 bc (b) abc 2 (c) ab 2 c (d) abc

Answer(c) ab2c

3.3) What are the properties of heating element ? (a) high resistance, high melting point (b) low resistance, high melting point (c) low resistance, high melting point (d) low resistance, low melting point

Answer (a) high resistance, high melting point

3.4) Calculate the heat produced when 96,000 coulombs of charge is transferred in one hour through a potential difference of 50 volts. (a) 4788 J (b) 4788 kJ (c) 478 kJ (d) 478 J

Answer (b) 4788 kJ

3.5) Which of the following characteristic is not suitable for a fuse wire ? (a) thin and short (b) low melting point (c) thick and short (d) high resistance

Answer (c) thick and short

Case Study 4

Substance through which charges cannot pass is called insulators. Glass, pure water, and all gases are insulators. Insulators are also called dielectrics. In insulators, the electrons are strongly bound to their atoms and cannot get themselves freed. Thus, free electrons are absent in insulators. Insulators can easily be charged by friction. This is due to the reason that when an electric charge is given to an insulator, it is unable to move freely and remains localized. But this does not mean that conductors cannot be charged. A metal rod can be charged by rubbing it with silk if it is held in a handle of glass or amber

4.1) Calculate the current in a wire if a 1500 C charge is passed through it in 5 minutes. (a) 2 A (b) 5 A (c) 3 A (d) 4 A

Answer (b) 5 A

4.2) Electrons and conventional current flows in : (a) The same direction (b) The opposite direction (c) Any direction (d) Can’t say

Answer (b) The opposite direction

4.3) If the current passing through a lamp is 5 A, what charge passes in 10 second ? (a) 0.5 C (b) 3 C (c) 5 C (d) 50 C

Answer (d) 50 C

4.4) One-coulomb charge is equivalent to the charge contained in : (a) 6.2 × 10 19 electrons (b) 2.6 × 10 18 electrons (c) 2.65 × 10 19 electrons (d) 6.25 × 10 18 electrons

Answer (d) 6.25 × 1018 electrons

4.5) When an electric lamp is connected to 12 V battery, it draws a current of 0.5 A. The power of the lamp is : (a) 0.5 W (b) 6 W (c) 12 W (d) 24 W

Answer (b) 6 W

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CBSE 10th Standard Science Subject Electricity Chapter Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams

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Cbse 10th standard science subject electricity case study questions 2021.

10th Standard CBSE

Final Semester - June 2015

case study based questions class 10 chapter electricity

= 2I = 3I	= 4I = 3I
= I = 3I	= 2I = I

Several resistors may be combined to form a network. The combination should have two end points to connect it with a battery or other circuit elements. When the resistances are connected in series, the current in each resistance is same but the potential difference is different in each resistor. When the resistances are connected in parallel, the voltage drop across each resistance is same but the current is different in each resistor. (i) The household circuits are connected in

(v) Two resistances 10 \(\begin{equation} \Omega \end{equation}\) and 3 \(\begin{equation} \Omega \end{equation}\) are connected in parallel across a battery. If there is a current of 0.2 A in 10 .Q resistor, the voltage supplied by battery is

The heating effect of current is obtained by transformation of electrical energy in heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser etc. all are based on the heating effect of current. (i) What are the properties of heating element? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point. (ii) What are the properties of electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point (iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 \(\begin{equation} \Omega \end{equation}\) , the amount of heat produced is

The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule. Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved. So for commercial purposes we use a bigger unit of electrical energy which is called kilowatt hour. 1 kilowatt-hour is equal to 3.6 x 106 joules of electrical energy. (i) The energy dissipated by the heater is E. When the time of operating the heater is doubled, the energy dissipated is

(ii) The power of a lamp is 60 W The energy consumed in 1 minute is

(iii) The electrical refrigerator rated 400 W operates 8 hours a day. The cost of electrical energy is \(\begin{equation} ₹ \end{equation}\) 5 per kWh. Find the cost of running the refrigerator for one day?

32	16
8	4

(iv) Calculate the energy transformed by a 5 A current flowing through a resistor of 2 \(\begin{equation} \Omega \end{equation}\) for 30 minutes?

(v) Which of the following is correct? (a) 1 watt hour = 3600 J (b) lkWh = 36x10 6 J (c) Energy (in kWh) = power (in W) x time (in hr) (d) \(\begin{equation} \text { Energy (in kWh) }=\frac{V(\text { volt }) \times I(\text { ampere }) \times t(\text { sec })}{1000} \end{equation}\)

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Cbse 10th standard science subject electricity case study questions 2021 answer keys.

(i) (c) : The equivalent resistance in the parallel combination is lesser than the least value, of the individual resistance. (ii) (b ): Resistance of each piece \(\begin{equation} =\frac{12}{3}=4 \Omega \end{equation}\) \(\begin{equation} \frac{1}{R_{P}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \Rightarrow R_{p}=\frac{4}{3} \Omega \end{equation}\) (iii) (a): All the three resistors are in parallel. \(\begin{equation} \frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{3}+\frac{1}{1}=\frac{1+2+6}{6}=\frac{9}{6} \end{equation}\) \(\begin{equation} R_{P}=\frac{6}{9}=\frac{2}{3} \Omega \end{equation}\) (iv) (a): Voltage is same across each resistance. So, I 1 x 5 = I 2 X 10 = 15 x I 3 I 1 = 2I 2 = 3I 3 (v) (d): All are in parallel. \(\begin{equation} \begin{array}{l} \frac{1}{R_{p}}=\frac{1}{12} \times 4=\frac{1}{3} \Rightarrow R_{p}=3 \Omega \\ I=\frac{3}{3}=1 \mathrm{~A} \end{array} \end{equation}\) So,current in each resistor \(\begin{equation} I^{\prime}=\frac{3}{12}=\frac{1}{4} \mathrm{~A} \end{equation}\)

(i) (b) (ii) (c): In series combination, resistance is maximum and in parallel combination, resistance is mcm. (iii) (c) : R 1 = r 1 + r 2 \(\begin{equation} \begin{array}{l} R_{2}=\frac{r_{1} r_{2}}{r_{1}+r_{2}} \\ \frac{R_{1}}{R_{2}}=\frac{\left(r_{1}+r_{2}\right)^{2}}{r_{1} r_{2}} \end{array} \end{equation}\) (iv) (c): In the given circuit, 3 \(\begin{equation} \Omega \end{equation}\) resistors are in series. R S = 3 + 3 = 6 \(\begin{equation} \Omega \end{equation}\) Now, R S and 6 \(\begin{equation} \Omega \end{equation}\) are parallel. \(\begin{equation} \frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3} \Rightarrow R_{p}=3 \Omega \end{equation}\) (v) (a): V = 0.2 x 10= 2 V So, total voltage supplied is same as 2 V.

(i) (b) (ii) (c) (iii) (a): Given: H = I 2 Rt \(\begin{equation} \mathrm{So}, H^{\prime}=(2 I)^{2} \cdot \frac{R}{2} t=2 H \end{equation}\) (iv) (b): Given: 1= 5 A, resistance = R. Let r be the new radius. Now,H=I 2 Rt Also H' = I' 2 R' t From (i) and (ii), \(\begin{equation} 5^{2} \times \rho \frac{L}{\pi r^{2}} t=10^{2} \times \rho \frac{L}{\pi r^{\prime 2}} \cdot t \end{equation}\) \(\begin{equation} \frac{25}{r^{2}}=\frac{100}{r^{\prime 2}} \Rightarrow \frac{r^{\prime}}{r}=2 \Rightarrow r^{\prime}=2 r \end{equation}\) (v) (c): Given: \(\begin{equation} I=0.5 \mathrm{~A}, R=10 \Omega, t=5 \mathrm{~min} \end{equation}\) \(\begin{equation} \begin{array}{l} H=I^{2} R t=0.5 \times 0.5 \times 10 \times 5 \times 60 \\ H=750 \mathrm{~J} \end{array} \end{equation}\)

(i) (a) : \(\begin{equation} E \propto t \end{equation}\) (ii) (c) : Given: P = 60W, t = 1 min E = 60 x 1 x 60 = 3600J (iii) (b) : Given: P = 400 \(\begin{equation} \Omega \end{equation}\) , t = 8 hour E = 400 x 8 = 3200Wh = 3.2kWh Cost = 3.2 x 5 = \(\begin{equation} ₹ \end{equation}\) 16 (iv) (a) : Given: I= 5 A, R = 2 \(\begin{equation} \Omega \end{equation}\) , t = 30 min E = I 2 Rt = 5 x 5 x 2 x 30 x 60 E = 90000J = 90 kJ (v) (a): 1watt hr = 3600J

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Class 10 Science: Case Study Chapter 12 Electricity PDF Download

Here we are providing you with Class 10 Science Chapter 12 Electricity Case Study Questions, by practicing these Case Study and Passage Based Questions will help you in your Class 10th Board Exam.

Case Study Chapter 12 Electricity

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

Answer: (b) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(a) doubled	(b) halved
(c) four times	(d) one fourth times

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(a) 4 times	(b) 2 times
(c) 6 times	(d) 8 times

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

(a) 250 J	(b) 5000J
(c) 750J	(d) 1000J

Answer: (c) 750J

Question 2:

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

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Case Study Chapter 12 Electricity

Please refer to Chapter 12 Electricity Case Study Questions with answers provided below. We have provided Case Study Questions for Class 10 Science for all chapters as per CBSE, NCERT and KVS examination guidelines. These case based questions are expected to come in your exams this year. Please practise these case study based Class 10 Science Questions and answers to get more marks in examinations.

Case Study Questions Chapter 12 Electricity

Case/Passage – 1

Two tungston lamps with resistances R1 and R2 respectively at full incandescence are connected first in parallel and then in series, in a lighting circuit of negaligible internal resistance. It is given that: R 1 > R 2 .

Question: Which lamp will glow more brightly when they are connected in parallel? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs

Question: Which lamp will glow more brightly when they are connected in series? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs

Question: If the lamp of resistance R 2 now burns out and the lamp of resistance R1 alone is plugged in, will the illumination increase or decrease? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) None

Question: If the lamp of resistance R 1 now burns out, how will the illumination produced change? (a) Net illumination will increase (b) Net illumination will decrease (c) Net illumination will remain same (d) Net illumination will reduced to zero

Question: Would physically bending a supply wire cause any change in the illumination? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) It is not possible to predict from the given datas

Case/Passage – 2

The rate at which electric energy is dissipated or consumed in an electric circuit. This is termed as electric power, P = IV, According to Ohm’s law V = IR We can express the power dissipated in the alternative forms P =I 2 R=V 2 /R

If 100W – 220V is written on the bulb then it means that the bulb will consume 100 joule in one second if used at the potential difference of 220 volts. The value of electricity consumed in houses is decided on the basis of the total electric energy used. Electric power tells us about the electric energy used per second not the total electric energy. The total energy used in a circuit = power of the electric circuit × time.

Question: Which of the following terms does not represent electrical power in a circuit? (a) I 2 R (b) IR 2 (c) VI (d) V 2 /R

Question: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in sereis and then in parallel in an electric circuit. The ratio of heat produced in series and in parallel combinations would be– (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

Question: In an electrical circuit, two resistors of 2Ω and 4Ω respectively are connected in series to a 6V battery. The heat dissipated by the 4Ω resistor in 5s will be (a) 5 J (b) 10 J (c) 20 J (d) 30 J

Question: In an electrical circuit three incandescent bulbs. A, B and C of rating 40 W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness? (a) Brightness of all the bulbs will be the same (b) Brightness of bulb A will be the maximum (c) Brightness of bulb B will be more than that of A (d) Brightness of bulb C will be less than that of B

Question: An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be– (a) 100 W (b) 75 W (c) 50 W (d) 25 W

Case/Passage – 3

Answer the following questions based on the given circuit.

Question: The equivalent resistance between points A and B is (a) 7Ω (b) 6Ω (c) 13Ω (d) 5Ω

Question: The potential drop across the 3Ω resistor is (a) 1 V (b) 1.5 V (c) 2 V (d) 3 V

Question: The current flowing through in the given circuit is (a) 0.5 A (b) 1.5 A (c) 6 A (d) 3 A

Case/Passage – 4

Answer the following questions based on the given circuit.

Question: The current through each resistor is (a) 1 A (b) 2.3 A (c) 0.5 A (d) 0.75 A

Question: The equivalent resistance between points A and B, is (a) 12 Ω (b) 36 Ω (c) 32 Ω (d) 24 Ω

Question: The potential drop across the 12Ω resistor is (a) 12 V (b) 6 V (c) 8 V (d) 0.5 V

Case/Passage – 5

Question: The equivalent resistance between points A and B (a) 6.2 Ω (b) 5.1 Ω (c) 13.33 Ω (d) 1.33 Ω

Question: The current through the 4.0 ohm resistor is (a) 5.6 A (b) 0.98 A (c) 0.35 A (d) 0.68 A

Question: The current through the battery is (a) 2.33 A (b) 3.12 A (c) 4.16 A (d) 5.19 A

Case/Passage – 6

Question: The total resistance of the circuit is (a) 2 Ω (b) 4 Ω (c) 1.5 Ω (d) 0.5 Ω

Question: The current flowing through 6Ω resistor is (a) 0.50 A (b) 0.75 A (c) 0.80 A (d) 0.25

Question: The current flowing through 0.5Ω resistor is (a) 1 A (b) 1.5 A (c) 3 A (d) 2.5 A

Case/Passage – 7

Ohm’s law gives the relationship between current flowing through a conductor with potential difference across it provided the physical conditions and temperature remains constant. The electric current flowing in a circuit can be measured by an ammeter. Potential difference is measured by voltmeter connected in parallel to the battery or cell. Resistances can reduce current in the circuit. A variable resistor or rheostat is used to vary the current in the circuit.

Question. Which type of conductor is represented by the graph given alongside?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What is the slope of graph in (i) equal to? (a) V (b) I (c) R (d) VI

Question. Which of the following is the factor on which resistance of a conductor does not depend? (a) Length (b) Area (c) Temperature (d) Pressur

Question. What type of conductor is represented by the following graph?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What type of conductors are represented by the following graph?

Study this table related to material and their resistivity and answer the questions that follow.

Question. Which of the following is used in transmission wires? (a) Cr (b) Al (c) Zn (d) Fe

Question. Which is the best conducting metal? (a) Cu (b) Ag (c) Au (d) Hg

Question. Which of the following is used as a filament in electric bulbs? (a) Nichrome (b) Tungsten (c) Manganese (d) Silver

Question. What is the range of resistivity in metals, good conductors of electricity? (a) 10–8 to 10–6 Wm (b) 10–6 to 10–4 Wm (c) 1010 to 1014 Wm (d) 1012 to 1014 Wm

Question. Which property of the alloy makes it useful in heating devices like electric iron, toasters, immersion rods, etc.? (a) Higher resistivity (b) Do not oxidise at low temperature (c) Do not reduce at high temperature (d) Oxidise at high temperature

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Class 10 Science Chapter 11 Case Based Questions - Electricity

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Case Study - 1

When electric current flows through the circuit this electrical energy is used in two ways, some part is used for doing work and remaining may be expended in the form of heat. We can see, in mixers after using it for long time it become more hot, fans also become hot after continuous use. This type of effect of electric current is called as heating effect of electric current. If I is the current flowing through the circuit then the amount of heat dissipated in that resistor will be H = VIt This effect was discovered by Joule, hence it is called as Joule’s law of heating. Also, we can write, H = I 2 Rt Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point. In case of electric circuit, this heating effect is used to protect the electric circuit from damage. The rate of doing work or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI The SI unit of electric power is watt.

Q1: What is the SI unit of electric energy? Ans: The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h. Q2: How heating effect works to protect electric circuit? Ans: In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

Q3: 1KW h = ? Ans: 1kW h = 3.6*10 6 joule Q4: If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb? Ans: Given that, V = 200V, I = 1A Then, P = VI = 200*1 = 200 J/s = 200 W

Case Study - 2

Resistance is the opposition offered by the conductor to the flow of electric current. When two or more resistors are connected in series then electric current through each resistor is same but the electric potential across each resistor will be different. If R1, R2 and R3 are the resistance connected in series then current through each resistor will be I but potential difference across each resistor is V1, V2 and V3 respectively. Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3 Again, IR = IR1 + IR2 + IR3 Thus, R = R1 + R2 + R3 Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit. Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively. Thus, total current through the circuit is the sum of current flowing through each resistor. I = I1 + I2 + I3 Again, V/R= V/R1 + V/R2 + V/R3 Thus, 1/R = 1/R1 + 1/R2 + 1/R3 Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

Q1: In which case the equivalent resistance is more and why? Ans: In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel. Since, 1/R = 1/R1 + 1/R2 + 1/R3 +…. Q2: In our home, which type of combination of electric devices is preferred? Why? Ans: At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices. Q3: If n resistors of resistance R are connected in parallel then what is the equivalent resistance? Ans: If n resistors of resistance R are connected in parallel then equivalent resistance is given by, 1/Re = 1/R + 1/R + 1/R +….n times 1/R Thus, 1/Re = n/R Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study - 3

We can see that, as the applied voltage is increased the current through the wire also increases. It means that, the potential difference across the terminals of the wire is directly proportional to the electric current passing through it at a given temperature. Thus, V= IR Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity. The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

Q1: What is SI unit of resistivity? Ans: The SI unit of resistivity is ohm meter. Q2: What is variable resistance? Ans: The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance. Q3: Why tungsten is used in electric bulbs? Ans: Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high. Q4: 1M ohm = ? Ans: 1M ohm = 10 6 ohm

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Home > Class 10 Science Subject-wise Materials

Class 10 Science Chapter 12 Electricity

Class 10 Chapter 12 Electricity as the name suggests, covers everything about electricity in detail. The constitution of electricity, the flow of electricity in the circuit, how electricity can be regulated, and much more. The chapter also includes Ohm’s law, resistors, and heating effects of electric circuits. The questions constitute 7 marks in the CBSE Class 10 exams. The inclusion of CBSE Electricity Chapter 12 is to help students create a strong foundation especially when students want to pursue the field of science and technology.

The understanding of concepts and topics included in the NCERT Chapter 12 can be done with the help of study materials like notes of electricity class 10 CBSE, question bank, mind maps, and support materials. Preparing the right study material can help in scoring good marks in the final examination.

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CBSE Class 10 Electricity Notes

Below we have provided the links to downloadable PDFs of class 10 ch 12 science notes and get an in-depth explanation and understanding of the chapter.

<red> ➜ <red> Class 10 Electricity Notes

📈 Trending: Class 10 CBSE Syllabus 2024-25

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CBSE Class 10 Electricity DoE Worksheet

Below, we have provided the links to downloadable PDFs of DoE Worksheets for Electricity Class 10 to practice more questions.

<red> ➜ <red> Worksheet 16

<red> ➜ <red> worksheet 17, <red> ➜ <red> worksheet 18, <red> ➜ <red> worksheet 19, <red> ➜ <red> worksheet 20, <red> ➜ <red> worksheet 21, cbse class 10 electricity experiential activities.

Below, we have provided the links to downloadable PDFs of Experiential Learning Activity for ch 12 class 10 Science to help students implement their acquired knowledge in the real world.

<red> ➜ <red> Electricity Experiential Activities

Cbse class 10 electricity important questions.

Below, we have provided Class 10 Science Important Questions that cover all the important questions in Electricity.

<red> ➜ <red> Electricity Important Questions (View)

Cbse class 10 electricity mind maps.

Below, we have provided Class 10 Science Mind maps that include mind maps of the related concepts in Electricity.

<red> ➜ <red> Electricity Mind Maps

Cbse class 10 electricity question bank.

Below, we have provided Class 10 Science Question Banks that cover every typology question with detailed explanations from various resources in one place

<red> ➜ <red> CBSE Question Bank PDF

<red> ➜ <red> kendriya vidyalaya question bank, cbse class 10 electricity support material.

Below, we have provided Class 10 Science Support Materials that cover Case Study-based questions from the various concepts explained in Science NCERT chapters.

<red> ➜ <red> Electricity Support Material

Why download these chapter-wise pdfs.

Science Class 10 Electricity chapter can include both objective and subjective questions related to Ohm’s law, SI unit of current, and magnetic effects of currents. The study materials are exam-centric and with the help of visual study materials like mind maps can help in connecting the knowledge they have acquired. With the right preparation, working strategically can help students build a strong base and score at least 7 marks in the final exams. The chapter-wise study materials are effective for both teachers and students.

Creating a study timetable and including these chapter-wise PDFs can help students prepare the chapters in a strategic and organized manner. Just allot sufficient time for understanding and revising the concepts.
Students wouldn’t have to juggle between numerous sites to find various study material that suits their learning style. The chapter-wise study material can be accessed in one place.
Download and browse these chapter-wise PDFs on any device in the comfort of your place by using an Internet connection.
After downloading these PDFs, students can get these educational materials printed and prepare accordingly.

How Can This Chapter-wise Material Help Students?

The Science Electricity chapter-wise materials can help in completing the chapter from the 10th NCERT textbook in addition to the extra study materials. Students may efficiently prepare for the chapter by downloading chapter notes, DoE worksheets, question banks, key questions, and a plethora of additional study resources.

Students could structure and focus on a certain less strong area of the subject at hand by using science class 10 electricity notes.
When it comes to making quick and efficient preparations or refining the flow of concepts under a certain topic that you might have overlooked, mind maps are a valuable tool.
The DoE worksheets and question banks may be used to study for every category of question that will be analyzed in the tenth board examinations. Once they have mastered the material, students can make a timetable and practice answering pertinent questions.
Among the Class 10 CBSE, important questions are the recurring questions and the notion of questions you should practice for the test. You may improve your chances of receiving better exam marks by rehearsing important questions.

Educators can use the additional materials and practice questions that Educart has provided to help students practice these topics completely. To download these PDFs, the user only has to verify themselves and click the link.

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Question 1 - Case Based Questions (MCQ) - Chapter 12 Class 10 - Electricity

Last updated at April 16, 2024 by Teachoo

Observe the graph and answer any four questions from (a) to (e). V-I graph for a conductor is as shown in figure.

Observe the graph and answer any four questions from.jpg

Question 1 (a)

(a) what do you infer from this graph, (i) v ∝ 1/i, (ii) v ∝ i 2, (iii) v ∝ i, (iv) v ∝ 1/i 2.

The V-I graph is a straight line graph passing through the origin.

This implies that V is directly proportional to I.

So, the correct answer is (iii ).

Question 1 (b)

Name the physical quantity represented by the slope of this graph , (i) current, (ii) resistance, (iii) potential difference, (iv) none of the above.

The slope of the V-I graph gives the resistance.

So, the correct answer is (ii).

Question 1 (c)

Ohm is the si unit of:, (i) potential difference, (iii) current, (iv) resistivity.

The SI unit of resistance is ohm .

Question 1 (d)

Which of the following law justify the above graph: , (i) faradays law, (ii) ohm’s law, (iii) faradays law, (iv) joule’s law.

Ohm’s law , states that the current passing through a conductor is directly proportional to the potential difference across its ends, provided the physical conditions like temperature, density etc. remain unchanged.

The V-I graph given above also states that current is directly proportional to potential difference.

Question 1 (e)

Resistance of a conductor depends on: , (i) length of conductor, (ii) area of cross-section, (iii) temperature, (iv) all of the above.

Factors affecting resistance of a conductor are:

Length of the conductor
Area of cross section
Nature of Material
Temperature

Since all the factors are mentioned in the options.

So, the correct answer is (iv).

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CBSE Class 10

CBSE Class 10 Science Chapter Wise Important Case Study Questions

Chapter wise important case study questions cbse class 10 science: cbse class 10 science board exam 2024 is just around the corner and students are working hard to score maximum marks. check these case study questions from class 10 science to ace your examination this year also download the solutions from the pdf attached towards the end. .

CBSE Class 10 Science Chapter Wise Important Case Study Questions: While the CBSE Board exam for Class 10 students are ongoing, the CBSE Class 10 Science board exam 2024 is to be held on March 2, 2024. With the exams just a few days away, CBSE Class 10th Board exam candidates are rushing to prepare the remaining syllabus, practising their weak portions, trying to revise the important questions from the past year papers, practise questions, etc.

Why are CBSE Class 10 Science Case Study Questions Important?

Section A : 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
Section B : 6 Very Short Answer type questions carrying 2 marks each. Answers to these questions should be in the range of 30 to 50 words.
Section C : 7 Short Answer type questions carrying 3 marks each. Answers to these questions should be in the range of 50 to 80 words.
Section D : 3 Long Answer type questions carrying 5 marks each. Answers to these questions should be in the range of 80 to 120 words.
Section E : 3 Case Based/ Source Based units of assessment (4 marks each) with sub-parts.

How to solve case study questions in CBSE Class 10 Science?

Read the case given and the associated questions carefully.
Read the questions attentively and analyse what they are asking.
Apply your subject knowledge and theories in the given case to decide what the correct answers should be.

1.A chemical reaction is a representation of chemical change in terms of symbols and formulae of reactants and products. There are various types of chemical reactions like combination, decomposition, displacement, double displacement, oxidation and reduction reactions. Reactions in which heat is released along with the formation of products are called exothermic chemical reactions. All combustion reactions are exothermic reactions.

(i) The massive force that pushes the rocket forward through space is generated due to the

(a) combination reaction

(b) decomposition reaction

(d) double displacement reaction

(ii) A white salt on heating decomposes to give brown fumes and yellow residue is left behind. The yellow residue left is of

(a) lead nitrate

(b) nitrogen oxide

(d) oxygen gas

(iii) Which of the following reactions represents a combination reaction?

(a) CaO (s) + H2O (l) → Ca (OH)2 (aq)

(b) CaCO3 (s) → CaO (s) + CO2(g)

(d) 2FeSO4(s) → Fe2O3 (s) +SO2(g) + SO3(g)

(iv) Complete the following statements by choosing correct type of reaction for X and Y.

Statement 1: The heating of lead nitrate is an example of ‘X’ reaction.

Statement 2: The burning of magnesium is an example of ‘Y’ reaction.

(a)X-Combination,Y-Decomposition

(b)X-Decomposition,Y-Combination

(c)X-Combination,Y-Displacement

(d) X- Displacement, Y-Decomposition

2.The earlier concept of oxidation and reduction is based on the addition or removal of oxygen or hydrogen elements so, in terms of oxygen and hydrogen, oxidation is addition of oxygen to a substance and removal of hydrogen from a substance. On the other hand, reduction is addition of hydrogen to a substance and removal of oxygen from a substance. The substance which gives oxygen to another substance or removes hydrogen from another substance in an oxidation reaction is known as oxidising agent, while the substance which gives hydrogen to another substance or removes oxygen from another substance in a reduction reaction is known as reducing agent. For example,

(i) A redox reaction is one in which

(a) both the substances are reduced

(b) both the substances are oxidised

(d) one substance is oxidised while the other is reduced.

(ii) In the reaction, H2S+Cl2⟶S+2HCl

(a) H2S is the reducing agent.

(b) HCl is the oxidising agent.

(d) Cl2 is the reducing agent.

(iii) Which of the following processes does not involve either oxidation or reduction?

(a) Formation of slaked lime from quicklime.

(b) Heating mercuric oxide.

(d) Formation of zinc from zinc blende.

(iv) Mg+CuO⟶MgO+Cu

Which of the following is wrong relating to the above reaction?

(a) CuO gets reduced

(b) Mg gets oxidised.

(d) It is a redox reaction.

3.A copper vessel gets tarnished due to formation of an oxide layer on its surface. On rubbing lemon on the vessel, the surface is cleaned, and the vessel begins to shine again. This is due to the fact that which reacts with the acid present in lemon to form a salt which is washed away with water. As a result, the layer of copper oxide is removed from the surface of the vessel and the shining surface is exposed.

1.Which of the following acids is present in lemon?

(a) Formic acid

(b) Acetic acid

(d) Hydrochloric acid

2.The nature of copper oxide is

d) amphoteric

3.Name the salt formed in the above reaction

a) copper carbonate

b) copper chloride

c)copper citrate

d) copper citrate

4.The phenomenon of copper getting tarnished is

a) corrosion

b) rancidity

c) displacement

d)none of these

4.Metals as we know, are very useful in all fields, industries in particular. Non-metals are no less in any way. Oxygen present in air is essential for breathing as well as for combustion. Non-metals form a large number of compounds which are extremely useful, e.g., ammonia, nitric acid, sulphuric acid, etc. Non-metals are found to exist in three states of matter. Only solid non-metals are expected to be hard however, they have low density and are brittle. They usually have low melting and boiling points and are poor conductors of electricity.

i.____________ is a non-metal but is lustrous

A.Phosphorus

ii.Which of the following is known as 'King of chemicals'?

C. Sulphuric acid

D. Nitric acid

iii.Which of the following non-metals is a liquid?

iv.Hydrogen is used

A.for the synthesis of ammonia

B. for the synthesis of methyl alcohol

C.nitrogenous fertilizers

D. all of these

5.Nisha observed that the bottoms of cooking utensils were turning black in colour while the flame of her stove was yellow in colour. Her daughter suggested cleaning the air holes of the stove to get a clean, blue flame. She also told her mother that this would prevent the fuel from getting wasted.

a) Identify the reasons behind the sooty flame arising from the stove.

b) Can you distinguish between saturated and unsaturated compounds by burning them? Justify your answer.

c) Why do you think the colour of the flame turns blue once the air holes of the stove are cleaned?

6.Blood transport food, Oxygen and waste materials in our bodies. It consists of plasma as a fluid medium. A pumping organ [heart] is required to push the blood around the body. The blood flows through the chambers of the heart in a specific manner and direction. While flowing throughout the body, blood exerts a pressure against the wall or a vessel.

Pulmonary artery
Pulmonary vein
Very narrow and have high resistance
Much wide and have low resistance
Very narrow and have low resistance
Much wide and have high resistance
It is a hollow muscular organ
It is four chambered having three auricles and one ventricle.
It has different chambers to prevent O2 rich blood from mixing with the blood containing CO2
Both A & C
Blood = Plasma + RBC + WBC + Platelets
Plasma = Blood – RBC
Lymph = Plasma + RBC
Serum = Plasma + RBC + WBC

7.A brain is displayed at the Allen Institute for Brain Science. The human brain is a 3-pound (1.4-kilogram) mass of jelly-like fats and tissues—yet it's the most complex of all known living structures The human brain is more complex than any other known structure in the universe. Weighing in at three pounds, on average, this spongy mass of fat and protein is made up of two overarching types of cells—called glia and neurons— and it contains many billions of each. Neurons are notable for their branch-like projections called axons and dendrites, which gather and transmit electrochemical signals. Different types of glial cells provide physical protection to neurons and help keep them, and the brain, healthy. Together, this complex network of cells gives rise to every aspect of our shared humanity. We could not breathe, play, love, or remember without the brain.

1)Animals such as elephants, dolphins, and whales actually have larger brains, but humans have the most developed cerebrum. It's packed to capacity inside our skulls and is highly folded. Why our brain is highly folded?

b) Learning

3)Which among these protects our brain?

a)Neurotransmitter

b) Cerebrospinal fluid

d) Grey matter

4.Ram was studying in his room. Suddenly he smells something burning and sees smoke in the room. He rushes out of the room immediately. Was Ram’s action voluntary or involuntary? Why?

8.Preeti is very fond of gardening. She has different flowering plants in her garden. One day a few naughty children entered her garden and plucked many leaves of Bryophyllum plant and threw them here and there in the garden. After few days, Preeti observed that new Bryophyllum plants were coming out from the leaves which fell on the ground.

1.What does the incident sited in the paragraph indicate?

(a). Bryophyllum leaves have special buds that germinate to give rise to new plant.

(b). Bryophyllum can propagate vegetatively through leaves.

(c). Bryophyllum is a flowering plant that reproduces only asexually

(d). Both (a) and (b).

2.Which of the following plants can propagate vegetatively through leaves like Bryophyllum?

3.Do you think any other vegetative part of Bryophyllum can help in propagation? If yes, then which part?

4.Which of the following plant is artificially propagated (vegetatively) by stem cuttings in horticultural practices?

(b)Snakeplant

(d)Water hyacinth

9.The growing size of the human population is a cause of concern for all people. The rate of birth and death in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on. Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of the population.

1) What are common signs of sexual maturation in boys?

a) Broadening of shoulders

b) Development of mammary glands

c) Broadening of waist

d) High pitch of voice

2) Common sign of sexual maturation in girls is

a) Low pitch voice

b) Appearance of moustache and beard

c) Development of mammary glands

d) Broadening of shoulders

3) Which contraceptive method changes the hormonal balance of the body?

b) Diaphragms

c) Oral pills

d) Both a) and b)

4) What should be maintained for healthy society?

a) Rate of birth and death rate

b) Male and female sex ratio

c) Child sex ratio

d) None of these

10.Pea plants can have smooth seeds or wrinkled seeds. One of the phenotypes is completely dominant over the other. A farmer decides to pollinate one flower of a plant with smooth seeds using pollen from a plant with wrinkled seeds. The resulting pea pod has all smooth seeds.

i) Which of the following conclusions can be drawn?

(1) The allele for smooth seeds is dominated over that of wrinkled seeds.

(2) The plant with smooth seeds is heterozygous.

(3) The plant with wrinkled seeds is homozygous.

(b) 1 and 2 only

(d) 1, 2 and 3

ii) Which of the following crosses will give smooth and wrinkled seeds in same proportion?

(a) RR X rr

(b) Rr X rr

(d) rr X rr

iii) Which of the following cross can be used to determine the genotype of a plant with dominant phenotype?

(a) RR X RR

(b) Rr X Rr

(d) RR X rr

iv) On crossing of two heterozygous smooth seeded plants (Rr), a total of 1000 plants were obtained in F1 generation. What will be the respective number of smooth and wrinkled seeds obtained in F1 generation?

(a) 750, 250

(b) 500, 500

(d) 950, 50

11.Food chains are very important for the survival of most species.When only one element is removed from the food chain it can result in extinction of a species in some cases.The foundation of the food chain consists of primary producers.Primary producers or autotrophs,can use either solar energy or chemical energy to create complex organic compounds,whereas species at higher trophic levels cannot and so must consume producers or other life that itself consumes producers. Because the sun’s light is necessary for photosynthesis,most life could not exist if the sun disappeared.Even so,it has recently been discovered that there are some forms of life,chemotrophs,that appear to gain all their metabolic energy from chemosynthesis driven by hydrothermal vents,thus showing that some life may not require solar energy to thrive.

1.If 10,000 J solar energy falls on green plants in a terrestrial ecosystem,what percentage of solar energy will be converted into food energy?

(d)It will depend on the type of the terrestrial plant

2.Matter and energy are two fundamental inputs of an ecosystem. Movement of

(a)Energy is by directional and matter is repeatedly circulating

(b)Energy is repeatedly circulating and matter is unidirectional

(c)Energy is unidirectional and matter is repeatedly circulating

(d)Energy is multidirectional and matter is bidirectional

3.Raj is eating curd/yoghurt. For this food intake in a food chain he should be considered as occupying

(a)First trophic level

(b)Second trophic level

(c)Third trophic level

(d)Fourth trophic level

4.Which of the following, limits the number of trophic levels in a food chain

(a)Decrease in energy at higher trophic levels

(b)Less availability of food

(c)Polluted air

5.The decomposers are not included in the food chain. The correct reason for the same is because decomposers

(a) Act at every trophic level at the food chain

(b) Do not breakdown organic compounds

(d) Release enzymes outside their body to convert organic material to inorganic forms

12.Shyam participated in a group discussion in his inter school competition on the practical application of light and was very happy to win an award for his school. That very evening his father gave treat to celebrate Shyam’s win. Shyam while sitting saw an image of a person sitting at his backside in his curved plate and could see that person’s mobile drop in the flower bed. Person was not aware until Shyam went and informed him. He thanked Shyam for his clever move.

a)From which side of his plate Shyam observed the incident –

i)outward curved

ii)inward curved

iii)plane surface

b)Part of plate from which Shyam observed the incident acted like a-

i)concave mirror

ii)convex mirror

iii)plane mirror

c)The nature of the size of the image formed in above situation is –

i)real, inverted and magnified

ii)same size , laterally inverted

iii)virtual, erect and diminished

iv)real , inverted and diminished

d)Magnification of the image formed by convex mirror is –

more than 1

iii)equal to 1

iv)less than 1

The location of image formed by a convex lens when the object is placed at infinity is

(a) at focus

When the object is placed at the focus of concave lens, the image formed is

(a)real and smaller

(b) virtual and smaller

The size of image formed by a convex lens when the object is placed at the focus ofconvex lens is

(a) highly magnified

(b) point in size

When the object is placed at 2F in front of convex lens, the location of image is

(b) between F and optical center

(d) none of the above

14.One of the wires in domestic circuits supply, usually with a red insulation cover, is called live wire. with black insulation is called neutral wire. The earth wire, which has insulation of green colour, is usually connected to a metal plate deep in the earth near the house appliances that has a metallic body. Overloading contact, in such a situation the current in the circuit abruptly increases. circuit prevents damage to the appliances and the circuit due to overloading.

1 When do we say that an electrical appliance

2 Mention the function of earth wire in electrical line

3 How is an electric fuse connected in a domestic circuit?

4 When overloading and short circuiting are said to occur?

5 What is a live wire?

15.Light of all the colours travel at the same speed in vacuum for all wavelengths. But in any transparent medium(glass or water), the light of different colours travels at different speeds for different wavelengths, which means that the refractive index of a particular medium is different for different wavelengths. As there is a difference in their speeds, the light of different colours bend through different angles. The speed of violet colour is maximum and the speed of red colour is minimum in glass so, the red light deviates least and violet colour deviates most. Hence, higher the wavelength of a colour of light, smaller the refractive index and less is the bending of light.

(i)Which of the following statements is correct regarding the propagation of Light of different colours of white light in air?

(a) Red light moves fastest.

(b) Blue light moves faster than green light.

(d) Yellow light moves with the mean speed as that of the red and the violet light.

(ii)Which of the following is the correct order of wavelength?

(a) Red> Green> Yellow

(b) Red> Violet> Green

(d) Red> Yellow> Orange

(iii)Which of the following is the correct order of speed of light in glass?

(a) Red> Green> Blue

(b) Blue> Green> Red

(d) Green> Red> Blue

(iv)Which colour has maximum frequency?

16.The region around a magnet where magnetism acts is represented by the magnetic field.The force of magnetism is due to moving charge or some magnetic material. Like stationary charges produce an electric field proportional to the magnitude of charge, moving charges produce magnetic fields proportional to the current. In other words, a current carrying conductor produces a magnetic field around it. The subatomic particles in the conductor, like the electrons moving in atomic orbitals, are responsible for the production of magnetic fields. The magnetic field lines around a straight conductor (straight wire) carrying current are concentric circles whose centres lie on the wire.

1)The magnetic field associated with a current carrying straight conductor is in anti- clockwise direction. If the conductor was held horizontally along east west direction,what is the direction of current through it?

2)Name and state the rule applied to determine the direction of magnetic field in a straight current carrying conductor.

3)Ramus performs an experiment to study the magnetic effect of current around a current carrying straight conductor with the help of a magnetic compass. He reports that

a)The degree of deflection of magnetic compass increases when the compass is moved away from the conductor.

b)The degree of deflection of the magnetic compass increases when the current through the conductor is increased.

Which of the above observations of the student appears to be wrong and why?

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NCERT Exemplar
Science Exemplar Class 10
Electricity

NCERT Exemplar Class 10 Science Solutions for Chapter 12 - Electricity

Ncert exemplar solutions class 10 science chapter 12 – free pdf download.

NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity are the study materials necessary for you to understand the questions that can be asked from the Class 10 Science Electricity chapter. It is crucial for students to get acquainted with this chapter in order to score excellent marks in their CBSE Class 10 examination. This solution provides answers to the questions provided in NCERT Class 10 Exemplar book. This page has answers to 18 MCQs, 11 short answer questions and 7 long answer questions.

To help students grasp all the concepts clearly and in-depth, we are offering free NCERT Exemplar for Class 10 Science Chapter 12 here. These exemplars will enable students to learn the correct answers to all the questions given at the end of the chapter. These NCERT Exemplars are prepared by experts and can be used by students as an effective learning tool to improve their conceptual understanding.

Take a closer look at Class 10 Science Chapter 12 NCERT Exemplar below.

Download the PDF of NCERT Exemplar for Class 10 Science Chapter 12 – Electricity

NCERT Exemplar solutions class 10 science Chapter 12 part 01

Access Answers to NCERT Exemplar Class 10 Science Chapter 12 – Electricity

Multiple choice questions.

1. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure12.1. The current recorded in the ammeter will be

ncert solutions class 10 science chapter 12 electricity fig 1

(a) maximum in (i)

(b) maximum in (ii)

(d) the same in all the cases

The answer is (d) the same in all the cases

Explanation:

There are no changes in any of the circuits, hence current will be same in all the circuits.

2. In the following circuits (Figure 12.2), the heat produced in the resistor or combination of resistors connected to a 12 V battery will be

ncert solutions class 10 science chapter 12 electricity fig 2

(a) same in all the cases

(b) minimum in case (i)

(d) maximum in case(iii)

The answer is (c) maximum in case(ii)

Explanation

Here two transistors are in series. In figure (iii) total resistance will be less than individual resistances as they are connected parallel. Higher resistance produces more heat hence option c) is the right answer.

3. Electrical resistivity of a given metallic wire depends upon

(a) its length

(b) its thickness

(d) nature of the material

The answer is (d) nature of the material

4. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly

Answer is (a) 10 20

n = 16 /1.6 x 10 -19

n = 10 x 10 19

n = 10 20 electrons

The number of electrons flowing is 10 20 electrons

5. Identify the circuit (Figure 12.3) in which the electrical components have been properly connected.

ncert solutions class 10 science chapter 12 electricity fig 3

The answer is (b) (ii)

6. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

The answer is (d) 1 Ω

Maximum resistance is obtained when resistors are connected in series.

7. What is the minimum resistance which can be made using five resistors each of 1/5 Ω?

Answer is (b) 1/25 Ω

Minimum resistance is obtained when resistors are connected parallel

1/R = 5 + 5 + 5 +5 +5= 25 Ω

8. The proper representation of the series combination of cells (Figure 12.4) obtaining maximum potential is

ncert solutions class 10 science chapter 12 electricity fig 5

The answer is (a) (i)

Here positive terminal of the next cell is adjacent to the negative terminal of the previous cell.

9. Which of the following represents voltage?

(a) \(\begin{array}{l}\frac{Work done}{Current\times Time}\end{array} \)

(b) Work done × Charge

(d) Work done × Charge × Time

10. A cylindrical conductor of length l and uniform area of crosssection A has resistance R. Another conductor of length 2l and resistance R of the same material has an area of cross-section

Answer is (c) 2A

When Length doubles

11. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively (Figure.12.5). Which of the following is true?

ncert solutions class 10 science chapter 12 electricity fig 6

(a) R1 = R2 = R3

(b) R1 > R2 > R3

(d) R2 > R3 > R1

The answer is (c) R3 > R2 > R1

Current flow is inversely proportional to resistance. Highest resistance will show less flow of current hence answer is c).

12. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

Answer is (c) 300 %

Heat generated by a resistor is directly proportional to the square of the current. Hence, when the current becomes double, dissipation of heat will multiply by 2 =4. This means there will be an increase of 300%.

13. The resistivity does not change if

(a) the material is changed

(b) the temperature is changed

(d) both material and temperature are changed

Answer is (c) the shape of the resistor is changed

14. In an electrical circuit, three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

(a) The brightness of all the bulbs will be the same

(b) The brightness of bulb A will be the maximum

(d) The brightness of bulb C will be less than that of B

Answer is (c) Brightness of bulb B will be more than that of A

Bulbs are connected in parallel so the resistance of combination would be less than the arithmetic sum of the resistance of all the bulbs. So. there will be no negative effect on the flow of current. As a result, bulbs would glow according to their wattage.

15. In an electrical circuit, two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be

Answer is (c) 20 J

Equivalent resistance of the circuit is R = 4+2 = 6Ω

current, I= V/R = 6/6= 1A

the heat dissipated by 4-ohm resistor is, H = I 2 Rt = 20J

16. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

The answer is (d) 5 A

Or 1000 w = 220v x I

I = \(\begin{array}{l}\frac{1000w}{220v}\end{array} \) = 4.54 A

17. Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have

(a) same current flowing through them when connected in parallel

(b) same current flowing through them when connected in series

(d) different p

The answer is (b) same current flowing through them when connected in series

In series combination current does not get divided into branches because resistor receives a common current.

18. Unit of electric power may also be expressed as

(a) volt-ampere

(b) kilowatt-hour

(d) joule second

The answer is (a) volt-ampere

Volt-ampere (VA) is the unit used for the apparent power in an electrical circuit. A watt-second (also watt-second, symbol W s or W. s) is a derived unit of energy equivalent to the joule. The joule-second is the unit used for Planck’s constant.

Short Answer Questions

19. A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

ncert solutions class 10 science chapter 12 electricity fig 7

20. Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?

ncert solutions class 10 science chapter 12 electricity fig 9

Current P= I 2 R

18W = I 2 x 2Ω

I 2 = 18W/ 2Ω

Maximum value of current passing through A is 3A.

Current through B = Current through C = 1/2 x Current through A

Current through B = Current through C = 1/2 x 3

Current through B = Current through C = 1.5 A

21. Should the resistance of an ammeter be low or high? Give reason.

Resistance of ammeter should be zero because ammeter should not affect the flow of current.

22. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4Ω resistors? Give reason.

ncert solutions class 10 science chapter 12 electricity fig 10

Total resistance for parallel combination of 40 resistors can be calculated as follows:

Thus, resistance of parallel combination is equal to resistance of resistors in series. So, potential difference across 20 resistance will be same as potential difference across the other two resistors which are connected in parallel.

23. How does use of a fuse wire protect electrical appliances?

Fuse wire has great resistance than the main wiring. When there is significant increase in the electric current. Fuse wire melts to break the circuit. This prevents damage of electrical appliance.

24. What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Property of the conductor which resists the flow of electric current is called resistivity. Resistance for a particular material is unique. Resistance is directly proportional to length of conductor and inversely proportional to current flow.

When length is doubled resistance becomes double and current flow reduces to half. This is the reason for the decrease in ammeter reading.

25. What is the commercial unit of electrical energy? Represent it in terms of joules.

Commercial unit of electrical energy is kilowatt/hr

1 kw/hr = 1 kW h

= 1000 W × 60 × 60s

= 3.6 × 10 6 J

26. A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

1) Let R be the resistance of the electric lamp. In series total resistance = 5 + R

1 = 10/5+R

2) V across Lamp + conductor = 10 V

V acoess Lamp = I × R = 1 * 5 = 5 Volt

27. Why is parallel arrangement used in domestic wiring?

Parallel arrangement used in domestic wiring because it provides the same potential difference across each electrical appliance.

28. B1 , B2 and B3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

What happens to the glow of the other two bulbs when the bulb B1 gets fused?
What happens to the reading of A1 , A2 , A3 and A when the bulb B2 gets fused?
How much power is dissipated in the circuit when all the three bulbs glow together?

i) Potential difference does not get divided in parallel circuit. Hence glowing of other bulbs will not get affected when bulb one is fused.

ii) Ammeter A shows a reading of 3A. This means each of the Al. A2, and A3 show IA reading.

iii) R= V/I = 4.5V/3A= 1.5Ω

Now P= I 2 R

= (3A) 2 x 1.5 Ω

Long Answer Questions

29. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.

(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

(Resistance of the bulbs in series will be three times the resistance of single bulb. Hence, the current in the series combination will be one-third compared to current in each bulb in parallel combination. The parallel combination bulbs will glow more brightly.

The bulbs in series combination will stop glowing as the circuit is broken and current is zero. However the bulbs in parallel combination shall continue to glow with the same brightness.

30. State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Ohm’s law states that at constant temperature potential difference (voltage) across an ideal conductor is proportional to the current through it.

ncert solutions class 10 science chapter 12 electricity fig 11

Verification of Ohm’s law

Set up a circuit as shown in Fig. consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)

First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given

Next, connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.

Repeat the above steps using three cells and then four cells in the circuit separately.

ncert solutions class 10 science chapter 12 electricity fig 12

31. What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.

Resistivity is an inherent property of a conductor which resists the flow of electric current. Resistivity of each material is unique. SI unit of resistance is Ωm.

Experiment to study the factors on which the resistance of conducting wire depends.

Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.

Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig. 12.4.

Observation:

It is observed that resistance depend on material of conductor

Length of conductor determines resistance

Resistance depends on area of cross-section.

ncert solutions class 10 science chapter 12 electricity fig 13

Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.

Now repeat the above step with the 10 W bulb in the gap XY. Are the ammeter readings different for different components connected in the gap XY? What do the above observations indicate?

You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

32. How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Collect three resistors R1, R2, R3 in series to make the circuit.
Use ammeter to see the changes observed in the current flow.
Remove R1 and take the reading of potential difference of R2 and R3
Remove R2 and take the reading of potential difference of R1 and R3

ncert solutions class 10 science chapter 12 electricity fig 14

Ammeter reading was the same in each case so it can be inferred that the current remains the same in the circuit. To cross-check one can place ammeter and different places and observe the current flow.

33. How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Take three resistors R 1 , R 2 and R 3 , connect them in parallel to make a circuit; as shown in the figure.

Use voltmeter to take reading of potential difference of three resistors in parallel combination.

Now, remove the resistor R1 and take the reacting of the potential difference of remaining resistors combination.

Then, remove the resistor R 2 , and take the reading of potential difference of remaining resistor.

ncert solutions class 10 science chapter 12 electricity fig 15

In each case Voltmeter reading was the same which shows that the same potential difference exists across three resistors connected in a parallel arrangement.

34. What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

According to Joules heating effect heat produces in a resistor is

Directly proportional to square of current for the given resistor.
Directly proportional to resistance for a given current,
Directly proportional to the time of current flowing through the resistor.

This can be expressed as

H is heating effect, I is electric current, R is resistance and t is time.

Experiment to demonstrate Joules law of heating

Take a water heating immersion rod and connect to a socket which is connected to the regulator. It Is important to recall that a regulator controls the amount of current flowing through a device.
Keep the pointer of regulator on the minimum and count the time taken by immersion rod to heat a certain amount of water.
Increase the pointer of the regulator to the next level. Count the time taken by immersion rod to heat the same amount of water.
Repeat above step for higher levels on regulator to count the time.

It is seen that with an increased amount of electric current, less time is required o heat the same amount of water. This shows Joule’s Law of Heating.

Application:

Electric toaster, oven, electric kettle and electric heater etc. work on the basis of leafing effect of current.

35. Find out the following in the electric circuit given in Figure 12.9

(a) Effective resistance of two 8 Ω resistors in the combination

(b) Current flowing through 4 Ω resistor

(d) Power dissipated in 4 Ω resistor (e) Difference in ammeter readings, if any

ncert solutions class 10 science chapter 12 electricity fig 16

NCERT Exemplar Class 10 Chapter 12 Electricity

Sometimes we might have wondered about what constitutes electricity or how does it flow in an electric circuit, or what controls and regulates the current through an electric circuit? In Chapter 12 Electricity, students will find answers to these questions. They will also learn about other topics like the heating effect of electric current and its applications, the circuit diagram, Ohm’s law , resistors and conductors, electrical potential and potential difference.

Topics covered in Class 10 NCERT Exemplar Solutions for Science Chapter 12 Electricity

Introduction
The potential difference – Definition of volt and voltmeter
Ohm’s law – Ohm and resistance
Factors on which the resistance of the conductor depends – Resistivity
Resistors in series – Total/resultant/overall and voltage across each resistor
Resistors in parallel
The advantage of parallel combination over the series combination
Heating effect of an electric circuit – Joule’s law of the heating effect of electric current, electric fuse and electric power.

With BYJU’S, students can excel in their studies and can score better marks in the board examination. Class 10 is an important stage of a student’s life, as it consists of topics which are necessary to understand thoroughly for future entrance exams. To help you grasp the concepts clearly, BYJU’S brings you notes , sample papers , and animation videos. For a customised learning experience, visit BYJU’S website or download BYJU’S – The Learning App.

Frequently Asked Questions on NCERT Exemplar Solutions for Class 10 Science Chapter 12

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CBSE Class 10 Electricity Important Questions with Solutions

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Electricity Class 10 Questions and Answers

Chapter 12 of Class 10 Science is about ‘Electricity’. One of the most fundamental elements of our society is electricity. Since the beginning of the industrial revolution, electricity has contributed to the development of our civilization by powering numerous businesses and industries. Today, life would be in complete disarray if we were to lose this energy source. Therefore, it’s crucial for students to learn ideas about how electricity functions at the molecular level and explore its applications. Chapter 12 ‘Electricity’ will teach students about the fundamentals of electricity, the movement of current, and the operation of circuits as a whole.

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Throughout our question bank of Chapter 12 Class 10 Science Important Questions, students are given important problems to complete so they will be familiar with the questions which are expected in their final exams.

Students should not memorise the answer of science problems since it requires an in-depth understanding of the concepts involved. For each question in our Important Questions Class 10 Science Chapter 12, there is a step-by-step explanation prepared by our in-house Science subject experts. . This makes it easy for students to have complete faith and trust, to revise concepts covered in class.

The questions are compiled from the NCERT textbook, NCERT exemplar, and other reference books for providing the best and most authentic source of question bank. Furthermore, there are questions from previous year’s exams as well. In order to set the foundation for the chapter, it could be beneficial for students to refer to the Important Questions Class 10 Science Chapter 12.

There is a need for students to improve their understanding of the art of writing answers. They can practice representational diagrams by referring to them from NCERT textbook and from our NCERT chapter-wise solutions. Students can access our entire collection of Science Class 10 Chapter 12 Important Questions by registering on the Extramarks’ website. Besides, students can review other study materials on our Extramarks websites, such as NCERT solutions, revision notes, and the previous year’s question papers.

CBSE Class 10 Science Important Questions

CBSE Class 10 Science Important Questions are also available for the following chapters:



1
2
3
4
5	Periodic Classification of Elements
6
7
8
9
10
11
12	Electricity
13
14
15
16

Important Electricity Questions with Answers Class 10 Chapter 12

Given below is a set of questionnaires and their answers from our question bank of Class 10 Science Chapter 12 Important Questions .

For the best preparation, students are advised to go through this set of Important questions Class 10 Science Chapter 12. They can register on our website to get access to it.

Question 1: Which of the following does not represent electrical power in a circuit?

Answer : b) IR 2

Explanation:

Electrical power is represented by the expression P = VI . (Equation 1)

According to Ohm’s law,

Putting the value of V in ( Equation 1), we get

P = ( IR ) × I

Similarly, from Ohm’s law,

Putting the value of I in (Equation 1),

P = V × V / R = V 2 / R

It is thus clear that the equation IR 2 does not represent electrical power in a circuit. Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit.

Question 2: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be _____.

Answer : (d) 25 W

This expression demonstrates how much energy the electric bulb consumes.

P = VI = V 2 /R

The given formula can be used to calculate the light bulb’s resistance:

Putting the values, we get

R = (220)2/100 = 484 Ω

The resistance generally does not change when the voltage supply is decreased. Consequently, the amount of electricity used can be determined as follows:

P = (110) 2 V/484 Ω = 25 W

As a result, the electric bulb uses 25 W of power when it is operating at 110 V.

Question 3: What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

Answer: (d) 1 Ω

Explanation: Resistance is maximum when resistors are connected in series.

R= 1/ 5 + 1/ 5 + 1/ 5 + 1/ 5 + 1/ 5

Question 4: If the current ‘I’ through a resistor is increased by 100% (assuming that the temperature remains unchanged), the approximate increase in power dissipated will be

Answer: (c) 300 %

Explanation: The amount of heat produced by a resistor is inversely proportional to the square of the current. Therefore, the loss of heat will multiply by 2=4 when the current doubles. Accordingly, there will be a 300% increase.

Question 5: A piece of wire of resistance R is cut into five equal parts. These parts are then arranged in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is _____.

Answer: d) 25

The resistance is divided into five halves, each of which has a resistance of R/5.

Since we are aware that each component is linked to the others in parallel, we can compute the equivalent resistance as follows:

R’ = 5/ R + 5/ R + 5/ R + 5/ R + 5/ R

R’ = ( 5 + 5+ 5+ 5+ 5)/ R = 25/ R

R R’ = 25

The ratio of R/R′ is 25.

Question 6: The correct representation of the series combination of cells (Figure 12.4) obtaining maximum potential is

Answer: (a)

A cell’s positive terminal needs to be connected to the neighbouring cell’s negative terminal. The appropriate cell combination is represented by case I.

Question 7: Two pieces of conducting wire of the same material and of equal lengths and the equal diameters are first connected in series and then changed to parallel in a circuit across the same potential difference. The ratio of heat produced in both series and parallel combinations would be _____.

Answer 7: (c)

Let Rs and Rp represent the wires’ respective equivalent resistances when linked in series and parallel.

The ratio of heat generated in the circuit is provided by

H s/ H p = ( V 2 / R s) t/( V 2 / R p)/ t = R p/ R s

The equivalent resistance (Rs) of resistors connected in series is R + R = 2R

The equivalent resistance (Rp) of resistors connected in parallel is 1/ R + 1/ R = 2/R

Hence, the estimated ratio of the heat produced in series and parallel combinations would be

H s/Hp = 2R/( R/2) = 1/ 4

Thus, the ratio of heat produced is 1:4.

Question 8: What is the minimum resistance which can be made using five resistors, each of 1/5 Ω?

Answer: (b) 1/25 Ω

Resistance is the minimum when resistors are connected in parallel

1/ R = 1/(1/5) + 1/(1/5) + 1/(1/5) + 1/(1/5) + 1/(1/5) = 25 Ω

Question 9: A person carries out an experiment and thus plots the V-I graph of three taken samples of nichrome wire with different resistances R 1, R 2 and R 3, respectively (Figure.12.5). Which one of the following is true?

(a) R 1 = R 2 = R 3

(b) R 1 > R 2 > R 3

(d) R 2 > R 3 > R 1

Answer 9: (c)

The graph’s slope is 1/R because the current ( I ) is plotted on the y-axis, and the potential difference ( V ) is plotted on the x-axis . It implies that the less resistance, the steeper the slope. R 1 will therefore be the minimum and R 3 the maximum.

Question 10: Two resistors of resistance 2 Ω and 4 Ω, when connected to a battery, will have

(a) the same potential difference across them when connected in series

(b) same current flows through them when connected in series

(d) different p

Answer: (b) same current flowing through them when connected in series

Since the resistor gets a common current in a series arrangement, the current is not split into branches.

Question 11: What does an electric circuit mean?

Answer: An electric circuit is a continuous, closed path or loop composed of electronic components through which an electric current flows. Conductors, cells, Switch, and Load are the components of a simple circuit.

Question 12: An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω resistances are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current that flows through it?

R 1 = 100 , R 2 = 50 , R 3 = 500

All the devices are in parallel, so

1/ R = 1/ R 1 + 1/ R 2 + 1/ R 3

1/ R = 1/ 100 + 1/ 50 + 1/ 500 = ( 5 + 10 + 1 )/500 = 16/ 500

R = 500/ 16 = 31.25

Current, through all the appliances

I = V/ R = 220 / 31.25 = 220 X 31.25 = 7.04 A

Now, if only electric iron is connected to the same source such that it takes as much current as all three appliances, i.e. I = 7.04 A, its resistance should be equal to 31.25 .

Question 13: How is the resistivity of alloys compared with those of pure metals from which they may have been formed?

Answer: An alloy often has a higher resistivity than the individual metals that make up the alloy.

Question 14: Write the relation between the resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it.

Answer: The relation between resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it can be represented as follows:

P = V 2 / R

Question 15: How does the use of a fuse wire protect electrical appliances?

Answer: Compared to the main wiring, the fuse wire has a high resistance. Whenever there is an abrupt surge in electric current, the circuit is broken by melting fuse wire. This keeps electrical equipment from being damaged.

Question 16: Why are copper wires used as connecting wires?

Answer: Copper wires are used as the connecting wires because, in the case of copper, the electrical resistivity for it is low. It is ductile, inexpensive and it is an excellent electrical conductor.

Question: Define the SI unit of current.

Answer: The SI unit of current is ampere. An ampere is defined by the flow of one coulomb of Charge per second.

Question 18: How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω?

Answer: In order to get 4 Ω, resistance 2 Ω should be connected in series with the parallel combination of 3 Ω and 6 Ω.

1/ R CD = 1/ 3 + 1/ 6 = ( 2 + 1)/ 6

= 3/ 6 = 1/ 2

R CD = 2 , R AB = 2

R AD = R AB + R CD

= 2 + 2 = 4

Therefore, the total resistance of the circuit is R= 4

(b) In order to get 1 , all three resistors should be connected in parallel as

1/ R = 1/ 2 + 1/ 3 + 1/ 6 = ( 3 + 2+ 1)/ 6 = 1

Therefore, the net equivalent resistance of the circuit is R = 1

Question 19: A rectangular block of iron has dimensions L x L x b. What will be the resistance of the block measured between the two square ends? Given p resistivity.

Answer: We have given that a rectangular block of iron has dimensions l x l x b. We need to find the resistance of the block measured between the two square ends.

The resistance is given by the below formula as follows :

L is length of block

A is area of cross section

In this case,

Length of the rectangular block is l and area of block is l x b. So, resistance of the block measured between the two square ends is :

R = p b/l 2

So, the resistance of the block measured between the two square ends इस R = pb/l 2

Question 20: Ammeter burn out when connected in parallel. Give reasons.

Answer: When a low resistance wire is connected in parallel, a huge quantity of current travels through it, causing it to be either burned out or short-circuited.

Question 21: Should the resistance of an ammeter be low or high? Give reason(s).

Answer: The resistance of an ammeter should be zero, as the ammeter should not affect the flow of current in a circuit.

Question 22: Why does the connecting rod of an electric heater not glow, but the heating element does?

Answer: As the resistance of the connecting rod is lower than that of the heating element, the connecting rod of an electric heater does not glow. Thus, the heating element produces more heat than the connecting cord, and it glows.

Question 23: The power of a lamp is 60 W. Find the energy in joules consumed by it in 1s.

Answer: Here, given the power of the lamp, P = 60 W time,

So, energy consumed = power x time = (60 x 1) J = 60 J

Question 24: A wire of resistivity ‘p’ is stretched to double its length. What will be its new resistivity?

Answer: When a wire of resistivity p is stretched to double its length, then the new resistivity tends to remain the same because resistivity depends on the nature of the material.

Question 25: What is the resistance of any connecting wire?

Answer: The resistance of the connecting wire made of a good conductor is extremely low and they are assumed to have zero resistance. So, less heat is produced in them and they can be easily used in connections.

Question 26: A number of n resistors each of resistance ‘R’ are first connected in series and then in parallel connection. What is the ratio of the total effective resistance of the circuit in series combination and parallel combination?

Answer : Total effective resistance of the circuit in series combination is R s = nR

And for parallel combination is R p = R/ n and

R s/ R p = nR/ R/ n

The ratio will be n 2 .

Question 27: Calculate the total number of electrons constituting one coulomb of charge.

The Charge of an electron = 1.6 × 10 -19 C.

According to the concept of charge quantisation,

Q = nqe, where we suppose ‘ n’ is the number of electrons and similarly ‘ qe’ is the Charge of the electron.

Substituting these values in the said equation, the number of electrons constituting one coulomb of Charge can be calculated as follows:

1C = n X 1.6 X 10 -19

n= 1 1.6 X 10 -19 = 6.25 X 10 18

Therefore, the number of electrons in one coulomb of Charge = 6. 25 × 10 18 .

Question 28: How much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts.

Here, V = 220V , R = 55

By Ohm’s law, V = IR

Therefore, 220 = 7 x 55 or I = 4A

The wattage of electric iron = Power

= V 2 R = (220) 2 55 = 880 W

Question 29: A current of 1 ampere flows in a circuit of series connection containing an electric lamp and a conductor of 5 Ω and connected to a 10 V battery. Calculate the resistance of the given electric lamp.

Therefore, if the resistance of 10 Ω is connected in parallel with this series combination, what type of change (if any) in current flowing through the 5 Ω conductor and potential difference across the lamp will take place? Give reasons.

Let R lamp represent the resistance of the lamp.

Current ( I ) = 1 A

Resistance of conductor (R conductor ) = 5 Ω

The potential difference of battery ( V) = 10 V

Given that the lamp and conductor are linked in series, the same amount of current 1 A will flow through them both.

Using Ohm’s law,

R net = V I

R net = 10 क्ष 1

We know, in series connection

R net = R lamp + R Conductor

10 = R lamp + 5

The potential difference across lamps,

R lamp = I x R lamp

= 1 x 5 = 5 V

When a resistor of 10 Ω resistor connected parallel to the series combination of lamp and conductor

( R net = 5 + 5 = 10 ) then the equivalent resistance,

1/ R eq = 1/ 10 + 1/ 10 = 2/ 10 = 1/ 5

Using Ohm’s law,

I’= V/ R eq

Equal distribution of current will occur in two parallel parts.

Thus, I’/2 = 1A current will pass through both the lamp and the resistor of 5 (because they are connected in series).

The potential difference across the lamp (R lamp = 5 ).

V’ lamp = 1×5 = 5 V

Therefore, the current flowing through the conductor of resistance 5 and the potential difference across the bulb won’t change.

Question 30: What is electrical resistivity? In a particular series electrical circuit comprising a resistor made up of a metallic wire, the ammeter generally reads 5 A. The previous reading of the ammeter decreases to half in case the length of the wire is doubled. Why?

Answer: Resistivity is a property of a conductor that prevents the flow of electric current. A specific material has a particular resistance. Resistance is inversely proportional to current flow and directly proportional to conductor length.

When the length is doubled, the resistance doubles and the current flow is reduced by half. This is what’s causing the ammeter value to drop.

Question 31: (i) List the three factors on which the resistance of a conductor depends.

(ii) Write the SI unit of resistivity.

(i) A conductor’s resistance is influenced by the following factors:

(1) Length of the conductor: The resistance (R) will increase as the conductor’s length (I) increases.

(2) Area of the cross-section of the conductor: (as the cross-sectional area of the conductor increases, the resistance decreases.

(3) Nature of conductor.

(ii) SI unit of resistivity is Ω m.

Question 32: An electric bulb which is connected to a 220 V generator and the current is 2.5 A. Calculate the power of the bulb.

Here, V= 220 V, I = 2.5 A

Given, Power of the bulb, P = VI = 220 × 2.5 W = 550 W

Question 33: Name a device that helps to maintain a potential difference across a conductor.

One of the devices that aid in maintaining a potential difference across a conductor is a battery, which can consist of one or more electric cells.

Question 34: What is the resistance of an ammeter?

An ammeter’s resistance generally is very minimal, and in an ideal ammeter, it is zero.

Question 35: What is the resistance of a voltmeter?

Answer: The resistance of a voltmeter is ideally infinite resistance.

Question 36: What is the commercial unit of electrical energy? Represent it in terms of joules.

The commercial unit of electrical energy is kilowatt/hr

1 kW/hr = 1 kW h

= 1000 W × 60 × 60s

= 3.6 × 10 6 J

Question 37: Explain two disadvantages of series arrangement for a household circuit.

The two drawbacks of series circuits for household wiring are:

If one electrical appliance in a series circuit stops functioning for any reason, the entire circuit will break, and all other electrical appliances will also stop functioning.
Because there is only one switch for every electrical device in a series circuit, they cannot be turned on or off independently.

Question 38: What is meant by the saying that the potential difference between two points is 1 V?

The potential difference between two points is 1V when 1 J of work is done to move a 1 C of Charge from one location to the other.

Question 39: Two equal wires of equal cross-sectional area, one of copper and the other of manganin , have the same resistance. Which one will be longer?

Using the equation, = RA I , where is the resistivity, R is the resistance, and A is the area.

Resistance of Copper wire = 1 l 1 A

Resistance of Manganin wire = 2 l 2 A

1 l 1 = 2 l 2 (As l is constant)

Since 1 <<< 2

So, l 1 >>>> l 2

I.e. Copper wire would be longer.

Question 40: Three equal resistances are connected in series and then in parallel. What will be the ratio of their change in resistances?

Answer 40: When connected in series, Resistance R series = R+R+R= 3R

When connected in parallel, Resistance R parallel = R/3

Ratio of change in resistances= R series R parallel = 3R R/3

Therefore, the ratio of change in resistances is 9:1

Question 41: State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

According to Ohm’s law, the potential difference (voltage) across an ideal conductor is proportional to the current flowing through it at a given temperature.

I.e. V/I = R

Verification of Ohm’s law

Make the circuit indicated in Fig., which consists of four 1.5 V cells, an ammeter, a voltmeter, and a nichrome wire of length XY, say, 0.5 m. (The metals nickel, chromium, manganese, and iron make up the alloy known as nichrome.)

Start by using a single cell as the circuit’s source. Take note of the ammeter’s reading for current (I) and the voltmeter’s reading (V) for the potential difference across the nichrome wire ‘XY’ in the circuit. Add them to the table provided.

Connect two batteries to the circuit next, and then note the ammeter and voltmeter readings for the current flowing through the nichrome wire and the potential difference across the nichrome wire values, respectively.

Use three cells in the circuit first, then four cells, and repeat the process above for each group of cells.

Question 42: If there are 3 x 10 11 electrons flowing through the filament of the bulb for two minutes. Find the current flowing through the circuit. Charge on one electron 1.6 x 10 19 C.

Using the equation, q = ne

= 3 x 10 11 x 1.6 x 10 – 19 C

= 4.8 x 10 8 C

= 4.8 x 10 8 2 x 60

= 4 x 10 7 A

The current flowing through the electric circuit is 4 x 10 7 A.

Question 43: Three resistors of 5 Ω, 10 Ω and 15 Ω are connected in series, and the entire combination is connected to a battery of 30 V. Ammeter and Voltmeter are connected in the circuit. Draw a circuit diagram to connect all the devices in the proper, correct order. What is the current flowing and potential difference across 10 Ω resistance?

Answer 43:

Given, Total resistance, R = R1 + R2 + R3 = 5 + 10 + 15 = 30Ω

Total potential difference, V = 30 volts

V = IR ⇒ I = V R = 30 30 = 1 ampere

∴ Current remains constant in this series,

∴ I1 = I2 = I3 = I;

I2 = 1amp;

As V2 = I2

R2 = 1 × 10 = 10 volts

∴ The potential difference across the 10 Ω is 10 volts.

Question 44: If an electric heater rated 800 W operates 6h/day. Find the Cost of energy to operate it for 30 days at ₹3.00 per unit of consumption.

Answer 44:

Here, the Power of the heater, P = 800 W;

Time, t = 6 hour/day;

No. of days, n = 30;

Cost per unit = ₹3.00;

Thus, Consumed in 1 day = 800 × 6 = 4800 Wh

And, Energy consumed in 30 days = 4800 × 30 = 144000 Wh

144000 1000 = kWh = 144 units

Now, the Cost of 1 unit = ₹3

Therefore, Cost of 144 units = 3 × 144 = ₹432

Question 45: What is the electrical resistivity of a given material? What is its unit? Discuss an experiment to study the factors on which the resistance of conducting wire depends.

Answer 45: Resistivity is an inherent property of a conductor that resists the flow of electric current. The resistivity of each material is unique.

The SI unit of resistance is Ω m.

Experiment to study the depending factors of the resistance of conducting wire.

A nichrome wire, a torch, a 10 W bulb, an ammeter (0–5 A range), a plug key, and some connecting wires are needed.

As illustrated in the figure, assemble the circuit by connecting four 1.5 V dry batteries in series with the ammeter, thereby leaving a gap XY in the circuit.

Observation:

Resistance depends on the length of the conductor, the material of the conductor, and the area of the cross-section.

Connecting the nichrome wire in the XY gap completes the circuit. Insert the key. The ammeter reading should be noted. From the plug, remove the key. [Remember: After measuring the current flowing through the circuit, always remove the key from the plug.]

Replace the nichrome wire in the circuit with the torch bulb, and then determine the current flowing through it by measuring the ammeter’s reading.

Repeat the previous process now using the 10 W bulb in the XY gap. Are there variations in the ammeter readings for the various components connected in the gap XY? What do the aforementioned observations suggest?

By leaving any material component in the gap, you are able to repeat this activity. Watch the ammeter values for each situation. Analyse the results.

Question 46: Calculate the resistance of a given metal wire of length 2m and area of cross-section 1.55 × 106 m² if the resistivity of the metal is taken to be 2.8 × 10-8 Ωm.

Answer 46: For the given metal wire,

Length, l= 2 m

Area of cross-section, A= 1.55 X 10 -6 m 2

Resistivity of the metal, p = 2.8 X 10 -8 m

Since, resistance, R = l A

So, R = ( 2.8 X 10 -8 X 2 1.55 X 10 -6 )

= 5.6 1.55 X 10 -2

= 3.6 X 10 -2

Therefore, R = 0.036

Question 47: When will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?

Resistance is represented by the equation,

ρ is the resistivity of the wire material,

l is the wire

A is the cross-sectional area of the wire.

It is clear from the equation that the resistance is inversely proportional to the area of the wire cross-section. Therefore, the resistance increases with wire thickness and vice versa. Therefore, a thick wire conducts current more readily than a thin wire.

Question 48: What is represented by joule/coulomb?

Answer 48: The potential difference is represented by the joule/coulomb.

Question 49: A nichrome wire of resistivity 100 W m and copper wire of resistivity 1.62 ohm -m of the same length and same area of the cross-section are connected in series, and current is passed through them. Why does the nichrome wire get heated first?

Answer 49: Looking at the equation

Q= I 2 (pL / A)t

Nichrome wire gets heated first because it has a higher resistance than copper wire.

Question 50: Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and thus can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.

Answer 50:

P = Potential difference

I = Current

R = Resistance

For resistor A,

18 = I 2 x 2

I 2 = 18 2

This is the maximum known current flowing through resistor A.

The maximum known current flowing through the resistors B and C, I’= 3 x 1 2 = 1.5 A.

Question 51: How will you infer, with the help of an experiment, that the same current flows through each and every part of the circuit containing three resistances in series connected to a battery?

You can collect three resistors, R1, R2, and R3, in series to make the circuit.
Then use an ammeter to observe the changes in the overall current flow.
You can remove R1 and take the readings of the potential difference of R2 and R3.
You can remove R2 and take the reading of the potential difference between R1 and R3.

Since the ammeter reading was the same in each case, it can be assumed that the circuit’s current is constant. One can set up an ammeter in several places and watch the current flow to double-check.

Question 52: Calculate the estimated resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance of 20 ohms.

Answer 52: 3.14 X (10 -4 ) 2 m 2

length of the wire, l = 1 m,

radius of the wire, r = 0.01 cm = 1 × 10 -4 m and

given resistance, R = 20Ω

R = l A , where is the resistivity of the material of the wire.

20 = l r 2 = 1 m 3.14 X (10 -4 ) 2 m 2

Therefore, = 6.28 X 10 -7 m

Question 53: A charge of 2 C moves between two plates, maintained at a potential difference of IV. What is the energy acquired by the Charge?

Answer 53: The energy acquired by the Charge, W = QV

Therefore, the energy acquired is 2 J.

Question 54: Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Ohm’s law is used to calculate how the current flow changes through an electrical component.

Ohm’s law states that

I = V/R, which gives the current.

The potential difference is now divided in half while maintaining the same resistance,

Let V’ = V/2 be the new voltage .

Let R’ = R be the new resistance, and the new amount of current be I’.

Ohm’s law is thus used to calculate the current change as shown below:

I’ = V’ R’ = ( V 2 ) R = 1 2 V R = 1 2

As a result, the electrical component’s current is reduced by half,keeping resistance constant.

Question 55: What is the overloading of an electrical circuit? Explain two possible causes due to which overloading might occur in any household circuit. Explain one precaution, if any, that should be taken to avoid the overloading of a domestic electric circuit.

Overloading: The power ratings of the appliances being utilised at a given moment determine the current flowing in household wiring. Electrical appliances with high power ratings take a tremendous amount of electricity from the circuit if too many of them are turned on at once. The overloading of the circuit takes place. . The copper wires in residential circuits get heated up due to extremely high temperatures and can immediately catch fire as a result of heavy currents running through them.

Precaution: As a result, overloading can seriously harm buildings and electrical equipment. To prevent these damages, a fuse with the appropriate rating must be used. Such a fuse wire will melt before the heated circuit wire’s temperature rises to a point where it breaks the circuit.

Question 56: What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

According to the Joule’s heating effect, the heat produced in a resistor is known to be

(i) Directly proportional to the square of current for the given resistor.
(ii) Directly proportional to the resistance for a given current,
(iii) Directly proportional to the time of current flowing through the resistor.

It can be expressed as H = I2Rt

‘H’ is the heating effect, ‘I’ is the electric current, ‘R’ is resistance, and ‘t’ is time.

Experiment to demonstrate Joule’s law of heating

Take an immersion rod for water heating and attach it to a regulator-connected socket. It’s crucial to keep in mind that a regulator regulates how much current flows through a gadget.
Keep the regulator’s pointer at the lowest setting and time how long it takes the immersion rod to heat a specific volume of water.
Increase the regulator’s pointer to the following level. Time the same quantity of water heating with an immersion rod.
To measure higher amounts of the regulator, repeat the previous step.

It has been observed that it takes less time to heat the same amount of water with an increasing electric current. This illustrates the Joule’s Law of Heating.

Application:

Electric appliances like toasters, ovens, kettles, and heaters operate using the leafing effect of current.

Question 57: Why are the coils of electric toasters and irons made of an alloy rather than any pure metal.Give reason(s).

Due to its high resistivity, an alloy has a substantially higher melting point than a pure metal. Alloys are resistant to melting when temperatures are high. As a result, alloys are utilised in heating devices like electric toasters and irons.

Question 58: Name a device that helps to maintain a potential difference across a conductor in a circuit. When do we say that the potential difference across a conductor is 1 volt? Calculate the amount of work done in shifting a charge of 2 coulombs from a point A to B having potential +10V and -5V, respectively.

Answer 58: In a circuit, a battery (or cell) aids in maintaining the potential difference across a conductor.

If 1 joule of labour is expended in transporting 1 coulomb of electrical charge from one location to the other, the potential difference between the two points is said to be 1 volt.

Given, Charge, Q = 2C

Potential at A = +10 V, Potential at B = -5V

Potential difference, (V) = +10 – (-5) = 10 + 5 = 15 volts

W = 15 × 2 = 30 J

Question 59: Which is a better conductor among iron and mercury?

Answer 59: Iron is a better conductor than mercury because the resistivity of mercury is more than the resistivity of iron.

Question 60: Which has more resistance, 100 W bulb or 60 W bulb?

Answer 60: As it is clearly known that R 1 P , the resistance of the 60 W bulb is more.

Question 61: Find the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 106 Ω, (b) 1 Ω, 103 Ω, and 106 Ω.

Answer 61: (a) When 1 Ω and 106 Ω when connected in parallel gives the 10 6 equivalent resistance as follows:

1 R = 1 1 + 1 10 6

R= 10 6 1+ 10 6 10 6 10 6 = 1

Therefore, the equivalent resistance is 1 Ω. 1+ 10 6

(b) When 1 Ω, 103 Ω, and 106 Ω are in parallel, the equivalent resistance is given by

1 R = 1 1 + 1 10 3 + 1 10 6

Solving, we get

R = 10 6 + 10 3 +1 10 6 = 1000000 1000001 = 0.999

Therefore, the equivalent resistance is 0.999 Ω.

Question 62: What are the benefits of connecting electrical devices in parallel with the battery instead of connecting them in series?

There is no voltage division among the appliances when the electrical devices are connected in parallel. The supply voltage is equal to the potential difference across the devices. Devices connected in parallel lower the circuit’s effective resistance as well.

Question 63: Why does the cord of an electric heater not glow while the heating element does?

Answer 63: An electric heater’s heating element is constructed from a high-resistance alloy. The heating element glows red and gets excessively hot when the electricity passes through it. Typically, copper or aluminium, which have low resistance, is used to make the rope. Consequently, the cord doesn’t glow.

Question 64: Discuss the heat generated while transferring 96000 coulombs of Charge in one hour through a potential difference of 50 V.

Answer 64: According to Joule’s law, the heat produced can be calculated as follows:

where assuming,

voltage, V = 50 V

I will be current

t be the time in seconds, 1 hour = 3600 seconds

The amount of current is calculated as follows:

Amount of Current = Amount of Charge Time flow of Charge

Substituting the value, we get

I = 96000 3600 = 26.66 A

Now, to find the heat generated

H = 50 X 26.66 X 3600 = 4.8 X 10 6 J

Therefore, the heat generated is 4.8 X 10 6 J

Question 65: An electric iron of resistance 20 Ω draws a current of 5 A. Calculate the heat developed in 30 s.

Answer 65: The Joule’s law of heating, which is represented by the equation, can be used to determine how the heat is produced as follows:

Putting the data in the above equation, we get,

H = 100 × 5 × 30 = 15,000 J

The amount of heat produced by the electric iron in 30 s is 15,000 J.

Question 66: What factors determine the rate at which energy is delivered by a current?

Answer 66: Electric power is the rate at which electric equipment uses electricity. Therefore, the power of the appliance is defined as the rate at which energy is delivered by a current.

Question 67: How is the connection of a voltmeter made in the circuit to measure the potential difference between two points?

Answer 67: The voltmeter should be connected in parallel to each of the two points in order to measure the voltage between any two points.

Question 68: Draw a circuit taking an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the resistor of 12 ohm. What would be the new reading in the ammeter and the voltmeter?

Answer 68:

Let us take the total resistance of the circuit = R

The equivalent resistance R is equal to the total resistance because all three resistors are connected in series.

R = 5 Ω + 8 Ω + 12 Ω = 25 Ω

Therefore,

V = 2V + 2V + 2V = 6V

I = V R = 6 25 = 0.24 A

The reading of the voltmeter across R’ = 12 Ω is

= 0.24 X 12 = 2.88 V

Question 69: Find the following in the electric circuit given in Figure 12.9

(a) Effective resistance of the two 8 Ω resistors in the given combination

(b) Current flowing through the resistor of 4 Ω

(d) Power dissipated in a resistor of 4 Ω (e) Difference in ammeter readings, if any

Answer 69: (i) Since two 8 resistors are in parallel, then their effective resistance R p is given by

1 R p = 1 R 1 + 1 R 2 = 1 8 + 1 8 = 1 4

(ii) Total resistance in the circuit

R = 4 + R p = 4 + 4 + 8

Current, through the circuit,

I = V R = 8 8 = 1 A

Thus, the current through the 4 resistor is 1 A as 4 and R p are in series and the same current flows through them.

(iii) Potential difference across 4 resistor is potential drop by the 4 resistor.

i.e. V = IR = 1 X 4 = 4 V

(iv) Power dissipated in 4 resistor

P = I 2 R = 1 2 X 4 = 4 W

(v) There is no difference in the reading of ammeters A 1 and A 2 as the same current flows through all elements in a series current.

Question 70: A copper wire having a diameter of 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of the wire to make its resistance 10 Ω? How much will the resistance change if the diameter is doubled?

Answer 70: The formula provides the resistance of a copper wire with a cross-sectional area of m2 and a length in metres.

The area of the cross-section of the wire is calculated as follows

A = ( Diameter 2 ) 2

Substituting the values in the formula, we get

l = RA = 10 X 3.14 X ( 0.0005 2 2 ) 1.6 X 10 -18 = 10 X 3.14 X 25 4 X 1.6 = 122.72 m

The new diameter of the wire is 1mm, or 0.001m when the wire’s diameter is doubled. Therefore, the resistance can be calculated as follows:

R = l A = 1.6 X 10 -18 X 122.72 m ( 0.001 2 ) 2 = 250.2 X 10 -2 = 2.5

The new resistance is 2.5 , and the wire’s length is 122.72 m.

Question 71: Difference features between Overloading and Short-circuiting in Domestic circuits

Answer 71:

Overloading: Overloading occurs when a circuit is used by too many electrical devices with high power ratings that are switched on simultaneously.

The copper wire used in domestic wiring heats up to an exceedingly high temperature as a result of an excessively high current running through the circuit, and a fire may subsequently ignite.

Short-circuiting: Short-circuiting is a direct result of touching bare live and neutral wires. Because the circuit’s resistance is so low in this situation, a lot of current passes through it, heating the wires to a high temperature and possibly igniting a wire. .

Question 72: When a battery of 12 V is connected across an unknown resistor, there is a current flow of 2.5 mA in the electric circuit. Find the resistance of the resistor.

Answer 72: Using Ohm’s Law, the resistor’s value can be determined as follows:

Putting the data in the equation, we get

R = 12 2.5 X 10 -8 = 4.8 X 10 3 = 4.8

Question 73: Three resistors of 10 Ω, 15 Ω and 5 Ω are connected in parallel. Find their equivalent resistance.

Here, R 1 = 10 , R 2 = 15 , R 3 = 5

In a parallel connection, equivalent resistance ( R eq ) is given by

1 R eq = 1 R 1 + 1 R 2 + 1 R 3

So, 1 R eq = 1 10 + 1 15 + 1 5

1 R eq = 3+ 2+6 30 = 11 30

Therefore, R eq = 30 11 = 2.73

Question 74: A battery of 9 V is connected in a series system with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. What quantity of current would flow through the 12 Ω resistor?

Answer 74: There is no existing division in a series connection. An equal amount of current travels across each resistor.

We apply Ohm’s law to determine the amount of current passing through the resistors.

Let’s

first determine the equivalent resistance in the manner described below:

R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

Using Ohm’s law,

I = V R = 9V 13.4 = 0.671 A

The current flowing across the 12 Ω resistor is 0.671 A.

Question 75: Suppose the resistance of an electrical component remains constant, and the potential difference across the two ends of the component decreases to half of its earlier value. What type of change will occur in the current through it?

Answer 75:

Knowing that,

If, V’ = V 2

If I’ = V ‘ R = V 2R = 1 2

As a result, an electrical component’s current reduces by half of what it was.

Question 76: Two same resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in both cases.

Answer 76:

Let the resistance of each resistor be R.

For series combination,

R s = R 1 + R 2

So, R s = R + R = 2R

For parallel combination,

1 R p = 1 R 1 + 1 R 2 or R p = R 1 R 2 R 1 + R 2

So, R p = R x R R + R = R 2

Required Ratio = R s R p = 2R R/2 = 4 : 1

Question 77: Explain the use of an electric fuse. What type of material is used for fuse wire and why?

Answer 77: Electric fuses guard against the very high electric current by blocking it from flowing into circuits and appliances. It is composed of a wire formed of a metal or alloy with an appropriate melting point, such as lead, copper, iron, or aluminium. The temperature of the fuse wire rises if a current more than the allowed amount runs through the circuit. The fuse wire melts, as a result, breaking the circuit.

Question 78: When an electric current flows through a conductor, it tends to become hot. Justify.List the factors on which the heat produced in a conductor depends. State Joule’s law of heating. How will the heat produced in an electric circuit be affected by this if the resistance in the circuit is doubled for the same current?

Answer 78: A conductor heats up when an electric current is carried through it. This is referred to as the current heating effect. The electrical energy is converted into heat energy to produce the heating effect of current. An electrical energy source is a cell or battery. The potential difference between the two terminals of the cell is created by the chemical reaction within, which causes the electrons to move and for current to flow through a resistor. The source must continue using up its energy. While maintaining the current, some of the source energy may be used for productive activity, while the remaining source energy may be used to generate heat.

Question 79: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer 79: Let ‘x’ be the number of resistors required.

The equivalent resistance of the resistor R in the parallel combination is given by

1 R = x X 1 176 = R = 176 x

Now, using Ohm’s law. The number of resistors can be calculated as follows:

Substituting the values, we get

176 x = V I

x = 176 X 5 220 = 4

The number of resistors required is 4.

Question 80: Two wires of the same material and same length have radii R and r. Compare their resistances.

Answer 80: Suppose R and r are resistances, then R = r as p and I are the same.

Question 81: Several electric bulbs designed and are supposed to be used on a 220 V electric supply line are rated 10 W. How many such lamps can be connected in parallel together across the two wires of a 220 V supply line if the maximum allowable current is 5 A?

Answer 81: The resistance of the bulb is calculated as follows:

R = (220) 2 10 = 4840

The resistance of x number of electric bulbs is calculated as follows:

R = V I = 220 5 = 44

The resistance of each electric bulb is 4840 .

The equivalent resistance of x bulbs is given by

1 R = 1 R 1 + 1 R 1 + 1 R 1 + ………up to x times

1 R = 1 R 1 X x

x = R1 R = 4840 44 = 110

Hence, 110 lamps can be connected together in parallel.

Question 82: A fuse wire melts at 5 A. If it is desired that the fuse wire of the same material melt at 10 A, then in your opinion whether the new fuse wire should be of a smaller or larger radius than the earlier one? Give reasons for your answer.

Answer 82: Let R be the resistance of the wire; the heat produced in the fuse at 5A is

H = (5) 2 R (H – I 2 R t)

Fue melts at (5) 2 R joules of heat

Let R’ be the resistance of the new wire

So, the heat produced in 1 second =(10) 2 R’

To prevent it from melting

(5)2 R = (10)2R’ or R’ = R 4

Therefore, the cross-sectional area of the new fuse wire is four times the first fuse.

Now, A = r 2 , so the new radius is twice as large as the old one. The new fuse wire, which is the same material and length as the old one, has a greater radius at 10 A.

Question 83: How many bulbs of 81 should be joined in parallel to draw a current of 2 A from a battery of 4V?

Answer 83:

Let n be the number of bulbs.

1/R = 1/R 1 + 1/R 2 +………………+ 1/R n = n 8

The number of bulbs is 4.

Question 84: When will current flow more easily: through a thick wire or a thin wire of the same material, when connected to the same electric source? Why?

Answer 84: A thick wire connected to the same source conducts current more easily than a short wire made of the same material. As thickness reduces, resistance increases.

Question 85: A hot plate of an electric oven connected to a 220 V supply line has two resistance coils such as A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What will be the currents in the three cases?

Case (i) When coils are used separately

By using Ohm’s law, we will be able to find the current flowing through each coil as follows:

I = 220 V 24 = 9.166 A

When used individually, each resistor allows 9.166 A of current to pass through it.

Case (ii) When the coils are connected in series

The total resistance is 24 Ω + 24 Ω = 48 Ω in the series circuit

The current flowing through this series circuit is calculated as follows:

I = V R = 220 V 48 = 4.58 A

Therefore, a current of 4.58 A will flow through the circuit in series.

Case (iii) When the coils are in parallel

connection,

the equivalent resistance is calculated as follows:

R = 24 X 24 24 + 24 = 576 48 = 12

By using Ohm’s law, the current flowing through the parallel circuit is given by

I = V R = 220 12 = 18.33 A

The current is 18.33 A in the parallel circuit.

Question 86: What happens to the current in a circuit if its resistance is doubled?

Answer 86: As current and resistance are inversely proportional, the current is reduced to half of its previous value.

Question 87: Let the resistance of a component of electric current remain constant while the potential difference across the two ends of the component decreases to half of its earlier value. What type of change will occur in the current through it?

Answer 87: A thick wire connected to the same source conducts current more easily than a short wire made of the same material. As thickness reduces, resistance increases.

Question 88: Compare the power consumed in the 2 Ω resistor in each of the following circuit conditions: (i) a 6 V battery in series connection with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel connection with 12 Ω and 2 Ω resistors.

Answer 88: (i) Since the resistors 1 Ω and 2 Ω are connected in series, and there is a 6 V potential difference, their equivalent resistance is given by 1 Ω + 2 Ω = 3 Ω. Using Ohm’s law, the following formula is used to determine the circuit’s current:

I = V R = 6 3 = 2 A

2 A current will flow across all the components in the circuit because there is no division of current in a circuit of series connection.

The power in the 2 resistor is calculated as follows:

P = I 2 R = (2) 2 X 2 = 8 W

Thus, the power consumed by the 2 Ω resistor is 8 W.

(ii) The voltage between the resistors stays constant when 12 and 2 resistors are linked in parallel. Given that a 2 Ω resistor has a 4 V voltage across it, we can use the formula below to determine how much power is used by the resistor: V 2 4 2

P = V 2 R = 4 2 2 = 8 W

The power consumed by the 2 Ω resistor is 8 W.

Question 89: Two cubes, A and B, are made of the same material. The side of B is thrice that of A. Find the ratio R A /R B .

The value of R A = L A and

R B = 3L 9 A

R A : R B = 3: 1

Question 90: What happens to the resistance of a circuit if the current through it is doubled?

Resistance is unchanged since the circuit’s resistance is independent of the current flowing through it.

Question 91: Two metallic wires, A and B, are connected. Wire A has lengths l and radius r, while B has lengths 2l and 2r. If both the wires are of the same material then find the ratio of total resistances of series combination and the resistance of wire A.?

Answer 91 . Here, the Resistance of metallic wire A, R 1 = l A

Resistance of metallic wire B , R 2 = 2l 4 r 2

The total resistance in series can be expressed as R = R 1 + R 2

= l A + 2l 4 r 2

= 3 l 2 r 2

The ratio of the total resistance (R) in series to the resistance of A (R1) is

R R 1 = 3 l 2 r 2 l r 2

The ratio of the total resistance (R) in series to the resistance of wire A is 3 2 .

Question 91: Illustrate how you would connect given three resistors, each of whose resistance is 6 Ω so that the combination has a resistance of (i) 9 Ω or (ii) 4Ω

Answer 91:

(i) When we connect R1 in series with the parallel combination of R2 and R3, as shown in Fig. (a).

The equivalent resistance is

R = R 1 + R 2 R 3 R 2 + R 3 = 6 + 6 X 6 6 + 6

= 6 + 3 = 9

(ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is

R = 12 x 6 12 + 6 = 72 18 = 4

Question 92: How does the resistance of a wire depend upon its radius?

Answer 92: As R 1 A

The resistance of the above-mentioned wire is directly proportional to its given radius.

Question 93: Which of the two uses more energy, a 250 W TV set in 1 hr or a 1200 W toaster in 10 minutes?

The total energy consumed by the electrical devices is represented by the equation

H = Pt , where the power of the appliance is P and t is the time

This formula is used to determine how much energy a TV with a 250 W power rating uses:

H = 250 W × 3600 seconds

= 9 × 105 J

In the same way, the energy consumed by a toaster with a 1200 W power rating is

H = 1200 W × 600 s = 7.2 × 105 J

From the part of the above calculation, it can be said that the energy consumed by the TV is greater than the toaster.

Question 94: Two wires are of the same length and same radius, but one of them is of copper, and the other is of iron. Which will have more resistance.

Since, R = 1 A

But A and I have the same value. It is absolutely determined by the resistivity; hence iron has a higher resistance.

Question 95: An electric heater of 8 Ω resistance draws 15 A of current from the service mains supply for 2 hours. Calculate the rate at which heat is produced in the given heater.

The rate at which the heat production takes place in the heater is thus calculated using the following formula

P = (15A) 2 × 8 Ω = 1800 watt

The electric heater produces heat at the rate of 1800 watt

Question 96: The resistance of a given wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10-8 ohm metre , find the estimated length of the wire.

Answer 96:

Here, r= 0.01 cm = 10 -4 m, p = 50 x 10 -8 m and R = 10

As, R = l A

Or, I = RA = R ( r 2 )

So, I = 10 50 X 10 -8 3.14 X (10 -4 ) 2

I = 0.628 m = 62.8 cm

Question 97: Explain the following.

Why is tungsten used almost exclusively for the filament of electric lamps?
Why are the conductors of electrical heating devices, like bread-toasters and electric irons, mostly made of an alloy rather than a pure metal?
Why is the series arrangement not used in domestic circuits?
How does the resistance of a wire vary with its area of cross-section?
Why are copper and aluminium wires usually employed for electricity transmission?
Tungsten has a very high resistance and melting point. This characteristic prevents it from burning easily when heated. At high temperatures, electric bulbs are operated. As a result, tungsten is a popular metal choice for electric lamp filaments.
Due to their high resistivity, alloys are used as the conductors in electric heating equipment like bread-toasters and electric irons. Because of its high resistance, it generates a lot of heat.
Because each component in the circuit only receives a tiny voltage as a result of the voltage being divided into a series circuit, when one component fails, the circuit is broken and none of the components work. . Because of this, domestic circuits do not employ series circuits.
The relationship between resistance and cross-sectional area is inversely proportional. This means the resistance decreases as the cross-sectional area increases and vice versa.
Copper and aluminium are frequently used for the transmission of electricity because they are effective conductors of electricity and have low resistance.

Question 98: How will you justify that the same potential difference (voltage) will exist across three resistors connected in a parallel arrangement to a battery?

You can take three resistors, R1, R2 and R3, and connect them in parallel to make a circuit, as shown in the figure.

Then use a voltmeter to take the reading of the potential difference of the three resistors in parallel combination.

Now, you can remove the resistor R1 and take the reading of the potential difference of the remaining resistors in combination.

Then, you can remove the resistor R2 and take the reading of the potential difference of the remaining resistor.

In each case, the Voltmeter reading appears to be the same, which shows that the same potential difference tends to exist across three resistors connected in a parallel arrangement.

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Q.1 An object is placed in front of a convex lens at a distance of

where f is the magnitude of the focal length of the lens. Prove that the magnification produced by the lens is n. Also find the two values of the object distances for which a convex lens of power 2.5 D will produce an image that is four times as large as the object.

Marks: 5 Ans

Given: Object distance , | u | = f f n ; Focal length = f; Image distance, v = ByLens’Formula , 1 f = 1 v 1 u Object distance is always negative. 1 f = 1 v + 1 f f n 1 v = 1 f 1 f f n 1 v = f f n f f f f n 1 v = 1 n f f n v= f f n 1 n Now, magnification, | m | = v u |m|=n Hence, proved. Now, u= f f n Butf = 1 2.5 m = 0.4 m ˆµ f = 1 P u = 0.4 0.4 4 ( ˆµ n=4 ) u= ( 0.4 0.1 ) u = 0.5mor 0.3 m

Q.2 An object is placed at 10cm in front of a concave mirror of focal length 15cm. Find the position, nature and size of the image.

Given u = 10 cm ; f = 15 cm ; v = ; m = 1 v + 1 u = 1 f 1 v + 1 10 = 1 15 1 v = 1 15 + 1 10 1 v = 2+3 30 = 1 30 v = 30 cm Positive sign indicates that the image is virtual and erect and form 30 cm behind the mirror.

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Chapter 1 - chemical reactions and equations.

Chapter 2 - Acids, Bases and Salts

Chapter 3 - metals and non-metals, chapter 4 - carbon and its compounds, chapter 5 - periodic classification of elements, chapter 6 - life processes, chapter 7 - control and coordination, chapter 8 - how do organisms reproduce, chapter 9 - heredity and evolution, chapter 10 - light reflection and refraction, chapter 11 - human eye and colourful world, chapter 13 - magnetic effects of electric current, chapter 14 - sources of energy, chapter 15 - our environment, chapter 16 - management of natural resources, faqs (frequently asked questions), 1. why is learning about electricity important for the students in school.

In order to be thorough with the most important commodities in modern Sciences , one must understand electricity. A fundamental understanding of electricity is also necessary for many technical occupations in order to create the technologies and goods that we use in our every -day lives. For this purpose, Extramarks has curated this important questions Class 10 Science Chapter 12

2. How has electricity made our life easier in present times?

Nowadays, electricity is almost necessary for our daily work functioning. It is utilized for residential purposes such as operating fans, computer, electric stoves, air conditioning, and lighting up rooms. It is used in factories to operate heavy machines. Electricity is also used to produce various products, including food, clothing, and paper.

3. Is Electricty Important chapter for CBSE Class 10 Exam?

Yes, electricity chapter is important for CBSE Class 10 Exam

4. Where can I get all study materials related to CBSE Class 10?

You can get all the study materials of CBSE Class 10 at Extramarks

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Electricity Class 10 Science Important Questions and Answers

Important Questions for Class 10 Science Chapter 12 Electricity covers each topic of the chapter. These questions aim at providing a better understanding of the chapter to the students and can be downloaded in PDF format. These important question bank help students in clearing their doubts so that they can score well in the exam.

While preparing for exams, students should practise these important questions of Class 10 Science to understand the concepts better. Solving important questions of Class 10 Science Chapter 12 will teach students time management skills and enhance their problem-solving skills. Also, students may come across a few of these questions in the board exam.

Case Study Questions Class 10 Science Magnetic Effects of Electric Current

Case study questions class 10 science chapter 13 magnetic effects of electric current.

CBSE Class 10 Case Study Questions Science Magnetic Effects of Electric Current. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Magnetic Effects of Electric Current.

CBSE Case Study Questions Class 10 Science Magnetic Effects of Electric Current

Case Study – 1

Answer- It states that if we stretch thumb, forefinger or the index finger and the middle finger in such a way that they are mutually perpendicular to each other then the thumb gives the direction of the motion or the force acting on conductor, index finger gives the direction of magnetic field and the middle finger gives the direction of current.

State your observation in the galvanometer.

Explanation- Electromagnetic induction is a process by changing a magnetic field in a conductor, which induces a current in another conductor placed in nearby.

Case study: 4

If we stretched the thumb, forefinger and middle finger of our left hand so that they are mutually perpendicular to each other. If the forefinger gives the direction of magnetic field and middle finger gives the direction of electric current then the thumb gives the direction of motion or the force acting on the conductor.

The domestic electric circuit consist of red insulated cover called as live wire, wire with black insulation called as neutral wire and the wire with green insulation is called as Earth wire. We know that fuse is connected in series with the circuit to prevent the damaging of electrical appliances and circuit from overloading. Overloading occurs when live wire and the neutral wire comes in direct contact with each other. Because of which current through the circuit increases suddenly. Also, overloading may occurs because of connecting many appliances to a single socket. The Earth wire which is green in colour is connected to a metal plate deep in the earth near the house. This type of safety measure is used in appliances like electric press, toaster, table fan, refrigerator etc. The Earth wire is gives low resistance conducting path for the electric current. In this way it protects us from severe electric shock.

4) What is the main purpose of using fuse in electric circuit?

4) Because of Joule’s heating effect the heat produced causes the fuse to melt and to break the circuit. And thereby protect the circuit and electric appliances.

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Class 10 Electricity and Magnetism Notes

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Direct Current and Alternating Current

Direct current.

Direct Current (DC) is the electric current whose polarity doesn't change with time. Such a current has a fixed magnitude and a fixed direction (polarity).

Sources of Direct Current are Batteries and DC generators.

The current-time graph of a Direct Current is shown below:

Alternating Current

Alternating Current (AC) is the electric current whose polarity changes with time. Such a current has a variable magnitude and a variable direction (polarity).

Sources of Alternating Current are AC generators and Dynamos. Alternating Current is used in Transformers.

The current-time graph of an Alternating Current is shown below:

Magnetic Effect of Electric Current

When an electric current flows through a conductor, it generates a magnetic field around it. This phenomenon is known as the magnetic effect of electric current and was first observed by Hans Christian Orsted in 1820.

Orsted's experiment demonstrated that a compass needle placed near a current-carrying wire deflects, indicating the presence of a magnetic field.

Magnetic Field Around a Current-Carrying Conductor and Solenoid

The magnetic field around a straight current-carrying conductor can be determined using Ampere's Circuital Law, which states that the line integral of the magnetic field B around any closed path is equal to μ 0 times the total current I passing through the enclosed area.

Mathematically,

\[ \rm \oint B \cdot dl = \mu_{o} I \]

For a straight conductor, the magnetic field at a distance r from the wire is:

\[ \rm B = \frac{\mu_{o} I}{2 \pi r} \]

where μ 0 is the permeability of free space (\( \rm 4 \pi \times 10^{-7} \) T m/A).

When the conductor is shaped into a coil or solenoid, the magnetic field becomes concentrated inside the coil. For a long solenoid with n turns per unit length carrying a current I, the magnetic field inside the solenoid is:

\[ \rm B = \mu_{o} n I \]

Magnetic Flux

Magnetic flux is the total magnetic field passing through a given area.

\[ \rm \phi = \vec{B} \cdot \vec{A} = B \cdot A \cdot \cos ( \theta ) \]

where B is the magnetic field strength, A is the area through which the field lines pass, and \( \rm ( \theta ) \) is the angle between the field lines and the perpendicular to the surface.

Motor Effect

It is the phenomenon where a current-carrying conductor placed within a magnetic field experiences a force.

\[ \rm F = I ( \vec{dl} \times \vec{B} ) = B I L \sin ( \theta ) \]

This force is described by Fleming's Left-Hand Rule, which states that if we position our left hand such that the thumb, forefinger, and middle finger are mutually perpendicular, the thumb indicates the direction of the force (motion), the forefinger indicates the magnetic field, and the middle finger indicates the current.

Electromagnetic Induction

Electromagnetic Induction is a current produced because of voltage production (electromotive force) due to a changing magnetic field.

Faraday's Law of Electromagnetic Induction

First law: It states that whenever there is a change in magnetic flux associated with a coil, EMF is induced in that coil.
Second law: It states that the magnitude of EMF induced in the coil is directly proportional to the rate of change of magnetic flux associated with that coil.

Mathematically, it can be expressed as

\[ \rm E = - \frac{d \phi}{dt} \]

where E is the induced EMF and \( \rm \frac{d \phi}{dt} \) is the rate of change of magnetic flux.

Dynamo and AC Generator

A dynamo is a device that converts mechanical energy into electrical energy using the principle of electromagnetic induction.

An AC generator operates on the same principle but specifically produces alternating current (AC).

In both devices, rotating a coil within a magnetic field induces an electric current in the coil.

Large Scale Sources of Electricity

Large-scale sources of electricity include power plants that utilize various forms of energy such as fossil fuels, nuclear reactions, and renewable sources like hydro, wind, and solar energy.

These plants often use turbines and generators to convert mechanical energy into electrical energy.

Alternating Current Generator

An alternating current generator, or AC generator, produces alternating current by rotating a coil within a magnetic field.

The rotation causes the direction of the induced current to reverse periodically, resulting in AC.

Transformer: Construction, Working Principle, and Types

A transformer is an electrical device that transfers electrical energy between two or more circuits through electromagnetic induction.

It consists of primary and secondary coils wound around a magnetic core.

The working principle is based on Faraday's Law of Induction, where an alternating current in the primary coil generates a changing magnetic field, which induces a current in the secondary coil.

Types of Transformers

Step-up transformer.

A step-up transformer increases the voltage from the primary coil to the secondary coil.

This type of transformer has more turns of wire on the secondary coil than on the primary coil.

Step-down Transformer

A step-down transformer decreases the voltage from the primary coil to the secondary coil.

This type of transformer has fewer turns of wire on the secondary coil than on the primary coil.

The relationship between the primary and secondary voltages and the number of turns in the coils is given by:

\[ \rm \frac{V_{S}}{V_{P}} = \frac{N_{S}}{N_{P}} \]

where: \( \rm V_{S} \) is the secondary voltage, \( \rm V_{P} \) is the primary voltage, \( \rm N_{S} \) is the number of turns in the secondary coil, \( \rm N_{P} \) is the number of turns in the primary coil.

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Case Study Questions for Class 10 Science Chapter 13 Magnetic Effect of Electric Current

Last modified on: 1 year ago
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Question 1:

A student wants to study the working of electric motor. He used a model of DC motor for electromagnetism as shown in figure.

He fixed the two ends of the coil to a pair of curved elastic metal strips. The metal strips are connected to the power supply with a rheostat.

(i) The direction of rotation of the coil, when viewed from the front by the student is (a) clockwise (b) anti clockwise (c) First half clockwise and other half anti-clockwise (d) First half anti-clockwise and other half clockwise

(ii) The student is still testing on the feasibility of using metal strips in the model. His observations are given below. I. As the current reverses in the coil for every half turn, the coil rotates in one direction. II. The speed of rotation of the motor is increased, if the value of current is increased. III. The direction of force, acting on the coil is given by Fleming’s left hand rule. IV. The coil continues its rotation in magnetic field even if there is no current in circuit. The correct observations made by him are (a) I, II and IV (b) II, III and IV (c) I, II and III (d) II and III

(iii) Commercial electric motors do not use (a) an electromagnet to rotate the armature. (b) effectively large number of turns of conducing wire in the current carrying coil (c) a permanent magnet to rotate the armature (d) a soft iron core on which the coil is wound

(iv) Which one of the following is true about electric motor? (a) It converts electrical energy into mechanical energy (b) It converts mechanical energy into electrical energy (c) It converts magnetic energy into electric energy (d) It converts electric energy into magnetic energy

(v) The direction of magnetic field at a place is coming out of the paper. A wire whose direction of current flow is as shown in the figure is placed there. In which direction is the force due to the magnetic field experienced by the wire?

(a) North-West direction (b) North direction (c) South-West direction (d) South-East direction

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Forest and Wildlife Resources Class 10 Case Study Social Science Geography Chapter 2

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Last Updated on September 3, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 10 social science. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 10 social science. In this article, you will find case study for CBSE Class 10 Social Science Geography Chapter 2 Forest and Wildlife Resources. It is a part of Case Study Questions for CBSE Class 10 Social Science Series.

	Forest and Wildlife Resources
	Case Study Questions
	Competency Based Questions
	CBSE
	10
	Social Science – Geography
	Contemporary India-II
	Resources and Development
	Class 10 Studying Students
	Yes
	Mentioned

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Table of Contents

Case Study Questions on Forest and Wildlife Resources Class 10

Read the following passage and answer the questions:

Nature worship is an age old tribal belief based on the premise that all creations of nature have to be protected. Such beliefs have preserved several virgin forests in pristine form called Sacred Groves (the forests of God and Goddesses). These patches of forest or parts of large forests have been left untouched by the local people and any interference with them is banned.

Certain societies revere a particular tree which they have preserved from time immemorial. The Mundas and the Santhal of Chota Nagpur region worship mahua (Bassia latifolia) and kadamba (Anthocaphalus cadamba) trees, and the tribals of Odisha and Bihar worship the tamarind (Tamarindus indica) and mango (Mangifera indica) trees during weddings. To many of us, peepal and banyan trees are considered sacred.

Indian society comprises several cultures, each with its own set of traditional methods of conserving nature and its creations. Sacred qualities are often ascribed to springs, mountain peaks, plants and animals which are closely protected. You will find troops of macaques and langurs around many temples. They are fed daily and treated as a part of temple devotees. In and around Bishnoi villages in Rajasthan, herds of blackbuck, (chinkara), nilgai and peacocks can be seen as an integral part of the community and nobody harms them.

Q. 1. How is nature worship an age old tribal belief ? Ans. Nature worship is an age old tribal belief as it is based on the promise that all creations of nature have to be protected. Such beliefs have preserved several virgin forests in pristine form called Sacred groves. These patches of forests, have been left untouched by the local people and any interference with them is banned.

Q. 2. Which tribal societies used to worship tress during weddings? Ans. The Mundas and the Santhal of Chota Nagpur region worship mahua (Bassia latifolia) and Kadamba trees. The tribes of Odisha and Bihar worship the tamarind and mango trees during weddings.

Q. 3. Name the animals that are treated as a part of temple devotees and the community. Ans. The animals that are treated as a part of temple devotees and the community are macaques and langurs while the herds of blackbuck, nilgai and peacocks can be seen as an integral part of community in and around Rajasthan.

Power Sharing Class 10 Case Study Social Science Political Science Chapter 1
Resources and Development Class 10 Case Study Social Science Geography Chapter 1
The Making of a Global World Class 10 Case Study Social Science History Chapter 3

Nationalism in India Class 10 Case Study Social Science History Chapter 2

The rise of nationalism in europe class 10 case study social science history chapter 1, topics from which case study questions may be asked.

Examine the importance of conserving forests and wild life and their interdependency in maintaining the ecology for the sustainable development of India.
Analyse the role of grazing and wood cutting in the development and degradation
Comprehends the reasons for conservation of biodiversity in India under sustainable development.

We humans along with all living organisms form a complex web of ecological system in which we are only a part of and very much dependent on this system for our own existence. Forests play a key role in the ecological system as these are also the primary producers on which all other living beings depend.

The famous Chipko movement in the Himalayas has not only successfully resisted deforestation in several areas but has also shown that community afforestation with indigenous species can be enormously successful.

Frequently Asked Questions (FAQs) on Forest and Wildlife Resources Class 10 Case Study

Q1: what are case study questions.

A1: Case study questions are a type of question that presents a detailed scenario or a real-life situation related to a specific topic. Students are required to analyze the situation, apply their knowledge, and provide answers or solutions based on the information given in the case study. These questions help students develop critical thinking and problem-solving skills.

Q2: How should I approach case study questions in exams?

A2: To approach case study questions effectively, follow these steps: Read the case study carefully: Understand the scenario and identify the key points. Analyze the information: Look for clues and relevant details that will help you answer the questions. Apply your knowledge: Use what you have learned in your course to interpret the case study and answer the questions. Structure your answers: Write clear and concise responses, making sure to address all parts of the question.

Q3: What are the benefits of practicing case study questions from your website?

A3: Practicing case study questions from our website offers several benefits: Enhanced understanding: Our case studies are designed to deepen your understanding of historical events and concepts. Exam preparation: Regular practice helps you become familiar with the format and types of questions you might encounter in exams. Critical thinking: Analyzing case studies improves your ability to think critically and make connections between different historical events and ideas. Confidence: Practicing with our materials can boost your confidence and improve your performance in exams.

Q4: What do you know about ‘Permanent forest estates’?

A4: Reserved and protected forests are also called as ‘Permanent forest estates’. These forest estates are maintained for the purpose of producing timber and other forest produce and for other protective reasons.

Q5: What is the main reason for the depletion of flora and fauna?

A5: Insensitivity to our environment is the main reason for the depletion of flora and fauna.

Q6: What is flora and fauna?

A6: Plants of particular region or period are referred to as flora. Species of animals of particular region or period are referred as fauna.

Q7: Why is it necessary to increase the area of forest in India?

A7: It is necessary to increase the area of forest in India due to the following reasons: (i) Forests play a key role in the ecological systems these are the primary producers on which all other living beings depend. (ii) Many forest dependent communities directly depends on them for food, drink, medicine, culture, spirituality, etc. (iii) Forest provide us timber. (iv) Forests also provide bamboo, wood for fuel, grass, charcoal, fruits, flowers, etc.

Q8: What is Joint Forest Management Programme? Which was the first state to adopt this programme?

A8: A programme which involves local communities in the management and restoration of degraded forests is called Joint Forest Management Programme. It involves local communities and land managed by forest department. Its major purpose is to protect the forests from encroachments, grazing, theft and fire and also to improve the forests in accordance with an approved Joint Forest Management Plan. This programme was first adopted in 1988 by the state of Odisha.

Q9: Which agency manages forests in India? Name three broad categories in which the forests are classified.

A9: The forests in India are owned and managed by the government through the Forest Department. They are classified under the following categories: (i) Reserved Forests (ii) Protected Forests (iii) Unclassed Forests

Q10: Explain the role of human in resource development.

A10: Human is at the centre of resource development. Actually all resources become resources only when they are put to use by humans. It is human who makes natural things usable with the help of technology. Had no technology been there, development would not have been possible. There are regions where natural resources are in abundance but the regions are not developed, e.g., Africa. But if humans are developed, they make the region developed with technology, e.g., Japan.

Q11: Are there any online resources or tools available for practicing “ Forest and Wildlife Resources” case study questions?

A11: We provide case study questions for CBSE Class 10 Social Science on our website . Students can visit the website and practice sufficient case study questions and prepare for their exams.

Forest and Wildlife Resources Class 10 Case Study Social Science Geography Chapter 2

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Case Study Questions Class 10 Science Electricity
CBSE Class 10 Case Study Questions Class 10 Science Chapter 12 Electricity. Electricity Case Study Questions with Solution. 100% FREE Exercise & Practice for CBSE, NCERT and ICSE ... In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks. ...
Case Study Questions Class 10 Science Chapter 12 Electricity
Case Study/Passage Based Questions. Question 1: The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire.
Electricity Class 10 Case Study Questions Science Chapter 12
Hello students, we are providing case study questions for class 10 science. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 10 science.
Case Study Questions for Class 10 Science Chapter 12 Electricity
Case Study Questions for Class 10 Science Chapter 12 Electricity. Question 1: The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule (as shown in figure). Actually, Joule represents a very small quantity of energy and ...
CBSE 10th Science Electricity Case Study Questions With Solution 2021
CBSE 10th Standard Science Subject Electricity Case Study Questions With Solution 2021. 10th Standard CBSE. Reg.No. : Science. Time : 00:30:00 Hrs. Total Marks : 16. The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow ...
CBSE 10th Science Electricity Case Study Questions 2021
CBSE 10th Standard Science Subject Electricity Case Study Questions 2021 Answer Keys. (i) (c) : The equivalent resistance in the parallel combination is lesser than the least value, of the individual resistance. (iii) (a): All the three resistors are in parallel. (iv) (a): Voltage is same across each resistance.
Class 10 Science: Case Study Chapter 12 Electricity PDF Download
Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity. Case Study/Passage Based Questions. Question 1: The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way ...
Case Study Chapter 12 Electricity
Please refer to Chapter 12 Electricity Case Study Questions with answers provided below. We have provided Case Study Questions for Class 10 Science for all chapters as per CBSE, NCERT and KVS examination guidelines. These case based questions are expected to come in your exams this year.
Class 10 Science Chapter 11 Case Based Questions
The Case Based Questions: Electricity is an invaluable resource that delves deep into the core of the Class 10 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
Electricity Case Study Based Questions Class 10
Join Telegram For Exclusive Content. Students who are studying in CBSE class 10 board, need to get the knowledge about the Electricity Case Study Based Questions. Case based questions are generally based on the seen passages from the chapter Electricity. Through solving the case based questions, students can understand each and every concept.
Top 15 Case Based Questions
Join Our Free Batch for Class 10: https://sankalpbharat.page.link/mFvZ Playlists & PDFs:- https://linktr.ee/sbfoundation In this video, ️ Class: 10th ️ Su...
Class 10 Science Chapter 12 Electricity
Class 10 Chapter 12 Electricity as the name suggests, covers everything about electricity in detail. The constitution of electricity, the flow of electricity in the circuit, how electricity can be regulated, and much more. ... Below, we have provided Class 10 Science Support Materials that cover Case Study-based questions from the various ...
Question 1
Answer: Factors affecting resistance of a conductor are: Since all the factors are mentioned in the options. So, the correct answer is (iv). Observe the graph and answer any four questions from (a) to (e). V-I graph for a conductor is as shown in figure.Question 1 (a) (a) What do you infer from this graph?
CBSE Class 10 Science Chapter Wise Important Case Study Questions
As mentioned above, Case Based Questions will carry a total of 12 marks i.e about 15 percent of the total CBSE Class 10 Science marks will come from the Case Study Based Question.
NCERT Exemplar Class 10 Science Solutions Chapter 12
This page has answers to 18 MCQs, 11 short answer questions and 7 long answer questions. To help students grasp all the concepts clearly and in-depth, we are offering free NCERT Exemplar for Class 10 Science Chapter 12 here. These exemplars will enable students to learn the correct answers to all the questions given at the end of the chapter.
Electricity Class 10 Questions and Answers
Question 27: Calculate the total number of electrons constituting one coulomb of charge. Answer: We know, The Charge of an electron = 1.6 × 10-19 C. According to the concept of charge quantisation, Q = nqe, where we suppose 'n' is the number of electrons and similarly 'qe' is the Charge of the electron.
Electricity Class 10 Important Questions and Answers
Answer: (a) Electric bulbs are generally filled with some inert gas like nitrogen or argon. This enables to prolong the life of the filament of electric bulb. (b) Here radius of wire r = 0.01 cm = 0.01 × 10 -2 m, resistance R = 10 Ω and resistivity ρ = 50 × 10 -8 Ω/m. 33. (a)Define electric power.
Case Study Questions for Class 10 Science Chapter 12 Electricity
Case Study Questions for Class 10 Science Chapter 12 Electricity - Free download as Word Doc (.doc / .docx), PDF File (.pdf), Text File (.txt) or read online for free.
Case Study Questions Class 10 Science
CBSE Case Study Questions Class 10 Science Magnetic Effects of Electric Current. Case Study - 1. Andre Marie Ampere suggested that a magnet must exert an equal and opposite force on a current carrying conductor, which was experimentally found to be true. But we know that current is due to charges in motion.
Electricity Class 10 Important Questions with Answers Science Chapter
Electricity Class 10 Important Questions Very Short Answer Type. Question 1. Write the relation between electric power (P) of a device with potential difference (Volt) across it and current (amp) flowing through it. (2013) Answer: The power (P) in watts is found by multiplying the potential difference (V) in volts by the current (I) in amperes.
CBSE Class 10 Physics Electricity Case Study Questions
These tests are unlimited in nature…take as many as you like. You will be able to view the solutions only after you end the test. TopperLearning provides a complete collection of case studies for CBSE Class 10 Physics Electricity chapter. Improve your understanding of biological concepts and develop problem-solving skills with expert advice.
Electricity Case Study Based Questions Class 10
Electricity Case Study: Here, Students can read Class 10 Electricity Case Based Questions with Answers in PDF File format. Apart from this, You can download Class 10 ...
Class 10 Electricity and Magnetism Notes
In both devices, rotating a coil within a magnetic field induces an electric current in the coil. Large Scale Sources of Electricity . Large-scale sources of electricity include power plants that utilize various forms of energy such as fossil fuels, nuclear reactions, and renewable sources like hydro, wind, and solar energy.
Heron's Formula Class 9 Case Study Questions Maths Chapter 10
Reading Time: 8 minutes Last Updated on September 8, 2024 by XAM CONTENT. Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board.
Case Study Questions for Class 10 Science Chapter 13 Magnetic Effect of
Case Study Questions for Class 10 Science Chapter 13 Magnetic Effect of Electric Current Question 1: A student wants to study the working of electric motor. He used a model of DC motor for electromagnetism as shown in figure. He fixed the two ends of the coil to a pair of curved elastic metal strips. … Continue reading Case Study Questions for Class 10 Science Chapter 13 Magnetic Effect of ...
Nationalism in India Class 10 Case Study Social Science History Chapter
A2: To approach case study questions effectively, follow these steps: Read the case study carefully: Understand the scenario and identify the key points. Analyze the information: Look for clues and relevant details that will help you answer the questions. Apply your knowledge: Use what you have learned in your course to interpret the case study and answer the questions.
Power Sharing Class 10 Case Study Social Science Political Science
Case Study Questions on Power Sharing Class 10. Questions. Passage 1: Read the following passage and answer the questions: Majoritarianism is a concept which signifies a belief that the majority community should be able to rule a country in whichever way it wants, by disregarding the wishes and needs of the minority.
Exponents and Powers Class 8 Case Study Questions Maths Chapter 10
Case Study Questions on Exponents and Powers. Questions. Passage 1: In a class science teacher give some information to all students about Solar system in following manner.
Resources and Development Class 10 Case Study Social Science Geography
A2: To approach case study questions effectively, follow these steps: Read the case study carefully: Understand the scenario and identify the key points. Analyze the information: Look for clues and relevant details that will help you answer the questions. Apply your knowledge: Use what you have learned in your course to interpret the case study and answer the questions.
Forest and Wildlife Resources Class 10 Case Study Social Science
A2: To approach case study questions effectively, follow these steps: Read the case study carefully: Understand the scenario and identify the key points. Analyze the information: Look for clues and relevant details that will help you answer the questions. Apply your knowledge: Use what you have learned in your course to interpret the case study and answer the questions.

Case Study Questions Class 10 Science Chapter 12 Electricity

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Class 10 Science Chapter 12 Electricity

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Question 1 - Case Based Questions (MCQ) - Chapter 12 Class 10 - Electricity

Observe the graph and answer any four questions from (a) to (e). V-I graph for a conductor is as shown in figure.

Question 1 (a)

Question 1 (b)

Question 1 (c)

Question 1 (d)

Question 1 (e)

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CBSE Class 10 Electricity Important Questions with Solutions

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Chapter 2 - Acids, Bases and Salts

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