WBCHSE Class 12 Physics Electric Energy And Power Multiple Choice Questions

Electric Energy And Power Multiple Choice Question And Answers

Question 1. Which quantity expresses the work done by an electrical machine?

  1. VI
  2. VIt
  3. I2R
  4. \(\frac{I^2 R t}{J}\)

Answer: 2. VIt

Question 2. A-s is a unit of

  1. Emf
  2. Energy
  3. Power
  4. Charge

Answer: 4. Charge

Question 3. Which one is the unit of power?

  1. A.s
  2. W.half
  3. \(\frac{\mathrm{A}^2}{\Omega}\)
  4. A².Ω

Answer: 4. A².Ω

Question 4. Two bulbs are marked 220V- 100 W and 110V-100W. The ratio of the resistances of the two bulbs is

  1. 1:4
  2. 1:2
  3. 2:1
  4. 4:1

Answer: 4. 4: 1

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 5. What is the power of the combination?

Class 12 Physics Unit 2 Current Electricity Chapter 3 Electric Energy And Power Multiple Choice Question And Answers Question 5

  1. 50W
  2. 20W
  3. 10W
  4. 8W

Answer: 4. 8W

Question 6. What is the power of the combination?

Class 12 Physics Unit 2 Current Electricity Chapter 3 Electric Energy And Power Multiple Choice Question And Answers Question 6

  1. 2.5 W
  2. 2.0W
  3. 0.5W
  4. 0.8W

Answer: 1. 2.5W

Question 7. A fuse wire of length Z and radius r is connected in series with a circuit. The safe current that can pass through the circuit is proportional to

  1. r3
  2. r3/2
  3. l-3/2
  4. l-1

Answer: 2. r3/2

Question 8. The ratio of the resistances of 100 W and 40 W bulbs of the same rated voltage is

  1. 2:5
  2. 5: 2
  3. 25:4
  4. 4:25

Answer: 1. 2: 5

Question 9. If power dissipated in the 9 H resistor in the circuit shown is 36 W, the potential difference across the 2Ω resistor is

Class 12 Physics Unit 2 Current Electricity Chapter 3 Electric Energy And Power Multiple Choice Question And Answers Question 9

  1. 4 V
  2. 8 V
  3. 10 V
  4. 2 V

Answer: 3. 10 V

Question 10. 1 BOT unit is equal to

  1. 3600 W
  2. 3600 J
  3. 3.6 x 106W
  4. 3.6 x 10-6 J

Answer: 4. 3.6 x 106 J

Question 11. BOT unit is a unit of

  1. Charge
  2. Energy
  3. Power
  4. Efficiency

Answer: 2. Energy

Question 12. 1 BOT unit is equal to

  1. lW.h
  2. 1000W.h
  3. \(\frac{1}{1000}\)
  4. \(\frac{1}{1000}\)V.A

Answer: 2. 1000W.h

Question 13. An electric conductor has a resistance R and its terminal potential difference is V. If charge Q passes through it in time t, then the amount of electrical energy transmitted is

  1. QV
  2. \(\frac{Q^2 R}{t}\)
  3. \(\frac{V^2 t}{R}\)
  4. \(\frac{Q V}{t}\)

Answer:

1. QV

2. \(\frac{Q^2 R}{t}\)

3. \(\frac{V^2 t}{R}\)

Question 14. Each of the two electric lamps has a voltage rating of V and a watt rating of P. If they are joined in series and are connected to a supply line of V volt, then,

  1. Current through each = \(\frac{P}{V}\)
  2. Current through each = \(\frac{P}{2V}\)
  3. Power consumed by each = \(\frac{P}{2}\)
  4. Power consumed by each = \(\frac{P}{4}\)

Answer:

2. Current through each = \(\frac{P}{2V}\)

4. Power consumed by each = \(\frac{P}{4}\)

Question 15. A 10 kn carbon resistor has a watt rating of 1 W, i.e., it may be damaged if the power consumed exceeds 1 W. Which of the following currents are safe for the resistor?

  1. 5mA
  2. 8mA
  3. 12mA
  4. 20mA

Answer:

1. 5mA

2. 8mA

Question 16. The powers consumed in the resistances \(\frac{R}{3}\), R and 2R are P1 P2 and P3 respectively. Then,

Class 12 Physics Unit 2 Current Electricity Chapter 3 Electric Energy And Power Multiple Choice Question And Answers Question 16

  1. P1<P2
  2. P1<P3
  3. P2 > P3
  4. P2<P3

Answer:

1. P1< P2

3. P2<P3

Question 17. A fuse wire has a length of l and a radius of r. The maximum safe. current through it is

  1. Proportional to r2
  2. Proportional to r3/2
  3. Inversely proportional to l
  4. Independent of l

Answer:

2. Proportional to r3/2

4. Independent of l

Question 18. For an electric lamp, ‘the voltage rating is V and the watt rating is P’- this means that the lamp will be brightest when its terminal potential difference is V volt, and in that condition, its power consumed will be P watt. The current through the’lamp is I = \(\frac{P}{V}\) and its resistance R = \(\frac{V}{l}\). A voltage higher than V is never applied between its terminals, as in that case the lamp may get damaged. On the other hand, the supply voltage may fall below V for different reasons say, the terminal voltage then is V'(V’ < V); the brightness of the lamp diminishes in this case. The resistance of the lamp, however, may be assumed to be unchanged, and it may be said that the current through the lamp, \(I^{\prime}=\frac{V^{\prime}}{R}\) and the power consumed, P’ = V’l’. The voltage rating of more than one lamp is V, and their watt ratings are P1, P2…. (suppose). When a voltage V is applied across a parallel combination of them, each lamp attains a terminal potential difference V. As a result, each lamp glows with its maximum brightness. On the other hand, when the same voltage V is applied at the two ends of their series combination, the lamps share the voltage V among themselves. As no lamp attains the rated terminal voltage V, none of them achieves the maximum brightness.

Class 12 Physics Unit 2 Current Electricity Chapter 3 Electric Energy And Power Multiple Choice Question And Answers Question 18

1. Two lamps have ratings of 200V-40W and 200V-60W respectively. The power consumed by their series combination, driven by a potential difference of 200 V, is

  1. 24 W
  2. 50 W
  3. 72 W
  4. 100W

Answer: 1. 24 W

2. Two lamps have ratings of 200V-40W and 200V-60W respectively. If the supply voltage across their parallel combination falls by 2% from 200 V, then the power consumed by the two lamps decreases by

  1. 1%
  2. 2%
  3. 4%
  4. 8%

Answer: 4%

3. A potential difference of 100 V is applied across the parallel combination of two electric bulbs rated at 100 V-100 W and 200 V-100 W respectively. The ratio of the power consumed by them would be

  1. 1:1
  2. 2:1
  3. 4:1
  4. 16:1

Answer: 4: 1

4. The power of an electric heater is P. This becomes P’ when the heater coil is cut to have a slightly shorter length. P’ would be related to P as

  1. P’ = P
  2. P’ > P
  3. P’ < P
  4. P’ ≠ P

Answer: P’ > P

Question 19. Let us take an electrical conductor in which the electrical energy supplied is entirely converted into heat. If, for the conductor, the terminal potential difference = V, the current through it =I, and its resistance = R, then the electrical energy consumed in time t is, W = I²Rt (from Ohm’s law, R = \(\frac{V}{I}\)). So, if the electrical and the heat energies both are expressed in joule, the heat developed in time t is H = PRt. However, if H is expressed in the conventional unit calorie, then from the law, W = JR, we may write, H = \(H=\frac{I^2 R t}{J}\) where J = mechanical equivalent of heat = 4.2 J- cal-1. The resistance R of a conducting wire depends on its material, its length l, and its area of cross-section A. The resistivity of the material of the conductor is, \(\rho=\frac{R A}{l}\) When more than one heat-producing conductor is kept in series in a circuit, the same current passes through each of them; but as their resistances are different in general, the terminal potential differences are also unequal. On the other hand, each conductor has the same terminal potential difference in a parallel combination; however, the rents through them are different.

1. The terminal potential difference and the currents through two conducting wires are both in the ratio 2: 1. The ratio of the rates of heat evolved in them is

  1. 1: 1
  2. 2: 1
  3. 4: 1
  4. 8: 1

Answer: 3. 4: 1

2. Heat is produced at the rate of 8 cal.s-1 in a uniform wire when its terminal potential difference is 10 V. What would be the rate in another wire of the same material, of the same length, but of half the diameter, for the same potential difference?

  1. 32 cal.s-1
  2. 16 cal.s-1
  3. 4 cal.s-1
  4. 2 cal.s-1

Answer: 2 cal.s-1

3. The first one of two wires, of the same material and of equal cross sections, is longer than the second. A current through their series combination produces heat in them at the rates h1 and h2, respectively. Then,

  1. h1 = h2
  2. h1>h2
  3. h1<h2
  4. h1 ≠ h2

Answer: h1>h2

Question 20. By how much will the power of an electric bulb decrease if the current drops by 0.5%?

  1. 0.25%
  2. 0.5%
  3. 1%
  4. 2%

Answer: 3. 1%

The resistance of an electric bulb can be assumed to be constant.

Power, P = I²R or, InP = 21nI + InR

∴ \(\frac{\Delta P}{P}=2 \frac{\Delta I}{I}\)

= 2 x 0.5 %

= 1%

The option 3 is correct

Question 21. Consider the circuit. The value of the resistance X for which the thermal power generated in it is practically independent of small variation of its resistance is

Class 12 Physics Unit 2 Current Electricity Chapter 3 Electric Energy And Power Multiple Choice Question And Answers Question 21

  1. X = R
  2. X = \(\frac{R}{3}\)
  3. X = \(\frac{R}{2}\)
  4. X = 2R

Answer: 3. X = \(\frac{R}{2}\)

⇒ \(i=\frac{E}{\left(R+\frac{R X}{R+X}\right)}[i=\text { main current in the circuit }]\)

⇒ \(V_{R X}=\frac{E \frac{R X}{R+X}}{\left(R+\frac{R X}{R+X}\right)}=\frac{E X}{(R+2 X)}\) [potential difference across two terminals of R and X]

⇒ \(\left.P_X=\frac{V_{R X}^2}{X}=\frac{E^2 X}{(R+2 X)^2} \text { [power consumed by } X\right]\)

∴ \(\frac{d P_X}{d X}=E^2 \frac{(R-2 X)}{(R+2 X)^3} \text { or, } d P_X=\frac{E^2(R-2 X)}{(R+2 X)^3} d X\)

Now if X = \(\frac{R}{2}\), then for any value of dX, dPx = 0

The option 3 is correct.

Question 22. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80, and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be

  1. 8A
  2. 10 A
  3. 12 A
  4. 14 A

Answer: 3. 12 A

15 x 40 + 5 x 100 + 5 x 80 + 1000

= 220 x I

∴ I = 11.36 A ≈ 12 A

The option 3 is correct.

Question The two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volts and the average resistance per km is 0.5Ω. The power loss in the wire is

  1. 19.2W
  2. 19.2 kW
  3. 19.2 J
  4. 12.2 kW

Answer: 2. 19.2 kW

Total resistance =0.5 X 150

= 75Ω

Total voltage drop = 8 x 150

= 1200 V

∴ Power dissipated

⇒ \(\frac{V^2}{R}=\frac{(1200)^2}{75}=19200 \mathrm{~W}=19.2 \mathrm{~kW}\)

The option 2. is correct.

Question 23. The charge flowing through a resistance R varies with time t as Q = at – bt², where a and b are positive constants. The total heat produced in R is

  1. \(\frac{a^3 R}{3 b}\)
  2. \(\frac{a^3 R}{2 b}\)
  3. \(\frac{a^3 R}{b}\)
  4. \(\frac{a^3 R}{6 b}\)

Answer: 4. \(\frac{a^3 R}{6 b}\)

Current passing through resistance R,

⇒ \(I=\frac{d Q}{d t}=a-2 b t\)

Now, I will be zero when \(a-2 b t=0 \text { or, } t=\frac{a}{2 b}\)

∴ Total heat produced in R

⇒ \(\int_0^t I^2 R d t=\int_0^{\frac{a}{2 b}}(a-2 b t)^2 R d t\)

⇒ \(\int_0^{\frac{a}{2 b}}\left(a^2 R+4 b^2 R t^2-4 a b R t\right) d t\)

⇒ \(\left[a^2 R t+\frac{4 b^2}{3} R t^3-\frac{4 a b R}{2} t^2\right]_0^{\frac{a}{2 b}}=\frac{a^3 R}{6 b}\)

The option 4 is correct.

Leave a Comment