Electrical Energy, Heating Effect of Current and Electrical Power – Electrical energy and power: NEET MCQs with Answers

13 free questions on Electrical Energy, Heating Effect of Current and Electrical Power – Electrical energy and power for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Electrical energy and power

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Question 1

Previous-year question
Which one of the following acts as a circuit protection device? [NEET 2019]
  1. conductor
  2. inductor
  3. switch
  4. fuse
Show answer
Answer: (D) fuse
Explanation: A fuse is a short piece of wire of low melting point and suitable rating placed in series with the circuit. If the current exceeds the safe value, the I²R heating melts the fuse wire and breaks the circuit, protecting the appliances and the wiring.

Question 2

Previous-year question
The charge flowing through a resistance R varies with time as Q = at − bt², where a and b are positive constants. The total heat produced in R is [NEET 2016]
  1. a³R/6b
  2. a³R/3b
  3. a³R/2b
  4. a³R/b
Show answer
Answer: (A) a³R/6b
Explanation: The current is I = dQ/dt = a − 2bt, which falls to zero at t = a/2b; the heating lasts only up to that instant. Then H = ∫₀^(a/2b) (a − 2bt)²R dt = R[−(a − 2bt)³/6b]₀^(a/2b) = a³R/6b.

Question 3

Previous-year question
Two cities are 150 km apart. Electric power is sent from one city to the other through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 Ω. The power loss in the wires is [AIPMT 2014]
  1. 19.2 W
  2. 19.2 kW
  3. 19.2 J
  4. 12.2 kW
Show answer
Answer: (B) 19.2 kW
Explanation: The current is fixed by the drop per kilometre: I = 8/0.5 = 16 A. The total line resistance is 150 × 0.5 = 75 Ω. Hence the power lost as heat is P = I²R = (16)² × 75 = 19200 W = 19.2 kW.

Question 4

A lamp is rated 50 W, 200 V and a small heater is rated 80 W, 100 V. Treating each resistance as constant, each device is connected in turn to the same 100 V supply. Which statement is correct?
  1. The lamp has the smaller resistance and draws the greater current.
  2. The lamp has the greater resistance and draws the smaller current.
  3. The lamp has the greater resistance and draws the greater current.
  4. The lamp has the smaller resistance and draws the smaller current.
Show answer
Answer: (B) The lamp has the greater resistance and draws the smaller current.
Explanation: From R = V²/P, the lamp has 200²/50 = 800 Ω and the heater has 100²/80 = 125 Ω. On the same 100 V supply the lamp draws 100/800 = 0.125 A and the heater draws 100/125 = 0.8 A. So the greater resistance goes with the smaller current.

Question 5

A cell of internal resistance 1 Ω is connected first to an external resistor of 1 Ω and then to an external resistor of 4 Ω. What percentage of the power generated by the cell reaches the external resistor in the two cases, respectively?
  1. 25% and 80%
  2. 50% and 20%
  3. 100% and 80%
  4. 50% and 80%
Show answer
Answer: (D) 50% and 80%
Explanation: The same current flows through the external resistance R and the internal resistance r. The cell generates I²(R + r), of which I²R reaches the external resistor, so the fraction is R/(R + r). This gives 1/2 = 50% for R = 1 Ω and 4/5 = 80% for R = 4 Ω.

Question 6

Three identical bulbs are connected to an ideal cell: bulb A is in the main line, in series with a parallel pair formed by bulbs B and C. The filament of bulb C now burns out, leaving an open gap in its branch. Assuming constant bulb resistance, what happens to bulbs A and B?
  1. A becomes dimmer and B becomes brighter, its power rising to 3/2 of the earlier value
  2. A becomes dimmer and B becomes brighter, its power rising to 9/4 of the earlier value
  3. A becomes brighter and B becomes dimmer, its power falling to 4/9 of the earlier value
  4. A stays unchanged and B becomes brighter, its power rising to 9/4 of the earlier value
Show answer
Answer: (B) A becomes dimmer and B becomes brighter, its power rising to 9/4 of the earlier value
Explanation: Let each bulb have resistance R and the supply be V. Before the failure the circuit resistance is 3R/2, so A carries 2V/3R and B carries V/3R. After it the resistance is 2R and both carry V/2R. A's current falls, so it dims, while B's current rises by a factor of 3/2, so its power rises by a factor of 9/4.

Question 7

Lamp A is rated 6 V, 3 W and lamp B is rated 12 V, 3 W. They are joined in series. What is the greatest supply voltage that can be applied across the pair without either lamp exceeding its rated current?
  1. 12 V
  2. 24 V
  3. 18 V
  4. 15 V
Show answer
Answer: (D) 15 V
Explanation: From R = V²/P, lamp A has R = 12 Ω and lamp B has R = 48 Ω. The rated currents, I = P/V, are 0.5 A and 0.25 A. In series both carry the same current, so it must not exceed the smaller rating, 0.25 A. The maximum supply is 0.25 × (12 + 48) = 15 V.
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Question 8

A 12 V battery of negligible internal resistance drives a 2 Ω resistor in series with a parallel section. One branch of this section is a 6 Ω resistor and the other is a 3 Ω resistor in series with a switch S. When S, initially closed, is opened, how do the powers dissipated in the 2 Ω resistor and in the 6 Ω resistor change?
  1. 2 Ω: 18 W to 4.5 W; 6 Ω: 6 W to 13.5 W
  2. 2 Ω: 4.5 W to 18 W; 6 Ω: 6 W to 13.5 W
  3. 2 Ω: 18 W to 4.5 W; 6 Ω: 13.5 W to 6 W
  4. 2 Ω: 18 W to 9 W; 6 Ω: 6 W to 13.5 W
Show answer
Answer: (A) 2 Ω: 18 W to 4.5 W; 6 Ω: 6 W to 13.5 W
Explanation: With S closed the parallel section is 2 Ω, the total is 4 Ω and the current is 3 A, so the 2 Ω resistor dissipates 18 W and the 6 Ω resistor, with 6 V across it, dissipates 6 W. With S open the total is 8 Ω and the current is 1.5 A through both resistors. The powers become 1.5² × 2 = 4.5 W and 1.5² × 6 = 13.5 W.

Question 9

A station sends 100 kW of power through a line of total resistance 2 Ω. Treating the line as a pure resistance, what is the power lost in the line when the power is sent at 1 kV and at 10 kV, respectively?
  1. 2 kW and 200 W
  2. 200 kW and 2 kW
  3. 20 kW and 2 kW
  4. 20 kW and 200 W
Show answer
Answer: (D) 20 kW and 200 W
Explanation: The line current is I = P/V, which is 100 A at 1 kV and 10 A at 10 kV. The loss is I²R, giving 10000 × 2 = 20 kW and 100 × 2 = 200 W. Raising the voltage ten times cuts the loss a hundred times, since for fixed power the loss varies as 1/V².

Question 10

A dc motor connected to a 220 V supply draws a steady current of 5 A. Its coil has a resistance of 4 Ω. Ignoring friction, what are the rate of heat production in the coil and the mechanical power output, respectively?
  1. 20 W and 1080 W
  2. 1000 W and 100 W
  3. 1100 W and 0 W
  4. 100 W and 1000 W
Show answer
Answer: (D) 100 W and 1000 W
Explanation: The electrical power input is VI = 220 × 5 = 1100 W. Heat in the coil follows Joule's law: I²R = 25 × 4 = 100 W. The remaining 1000 W is converted to mechanical work, because a motor is not a purely resistive element.

Question 11

A tungsten filament lamp runs on a constant supply voltage. With long use, tungsten evaporates unevenly and the filament becomes slightly thinner at one spot. Compared with the rest of the filament, this thin spot
  1. carries the same current but produces more heat per unit length, so it runs hotter and evaporates faster
  2. carries a larger current because the same charge is squeezed through a smaller area
  3. carries a smaller current because its resistance is higher, so it runs cooler
  4. carries the same current but produces less heat per unit length, so it runs cooler and slowly recovers
Show answer
Answer: (A) carries the same current but produces more heat per unit length, so it runs hotter and evaporates faster
Explanation: All parts of the filament are in series, so the current is the same everywhere. A thinner section has a higher resistance per unit length, and heat produced per unit length is I² times this resistance. The spot therefore becomes hotter, evaporates faster and thins further, until the filament breaks there.

Question 12

Two filament bulbs are rated 100 W, 110 V and 100 W, 220 V. They are joined in series across a 330 V supply. Assuming their resistances stay constant, which is correct?
  1. Each bulb runs at 50 W
  2. The 110 V bulb is overloaded at 144 W, and the 220 V bulb runs at 36 W
  3. The 220 V bulb is overloaded at 144 W, and the 110 V bulb runs at 36 W
  4. Each bulb runs at its rated 100 W, since the rated voltages add up to the supply voltage
Show answer
Answer: (C) The 220 V bulb is overloaded at 144 W, and the 110 V bulb runs at 36 W
Explanation: From R = V²/P, the resistances are 121 Ω and 484 Ω, so the total is 605 Ω and the current is 330/605 = 6/11 A. The voltages are 66 V across the 110 V bulb and 264 V across the 220 V bulb. The powers are 66²/121 = 36 W and 264²/484 = 144 W, so the higher-voltage bulb gets more than its rated power. Series voltages divide by resistance, not by the rated voltages.

Question 13

Four identical cells in series, each of emf 2.5 V and internal resistance 0.25 Ω, are charged from a 60 V d.c. supply. A series resistor is chosen so that the charging current is 5 A. What fraction of the power delivered by the supply is stored as chemical energy in the cells?
  1. 1/6
  2. 1/5
  3. 5/6
  4. 1/12
Show answer
Answer: (A) 1/6
Explanation: The cells give a total emf of 10 V and a total internal resistance of 1 Ω. The supply delivers 60 × 5 = 300 W. Chemical energy is stored at the rate EI = 10 × 5 = 50 W, and the remaining 250 W appears as heat in the series resistor and the internal resistance. The stored fraction is 50/300 = 1/6.

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