Cells, Its Combination and Kirchhoff's Law – Internal resistance: NEET MCQs with Answers

15 free questions on Cells, Its Combination and Kirchhoff's Law – Internal resistance for NEET (UG). Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Internal resistance, potential difference and emf of a cell, a combination of cells in series and parallel. Kirchhoff’s laws and their applications.

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Question 1

A cell of emf 8 V and internal resistance 1 Ω supplies a network of a 4 Ω resistor and a 12 Ω resistor joined in parallel. What are the current drawn from the cell and the total power given out by the cell, including the power lost inside it?
  1. 2 A and 12 W
  2. 2 A and 16 W
  3. 2.67 A and 21.3 W
  4. 1 A and 8 W
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Answer: (B) 2 A and 16 W
Explanation: The parallel network has resistance 4 × 12/16 = 3 Ω, and adding the internal resistance gives 4 Ω. The current is 8/4 = 2 A. The total power given out by the cell is εI = 8 × 2 = 16 W, of which 12 W goes to the network and 4 W is lost inside the cell.

Question 2

A cell of emf 6 V and internal resistance 1 Ω and a cell of emf 4 V and internal resistance 2 Ω are joined between two terminals A and B. The positive terminal of the first and the negative terminal of the second are at A; the other two terminals are at B. With no load, what are the equivalent emf and equivalent internal resistance between A and B?
  1. 8/3 V and 2/3 Ω
  2. 10 V and 3 Ω
  3. 2 V and 3 Ω
  4. 16/3 V and 2/3 Ω
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Answer: (A) 8/3 V and 2/3 Ω
Explanation: For cells across the same two points, 1/r_eq = 1/r₁ + 1/r₂ and ε_eq/r_eq = ε₁/r₁ + ε₂/r₂, with the sign of the emf reversed for a cell connected the opposite way. So r_eq = 2/3 Ω and ε_eq/r_eq = 6/1 − 4/2 = 4 V/Ω, giving ε_eq = 8/3 V. As a check, a current of (6 + 4)/3 = 10/3 A circulates through the two cells, and the voltage between A and B is 6 − 10/3 = 8/3 V.

Question 3

A cell of emf 6 V and internal resistance 1 Ω is joined in parallel with a cell of emf 3 V and internal resistance 2 Ω, positive terminal to positive terminal and negative to negative. A resistor of 4/3 Ω is connected across the pair. What is the current in the resistor, and what is the 3 V cell doing?
  1. 2.5 A; the 3 V cell takes in a small current at its positive terminal and is being charged
  2. 2.5 A; the 3 V cell supplies one-third of the current in the resistor
  3. 2.25 A; both cells send current into the resistor
  4. 2.08 A; both cells send current into the resistor
Show answer
Answer: (A) 2.5 A; the 3 V cell takes in a small current at its positive terminal and is being charged
Explanation: The combination has ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) = (12 + 3)/3 = 5 V and r_eq = r₁r₂/(r₁ + r₂) = 2/3 Ω. The current is therefore 5/(2/3 + 4/3) = 2.5 A, and the voltage across the resistor is 10/3 V. This exceeds 3 V, so the second cell carries (3 − 10/3)/2 = −1/6 A. The minus sign means current enters its positive terminal, so the first cell supplies 8/3 A.

Question 4

An ideal 12 V cell is connected across a 4 Ω resistor to form one closed loop, and an ideal 18 V cell is connected across a 9 Ω resistor to form a second closed loop. A 6 Ω resistor then joins one point of the first loop to one point of the second loop, and there is no other connection between the loops. What are the currents in the 4 Ω, 9 Ω and 6 Ω resistors respectively?
  1. 30/19 A in each resistor
  2. 3 A, 2 A and zero
  3. 3 A, 2 A and 5 A
  4. 3 A, 2 A and 1 A
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Answer: (B) 3 A, 2 A and zero
Explanation: Any charge entering a loop through the 6 Ω resistor would have no second path by which to return, so the junction rule makes the current in that resistor zero. It is not part of any closed loop. Each loop then behaves independently, giving 12/4 = 3 A and 18/9 = 2 A.

Question 5

A healthy battery of emf 12 V and internal resistance 0.02 Ω is connected by thick leads of negligible resistance to a run-down battery of emf 10 V and internal resistance 0.03 Ω, positive terminal to positive terminal and negative to negative. No load is connected. What are the current in the leads and the common terminal voltage?
  1. 440 A and 3.2 V
  2. 40 A and 11.2 V
  3. 40 A and 10.8 V
  4. 100 A and 11.0 V
Show answer
Answer: (B) 40 A and 11.2 V
Explanation: With like terminals joined, the two emfs oppose each other around the loop, so the net emf is 12 − 10 = 2 V. The loop resistance is 0.02 + 0.03 = 0.05 Ω, giving I = 2/0.05 = 40 A. The healthy battery discharges, so its terminal voltage is 12 − 40 × 0.02 = 11.2 V. The run-down battery is being charged, and 10 + 40 × 0.03 = 11.2 V agrees.

Question 6

A car battery has emf 12 V and internal resistance 0.02 Ω. With only the headlamps on, it supplies 5 A. When the starter motor is switched on as well, the total current from the battery rises to 100 A. Treating the headlamp resistance as constant, by about what percentage does the headlamp power fall?
  1. About 29%
  2. About 50%
  3. About 3%
  4. About 16%
Show answer
Answer: (A) About 29%
Explanation: Terminal voltage is V = E − Ir, so it is 12 − 5 × 0.02 = 11.9 V with the lamps alone and 12 − 100 × 0.02 = 10 V with the starter on. For a fixed resistance, power is proportional to V². The new power is (10/11.9)² ≈ 0.71 of the old value, a fall of about 29%. The large current through the internal resistance is what dims the lamps.

Question 7

An ideal voltmeter across a cell reads 2.0 V with nothing else connected. When a 9 Ω resistor is joined across the cell, the voltmeter reads 1.8 V. What are the internal resistance of the cell and the power wasted inside it?
  1. 1 Ω; 0.04 W
  2. 10 Ω; 0.4 W
  3. 0.5 Ω; 0.02 W
  4. 1 Ω; 0.36 W
Show answer
Answer: (A) 1 Ω; 0.04 W
Explanation: The open-circuit reading is the emf, 2.0 V. With the resistor connected, the current is 1.8/9 = 0.2 A. The drop inside the cell is 2.0 − 1.8 = 0.2 V, so r = 0.2/0.2 = 1 Ω. The power wasted inside is I²r = 0.04 W.
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Question 8

A 12 V battery has an internal resistance of 0.4 Ω. What is the potential difference across its terminals in these three situations, taken in order: delivering a current of 5 A to a load, being charged with a current of 5 A, and with no current flowing?
  1. 12 V, 14 V, 10 V
  2. 10 V, 14 V, 12 V
  3. 10 V, 10 V, 12 V
  4. 14 V, 10 V, 12 V
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Answer: (B) 10 V, 14 V, 12 V
Explanation: While delivering current, V = ε − Ir = 12 − 5 × 0.4 = 10 V. While being charged, current is forced through the cell against its emf, so V = ε + Ir = 14 V. With no current, there is no drop across the internal resistance, so the terminal voltage equals the emf, 12 V.

Question 9

An ideal 12 V cell drives a series circuit of a 4 Ω resistor and an 8 Ω resistor, and X is the junction between them. The other end of the 8 Ω resistor is connected to the negative terminal of the cell. An extra branch, made of an open switch S, a 2 Ω resistor and an ideal 8 V cell in series, is joined across the 8 Ω resistor, with the positive terminal of the 8 V cell towards X. What happens to the power dissipated in the 4 Ω resistor when S is closed?
  1. It falls by 3 W
  2. It rises by 5 W
  3. It rises by 2 W
  4. It stays at 4 W
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Answer: (D) It stays at 4 W
Explanation: With S open, the current is 12/12 = 1 A, so the potential difference across the 8 Ω resistor is 8 V with X at the higher potential. The emf of the extra branch exactly balances this difference, so no current flows through that branch when S is closed. The main current remains 1 A, and the power in the 4 Ω resistor stays at 1² × 4 = 4 W.

Question 10

A cell of emf 6 V and internal resistance 1 Ω supplies a 6 Ω resistor and a 3 Ω resistor joined in parallel with each other. What fraction of the total power produced by the cell is lost inside the cell?
  1. 1/4
  2. 2/3
  3. 1/2
  4. 1/3
Show answer
Answer: (D) 1/3
Explanation: The parallel combination has a resistance of (6 × 3)/(6 + 3) = 2 Ω, so the current is 6/(2 + 1) = 2 A. The total power produced is 6 × 2 = 12 W. The power lost in the internal resistance is 2² × 1 = 4 W, which is 4/12 = 1/3 of the total.

Question 11

A single closed loop is traced in this order: point P, an ideal 18 V cell entered at its negative terminal and left at its positive terminal, a 4 Ω resistor, point Q, an ideal 6 V cell entered at its positive terminal and left at its negative terminal, an 8 Ω resistor, and back to P. Which point is at the higher potential, and by how much?
  1. Q is higher by 22 V
  2. P is higher by 14 V
  3. Q is higher by 14 V
  4. Q is higher by 12 V
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Answer: (C) Q is higher by 14 V
Explanation: The cells oppose each other, so the net emf is 18 − 6 = 12 V and the current is 12/(4 + 8) = 1 A in the sense of the 18 V cell. Going from P to Q, the potential rises by 18 V across the cell and falls by 1 × 4 = 4 V across the resistor. Hence V_Q − V_P = 14 V. The other path gives the same result: from Q, a 6 V drop and an 8 V drop bring the potential down by 14 V to P.

Question 12

Which one of the following changes lowers the internal resistance of an electrolytic cell?
  1. Using the cell for a long time
  2. Moving the electrodes farther apart
  3. Cooling the electrolyte
  4. Increasing the area of the electrodes in contact with the electrolyte
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Answer: (D) Increasing the area of the electrodes in contact with the electrolyte
Explanation: The electrolyte between the electrodes acts as a conductor, so a larger area gives it a wider path and a lower resistance. A greater electrode separation lengthens that path and raises the resistance. Cooling makes the ions less mobile, and internal resistance also grows as a cell ages with use.

Question 13

The external resistance connected to a cell is trebled from R to 3R. As a result, the potential difference across the terminals of the cell rises from 6 V to 9 V. What is the emf of the cell?
  1. 10.5 V
  2. 15 V
  3. 18 V
  4. 12 V
Show answer
Answer: (D) 12 V
Explanation: The terminal voltage is V = ER/(R + r). Let x = r/R. Then E = 6(1 + x) and E = 9(1 + x/3) = 9 + 3x. Equating gives 6 + 6x = 9 + 3x, so x = 1 and E = 12 V.

Question 14

Five identical cells are joined in series and connected to a rheostat. As the rheostat is varied, the terminal voltage of the combination falls linearly with the current, from 7.5 V at negligible current to 3.0 V at 1.5 A. What are the emf of each cell and the current at which the power delivered to the rheostat is maximum?
  1. 1.5 V and 0.75 A
  2. 1.5 V and 2.5 A
  3. 1.5 V and 1.25 A
  4. 7.5 V and 1.25 A
Show answer
Answer: (C) 1.5 V and 1.25 A
Explanation: At negligible current the terminal voltage equals the total emf, so 5E = 7.5 V and E = 1.5 V. The slope of the line gives the total internal resistance: (7.5 − 3.0)/1.5 = 3 Ω. Power in the rheostat is maximum when its resistance equals 3 Ω, so I = 7.5/(3 + 3) = 1.25 A.

Question 15

Two cells, each of emf E, have internal resistances 1.5 Ω and 4 Ω. They are joined in series, aiding each other, and connected to an external resistor R. For what value of R does the terminal potential difference of one cell become zero, and which cell is it?
  1. 4 Ω, the cell of internal resistance 1.5 Ω
  2. 2.5 Ω, the cell of internal resistance 4 Ω
  3. 2.5 Ω, the cell of internal resistance 1.5 Ω
  4. 5.5 Ω, the cell of internal resistance 4 Ω
Show answer
Answer: (B) 2.5 Ω, the cell of internal resistance 4 Ω
Explanation: The current is I = 2E/(R + 5.5). For a cell of internal resistance r, V = E − Ir = 0 requires 2r = R + 5.5, so R = 2r − 5.5. With r = 4 Ω this gives R = 2.5 Ω, while r = 1.5 Ω would need a negative R, which is impossible. Only the cell with the larger internal resistance can have zero terminal voltage.

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