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Cells, emf, internal resistance and cell combinations: NEET MCQs with answers

74 questions on this topic of Current Electricity (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. When a resistor of 5 Ω is connected across the terminals of a cell, a current of 0.5 A flows. When the external resistor is replaced by a 11 Ω resistor, the current decreases to 0.25 A. The electromotive force (emf) and internal resistance of the cell are:

  1. 2.5 V and 0.5 Ω
  2. 3.0 V and 1.0 Ω
  3. 4.0 V and 2.0 Ω
  4. 3.5 V and 1.5 Ω
Show answer and explanation

Answer: (B) 3.0 V and 1.0 Ω

Using E = I(R + r): E = 0.5(5 + r) and E = 0.25(11 + r). Equating the two gives 0.5(5 + r) = 0.25(11 + r) => 2(5 + r) = 11 + r => 10 + 2r = 11 + r => r = 1.0 Ω. Substituting r = 1 Ω into the first equation gives E = 0.5(6) = 3.0 V.

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2. Two cells having the same emf E but different internal resistances r₁ and r₂ are connected in series with an external resistance R. If the potential difference across the terminals of the first cell is zero, the value of R is:

  1. (r₁ + r₂) / 2
  2. r₁ + r₂
  3. r₁ - r₂
  4. √(r₁ r₂)
Show answer and explanation

Answer: (C) r₁ - r₂

The current in the circuit is I = 2E / (R + r₁ + r₂). The terminal potential difference across the first cell is V₁ = E - I r₁ = 0, which gives E = I r₁. Substituting I gives E = [2E / (R + r₁ + r₂)] × r₁, leading to R + r₁ + r₂ = 2r₁, hence R = r₁ - r₂.

hard

3. Two cells of emf 4 V and 2 V, having internal resistances 2 Ω and 1 Ω respectively, are connected in parallel such that the positive terminal of one is connected to the negative terminal of the other. The equivalent emf of this combination is:

  1. 1 V
  2. 2 V
  3. 0 V
  4. 3 V
Show answer and explanation

Answer: (C) 0 V

When connected with opposite polarities in parallel, the equivalent emf formula is E_eq = (E₁/r₁ - E₂/r₂) / (1/r₁ + 1/r₂). Substituting the values: E_eq = (4/2 - 2/1) / (1/2 + 1/1) = (2 - 2) / (1.5) = 0 V.

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4. A cell of emf 12 V and internal resistance 0.5 Ω is connected to an external 15 Ω resistor. The terminal potential difference is:

  1. 11.6 V
  2. 0.8 V
  3. 12 V
  4. 0.387 V
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Answer: (A) 11.6 V

I = E/(R + r) = 12/15.5 A. V = IR = 12 × 15/15.5 = 11.6 V.

easy

5. Which of the following statements is INCORRECT regarding an electrochemical cell?

  1. The emf of a cell is independent of the quantity of electrolyte present inside it.
  2. The terminal potential difference across a cell is always strictly less than its emf under all operating conditions.
  3. The internal resistance of a cell increases with an increase in the concentration of the electrolyte.
  4. The internal resistance of a cell decreases with an increase in temperature of the electrolyte.
Show answer and explanation

Answer: (B) The terminal potential difference across a cell is always strictly less than its emf under all operating conditions.

When a cell is being charged by an external source, current enters through the positive terminal. Under this condition, the terminal potential difference across the cell is V = E + Ir, which is greater than its emf E. Hence, statement C is incorrect.

hard

6. A student connects three identical cells in series across an external resistor of 10 Ω, measuring a current of 0.6 A. When five identical cells of the same type are connected in series across the same 10 Ω resistor, the current becomes 0.9 A. The emf of each individual cell is:

  1. 1.8 V
  2. 2.4 V
  3. 3.0 V
  4. 1.2 V
Show answer and explanation

Answer: (B) 2.4 V

Let E be the emf and r be the internal resistance of each cell. In the first case: 3E = 0.6(10 + 3r) = 6 + 1.8r. In the second case: 5E = 0.9(10 + 5r) = 9 + 4.5r. Multiplying the first equation by 5 gives 15E = 30 + 9r, and the second by 3 gives 15E = 27 + 13.5r. Equating these: 30 + 9r = 27 + 13.5r => 3 = 4.5r => r = 3/4.5 = 2/3 Ω. Substituting r into the first equation: 3E = 6 + 1.8(2/3) = 6 + 1.2 = 7.2 V => E = 2.4 V.

medium

7. A cell of emf 6 V and internal resistance 1 Ω is connected to an external 12 Ω resistor. The terminal potential difference is:

  1. 5.54 V
  2. 6 V
  3. 0.5 V
  4. 0.462 V
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Answer: (A) 5.54 V

I = E/(R + r) = 6/13 A. V = IR = 6 × 12/13 = 5.54 V.

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8. A cell of emf 6 V and internal resistance 0.5 Ω is connected to an external 7 Ω resistor. The terminal potential difference is:

  1. 0.4 V
  2. 0.857 V
  3. 6 V
  4. 5.6 V
Show answer and explanation

Answer: (D) 5.6 V

I = E/(R + r) = 6/7.5 A. V = IR = 6 × 7/7.5 = 5.6 V.

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9. Two non-ideal cells with emfs E₁ = 4 V, E₂ = 2 V and internal resistances r₁ = 1 Ω, r₂ = 2 Ω are connected in parallel with their positive terminals connected together. The equivalent emf of this combination is:

  1. 2.67 V
  2. 3.00 V
  3. 3.33 V
  4. 3.50 V
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Answer: (C) 3.33 V

For cells in parallel with aiding polarities, equivalent emf is E_eq = (E₁/r₁ + E₂/r₂) / (1/r₁ + 1/r₂) = (4/1 + 2/2) / (1/1 + 1/2) = (4 + 1) / (3/2) = 5 / 1.5 = 10/3 V ≈ 3.33 V.

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10. A cell of emf 9 V and internal resistance 2 Ω is connected to an external 19 Ω resistor. The terminal potential difference is:

  1. 9 V
  2. 8.14 V
  3. 0.474 V
  4. 0.857 V
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Answer: (B) 8.14 V

I = E/(R + r) = 9/21 A. V = IR = 9 × 19/21 = 8.14 V.

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11. A cell of emf 1.5 V and internal resistance 1 Ω is connected to an external 13 Ω resistor. The terminal potential difference is:

  1. 1.39 V
  2. 1.5 V
  3. 0.115 V
  4. 0.107 V
Show answer and explanation

Answer: (A) 1.39 V

I = E/(R + r) = 1.5/14 A. V = IR = 1.5 × 13/14 = 1.39 V.

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12. A cell of emf 1.5 V and internal resistance 1 Ω is connected to an external 16 Ω resistor. The terminal potential difference is:

  1. 0.0882 V
  2. 1.41 V
  3. 0.0938 V
  4. 1.5 V
Show answer and explanation

Answer: (B) 1.41 V

I = E/(R + r) = 1.5/17 A. V = IR = 1.5 × 16/17 = 1.41 V.

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