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Kirchhoff’s laws, Wheatstone bridge and metre bridge: NEET MCQs with answers

45 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. Five identical resistors, each of resistance R = 10 Ω, are connected to form a bridge network ABCDA. Resistors are connected along AB, BC, CD, DA and a central resistor is connected across BD. A cell of emf 15 V and negligible internal resistance is connected across the terminals A and C. What is the current delivered by the cell?

  1. 1.5 A
  2. 3.0 A
  3. 0.75 A
  4. 2.0 A
Show answer and explanation

Answer: (A) 1.5 A

The network forms a balanced Wheatstone bridge since the ratio of resistances in the adjacent arms is equal (R/R = R/R). Therefore, no current flows through the central resistor connected across BD. The equivalent resistance between A and C is (R + R) in parallel with (R + R), which equals 2R / 2 = R = 10 Ω. The current delivered by the cell is I = V / R_eq = 15 V / 10 Ω = 1.5 A.

medium

2. A Wheatstone bridge with arms P, Q, R and S is balanced (no galvanometer current) when:

  1. P/Q = R/S
  2. P + Q = R + S
  3. PQ = RS always
  4. P = Q = R = S only
Show answer and explanation

Answer: (A) P/Q = R/S

At balance, the potentials at the galvanometer terminals are equal, giving P/Q = R/S.

NCERT Class 12

hard

3. A circuit consists of two parallel branches connected across terminals A and B. Branch 1 contains a cell of emf E₁ = 12 V with internal resistance r₁ = 2 Ω. Branch 2 contains a cell of emf E₂ = 6 V with internal resistance r₂ = 1 Ω connected in the same polarity (positive terminals connected together). The potential difference V_A - V_B across the parallel combination is:

  1. 7.5 V
  2. 9.0 V
  3. 10.0 V
  4. 8.0 V
Show answer and explanation

Answer: (D) 8.0 V

Using Millman's theorem or Kirchhoff's laws for two parallel sources, the equivalent emf (which equals the terminal potential difference on open circuit) is E_eq = (E₁/r₁ + E₂/r₂) / (1/r₁ + 1/r₂). Substituting the given values gives E_eq = (12/2 + 6/1) / (1/2 + 1/1) = (6 + 6) / (1.5) = 12 / 1.5 = 8.0 V.

medium

4. In a metre bridge experiment, resistances P and Q are in the left and right gaps, and the null point is obtained at 60 cm from the left end. What resistance must be connected in parallel with P so that the balance point shifts to the exact center (50 cm) of the wire?

  1. Q
  2. 3Q
  3. 2Q
  4. 1.5Q
Show answer and explanation

Answer: (B) 3Q

From the initial setup, P / Q = 60 / 40 = 3 / 2, so P = 1.5 Q. To bring the balance point to 50 cm, the effective resistance in the left gap must equal Q. If resistance R is shunted across P, (P × R) / (P + R) = Q. Substituting P = 1.5 Q: (1.5 Q × R) / (1.5 Q + R) = Q ⇒ 1.5 R = 1.5 Q + R ⇒ 0.5 R = 1.5 Q ⇒ R = 3 Q.

easy

5. Kirchhoff's junction rule is based on:

  1. Conservation of energy
  2. Conservation of charge
  3. Conservation of momentum
  4. Ohm's law
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Answer: (B) Conservation of charge

ΣI = 0 at junction. Current in = Current out. Based on conservation of charge.

Quick tip: Junction rule = charge conservation. Loop rule = energy conservation.

NCERT Class 12, Chapter 3, page 93

hard

6. In the circuit loop shown conceptually, two cells of emf 4 V and 8 V with internal resistances 1 Ω and 2 Ω respectively are connected in opposition in a single closed loop with an external resistor of 5 Ω. What is the potential difference across the terminals of the 4 V cell?

  1. 4.0 V
  2. 3.5 V
  3. 4.5 V
  4. 5.0 V
Show answer and explanation

Answer: (C) 4.5 V

Since the cells are in opposition, the net emf is E_net = 8 - 4 = 4 V. The total circuit resistance is R_total = 1 + 2 + 5 = 8 Ω. The current in the loop is I = 4 / 8 = 0.5 A, driven in the direction of the 8 V cell. Current therefore enters the positive terminal of the 4 V cell (it is being charged). Thus, the terminal potential difference across the 4 V cell is V = E + I · r = 4 + (0.5 × 1) = 4.5 V.

medium

7. In a metre bridge, the balance point is at 40 cm from one end with a known resistance of 30 Ω in the right gap. The unknown resistance in the left gap is:

  1. 12 Ω
  2. 20 Ω
  3. 45 Ω
  4. 30 Ω
Show answer and explanation

Answer: (B) 20 Ω

At balance R/S = l/(100 - l): R/30 = 40/60, so R = 20 Ω.

NCERT Class 12

easy

8. In a balanced Wheatstone bridge, the cell of emf E and internal resistance r is connected between terminals A and C, while a galvanometer of resistance G is connected between B and D. If the positions of the cell and the galvanometer are interchanged, which of the following statements is correct?

  1. The condition for balance changes and the bridge is no longer balanced.
  2. The bridge remains balanced, but the current drawn from the cell must become zero.
  3. The bridge remains balanced and no current flows through the galvanometer.
  4. The sensitivity of the bridge becomes infinitely large.
Show answer and explanation

Answer: (C) The bridge remains balanced and no current flows through the galvanometer.

The conjugate arms of a Wheatstone bridge can be interchanged without disturbing the balance condition. When the positions of the galvanometer and battery are swapped, the null condition (no current through the galvanometer) remains satisfied.

easy

9. At a junction in a circuit, four wires meet. Wire 1 carries a current of 2.5 A towards the junction, Wire 2 carries 1.8 A away from the junction, and Wire 3 carries 3.2 A towards the junction. The magnitude and direction of the current in Wire 4 is:

  1. 3.9 A towards the junction
  2. 3.9 A away from the junction
  3. 0.5 A away from the junction
  4. 7.5 A towards the junction
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Answer: (B) 3.9 A away from the junction

By Kirchhoff's junction rule, the total incoming current equals the total outgoing current: Σ I_in = Σ I_out. Here, 2.5 A + 3.2 A = 1.8 A + I₄, which gives 5.7 A = 1.8 A + I₄, so I₄ = 3.9 A. Since the incoming current exceeds the outgoing current, Wire 4 must carry 3.9 A away from the junction.

easy

10. Kirchhoff's first law (junction rule) and second law (loop rule) for electrical networks are direct consequences of respectively:

  1. Conservation of electric charge and conservation of energy
  2. Conservation of energy and conservation of electric charge
  3. Conservation of linear momentum and conservation of electric charge
  4. Conservation of electric charge and conservation of linear momentum
Show answer and explanation

Answer: (A) Conservation of electric charge and conservation of energy

Kirchhoff's junction rule states that total current entering a junction equals total current leaving, which is based on the law of conservation of charge. Kirchhoff's loop rule states that the algebraic sum of changes in potential around any closed loop is zero, which is based on the law of conservation of energy.

easy

11. In the network shown, three conductors carry currents towards a junction J: I₁ = 3 A through branch 1, I₂ through branch 2, and I₃ = 4 A through branch 3. Two conductors carry currents away from the junction: I₄ = 5 A through branch 4, and I₅ = 6 A through branch 5. What is the magnitude and direction of current I₂?

  1. 4 A away from the junction
  2. 8 A away from the junction
  3. 4 A towards the junction
  4. 8 A towards the junction
Show answer and explanation

Answer: (C) 4 A towards the junction

According to Kirchhoff's first law (junction rule), the sum of currents entering a junction must equal the sum of currents leaving it: Σ I_in = Σ I_out. Assuming I₂ enters the junction: I₁ + I₂ + I₃ = I₄ + I₅. Substituting the known values: 3 A + I₂ + 4 A = 5 A + 6 A, which gives 7 A + I₂ = 11 A. Thus, I₂ = 4 A, directed towards the junction.

easy

12. Why is the Wheatstone bridge method inherently more accurate than the simple ammeter-voltmeter method for measuring an unknown resistance?

  1. It uses direct current instead of alternating current
  2. The connecting wires in a Wheatstone bridge have zero resistance
  3. It is a null-deflection method, so the measurement does not depend on the calibration or internal resistance of the detector
  4. It can measure dynamic resistance of semiconductor diodes accurately
Show answer and explanation

Answer: (C) It is a null-deflection method, so the measurement does not depend on the calibration or internal resistance of the detector

The Wheatstone bridge operates on the null-deflection principle, where no current flows through the galvanometer at balance. Consequently, the measurement is unaffected by the internal resistance or calibration of the galvanometer, eliminating measurement loading errors present in ammeter-voltmeter methods.

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