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Conversion of galvanometer to ammeter and voltmeter: NEET MCQs with answers

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

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easy

1. Which of the following statements is INCORRECT regarding ammeters and voltmeters?

  1. A voltmeter is always connected in parallel across the circuit element whose potential difference is to be measured.
  2. An ammeter is connected in series with the branch whose current is to be measured.
  3. The effective resistance of a voltmeter is always greater than the resistance of the galvanometer from which it is made.
  4. The effective resistance of an ammeter is always greater than the resistance of the galvanometer from which it is made.
Show answer and explanation

Answer: (D) The effective resistance of an ammeter is always greater than the resistance of the galvanometer from which it is made.

An ammeter is formed by connecting a very small shunt resistance in parallel with the galvanometer. Therefore, the equivalent resistance Req = (G × S) / (G + S) is strictly less than both G and S.

medium

2. A galvanometer of resistance 25 Ω gives full-scale deflection at 2 mA. The series resistance needed to convert it into a voltmeter of range 0 to 10 V is:

  1. 0.005 Ω
  2. 4975 Ω
  3. 5000 Ω
  4. 5025 Ω
Show answer and explanation

Answer: (B) 4975 Ω

At full scale the whole meter carries Ig, so G + R = V/Ig. R = V/Ig − G = 10/0.002 − 25 = 4975 Ω.

hard

3. A galvanometer of resistance 20 Ω gives full-scale deflection at 1 mA. The shunt resistance needed to convert it into an ammeter of range 0 to 10 A is:

  1. 0.004 Ω
  2. 2 × 10⁵ Ω
  3. 0.002 Ω
  4. 0.02 Ω
Show answer and explanation

Answer: (C) 0.002 Ω

The shunt carries I − Ig: S = IgG/(I − Ig) = 0.001 × 20/9.999 = 0.002 Ω.

medium

4. A galvanometer of resistance 100 Ω gives full-scale deflection at 5 mA. The series resistance needed to convert it into a voltmeter of range 0 to 20 V is:

  1. 3900 Ω
  2. 4100 Ω
  3. 4000 Ω
  4. 0.025 Ω
Show answer and explanation

Answer: (A) 3900 Ω

At full scale the whole meter carries Ig, so G + R = V/Ig. R = V/Ig − G = 20/0.005 − 100 = 3900 Ω.

easy

5. When a shunt of resistance S is connected across a galvanometer, the fraction of the total current passing through the galvanometer coil is:

  1. G / (S + G)
  2. S / G
  3. (S + G) / S
  4. S / (S + G)
Show answer and explanation

Answer: (D) S / (S + G)

Since the galvanometer and shunt are connected in parallel, the potential difference across both is the same: I_g × G = I_s × S = (I - I_g) × S. Expanding gives I_g(G + S) = I × S, which yields the fraction I_g / I = S / (S + G).

hard

6. A galvanometer of resistance 20 Ω gives full-scale deflection at 2 mA. The shunt resistance needed to convert it into an ammeter of range 0 to 5 A is:

  1. 49980 Ω
  2. 50000 Ω
  3. 0.008 Ω
  4. 0.08 Ω
Show answer and explanation

Answer: (C) 0.008 Ω

The shunt carries I − Ig: S = IgG/(I − Ig) = 0.002 × 20/4.998 = 0.008 Ω.

medium

7. A galvanometer of resistance 50 Ω gives full-scale deflection at 10 mA. The series resistance needed to convert it into a voltmeter of range 0 to 5 V is:

  1. 500 Ω
  2. 550 Ω
  3. 450 Ω
  4. 0.1 Ω
Show answer and explanation

Answer: (C) 450 Ω

At full scale the whole meter carries Ig, so G + R = V/Ig. R = V/Ig − G = 5/0.01 − 50 = 450 Ω.

medium

8. A galvanometer has 50 divisions and its figure of merit is 2 × 10⁻⁴ A/division. If its internal resistance is 60 Ω, the resistance that must be connected in series to convert it into a voltmeter of range 0 to 15 V is:

  1. 1500 Ω
  2. 1440 Ω
  3. 1460 Ω
  4. 1560 Ω
Show answer and explanation

Answer: (B) 1440 Ω

Full-scale deflection current I_g = (number of divisions) × (figure of merit) = 50 × (2 × 10⁻⁴ A) = 10⁻² A = 10 mA. The series resistance R = (V / I_g) - G = (15 / 0.01) - 60 = 1500 - 60 = 1440 Ω.

medium

9. A galvanometer of resistance 40 Ω gives full-scale deflection at 2 mA. The series resistance needed to convert it into a voltmeter of range 0 to 50 V is:

  1. 25000 Ω
  2. 24960 Ω
  3. 25040 Ω
  4. 0.0016 Ω
Show answer and explanation

Answer: (B) 24960 Ω

At full scale the whole meter carries Ig, so G + R = V/Ig. R = V/Ig − G = 50/0.002 − 40 = 24960 Ω.

medium

10. A galvanometer of resistance 25 Ω gives full-scale deflection at 5 mA. The series resistance needed to convert it into a voltmeter of range 0 to 20 V is:

  1. 4025 Ω
  2. 3975 Ω
  3. 4000 Ω
  4. 0.00625 Ω
Show answer and explanation

Answer: (B) 3975 Ω

At full scale the whole meter carries Ig, so G + R = V/Ig. R = V/Ig − G = 20/0.005 − 25 = 3975 Ω.

medium

11. A galvanometer of resistance 40 Ω gives full-scale deflection at 1 mA. The series resistance needed to convert it into a voltmeter of range 0 to 20 V is:

  1. 20000 Ω
  2. 0.002 Ω
  3. 19960 Ω
  4. 20040 Ω
Show answer and explanation

Answer: (C) 19960 Ω

At full scale the whole meter carries Ig, so G + R = V/Ig. R = V/Ig − G = 20/0.001 − 40 = 19960 Ω.

hard

12. A galvanometer of resistance 50 Ω gives full-scale deflection at 5 mA. The shunt resistance needed to convert it into an ammeter of range 0 to 10 A is:

  1. 99900 Ω
  2. 0.25 Ω
  3. 0.025 Ω
  4. 1 × 10⁵ Ω
Show answer and explanation

Answer: (C) 0.025 Ω

The shunt carries I − Ig: S = IgG/(I − Ig) = 0.005 × 50/9.995 = 0.025 Ω.

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