Moving Coil Galvanometer: NEET MCQs with Answers

15 free questions on Moving Coil Galvanometer 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: Moving coil galvanometer, its sensitivity, and conversion to ammeter and voltmeter.

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Question 1

Previous-year question
The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is [NEET 2018]
  1. 40 Ω
  2. 25 Ω
  3. 500 Ω
  4. 250 Ω
Show answer
Answer: (D) 250 Ω
Explanation: Voltage sensitivity is current sensitivity divided by the galvanometer resistance, so G = (current sensitivity)/(voltage sensitivity) = (5 div/mA)/(20 div/V) = (5000 div/A)/(20 div/V) = 250 Ω.

Question 2

Previous-year question
A galvanometer having a resistance of 8 Ω is shunted by a wire of resistance 2 Ω. If the total current is 1 A, the part of it passing through the shunt will be [AIPMT 1998]
  1. 0.25 A
  2. 0.8 A
  3. 0.2 A
  4. 0.5 A
Show answer
Answer: (B) 0.8 A
Explanation: In a parallel combination the current divides in inverse proportion to resistance, so I_S = I × G/(G + S) = 1 × 8/(8 + 2) = 0.8 A. The remaining 0.2 A flows through the galvanometer.

Question 3

Previous-year question
A galvanometer acting as a voltmeter will have [AIPMT 2004, 2002]
  1. a low resistance in series with its coil
  2. a high resistance in parallel with its coil
  3. a high resistance in series with its coil
  4. a low resistance in parallel with its coil
Show answer
Answer: (C) a high resistance in series with its coil
Explanation: A voltmeter must draw negligible current from the circuit it measures, so its total resistance must be large. This is achieved by connecting a high resistance in series with the galvanometer coil. (In contrast, an ammeter uses a low resistance shunt in parallel.)

Question 4

Previous-year question
A galvanometer of 50 Ω resistance has 25 divisions. A current of 4 × 10⁻⁴ A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V, it should be connected with a resistance of [AIPMT 2004, 2002]
  1. 2450 Ω in series
  2. 2500 Ω in series
  3. 245 Ω in series
  4. 2550 Ω in series
Show answer
Answer: (A) 2450 Ω in series
Explanation: Full-scale current is I_g = 25 × 4 × 10⁻⁴ = 10⁻² A. For a voltmeter, V = I_g(R_e + R_g), so R_e = V/I_g − R_g = 25/10⁻² − 50 = 2500 − 50 = 2450 Ω, to be connected in series.

Question 5

Previous-year question
The resistance of an ammeter is 13 Ω and its scale is graduated for a current up to 100 A. After an additional shunt has been connected to this ammeter, it becomes possible to measure currents up to 750 A by this meter. The value of the shunt resistance is [AIPMT 2007]
  1. 20 Ω
  2. 2 Ω
  3. 0.2 Ω
  4. 2 kΩ
Show answer
Answer: (B) 2 Ω
Explanation: The potential difference across the ammeter and shunt is the same: i_aR = (i − i_a)S. Hence S = i_aR/(i − i_a) = (100 × 13)/(750 − 100) = 1300/650 = 2 Ω.

Question 6

Previous-year question
A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be [AIPMT 2008]
  1. 5050 Ω
  2. 5550 Ω
  3. 6050 Ω
  4. 4450 Ω
Show answer
Answer: (D) 4450 Ω
Explanation: Initially I = 3/(50 + 2950) = 1 mA gives 30 divisions, so 20 divisions corresponds to I′ = (20/30) × 1 = 2/3 mA. The total resistance needed is 3/(2/3 × 10⁻³) = 4500 Ω. Subtracting the galvanometer resistance, the series resistance is 4500 − 50 = 4450 Ω.

Question 7

Previous-year question
A galvanometer having a coil resistance of 60 Ω shows full-scale deflection when a current of 1.0 A passes through it. It can be converted into an ammeter reading currents up to 5.0 A by [AIPMT 2009]
  1. putting a resistance of 15 Ω in series
  2. putting a resistance of 240 Ω in series
  3. putting a resistance of 15 Ω in parallel
  4. putting a resistance of 240 Ω in parallel
Show answer
Answer: (C) putting a resistance of 15 Ω in parallel
Explanation: A shunt is connected in parallel: S = I_gG/(I − I_g) = (1.0 × 60)/(5.0 − 1.0) = 60/4 = 15 Ω. Hence a 15 Ω resistance is placed in parallel with the galvanometer.
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Question 8

Previous-year question
A galvanometer has a coil of resistance 100 ohm and gives a full-scale deflection for a current of 30 mA. It is to be used as a voltmeter of 30 V range. The resistance required to be added in series will be [AIPMT 2010]
  1. 900 Ω
  2. 1800 Ω
  3. 500 Ω
  4. 1000 Ω
Show answer
Answer: (A) 900 Ω
Explanation: For a voltmeter, V = I_g(R + G), so R = V/I_g − G = 30/(30 × 10⁻³) − 100 = 1000 − 100 = 900 Ω, connected in series.

Question 9

Previous-year question
A galvanometer of resistance G is shunted by a resistance of S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is [AIPMT 2011]
  1. S²/(S + G)
  2. SG/(S + G)
  3. G²/(S + G)
  4. G/(S + G)
Show answer
Answer: (C) G²/(S + G)
Explanation: The unshunted branch had resistance G. After shunting, the parallel combination is GS/(G + S). To restore the original resistance, a series resistance S′ is added such that GS/(G + S) + S′ = G, giving S′ = G − GS/(G + S) = G²/(G + S).

Question 10

Previous-year question
A millivoltmeter of 25 mV range is to be converted into an ammeter of 25 A range. The value (in ohm) of the necessary shunt will be [AIPMT 2012]
  1. 0.001
  2. 0.01
  3. 1
  4. 0.05
Show answer
Answer: (A) 0.001
Explanation: The voltage across the parallel combination equals the full-scale reading of the millivoltmeter, 25 mV. Since almost the whole current passes through the shunt, S ≈ 25 mV/25 A = 25 × 10⁻³/25 = 0.001 Ω.

Question 11

Previous-year question
In an ammeter, 0.2% of the main current passes through the galvanometer. If the resistance of the galvanometer is G, the resistance of the ammeter will be [AIPMT 2014]
  1. G/499
  2. 499G/500
  3. G/500
  4. 500G/499
Show answer
Answer: (C) G/500
Explanation: With I_g = 0.002I, the shunt carries 0.998I. From I_gG = (I − I_g)S, S = 0.002G/0.998 = G/499. The ammeter resistance is the parallel combination: R = GS/(G + S) = G(G/499)/[G(1 + 1/499)] = G/500.

Question 12

Previous-year question
The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is [NEET 2018]
  1. 250 Ω
  2. 25 Ω
  3. 40 Ω
  4. 500 Ω
Show answer
Answer: (A) 250 Ω
Explanation: Current sensitivity is I_S = θ/I = NAB/k and voltage sensitivity is V_S = θ/V = NAB/kR_G. Dividing, I_S/V_S = R_G. Here I_S = 5 div/mA = 5 × 10³ div/A and V_S = 20 div/V, so R_G = 5000/20 = 250 Ω.

Question 13

A student wants to study a 100 Ω resistor connected to a 12 V battery of negligible internal resistance. By mistake, a voltmeter of resistance 20 kΩ is joined in series with the resistor, and an ammeter of resistance 0.1 Ω is joined in parallel with the resistor. What will the instruments read, approximately?
  1. Voltmeter 0 V; ammeter 120 A, so that the ammeter burns out
  2. Voltmeter 12 V; ammeter 120 mA
  3. Voltmeter 12 V; ammeter 0.6 mA, with no harm to either instrument
  4. Voltmeter 6 V; ammeter 0.3 mA
Show answer
Answer: (C) Voltmeter 12 V; ammeter 0.6 mA, with no harm to either instrument
Explanation: The ammeter and resistor in parallel have a combined resistance of about 0.1 Ω, which is tiny compared with the 20 kΩ voltmeter in series. The total resistance is therefore about 20 kΩ, and the current is 12 V ÷ 20 kΩ = 0.6 mA. Nearly all of this passes through the ammeter. Almost all of the 12 V appears across the voltmeter, and the small current harms neither instrument.

Question 14

A moving coil galvanometer has concave pole pieces and a fixed cylindrical soft iron core inside its coil. Suppose the core is removed and the pole pieces are replaced by flat ones that give a uniform field in the gap. Compared with the original instrument, what would be the result?
  1. The coil would feel no torque in any position
  2. The torque would depend on the coil's orientation, so deflection would not be proportional to current, and the field through the coil would be weaker
  3. The restoring torque of the spring would vanish, so the coil would turn through the same large angle for any current
  4. The scale would stay linear and the instrument would be more sensitive
Show answer
Answer: (B) The torque would depend on the coil's orientation, so deflection would not be proportional to current, and the field through the coil would be weaker
Explanation: The concave poles and the soft iron core make the field radial, so the plane of the coil stays along the field at every deflection. The torque is then NIAB, and the deflection is proportional to current. In a uniform field the torque varies with the coil's angle, so the scale is no longer linear. The soft iron core also strengthens the field, so removing it weakens the field.

Question 15

The coil of a moving coil galvanometer is rewound with twice the number of turns of the same wire. The coil area, the magnetic field and the spring are unchanged, and the coil resistance doubles. What happens to the current sensitivity and the voltage sensitivity?
  1. Both double
  2. Current sensitivity doubles; voltage sensitivity does not change
  3. Current sensitivity does not change; voltage sensitivity doubles
  4. Current sensitivity doubles; voltage sensitivity is halved
Show answer
Answer: (B) Current sensitivity doubles; voltage sensitivity does not change
Explanation: Current sensitivity is φ/I = NAB/k, so doubling N doubles it. Voltage sensitivity is φ/V = NAB/(kR). Here N and R both double, so it stays the same. Adding turns therefore does not always make a galvanometer more sensitive to voltage.

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