Ammeter: galvanometer + low-resistance shunt in parallel. Voltmeter: galvanometer + high resistance in series.
-- NCERT Class 12 Physics, Ch. 4, p. 130Galvanometer Sensitivity
8 MCQs9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismOfficial key: NTA-verifiedLast updated: 8 Oct 2026
Try this first
In a moving coil galvanometer, what is the current sensitivity?
- A.The current needed per unit deflection.
- B.The torque produced per unit current.
- C.The deflection produced per unit current.
- D.The deflection produced per unit voltage.
Tap to see the answer
Answer: C. C is correct: NCERT defines the current sensitivity as the deflection per unit current, φ/I = NAB/k (NCERT Class 12 Physics, Chapter 4, page 130).
A is wrong: A is wrong because it inverts the definition; sensitivity is deflection per unit current, so a larger value means a larger deflection for the same current.
B is wrong: B is wrong because torque per unit current is NAB, not the sensitivity; the sensitivity also divides by the torsional constant k.
D is wrong: D is wrong because deflection per unit voltage is the voltage sensitivity, a different quantity.
Galvanometer Sensitivity, explained for NEET
The trap is "more turns, better galvanometer." Doubling the number of turns N does double the current sensitivity, but it does not raise the voltage sensitivity, and a question that asks for the voltage sensitivity is built to catch the aspirant who stops after the first half.
In the moving coil galvanometer a spring supplies a counter torque kφ that balances the magnetic torque NIAB, where k is the torsional constant, the restoring torque per unit twist. The steady deflection is φ = (NAB/k) I (NCERT Class 12 Physics, Chapter 4, page 130). The bracket NAB/k is a constant for a given galvanometer.
The current sensitivity is defined as the deflection per unit current, φ/I = NAB/k, and a convenient way to raise it is to increase N (NCERT Class 12 Physics, Chapter 4, page 130). The voltage sensitivity is the deflection per unit voltage, φ/V = (NAB/k)(1/R), where R is the resistance of the circuit the voltage drives the current through (NCERT Class 12 Physics, Chapter 4, page 131).
NCERT then makes the point that a higher current sensitivity does not necessarily mean a higher voltage sensitivity. If N becomes 2N, the current sensitivity doubles, but the wire is twice as long, so the resistance also roughly doubles, and φ/V stays unchanged (NCERT Class 12 Physics, Chapter 4, page 131).
The same pages explain why the bare galvanometer is not used as an ammeter or voltmeter: it gives full-scale deflection for a current of the order of microamperes. An ammeter needs a small shunt in parallel; a voltmeter needs a large resistance in series (NCERT Class 12 Physics, Chapter 4, pages 130 and 131).
Watch-out: write which sensitivity the question wants before touching N, k or R. Current sensitivity depends on NAB/k alone; voltage sensitivity also divides by the resistance.
How do you solve a Galvanometer Sensitivity question? A worked example
- 1
Given
A galvanometer coil has N = 100 turns, area A = 2.0 × 10⁻⁴ m², field B = 0.20 T in the gap, and a spring of torsional constant k = 8.0 × 10⁻⁸ N m rad⁻¹. The coil resistance is 25 Ω and the field is radial.
- 2
Required
The current sensitivity in rad per A, the deflection for a current of 10 μA, and the voltage sensitivity in rad per V.
- 3
Concept
In equilibrium kφ = NIAB, so φ = (NAB/k) I. The current sensitivity is φ/I = NAB/k, and the voltage sensitivity is φ/V = (NAB/k)(1/R) (NCERT Class 12 Physics, Chapter 4, pages 130 and 131).
- 4
Formula
φ/I = NAB/k; φ = (φ/I) × I; φ/V = (φ/I) / R.
- 5
Substitution
φ/I = (100 × 2.0 × 10⁻⁴ × 0.20) / (8.0 × 10⁻⁸); φ = (φ/I) × (10 × 10⁻⁶ A); φ/V = (φ/I) / 25.
- 6
Calculation
NAB = 100 × 2.0 × 10⁻⁴ × 0.20 = 4.0 × 10⁻³ N m A⁻¹, so φ/I = 4.0 × 10⁻³ / 8.0 × 10⁻⁸ = 5.0 × 10⁴ rad A⁻¹. Then φ = 5.0 × 10⁴ × 1.0 × 10⁻⁵ = 0.50 rad. The voltage sensitivity is φ/V = 5.0 × 10⁴ / 25 = 2.0 × 10³ rad V⁻¹. No physical constants are needed.
- 7
Final answer
Current sensitivity 5.0 × 10⁴ rad per A, deflection 0.50 rad for 10 μA, voltage sensitivity 2.0 × 10³ rad per V.
- 8
Common trap
Assuming that rewinding the coil with more turns raises both sensitivities. The resistance rises with the length of wire, so the voltage sensitivity can stay where it was.
- 9
Similar NEET-style question
The same galvanometer is rewound with 200 turns of the same wire, so the coil resistance becomes 50 Ω; A, B and k are unchanged. What is the new voltage sensitivity? (Answer: the turns go from 100 to 200, a factor 200/100 = 2, so φ/I = 2 × 5.0 × 10⁴ = 1.0 × 10⁵ rad A⁻¹. The resistance also doubles, 50/25 = 2, so φ/V = 1.0 × 10⁵ / 50 = 2.0 × 10³ rad V⁻¹, the same as before.)
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Can you answer these Galvanometer Sensitivity MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In a moving coil galvanometer, what is the current sensitivity?
Show answer and why every option is right or wrong
Answer: C. C is correct: NCERT defines the current sensitivity as the deflection per unit current, φ/I = NAB/k (NCERT Class 12 Physics, Chapter 4, page 130).
Why A is wrong: A is wrong because it inverts the definition; sensitivity is deflection per unit current, so a larger value means a larger deflection for the same current.
Why B is wrong: B is wrong because torque per unit current is NAB, not the sensitivity; the sensitivity also divides by the torsional constant k.
Why D is wrong: D is wrong because deflection per unit voltage is the voltage sensitivity, a different quantity.
MCQ 2Easy RecallPractice
In equilibrium the spring's counter torque kφ balances the magnetic torque NIAB on the coil. Which expression gives the steady deflection φ?
Show answer and why every option is right or wrong
Answer: A. A is correct: setting kφ = NIAB gives φ = (NAB/k) I, which NCERT numbers (4.26) (NCERT Class 12 Physics, Chapter 4, page 130).
Why B is wrong: B is wrong because it puts the torsional constant in the numerator; a stiffer spring (larger k) must give a smaller deflection.
Why C is wrong: C is wrong because it multiplies by k; k appears on the restoring side of kφ = NIAB, so it divides.
Why D is wrong: D is wrong because it moves the area A and the field B into the denominator; the magnetic torque NIAB is proportional to both.
MCQ 3Easy RecallPractice
Which change does NCERT give as a convenient way for a manufacturer to increase the current sensitivity of a galvanometer?
Show answer and why every option is right or wrong
Answer: D. D is correct: the current sensitivity is NAB/k, and NCERT notes that increasing the number of turns N is a convenient way to raise it (NCERT Class 12 Physics, Chapter 4, page 130).
Why A is wrong: A is wrong because fewer turns lower NAB and so lower the sensitivity.
Why B is wrong: B is wrong because a larger k appears in the denominator of NAB/k and reduces the deflection per unit current.
Why C is wrong: C is wrong because a smaller B lowers NAB, so the sensitivity falls.
MCQ 4Direct ApplicationPractice
A galvanometer has a current sensitivity of 4.0 divisions per mA. The spring is replaced by one with half the torsional constant; N, A and B are unchanged. What is the new current sensitivity?
Show answer and why every option is right or wrong
Answer: B. B is correct: the current sensitivity is NAB/k, so halving k doubles it, 2 × 4.0 = 8.0 divisions per mA (NCERT Class 12 Physics, Chapter 4, page 130).
Why A is wrong: A is wrong because it halves the sensitivity, treating k as if it sat in the numerator.
Why C is wrong: C is wrong because k appears in NAB/k, so changing the spring changes the sensitivity.
Why D is wrong: D is wrong because it squares the factor; the sensitivity is inversely proportional to k, not to k².
MCQ 5Direct ApplicationPractice
A galvanometer coil has 100 turns and a current sensitivity of 6.0 divisions per mA. It is rewound with 150 turns; A, B and k are unchanged. What is the new current sensitivity?
Show answer and why every option is right or wrong
Answer: D. D is correct: the current sensitivity NAB/k is proportional to N, so it becomes 6.0 × (150/100) = 9.0 divisions per mA (NCERT Class 12 Physics, Chapter 4, page 130).
Why A is wrong: A is wrong because 6.0 × 100/150 = 4.0 inverts the ratio of turns.
Why B is wrong: B is wrong because the current sensitivity does change with N; only the voltage sensitivity can stay unchanged, and only when the resistance also scales.
Why C is wrong: C is wrong because 6.0 × 1.5² = 13.5 squares the ratio; NAB/k is linear in N.
MCQ 6Direct ApplicationPractice
A galvanometer with a current sensitivity of 5.0 divisions per mA is connected so that the total resistance in its circuit is 50 Ω. What is its voltage sensitivity?
Show answer and why every option is right or wrong
Answer: A. A is correct: the voltage sensitivity is the current sensitivity divided by R. 5.0 divisions per mA is 5000 divisions per A, and 5000 / 50 = 100 divisions per V (NCERT Class 12 Physics, Chapter 4, page 131).
Why B is wrong: B is wrong because 5.0 × 50 = 250 multiplies by the resistance; φ/V = (NAB/k)(1/R) divides by it.
Why C is wrong: C is wrong because it takes the voltage sensitivity equal to the current sensitivity, ignoring the 1/R factor.
Why D is wrong: D is wrong because 5.0 / 50 = 0.10 leaves the current sensitivity in mA instead of converting to A first.
MCQ 7CalculationPractice
A manufacturer doubles the number of turns of a galvanometer coil using the same wire, so the coil resistance also doubles. A, B and k are unchanged. What happens to the two sensitivities?
Show answer and why every option is right or wrong
Answer: C. C is correct: with N → 2N the current sensitivity NAB/k doubles, but R → 2R as well, so (NAB/k)(1/R) is unchanged (NCERT Class 12 Physics, Chapter 4, page 131).
Why A is wrong: A is wrong because it counts the doubling of N in the voltage sensitivity but ignores that the resistance doubles too.
Why B is wrong: B is wrong because it swaps the two sensitivities: it is the current sensitivity that doubles.
Why D is wrong: D is wrong because the doubling of N in the numerator cancels the doubling of R in the denominator; it does not leave a net halving.
MCQ 8CalculationPractice
A galvanometer has a current sensitivity of 8.0 divisions per mA and a coil resistance of 40 Ω, and the voltage is applied directly across the coil. The number of turns is increased by 50 per cent, and the coil resistance rises in the same proportion. What is the new voltage sensitivity?
Show answer and why every option is right or wrong
Answer: B. B is correct: the original voltage sensitivity is 8000 / 40 = 200 divisions per V. The new current sensitivity is 12.0 divisions per mA and the new resistance is 60 Ω, so 12 000 / 60 = 200 divisions per V, unchanged (NCERT Class 12 Physics, Chapter 4, page 131).
Why A is wrong: A is wrong because 200 × 1.5 = 300 scales the voltage sensitivity with N and forgets that the resistance rises by the same factor.
Why C is wrong: C is wrong because 200 / 1.5 = 133 divides by the factor twice over; the factors cancel instead.
Why D is wrong: D is wrong because 12.0 / 60 = 0.20 leaves the current sensitivity in divisions per mA instead of converting to divisions per A.
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What to remember before solving Galvanometer Sensitivity questions
5 NCERT lines
The magnetic torque NIAB tends to rotate the coil. A spring Sp provides a counter torque kφ that balances the magnetic torque NIAB; resulting in a steady angular deflection φ. In equilibrium kφ = NI AB where k is the torsional constant of the spring; i.e. the restoring torque per unit twist. The deflection φ is indicated on the scale by a pointer attached to the spring. We have φ = NAB k I (4.26)
-- NCERT Class 12 Physics, Ch. 4, p. 130We define the current sensitivity of the galvanometer as the deflection per unit current. From Eq. (4.26) this current sensitivity is, NAB I k φ = (4.27) A convenient way for the manufacturer to increase the sensitivity is to increase the number of turns N.
-- NCERT Class 12 Physics, Ch. 4, p. 130We define the voltage sensitivity as the deflection per unit voltage. From Eq. (4.26), φ V NAB k I V NAB k R = = 1 (4.28)
-- NCERT Class 12 Physics, Ch. 4, p. 131Key Fact
Doubling N doubles current sensitivity but leaves voltage sensitivity unchanged (R also doubles)
An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. Let us take Eq. (4.27) which provides a measure of current sensitivity. If N → 2N, i.e., we double the number of turns, then 2 I I φ φ → Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Eq. (4.28), N →2N, and R →2R, thus the voltage sensitivity, V V φ φ → remains unchanged.
-- NCERT Class 12 Physics, Ch. 4, p. 131Galvanometer Sensitivity: NEET previous year questions (PYQs) with answers
18 questions in Magnetic Effects of Current and Magnetism
No question in our NEET 2020–2025 set targets this topic directly.
All 18 past-paper questions from Magnetic Effects of Current and Magnetism →
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
Sources
NCERT refs: Class 12 Physics Chapter 4, p.130 | Class 12 Physics Chapter 4, p.131
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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