PhysicsCBSE Class 12 · Moving Charges and Magnetism

Moving Coil Galvanometer Class 12: Working, Sensitivity and Conversion

Updated 15 October 2026 · Fact-checked

In a moving coil galvanometer, a current-carrying coil in a radial magnetic field feels a torque NIAB, balanced by a spring's torque kφ, so the deflection φ = (NAB/k)I is proportional to the current. A small shunt in parallel converts it into an ammeter; a large resistance in series converts it into a voltmeter.

Understand Moving Coil Galvanometer

Every analogue ammeter and voltmeter is built around the same device: a coil that turns when current flows through it in a magnetic field. The torque on a current loop, $τ = N I A B sin θ$ , is put to work.

Construction. A rectangular coil of many turns, wound on a light frame, can rotate about a fixed axis between the concave pole pieces of a permanent magnet. A soft iron cylinder sits inside the coil. A spring (or a fine suspension wire) provides a restoring torque, and a pointer moves over a scale.

Why a radial field. The concave pole pieces and the soft iron core make the field radial: whatever the coil's position, the plane of the coil is parallel to the field lines. So $sin θ = 1$ always, and the deflecting torque is simply $τ = N I A B$ . The soft iron core also strengthens the field, increasing sensitivity.

Working. The deflecting torque NIAB is opposed by the spring's restoring torque $k ϕ$ , where k is the torsional constant and $ϕ$ the deflection. In equilibrium $k ϕ = N I A B$ , so $ϕ = (\frac{N A B}{k}) I$ . The deflection is proportional to the current, so the scale is linear (evenly spaced).

Sensitivity.

Current sensitivity: deflection per unit current, $\frac{ϕ}{I} = \frac{N A B}{k}$ (unit rad/A or div/A).
Voltage sensitivity: deflection per unit voltage, $\frac{ϕ}{V} = \frac{ϕ}{I G} = \frac{N A B}{k G}$ , where G is the galvanometer's resistance.

To increase current sensitivity, increase N, A or B, or use a weaker spring. But doubling N also roughly doubles the length of wire and so G, so the voltage sensitivity stays about the same.

Galvanometer as an ammeter. A galvanometer gives full-scale deflection for a small current $I_{g}$ (micro- or milliampere), and its resistance G is not negligible. To measure a larger current I, a small resistance S, a shunt, is connected in parallel. At full scale, $I_{g}$ goes through the coil and $I - I_{g}$ through the shunt; the two have the same voltage: $I_{g} G = (I - I_{g}) S$ , so $S = \frac{I _{g} G}{I - I _{g}}$ . The ammeter's resistance $\frac{GS}{G + S}$ is less than S, close to zero, which is what an ammeter in series needs so as not to change the current it measures.

Galvanometer as a voltmeter. To measure up to a voltage V, a large resistance R is connected in series: $V = I_{g} (G + R)$ , so $R = \frac{V}{I _{g}} - G$ . The voltmeter's total resistance $G + R$ is high, so connected in parallel it draws very little current and hardly disturbs the circuit.

An ideal ammeter has zero resistance and an ideal voltmeter infinite resistance; real meters come close, and the small errors they cause can be estimated with Ohm's law.

Moving Coil Galvanometer: formulas and key facts

Deflecting torque in a radial field: $τ = N I A B$; Radial field keeps the plane of the coil parallel to B.
Deflection: $ϕ = (\frac{N A B}{k}) I$; k = torsional constant of the spring.
Current sensitivity: $\frac{ϕ}{I} = \frac{N A B}{k}$; Unit rad/A or div/A.
Voltage sensitivity: $\frac{ϕ}{V} = \frac{N A B}{k G}$; G = galvanometer resistance.
Shunt for an ammeter of range I: $S = \frac{I _{g} G}{I - I _{g}}$; Connected in parallel; ammeter resistance $\frac{GS}{G + S}$ .
Series resistance for a voltmeter of range V: $R = \frac{V}{I _{g}} - G$; Connected in series; voltmeter resistance G + R.

How to solve Moving Coil Galvanometer questions

Use this method for galvanometer sensitivity and conversion questions.

1List G and the full-scale current $I_{g}$ (convert mA or μA to A).
2Ammeter: shunt in parallel. Write $I_{g} G = (I - I_{g}) S$ and solve for S.
3Voltmeter: resistance in series. Write $V = I_{g} (G + R)$ and solve for R.
4Check: the ammeter's total resistance should be tiny; the voltmeter's should be large.
5For sensitivity changes, write $\frac{N A B}{k}$ and $\frac{N A B}{k G}$ , and remember G ∝ N if the same wire is used for more turns.
6For two-range problems, write one equation per range and divide them to cancel $I_{g}$ .
7State where the extra resistance goes (parallel or series) in words; it earns a mark.

Moving Coil Galvanometer in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Converting a meter movement

Suppose a multimeter is built around a galvanometer of resistance 100 Ω that gives full-scale deflection at 100 μA.

(a) What shunt makes it a 0-10 mA ammeter? (b) What series resistance makes it a 0-20 V voltmeter? (c) The voltmeter's quality is often quoted in ohm per volt. Find it.

Try it first, then show the solutionSolution

(a) $S = \frac{I _{g} G}{I - I _{g}} = \frac{1 0 ^{- 4} \times 100}{0.01 - 0.0001} = \frac{0.01}{0.0099} \approx 1.01$ Ω, in parallel.
(b) $R = \frac{V}{I _{g}} - G = \frac{20}{1 0 ^{- 4}} - 100 = 199900$ Ω, in series.
(c) Total resistance per volt $= \frac{1}{I _{g}} = 10000$ Ω/V.

Answer: (a) about 1.01 Ω in parallel; (b) 199 900 Ω in series; (c) 10 kΩ/V.

Memory hook: Ammeter: tiny shunt beside it. Voltmeter: huge resistor behind it.

Why a galvanometer needs protection

Imagine someone connects a bare galvanometer (G = 50 Ω, full scale 1 mA) straight across a 1.5 V cell to 'measure its voltage'.

(a) What current would flow? (b) What would happen to the meter? (c) What is the smallest series resistance that would keep the current within range?

Try it first, then show the solutionSolution

$I = \frac{1.5}{50} = 0.03$ A = 30 mA (ignoring the cell's internal resistance).
That is 30 times full scale: the coil would overheat and the pointer would slam against the stop, damaging the meter.
$R = \frac{V}{I _{g}} - G = \frac{1.5}{1 0 ^{- 3}} - 50 = 1450$ Ω.

Answer: (a) 30 mA; (b) it would be damaged; (c) 1450 Ω in series.

Memory hook: A galvanometer is delicate: never connect it straight across a supply.

Tricks to remember Moving Coil Galvanometer

Short memory aids. Say each one aloud once, then use it on a question the same day.

Shunt Sits Side by side
Shunt in parallel for an ammeter; a big Resistor in a Row (series) for a voltmeter.
Use it when: Getting the connections right.
NAB over k
Current sensitivity = NAB/k; divide by G for voltage sensitivity.
Use it when: Writing sensitivity formulas.
Radial = always parallel
In a radial field the coil's plane is always along B, so sin θ = 1 and the scale is linear.
Use it when: Explaining the radial field.
More turns, more G
Doubling turns doubles current sensitivity but also G, so voltage sensitivity stays put.
Use it when: Assertion-reason questions.

Moving Coil Galvanometer: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Converting into an ammeter and a voltmeter · 3 marks · Short answerBoard level

A galvanometer of resistance 50 Ω gives full-scale deflection for 2 mA. Find the resistance needed, and how it is connected, to convert it into (a) an ammeter of range 0-2 A and (b) a voltmeter of range 0-10 V.

Try it first, then show the solutionSolution

(a) Shunt in parallel: $S = \frac{I _{g} G}{I - I _{g}} = \frac{2 \times 1 0 ^{- 3} \times 50}{2 - 0.002} = \frac{0.1}{1.998} \approx 0.050$ Ω. (1½)
(b) Series resistance: $R = \frac{V}{I _{g}} - G = \frac{10}{2 \times 1 0 ^{- 3}} - 50 = 4950$ Ω. (1½)

Answer: (a) about 0.050 Ω in parallel; (b) 4950 Ω in series.

Example 2: Derivation: deflection and sensitivity · 3 marks · Short answerBoard level

Explain the working of a moving coil galvanometer. Show that the deflection is proportional to the current, and define its current and voltage sensitivity. Why is the field made radial?

Try it first, then show the solutionSolution

The coil (N turns, area A) in a field B carrying current I feels a torque $N I A B sin θ$ . The concave pole pieces and soft iron core make the field radial, so the coil's plane is always parallel to B and $τ = N I A B$ . (1)
The spring provides a restoring torque $k ϕ$ . At equilibrium $k ϕ = N I A B$ , so $ϕ = \frac{N A B}{k} I \propto I$ , giving a linear scale. (1)
Current sensitivity $\frac{ϕ}{I} = \frac{N A B}{k}$ ; voltage sensitivity $\frac{ϕ}{V} = \frac{N A B}{k G}$ . (1)

Answer: φ = (NAB/k)I; current sensitivity NAB/k; voltage sensitivity NAB/kG; radial field gives τ = NIAB at every position.

Example 3: Doubling the turns · 3 marks · Short answerHarder board level

The number of turns of a galvanometer coil is doubled using the same kind of wire, so its resistance also doubles. Find the factor by which (a) the current sensitivity and (b) the voltage sensitivity change. (c) Suggest a way to increase current sensitivity without changing G much.

Try it first, then show the solutionSolution

(a) Current sensitivity $\frac{N A B}{k} \propto N$ : doubles. (1)
(b) Voltage sensitivity $\frac{N A B}{k G}$ : N and G both double, so it is unchanged. (1)
(c) Use a stronger magnet (larger B) or a spring with a smaller torsional constant k. (1)

Answer: (a) doubles; (b) unchanged; (c) increase B or reduce k.

Example 4: Sensitivity from coil data · 3 marks · Short answerHarder board level

A galvanometer coil has 100 turns, each 1 cm × 2 cm, in a radial field of 0.1 T. The torsional constant of the spring is $1 0^{- 6}$ N m/rad and the coil resistance is 20 Ω. Find (a) the current sensitivity, (b) the deflection for 0.1 mA and (c) the voltage sensitivity.

Try it first, then show the solutionSolution

(a) $\frac{ϕ}{I} = \frac{N A B}{k} = \frac{100 \times 2 \times 1 0 ^{- 4} \times 0.1}{1 0 ^{- 6}} = 2000$ rad/A. (1)
(b) $ϕ = 2000 \times 1 0^{- 4} = 0.2$ rad, about 11.5°. (1)
(c) $\frac{ϕ}{V} = \frac{2000}{20} = 100$ rad/V. (1)

Answer: (a) 2000 rad/A; (b) 0.2 rad ≈ 11.5°; (c) 100 rad/V.

Example 5: Finding G and Ig from two voltmeter ranges · 5 marks · Long answerHOTS / competency

A galvanometer becomes a voltmeter of range 0-2 V with a series resistance of 1980 Ω, and of range 0-5 V with a series resistance of 4980 Ω.
(a) Find the resistance of the galvanometer and its full-scale current. (3)
(b) What shunt would convert it into an ammeter of range 0-1 A? (2)

Try it first, then show the solutionSolution

(a) $2 = I_{g} (G + 1980)$ and $5 = I_{g} (G + 4980)$ . (1)
Dividing: $\frac{G + 4980}{G + 1980} = 2.5$ , so $G + 4980 = 2.5 G + 4950$ and $G = 20$ Ω. (1)
$I_{g} = \frac{2}{2000} = 1 0^{- 3}$ A = 1 mA. (1)
(b) $S = \frac{I _{g} G}{I - I _{g}} = \frac{1 0 ^{- 3} \times 20}{1 - 0.001} = \frac{0.02}{0.999} \approx 0.020$ Ω, in parallel. (2)

Answer: (a) G = 20 Ω, I_g = 1 mA; (b) about 0.020 Ω in parallel.

Example 6: Effect of meter resistance on readings · 4 marks · Case-basedHarder board level

Read the passage and answer.

A 3 V cell (negligible internal resistance) drives current through a 10 Ω resistor. In a second circuit, two 100 Ω resistors are in series across a 6 V supply.
(a) An ammeter of resistance 0.5 Ω is put in series in the first circuit. What does it read, and what is the true current without it? (1)
(b) Find the percentage error. (1)
(c) A voltmeter of resistance 1000 Ω is connected across one 100 Ω resistor in the second circuit. What does it read, and what is the true voltage? (1)
(d) What do (a) to (c) show about ideal meters? (1)

Try it first, then show the solutionSolution

(a) Reading $= \frac{3}{10.5} \approx 0.286$ A; true current $= \frac{3}{10} = 0.3$ A. (1)
(b) Error $= \frac{0.3 - 0.286}{0.3} \times 100 \approx 4.8%$ . (1)
(c) $100 ∥ 1000 \approx 90.9$ Ω; reading $= 6 \times \frac{90.9}{190.9} \approx 2.86$ V; true voltage = 3 V. (1)
(d) An ideal ammeter should have zero resistance and an ideal voltmeter infinite resistance, so that inserting them does not change what they measure. (1)

Answer: (a) about 0.286 A against 0.3 A; (b) about 4.8%; (c) about 2.86 V against 3 V; (d) ammeters need very low and voltmeters very high resistance.

Where students lose marks in Moving Coil Galvanometer

The slips examiners see most often, and what to write instead.

Mistake: Connecting the shunt in series or the voltmeter resistance in parallel.
Mixing up the two conversions.
Do this: Ammeter: shunt in parallel (bypass). Voltmeter: resistance in series (limit current).
Usually costs: 1 mark, and the numerical often goes wrong too.
Mistake: Using S = I_gG/I instead of I_gG/(I − I_g).
Forgetting that only I − I_g goes through the shunt.
Wrong working
S = 0.002 × 50/2 = 0.05 Ω
Right working
S = 0.002 × 50/(2 − 0.002) ≈ 0.050 Ω (here close, but not in general)
Do this: Write the current-sharing equation $I_{g} G = (I - I_{g}) S$ first.
Usually costs: 1 mark when I_g is not tiny compared with I.
Mistake: Forgetting to subtract G in R = V/I_g − G.
Thinking the whole V/I_g is the added resistance.
Do this: The galvanometer's own G is already in series; add only the difference.
Usually costs: The final mark.
Mistake: Saying a radial field makes the torque zero.
Confusing 'plane parallel to B' with 'm parallel to B'.
Do this: In a radial field the plane of the coil is along B, so m ⊥ B and the torque is the maximum, NIAB.
Usually costs: 1 mark.
Mistake: Leaving μA or mA unconverted.
Rushing the substitution.
Do this: Convert to ampere before using any formula.
Usually costs: The answer mark.

Exam tips

Draw a neat labelled diagram for the working question: coil, concave poles, soft iron core, spring and pointer.
Write the equilibrium condition kφ = NIAB as a separate line; it is the key step.
For conversions, show the circuit with the shunt or series resistor drawn in.
Board papers have asked galvanometer sensitivity and conversion for 2 to 5 marks in recent years (for example 2023 55/5/1 Q29, 2025 55/7/1 Q32 and 2026 55/1/1 Q29).
Explain why increasing current sensitivity need not increase voltage sensitivity: the same idea underlies assertion-reason items on sensitivity.

Practice questions on Moving Coil Galvanometer

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
To convert a galvanometer into an ammeter, one connects
(a) a low resistance in series (b) a low resistance in parallel (c) a high resistance in series (d) a high resistance in parallel
Hint
The extra current must bypass the coil.
Answer
(b) a low resistance in parallel.
Q2 · 2 marks · Very short answerBoard level
Why are the pole pieces of a moving coil galvanometer made concave, with a soft iron core in the coil?
Hint
Think of sin θ in the torque.
Answer
They make the field radial, so the plane of the coil is always parallel to the field and the torque is NIAB at every position. Then φ ∝ I and the scale is linear. The soft iron core also concentrates the field, making B (and the sensitivity) larger.
Q3 · 1 mark · Very short answerBoard level
Define current sensitivity of a galvanometer and write its SI unit.
Hint
Deflection per unit ...
Answer
Deflection per unit current, $\frac{ϕ}{I} = \frac{N A B}{k}$ ; unit rad/A (or division per ampere).
Q4 · 2 marks · Very short answerHarder board level
A galvanometer of resistance 60 Ω gives full-scale deflection at 5 mA. Find the shunt needed to make it a 0-3 A ammeter, and the resistance of the ammeter.
Hint
S = I_gG/(I − I_g).
Answer
$S = \frac{0.005 \times 60}{2.995} \approx 0.1002$ Ω. Ammeter resistance $= \frac{60 \times 0.1002}{60.1002} \approx 0.1$ Ω.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Increasing the number of turns of a galvanometer coil may not increase its voltage sensitivity.
Reason (R): Increasing the number of turns also increases the resistance of the coil in the same proportion.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Voltage sensitivity = NAB/kG.
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
A galvanometer of resistance 40 Ω and full-scale current 5 mA is made into a 0-10 V voltmeter. What extra series resistance would make its range 0-20 V?
Hint
Find R for 10 V first.
Answer
For 10 V: $R = \frac{10}{0.005} - 40 = 1960$ Ω. For 20 V: total $\frac{20}{0.005} = 4000$ Ω, so R must be 3960 Ω: add 2000 Ω more in series.
Q7 · 2 marks · Very short answerHOTS / competency
A shunt of resistance $\frac{G}{9}$ is connected across a galvanometer of resistance G. What fraction of the total current passes through the galvanometer, and by what factor does the range increase?
Hint
Current divides in inverse ratio of resistances.
Answer
Fraction through G $= \frac{S}{S + G} = \frac{G /9}{G /9 + G} = \frac{1}{10}$ . The range becomes 10 times.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Deflection of a galvanometer in terms of current?Show answer
φ = (NAB/k) I.
2.Current sensitivity?Show answer
NAB/k.
3.Voltage sensitivity?Show answer
NAB/(kG).
4.Shunt for an ammeter of range I?Show answer
S = I_gG/(I − I_g), in parallel.
5.Series resistance for a voltmeter of range V?Show answer
R = V/I_g − G.
6.Why is the field radial?Show answer
So that the torque is NIAB at every position and the scale is linear.

Moving Coil Galvanometer: frequently asked questions

Why can't a galvanometer be used directly as an ammeter?

It gives full-scale deflection for a very small current (micro- or milliampere) and would be damaged by larger currents. Its resistance is also not negligible, so in series it would change the current being measured.

Why does an ammeter have low resistance and a voltmeter high resistance?

An ammeter goes in series, so it must not add resistance to the circuit. A voltmeter goes in parallel, so it must draw almost no current, which needs a very high resistance.

What is the role of the soft iron core?

It makes the field radial (along with the concave pole pieces) and increases its strength, which increases the sensitivity of the galvanometer.

Can voltage sensitivity be increased by adding turns?

Not usually. More turns increase current sensitivity, but the extra wire raises the coil's resistance in the same proportion, so NAB/(kG) stays about the same.

Is the galvanometer scale linear?

Yes. Because the field is radial, the torque is NIAB at every position, and balancing it with kφ gives φ ∝ I, so equal currents give equal steps of deflection.