PhysicsISC Class 12 · Moving Charges and Magnetism

Moving Coil Galvanometer: Theory, Sensitivity and Conversion

Updated 15 October 2026 · Fact-checked

In a moving coil galvanometer, a coil of N turns and area A in a radial field B feels a torque NIAB, balanced by the spring's restoring torque kφ, so $I = \frac{k}{NAB}\phi$: deflection is proportional to current. A small shunt in parallel makes it an ammeter; a large resistance in series makes it a voltmeter.

Understand Moving Coil Galvanometer and Its Conversion

Torque on a current loop. A rectangular coil of N turns, area A, carrying current I in a uniform field $B$ feels a torque
$τ = m \times B, τ = N I A B sin φ,$
where $m = N I A$ is the magnetic moment and φ the angle between the normal to the coil and $B$ . Derivation: the two sides parallel to the axis of rotation (length l) carry opposite currents and feel equal, opposite forces $F = N B I l$ a perpendicular distance $b sin φ$ apart, forming a couple $τ = N B I l b sin φ = N I A B sin φ$ . The other two sides feel forces along the axis that cancel.

Construction. A rectangular coil of fine insulated copper wire wound on a light non-magnetic frame, pivoted (or suspended) between the concave pole pieces of a strong permanent magnet, with a soft iron cylinder fixed inside the coil. A pair of hair springs (or the suspension fibre) provides the restoring torque and leads current in and out; a pointer moves over a scale.

Radial field. The concave poles and the soft iron core make the field radial: the plane of the coil is always parallel to the field lines whatever the deflection. So φ = 90° at every position and the deflecting torque is always the maximum, $τ = N I A B$ . The soft iron also concentrates the field, making B larger.

Theory. At equilibrium, deflecting torque = restoring torque:
$N I A B = k ϕ \Rightarrow I = \frac{k}{N A B} ϕ = G_{k} ϕ,$
where k is the torsion constant (N m rad⁻¹) and $G_{k} = \frac{k}{N A B}$ is the galvanometer constant. Deflection is proportional to current, so the scale is linear. Note: a circular coil does not produce a uniform field around itself; the uniform (radial) field comes from the magnet. A 2024 ISC assertion-reason item tested exactly this.

Sensitivity.

Current sensitivity = deflection per unit current: $\frac{ϕ}{I} = \frac{N A B}{k}$ (rad A⁻¹, or divisions per µA).
Voltage sensitivity = deflection per unit p.d.: $\frac{ϕ}{V} = \frac{N A B}{k G}$ , where G is the coil resistance.

Current sensitivity rises with more turns, larger area, stronger field and a weaker spring (small k). Doubling the turns of the same wire also doubles G, so it doubles current sensitivity but leaves voltage sensitivity unchanged.

Conversion into an ammeter. A galvanometer gives full-scale deflection at a small current $I_{g}$ . To measure up to I, connect a small shunt S in parallel, carrying the extra current $I - I_{g}$ . Equal p.d.s: $I_{g} G = (I - I_{g}) S$ , so
$S = \frac{I _{g} G}{I - I _{g}} .$
The ammeter's resistance $\frac{GS}{G + S}$ is very small, as an ammeter (in series) should be.

Conversion into a voltmeter. To measure p.d.s up to V, connect a large resistance R in series: $V = I_{g} (G + R)$ , so
$R = \frac{V}{I _{g}} - G .$
The voltmeter's resistance $G + R$ is large, as a voltmeter (in parallel) should be, so that it draws little current.

Why not use the bare galvanometer? It would be damaged by large currents, and as a voltmeter its low resistance would disturb the circuit.

Moving Coil Galvanometer and Its Conversion: formulas and key facts

Torque on a coil: $τ = N I A B sin φ, τ = m \times B$; φ between the normal and B.
Radial field: $τ = N I A B$; Plane of the coil always along B.
Galvanometer equation: $I = \frac{k}{N A B} ϕ$; Deflection ∝ current.
Current sensitivity: $\frac{ϕ}{I} = \frac{N A B}{k}$; rad A⁻¹.
Voltage sensitivity: $\frac{ϕ}{V} = \frac{N A B}{k G}$; rad V⁻¹.
Ammeter shunt: $S = \frac{I _{g} G}{I - I _{g}}$; Small, in parallel.
Voltmeter resistance: $R = \frac{V}{I _{g}} - G$; Large, in series.

How to solve Moving Coil Galvanometer and Its Conversion questions

Conversions all come from "the galvanometer branch carries $I_{g}$ at full scale".

1Note G and the full-scale current $I_{g}$ (convert mA or µA to A).
2Ammeter: the shunt carries $I - I_{g}$ with the same p.d. as the galvanometer; solve $I_{g} G = (I - I_{g}) S$ .
3Voltmeter: the full range V drives $I_{g}$ through $G + R$ ; solve $V = I_{g} (G + R)$ .
4State how the resistance is connected (parallel shunt or series resistance) with its value and unit.
5For sensitivity questions, write the formula and see how each factor changes.
6Check: an ammeter's resistance is small; a voltmeter's is large.

Moving Coil Galvanometer and Its Conversion 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.

Shunt for a high-current range

Suppose an analogue multimeter uses a movement with G = 100 Ω and a full-scale current of 50 µA, and needs a 0–10 A range.

Find the shunt resistance.

Try it first, then show the solutionSolution

$S = \frac{I _{g} G}{I - I _{g}} = \frac{50 \times 1 0 ^{- 6} \times 100}{10 - 0.00005}$ .
$S \approx 5.0 \times 1 0^{- 4}$ Ω: a short, thick strip of metal.

Answer: About 0.5 mΩ, in parallel with the movement.

Memory hook: Ammeter: tiny resistance, side by side.

Voltmeter resistance and "ohms per volt"

Imagine an analogue voltmeter whose movement has a full-scale current of 50 µA.

What is the total resistance of the meter on its 10 V range, and why is that useful?

Try it first, then show the solutionSolution

$G + R = \frac{V}{I _{g}} = \frac{10}{50 \times 1 0 ^{- 6}} = 2 \times 1 0^{5}$ Ω.
That is 20 000 Ω for every volt of range; a high resistance draws little current and hardly disturbs the circuit.

Answer: 200 kΩ on the 10 V range.

Memory hook: Voltmeter: huge resistance, in series with the coil.

Tricks to remember Moving Coil Galvanometer and Its Conversion

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

Ammeter: Small Shunt Side-by-side
Three S's: small, shunt, side by side (parallel).
Use it when: Ammeter conversion.
Voltmeter: Very big, in-line
Large resistance in series.
Use it when: Voltmeter conversion.
NAB over k
Current sensitivity = $\frac{N A B}{k}$ ; divide by G for voltage sensitivity.
Use it when: Sensitivity questions.
Radial = always 90°
Radial field keeps sin φ = 1, so deflection ∝ current.
Use it when: Explaining the radial field.

Moving Coil Galvanometer and Its Conversion: solved examples

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

Example 1: Purpose of the radial field · 1 mark · MCQBoard level

The magnetic field in a moving coil galvanometer is made radial so that:
(a) the coil does not rotate
(b) the deflecting torque is always NIAB and the scale is linear
(c) the coil's resistance is reduced
(d) no spring is needed

Try it first, then show the solutionSolution

In a radial field the plane of the coil is always parallel to B, so sin φ = 1 at every deflection.

Answer: (b).

Example 2: Galvanometer to ammeter · 2 marks · Short answerBoard level

A galvanometer of resistance 40 Ω gives full-scale deflection for 5 mA. How can it be converted into an ammeter of range 0–1 A? Find the resistance of the ammeter.

Try it first, then show the solutionSolution

Shunt in parallel: $S = \frac{0.005 \times 40}{1 - 0.005} \approx 0.201$ Ω. (1)
Ammeter resistance $\frac{40 \times 0.201}{40.201} \approx 0.20$ Ω. (1)

Answer: A 0.201 Ω shunt in parallel; ammeter resistance ≈ 0.2 Ω.

Example 3: Galvanometer to voltmeter · 2 marks · Short answerHarder board level

A galvanometer of resistance 30 Ω gives full-scale deflection for 2 mA. How can it be converted into a voltmeter of range 0–15 V? Find the voltmeter's resistance.

Try it first, then show the solutionSolution

$R = \frac{15}{0.002} - 30 = 7470$ Ω, in series. (1)
Voltmeter resistance $G + R = 7500$ Ω. (1)

Answer: 7470 Ω in series; 7500 Ω in all.

Example 4: Derivation: theory of the galvanometer · 4 marks · Long answerHarder board level

Explain why the pole pieces are concave and a soft iron core is used, and show that the deflection of a moving coil galvanometer is proportional to the current. Define its current sensitivity.

Try it first, then show the solutionSolution

Concave poles and a soft iron core make the field radial and strong; the plane of the coil is always parallel to B. (1)
Deflecting torque: $τ = N I A B sin 90° = N I A B$ . (1)
Restoring torque of the springs: $k ϕ$ . At equilibrium $N I A B = k ϕ$ , so $ϕ = \frac{N A B}{k} I$ , i.e. $ϕ \propto I$ . (1)
Current sensitivity: deflection per unit current, $\frac{ϕ}{I} = \frac{N A B}{k}$ . (1)

Answer: $ϕ = \frac{N A B}{k} I$ ; current sensitivity $\frac{N A B}{k}$ .

Example 5: Sensitivity numbers · 3 marks · Short answerHarder board level

A galvanometer coil has 120 turns of area 3 cm² in a radial field of 0.2 T; the springs have $k = 3.6 \times 1 0^{- 5}$ N m rad⁻¹ and the coil resistance is 25 Ω. Find the current sensitivity, the voltage sensitivity and the deflection for 0.5 mA.

Try it first, then show the solutionSolution

$\frac{N A B}{k} = \frac{120 \times 3 \times 1 0 ^{- 4} \times 0.2}{3.6 \times 1 0 ^{- 5}} = 200$ rad A⁻¹ (0.2 rad per mA). (1)
Voltage sensitivity $= \frac{200}{25} = 8$ rad V⁻¹. (1)
$ϕ = 200 \times 5 \times 1 0^{- 4} = 0.1$ rad ≈ 5.7°. (1)

Answer: 200 rad A⁻¹; 8 rad V⁻¹; 0.1 rad.

Example 6: A multi-range meter · 4 marks · Case-basedHOTS / competency

Suppose a student builds a meter from a galvanometer with G = 50 Ω and $I_{g}$ = 1 mA.
(i) Find the series resistance for a 0–3 V range. (1 mark)
(ii) Find the extra series resistance to add for a 0–15 V range. (1 mark)
(iii) Find the shunt for a 0–1 A range. (1 mark)
(iv) Find the resistance of the meter on each of the three ranges. (1 mark)

Try it first, then show the solutionSolution

(i) $R_{1} = \frac{3}{0.001} - 50 = 2950$ Ω.
(ii) Total for 15 V: $\frac{15}{0.001} - 50 = 14950$ Ω, so add $12000$ Ω after $R_{1}$ .
(iii) $S = \frac{0.001 \times 50}{1 - 0.001} \approx 0.0501$ Ω.
(iv) 3 V range: 3000 Ω; 15 V range: 15 000 Ω; 1 A range: ≈ 0.050 Ω.

Answer: (i) 2950 Ω; (ii) 12 kΩ more; (iii) ≈ 0.050 Ω; (iv) 3 kΩ, 15 kΩ and 0.05 Ω.

Where students lose marks in Moving Coil Galvanometer and Its Conversion

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

Mistake: Misreading the assertion-reason item
In ISC 2024 candidates chose (a), not seeing that a circular coil does not give a uniform field around itself.
Do this: Check each statement separately for truth before asking whether R explains A.
Usually costs: 1 mark.
Mistake: Wrong conversion formula
The 2025 examiners noted candidates who misread the question, used a wrong formula or gave an answer without calculation.
Wrong working
Shunt for a voltmeter: $S = \frac{I _{g} G}{I - I _{g}}$ .
Right working
Series resistance: $R = \frac{V}{I _{g}} - G = 9980$ Ω.
Do this: Voltmeter: series, large. Ammeter: parallel, small.
Usually costs: 2 to 3 marks.
Mistake: No unit or no connection stated
Also in 2025: the value of R was given without Ω or without saying "in series".
Do this: Answer in the form "connect … Ω in series (or parallel)".
Usually costs: 1 mark.
Mistake: Forgetting to subtract G
Using $R = \frac{V}{I _{g}}$ gives the total meter resistance, not the added resistance.
Do this: Subtract the galvanometer's own resistance.
Usually costs: 1 mark.

Exam tips

Draw the circuit for each conversion: shunt across the galvanometer, or resistance in line with it.
Convert mA and µA to A before substituting.
Mention both the concave pole pieces and the soft iron core when explaining the radial field.
Give the final ammeter or voltmeter resistance if the question asks for it.

Practice questions on Moving Coil Galvanometer and Its Conversion

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
An ammeter is made from a galvanometer by connecting:
(a) a high resistance in series
(b) a low resistance in parallel
(c) a high resistance in parallel
(d) a low resistance in series
Hint
Most current must bypass the coil.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
A galvanometer of resistance 50 Ω gives full-scale deflection at 2 mA. How can it be converted into a voltmeter of range 0–20 V? (A variation on ISC 2025.)
Hint
$V = I_{g} (G + R)$ .
Answer
Connect $R = \frac{20}{0.002} - 50 = 9950$ Ω in series with it.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The deflection of a moving coil galvanometer is directly proportional to the current through it.
Reason (R): The radial field keeps the plane of the coil parallel to the field, so the deflecting torque is NIAB at every deflection.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Balance NIAB with kφ.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
An ammeter of range 0–1 A has a resistance of 0.1 Ω. What shunt extends its range to 0–5 A?
Hint
Treat the ammeter as the "galvanometer".
Answer
$S = \frac{1 \times 0.1}{5 - 1} = 0.025$ Ω.
Q5 · 1 mark · Very 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. What happens to its current and voltage sensitivities?
Hint
$\frac{N A B}{k}$ and $\frac{N A B}{k G}$ .
Answer
Current sensitivity doubles; voltage sensitivity is unchanged.
Q6 · 2 marks · Short answerHOTS / competency
A voltmeter of range 0–3 V has a resistance of 3000 Ω. What must be added, and how, to make it read up to 15 V? What is its new resistance?
Hint
Full-scale current = 1 mA.
Answer
$I_{g} = 1$ mA; total needed 15 000 Ω, so add 12 000 Ω in series; new resistance 15 kΩ.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a galvanometer with G = 60 Ω and full-scale current 2 mA is to be used in a laboratory.
(i) Find the series resistance for a 0–6 V voltmeter. (1 mark)
(ii) Find the shunt for a 0–3 A ammeter. (1 mark)
(iii) The voltmeter is used across a 100 kΩ resistor. Why will it give a noticeably low reading? (1 mark)
(iv) How could the galvanometer be made more sensitive? (1 mark)
Hint
Compare the voltmeter's resistance with 100 kΩ.
Answer
(i) $\frac{6}{0.002} - 60 = 2940$ Ω. (ii) $\frac{0.002 \times 60}{2.998} \approx 0.040$ Ω. (iii) Its resistance (3 kΩ) is much smaller than 100 kΩ, so in parallel it lowers the combined resistance and the p.d. across it. (iv) More turns, larger area, stronger field or weaker springs (larger $\frac{N A B}{k}$ ).

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.Torque on a coil in a radial field?Show answer
$τ = N I A B$ .
2.Define current sensitivity.Show answer
Deflection per unit current: $\frac{ϕ}{I} = \frac{N A B}{k}$ .
3.Shunt for an ammeter?Show answer
$S = \frac{I _{g} G}{I - I _{g}}$ , in parallel.
4.G = 10 Ω, $I_{g}$ = 1 mA: series resistance for 0–5 V?Show answer
4990 Ω.
5.Why is a soft iron core used?Show answer
To make the field radial and to concentrate (strengthen) it.

Moving Coil Galvanometer and Its Conversion: frequently asked questions

Why can an ammeter not be connected in parallel?

Its resistance is tiny, so in parallel it would act almost as a short circuit, drawing a large current that could damage it and change the circuit.

Does increasing current sensitivity always increase voltage sensitivity?

No. Voltage sensitivity is current sensitivity divided by the coil resistance. Adding turns increases both N and G, so voltage sensitivity can stay the same.

What is the galvanometer constant?

It is \frac{k}{NAB}, the current needed per unit deflection; its reciprocal is the current sensitivity.

What does the ISC 2027 syllabus require here?

Construction, principle, working and theory (I = k\phi) of the moving coil galvanometer, its current and voltage sensitivity, the shunt, and conversion into an ammeter and a voltmeter of given range.

How was this tested recently?

In 2025 a 3-mark question asked to convert a galvanometer into a 0–10 V voltmeter; in 2024 a 1-mark assertion-reason item tested why the coil deflects.