A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works on the fact that a current-carrying coil experiences a deflecting torque when placed in a magnetic field. This deflection in the coil can be measured and it is related to the current flowing in the coil, the number of turns in the coil, area of the coil and the magnetic field. A hair spring attached to the coil provides a counter torque and helps in measuring the deflection. A galvanometer can be converted to an ammeter or a voltmeter of desired range by using suitable resistances. (I) The torque on the coil remains constant irrespective of the coil's orientation during rotation due to (A) use of soft iron core which increases the magnetic field. (B) radial magnetic field (C) hair spring which provides the counter torque (D) eddy current in the iron core which causes damping. (II) The best way to increase current sensitivity of a galvanometer is by (A) increasing number of turns of the coil (B) increasing area of coil and magnetic field strength (C) decreasing area of coil and magnetic field strength (D) increasing torsional constant of the hair spring (III) A moving coil galvanometer has a coil with area of cross-section $4.0\times10^{-3}\ \text{m}^2$ and number of turns $50$. The coil is rotating in a magnetic field of $0.25\ \text{T}$. The torque acting on the coil when a current of $5\ \text{A}$ passes through it is (A) $1.0\ \text{N m}$ (B) $2.0\ \text{N m}$ (C) $0.50\ \text{N m}$ (D) $0.25\ \text{N m}$ OR A galvanometer coil has a resistance of $15\ \Omega$ and the meter shows full scale deflection for a current of $3\ \text{mA}$. The value of resistance required to convert it into a voltmeter of range $(0-12\ \text{V})$ is (A) $4015\ \Omega$ (B) $3985\ \Omega$ (C) $415\ \Omega$ (D) $385\ \Omega$ (IV) A galvanometer with coil of resistance $20\ \Omega$ shows full scale deflection for a current of $5\ \text{mA}$. To convert it into an ammeter of range $(0-10\ \text{A})$, a resistance of (A) $0.05\ \Omega$ should be connected in series with it. (B) $0.05\ \Omega$ should be connected in parallel with it. (C) $0.01\ \Omega$ should be connected in parallel with it. (D) $0.01\ \Omega$ should be connected in series with it.

A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works

Question

4.0\times10^{-3}\ \text{m}^2

(OR)

15\ \Omega

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Part (a)Concept understanding — Moving Coil Galvanometer

Moving Coil Galvanometer: From Intuition to Formula

Imagine you have a tiny, lightweight coil of wire, suspended so it can rotate freely. If you pass a current through it, that coil becomes an electromagnet. Now place it between the poles of a strong permanent magnet. The coil will try to twist — it experiences a torque. The bigger the current, the harder it twists. That is the entire physical idea behind a moving coil galvanometer: use a current to produce a rotation, and measure the rotation to know the current.

But a freely spinning coil would just keep turning. To get a useful measurement, you need something that opposes that rotation — a restoring force that grows as the coil turns further. That is the job of a spring (usually a fine phosphor-bronze strip called a torsion fibre). The spring twists as the coil rotates, producing a restoring torque that exactly balances the magnetic torque at some angle. That equilibrium angle is your reading.

The Radial Magnetic Field — The Key Trick

Here is the clever part. If the magnetic field were uniform and the coil rotated out of alignment, the torque would change with angle — making the scale non-linear. To avoid that, the poles of the magnet are shaped into concave cylindrical surfaces, and a soft iron cylinder is placed inside the coil. This creates a radial magnetic field: the field lines always point radially outward (or inward), so the plane of the coil is always parallel to the field as it rotates.

Important

In a radial field, the magnetic torque on the coil is independent of the coil's angular position. The torque depends only on the current.

That is what makes the deflection directly proportional to current — a linear scale.

The Physics in Equations

Let the coil have $N$ turns, each of area $A$ . A current $I$ flows through it. The magnetic field strength is $B$ (radial). The torque due to the magnetic field on a single turn is:

\tau_m = N I A B

This is because the force on each vertical side of the coil is $I L B$ (where $L$ is the length of the side), and the lever arm is the width of the coil, so the product gives $I \times (\text{area}) \times B$ per turn.

The spring provides a restoring torque proportional to the twist angle $\theta$ :

\tau_s = k \theta

where $k$ is the torsion constant of the spring (unit: N·m/rad).

At equilibrium, the two torques balance:

N I A B = k \theta

So the deflection is:

\theta = \frac{N A B}{k} \, I

The quantity $\frac{N A B}{k}$ is called the current sensitivity of the galvanometer. It tells you how many radians of deflection you get per ampere of current.

\theta = \left( \frac{N A B}{k} \right) I

What This Means for a Student

Larger $N$ , $A$ , or $B$ makes the galvanometer more sensitive — more deflection for the same current. …

Part (b)Concept understanding — Galvanometer to Voltmeter Conversion

Galvanometer to Voltmeter Conversion

A moving-coil galvanometer deflects fully only at a small current $I_g$ (its full-scale deflection current) and has internal resistance $G$ . To use it as a voltmeter — reading much larger voltages $V$ — a large resistance $R_s$ (the multiplier) is connected in series with the galvanometer coil. This series resistance limits the current to exactly $I_g$ when the full-scale voltage $V$ is applied, so the needle deflects fully and the scale is recalibrated to read volts instead of amperes.

The Conversion Formula

The galvanometer and $R_s$ together form a series circuit of total resistance $R_s + G$ . Applying $V$ across this series combination drives a current

I = \frac{V}{R_s + G}

We want this current to equal $I_g$ exactly when $V$ is the maximum (full-scale) voltage, so $I_g = \dfrac{V}{R_s+G}$ , which rearranges to:

R_s = \frac{V}{I_g} - G

Since the coil's deflection is proportional to the current through it, and that current is proportional to the applied voltage (Ohm's law), the resulting scale is linear in $V$ : equal voltage steps give equal angular deflections.

Why series, not parallel?

A voltmeter must be connected across the component whose voltage is being measured, without diverting current away from it — so it should draw as little current as possible, meaning its own resistance must be as large as possible. Adding $R_s$ in series does exactly that: it raises the meter's total resistance to $R_s + G$ , which is deliberately made large. (This is the opposite requirement to an ammeter, which sits in the current path and needs the smallest possible resistance — achieved there with a small shunt in parallel, not a large resistor in series.)

Worked Example

For $I_g = 1\ \text{mA}$ , $G = 50\ \Omega$ , converting to a $0$ – $10\ \text{V}$ voltmeter:

R_{\text{total}} = \frac{V}{I_g} = \frac{10}{0.001} = 10{,}000\ \Omega \quad\Longrightarrow\quad R_s = 10{,}000 - 50 = 9{,}950\ \Omega

A $9.95\ \text{k}\Omega$ resistor in series gives full-scale deflection at exactly $10\ \text{V}$ .

Watch out

Do not forget to subtract $G$ from $V/I_g$ . For most galvanometers $G$ is small next to $R_s$ , but in precision work it matters.

Key Properties

High input resistance: $R_v = R_s + G$ is large (kΩ to MΩ), so the voltmeter draws minimal current and barely disturbs the circuit it measures.
Linear scale: deflection ∝ current ∝ voltage. …

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Shared: (I) (B) a radial field keeps the torque constant; (II) (A) more turns raise current sensitivity. Part (a): (III) $\tau=NBIA=0.25\ \text{N·m}$ — (D). Part (b): (III OR) a $3985\ \Omega$ series resistor makes a 12 V voltmeter — (B); (IV) a $0.01\ \Omega$ parallel shunt makes a 10 A ammeter — (C).

A moving-coil galvanometer's deflecting torque is $\tau=NBIA\sin\theta$ , balanced by the hair-spring torque $k\phi$ . The design makes $\sin\theta=1$ always, so $\tau=NBIA$ and the deflection $\phi=\dfrac{NBA}{k}I$ is proportional to the current.

Part (a)

(I) Why the torque is constant

A cylindrical soft-iron core with concave pole pieces produces a radial field, always in the plane of the coil, so $\theta=90^\circ$ at every orientation and $\tau=NBIA$ is independent of the deflection angle. The core also boosts $B$ (option A), the hair spring gives the restoring torque (C), and eddy currents give damping (D) — none of these keeps the deflecting torque constant. Answer (B).

(II) Increasing current sensitivity

Current sensitivity is $S_I=\dfrac{\phi}{I}=\dfrac{NBA}{k}$ . Increasing $N$ , $B$ or $A$ , or decreasing $k$ , raises it; the standard, practical way is to increase the number of turns $N$ . Answer (A).

(III) Torque on the coil

With $N=50$ , $B=0.25\ \text{T}$ , $I=5\ \text{A}$ , $A=4.0\times10^{-3}\ \text{m}^2$ :

$\tau=NBIA=50\times0.25\times5\times4.0\times10^{-3}=0.25\ \text{N·m}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set 55/3/11 markMCQ
$Q. Assertion (A) : The cylindrical soft iron core in a moving coil galvanometer only makes the magnetic field radial and does not affect the strength of the magnetic field. Reason (R) : In a moving coil galvanometer, the plane of the coil is always perpendicular to the magnetic field. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.$
›Reveal solutionSolution

The soft iron core makes the field radial AND, being ferromagnetic, concentrates the magnetic flux — it increases the field strength, so the Assertion is false. In a radial field the plane of the coil is always parallel to the field lines (the coil's normal is perpendicular to $\vec B$ ), so the Reason is also false. The correct option is (D).

The question tests two separate facts about the moving coil galvanometer: what the cylindrical soft iron core actually does, and how the coil sits relative to the magnetic field.

1. Role of the soft iron core — is the Assertion true?

The core, together with the concave pole pieces, shapes the field in the air gap so that it is radial: at every angular position of the coil, the field lines point along the radius. This makes the deflecting torque independent of the coil's position, which is what gives the galvanometer its linear scale $\theta \propto I$ .

But that is not all the core does. Soft iron is ferromagnetic, with a very high relative permeability ( $\mu_r \gg 1$ ). It provides a low-reluctance path for the magnetic flux, so the flux crowds through the core and the field strength $B$ in the narrow air gap becomes much larger than it would be without the core. The claim that the core "only makes the field radial and does not affect the strength" is therefore false — the Assertion is false.

2. Orientation of the coil — is the Reason true?

In the radial field the field lines run along the radius, and the plane of the rectangular coil (tangential to the cylindrical core) always contains those field lines. So the plane of the coil is always parallel to the magnetic field — equivalently, the coil's normal is perpendicular to $\vec B$ in every position. That is exactly what keeps the torque at its maximum value throughout the rotation: …
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CBSE 2026Set ANNUAL1 mark
$Q. The deflection per unit voltage in a voltmeter is called ______.$
›Reveal solutionSolution

This ratio (deflection / voltage) is called the voltage sensitivity of the instrument.

A voltmeter is a galvanometer connected in series with a high resistance. Its voltage sensitivity is defined as the deflection theta produced per unit voltage V applied across it, i.e. voltage sensitivity = th …

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CBSE 2025Set D1 markMCQ
$Q. If the number of turns is increased in any moving coil galvanometer, then its sensitivity (A) increases (B) decreases (C) remains unchanged (D) may increase or may decrease$
›Reveal solutionSolution

A galvanometer's current sensitivity is (NAB/k), directly proportional to the number of turns N, so more turns → higher sensitivity.

The deflection of a moving-coil galvanometer is

$\theta = \frac{NAB}{k}\,I$

so its current sensitivity is

$\frac{\theta}{I} = \frac{NAB}{k}$

…

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CBSE 2024Set 55/5/11 markMCQ
$50\,\Omega$
›Reveal solutionSolution

A galvanometer becomes a voltmeter by adding a high resistance in series; the same full-scale deflection current must flow in both configurations, so we first find $I_g$ from the $2\,\text{V}$ setup, then calculate the series resistance needed for $10\,\text{V}$ . The answer is (C) $5.2\,\text{k}\Omega$ .

Why a galvanometer needs a series resistor to measure voltage

A galvanometer is fundamentally a current-measuring device. Its coil deflects fully when a specific current $I_g$ (the full-scale deflection current) flows through it. To measure voltage instead, we exploit Ohm's law: place a large resistance $R$ in series with the galvanometer so that when the desired maximum voltage $V$ appears across the combination, exactly $I_g$ flows through the circuit.

The total resistance $(G + R)$ and the voltage $V$ are related by:
$V = I_g (G + R)$
where $G$ is the galvanometer resistance. The key insight is that $I_g$ and $G$ are fixed properties of the galvanometer—they don't change when we reconfigure the circuit.

Step-by-step solution

1. Extract the full-scale deflection current from the first configuration

When the galvanometer is converted to a $(0\text{–}2\,\text{V})$ voltmeter using $R_1 = 1.0\,\text{k}\Omega = 1000\,\Omega$ in series:
$2 = I_g (50 + 1000)$ $I_g = \frac{2}{1050} = \frac{1}{525}\,\text{A}$
This current is an intrinsic property of the galvanometer. Any voltmeter configuration must respect this same $I_g$ for full-scale deflection.

2. Determine the series resistance for the $(0\text{–}10\,\text{V})$ range

For the new voltmeter to read up to $10\,\text{V}$ , the same current $I_g$ must flow when $10\,\text{V}$ is applied. Let the required series resistance be $R_2$ :

$10 = I_g (G + R_2)$ …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. In any electric circuit, galvanometer in its original form is used to -$
(a) detect the current
(b) measure the current
(c) measure the voltage
(d) measure the resistance
›Reveal solutionSolution

A galvanometer in its basic form is a sensitive current-detecting device, not a calibrated measuring instrument.

A galvanometer is a sensitive instrument used to detect the presence (and direction) of a small current in a circuit through the deflection of a coil/needle. In its original form it is not calibrated to read numerical values of current, voltage o …

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CBSE 2024Set A1 markMCQ
$Q. The value of current obtained in a moving coil galvanometer is proportional to (A) deflection (θ) (B) resistance (R) (C) magnetic field (B) (D) none of these$
›Reveal solutionSolution

A moving-coil galvanometer is linear: I ∝ θ (the deflection).

In a moving-coil galvanometer, the current-carrying coil in the radial magnetic field experiences a deflecting torque $NBIA$ , balanced by the restoring torque $k\theta$ of the suspension:

$NBIA = k\theta \;\Rightarrow\; I = \frac{k}{NBA}\,\theta.$

…

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CBSE 2024Set ANNUAL1 mark
$Q. To convert a galvanometer into a voltmeter a resistance of __________ value is connected in series to it.$
›Reveal solutionSolution

Adding a large series resistance limits the current through the galvanometer coil and extends the range so it can measure large potential differences directly.

A galvanometer is a sensitive instrument that gives full-scale deflection for only a very small current, and it has some small internal (coil) resistance $G$ . To use it as a voltmeter capable of reading a potential difference V, a high resistance R is connected in series with it, chosen so that:

$V = I_g (G + R)$

…

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CBSE 2024Set ANNUAL1 mark
$Q. Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer ?$
›Reveal solutionSolution

The soft-iron core makes the field radial, ensuring the deflecting torque (and hence the scale) is uniform.

In a moving-coil galvanometer, concave pole pieces together with a cylindrical soft-iron core placed inside the coil make the magnetic field radial — i.e. $\vec B$ is always along the plane of the coil, no matter what angle the coil has turned through. Because of this, the angle between the field and the normal to the coil stays 90° throughout the motion, so the deflecting torque $\tau = NBIA$ has no $\sin\theta$ dependence and stays proportional to the current $I$ alone. This gives a uniform torque for a given current at every deflection, so the pointer's deflection is directly proportional to the cu …

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CBSE 2023Set B1 mark
$Q. How is Galvanometer changed into a voltmeter?$
›Reveal solutionSolution

Connecting a high resistance in series with a galvanometer converts it into a voltmeter of the desired range.

A galvanometer alone can only detect small currents and has a small, fixed full-scale voltage rating (since its coil has resistance $G$ ). To use it as a voltmeter capable of measuring a chosen larger range of potential difference $V$ , a high resistance $R$ is connected in series with the galvanometer coil, so that at full-scale deflection current $I_g$ , $V = I_g(G+R)$ , giving $R = \dfrac{V}{I_g} - G$ . This series resistance both extends the voltage range and ensures the …

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CBSE 2023Set ANNUAL1 mark
$Q. What happens to the voltage sensitivity of the galvanometer when the current sensitivity of a moving coil galvanometer is doubled by doubling the number of turns of the coil?$
›Reveal solutionSolution

Doubling the turns doubles both the current sensitivity and the coil's resistance, leaving voltage sensitivity the same.

Current sensitivity of a galvanometer is $I_s = \dfrac{NBA}{k}$ , which is directly proportional to the number of turns $N$ ; doubling $N$ doubles $I_s$ , as stated. Voltage sensitivity is defined as $V_s = \dfrac{I_s}{R} = \dfrac{NBA}{kR}$ , where $R$ is the resistance of the galvanometer coil. Since doubling the number of turns (with the same wire, same coil geometry) also roughly doubles the length of wire used and hence doubles the coil's resistance $R$ , the factor of 2 in the numerator ( $N$ ) is cancelled by the factor of 2 in the den …

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CBSE 2022Set I1 markMCQ
$Q. Which of the following is correct for galvanometer constant? (A) C/(NAB) (B) C(B)/(NA) (C) C(A/(NB)) (D) C(N/(AB))$
›Reveal solutionSolution

Galvanometer constant $k = \dfrac{C}{NAB}$ .

In a moving-coil galvanometer, the current $I$ produces a torque $NIAB$ that is balanced at equilibrium by the restoring torque $C\theta$ of the suspension (torsional constant $C$ ):

$NIAB = C\theta \;\Rightarrow\; I = \left(\frac{C}{NAB}\right)\theta.$

The proportionality factor relating current to deflection is the galvanometer constant …

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CBSE 2022Set ANNUAL1 mark
$Q. How can a galvanometer be converted into a voltmeter?$
›Reveal solutionSolution

A galvanometer is converted into a voltmeter by putting a large resistance in series with it, so that most of the applied voltage drops across this resistance rather than across the (delicate, low-current) galvanometer coil.

A galvanometer of coil resistance $G$ gives full-scale deflection for a small current $I_g$ . To use it to measure a larger potential difference $V$ across some element, we connect a high resistance $R$ in series with the galvanometer. The combination (galvanometer + R) is then connected across the points whose voltage is to be measured. The value of $R$ needed for the voltmeter to read up to $V$ at full-scale deflection is found from: …

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Part (a)

(I) Why the torque is constant

(II) Increasing current sensitivity

(III) Torque on the coil

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