State the principle of a moving coil galvanometer. Explain its working and obtain the expression for the defle

Question

Q. State the principle of a moving coil galvanometer. Explain its working and obtain the expression for the deflection produced due to the current passed through the coil. Define current sensitivity.

(OR)

Explain how a galvanometer can be converted into an ammeter of a given range. Derive an expression for shunt resistance and current for full scale deflection. Find the effective resistance of the ammeter.

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★

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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 Ammeter Conversion (Shunt)

Galvanometer to Ammeter Conversion (Shunt)

A moving-coil galvanometer carries only a tiny full-scale current $I_g$ (and has resistance $G$ ), so on its own it can measure at most $I_g$ . To read a much larger current $I$ , we connect a small resistance $S$ (the shunt) in parallel with the galvanometer. The shunt diverts most of the current, letting only $I_g$ pass through the coil.

Key idea: the galvanometer and shunt are in parallel, so they share the same potential difference:

I_g\,G = (I - I_g)\,S \quad\Longrightarrow\quad S = \frac{I_g}{I - I_g}\,G

Because $I \gg I_g$ , the shunt $S$ is very small, and the combined resistance of the ammeter is even smaller than $S$ — ideal, since an ammeter must not disturb the circuit it measures. …

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Part (a): A moving-coil galvanometer balances the magnetic torque $NIAB$ against the spring torque $k\theta$ , giving $\theta=\dfrac{NAB}{k}I$ and current sensitivity $\dfrac{NAB}{k}$ . Part (b): A shunt $S=\dfrac{I_g G}{I-I_g}$ in parallel converts it to an ammeter of range $I$ ; its effective resistance is $R_A=\dfrac{I_g G}{I}$ .

Part (a) — Moving-coil galvanometer

Principle. When a current-carrying coil is placed in a magnetic field it experiences a torque proportional to the current; measuring the resulting deflection measures the current.

Construction and working. A rectangular coil of $N$ turns and area $A$ is suspended between the poles of a permanent magnet with a cylindrical soft-iron core, which makes the field radial. As a result the plane of the coil is always parallel to $\vec B$ , so each side always feels the maximum force and the deflecting torque is

\tau_d=NIAB\qquad(\text{no }\sin\theta\text{ factor, thanks to the radial field}).

The suspension fibre (torsion constant $k$ ) supplies a restoring torque $\tau_r=k\theta$ . At equilibrium $\tau_d=\tau_r$ :

NIAB=k\theta\ \Rightarrow\ \theta=\frac{NAB}{k}\,I.

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

The deflection is directly proportional to the current — a linear (uniform) scale.

Current sensitivity is the deflection produced per unit current:

$S_I=\frac{\theta}{I}=\frac{NAB}{k}.$ …

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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 24 on this concept.

CBSE 2026Set 55/2/11 markMCQ
$27\ \Omega$
›Reveal solutionSolution
A galvanometer is converted to an ammeter by connecting a small shunt resistor in parallel. Using the current division rule, the full-scale deflection current of the galvanometer is found to be $1\ \text{mA}$ , which corresponds to option (C).

When a galvanometer is converted into an ammeter, a small resistance (shunt) is connected in parallel with it. The purpose is to allow most of the current to bypass the delicate galvanometer coil, so only a small fraction passes through the meter itself. The galvanometer shows full-scale deflection when the current through its coil reaches its maximum rated value, say $I_g$ . The shunt carries the remaining current.

Here, the galvanometer resistance is $G = 27\ \Omega$ , the shunt resistance is $S = 3\ \Omega$ , and the ammeter range is $0$ to $10\ \text{mA}$ — meaning the total current that produces full-scale deflection in the ammeter is $I = 10\ \text{mA}$ .

Let’s work through the reasoning step by step.

Understand the parallel connection In an ammeter, the galvanometer and shunt are in parallel. So the voltage across both is the same. If $I_g$ is the current through the galvanometer at full deflection, and $I_s$ is the current through the shunt, then:

$I_g \cdot G = I_s \cdot S$
Also, the total current entering the ammeter is:
$I = I_g + I_s$

Express $I_s$ in terms of $I_g$ From the voltage equality:

$I_s = I_g \cdot \frac{G}{S}$
Substitute into the total current equation:
$I = I_g + I_g \cdot \frac{G}{S} = I_g \left(1 + \frac{G}{S}\right)$

Plug in the given values $G = 27\ \Omega$ , $S = 3\ \Omega$ , $I = 10\ \text{mA}$ :

$10 = I_g \left(1 + \frac{27}{3}\right) = I_g \left(1 + 9\right) = I_g \cdot 10$
Therefore:

$I_g = \frac{10}{10} = 1\ \text{mA}$ …
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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 2025Set 55/6/11 markMCQ
$Q. A galvanometer can be converted into an ammeter of desired range by connecting a: (A) small resistance in series (B) large resistance in series (C) small resistance in parallel (D) large resistance in parallel$
›Reveal solutionSolution
To convert a galvanometer into an voltmeter, a large resistance is connected in series with it. For an ammeter, a small resistance is connected in parallel. The question asks for ammeter conversion, so the correct choice is (C) small resistance in parallel.

The key idea is that a galvanometer is a sensitive current-measuring device that deflects fully for a small current (its full-scale deflection current, $I_g$ ). To measure larger currents (as an ammeter does), we need to bypass most of the current around the galvanometer coil, protecting it from burning out. This is done by connecting a shunt — a small resistance — in parallel.

Why parallel? Because a parallel path divides the current. The galvanometer still sees only $I_g$ at full deflection, while the shunt carries the excess current $(I - I_g)$ . The shunt resistance $S$ is chosen so that at the desired maximum current $I$ , exactly $I_g$ flows through the galvanometer. Since the voltage across parallel branches is equal:
$I_g \cdot G = (I - I_g) \cdot S$
where $G$ is the galvanometer resistance. Solving:
$S = \frac{I_g G}{I - I_g}$
For a large range ( $I \gg I_g$ ), $S$ becomes very small — hence a small resistance in parallel.

Watch out

A common mistake is confusing ammeter and voltmeter conversion. For a voltmeter, you add a large series resistance to limit voltage. For an ammeter, you add a small parallel resistance to shunt current. Mixing them up leads to wrong answers.

Now, let's work through the reasoning step by step:

Understand the galvanometer's limitation: A galvanometer is essentially a sensitive moving-coil meter with resistance $G$ (typically 10–100 $\Omega$ ) and full-scale deflection current $I_g$ (often a few mA). It cannot handle large currents directly — passing a large current through it would permanently damage the coil.

Goal of an ammeter: An ammeter must measure a wide range of currents (say 0–1 A or more) while offering very low resistance to the circuit, so it doesn't disturb the current being measured. The galvanometer alone has too high a resistance and too low a current capacity.

Why parallel (shunt) works: Connecting a small resistance $S$ in parallel creates a current divider. At full-scale deflection, the total current $I$ entering the ammeter splits: $I_g$ through the galvanometer and $(I - I_g)$ through the shunt. The shunt "steals" the excess current. The parallel combination also reduces the overall ammeter resistance to $\frac{G S}{G+S}$ , which is very small — ideal for an ammeter.

Derive the shunt value: Using the voltage equality across parallel branches: …
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CBSE 2025Set D1 markMCQ
$Q. When an ammeter is shunted then its measurement limit (A) increases (B) decreases (C) remains unchanged (D) none of these$
›Reveal solutionSolution

A shunt is a small resistance placed in parallel with the ammeter; it bypasses most of the current so the meter reads only a fraction, extending (increasing) its range.

An ammeter (galvanometer) can carry only a small full-scale current $I_g$ . To measure a larger current I, a low resistance shunt S is connected in parallel:

$S = \frac{I_g}{I - I_g}\,G$

…

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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 2025Set ANNUAL1 mark
$Q. To convert a galvanometer into an ammeter ____________ is connected in parallel to it.$
›Reveal solutionSolution

A galvanometer becomes an ammeter by adding a low-resistance shunt in parallel, so most of the current bypasses the sensitive galvanometer coil.

A galvanometer is a sensitive device with a resistance $G$ that can only safely carry a small current $I_g$ for full-scale deflection. To measure large currents, a small resistance called a shunt ( $S$ ) is connected in parallel with the galvanometer. Most of the current flows through the low-resistance shunt, and only the small fraction $I_g$ flows through the galvanom …

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CBSE 2024Set 55/1/11 markMCQ
$G$
›Reveal solutionSolution

When a galvanometer is converted to an ammeter using a shunt, only 0.1% of the total current flows through the galvanometer coil. Using the parallel-resistance formula and the current-division condition, the net resistance of the ammeter is $\boxed{\frac{G}{1000}}$ .

Why a shunt converts a galvanometer into an ammeter

A galvanometer is a sensitive current-measuring device with high resistance $G$ that can only handle a small current $I_g$ before its coil deflects fully. To measure larger currents, we place a low-resistance shunt $S$ in parallel with the galvanometer. Most of the current bypasses the galvanometer through this shunt, while a small fraction flows through the coil to produce the deflection.

The ammeter's effective resistance is the parallel combination of $G$ and $S$ , which must be very small so that inserting the ammeter into a circuit doesn't significantly alter the current being measured.

Step-by-step solution

1. Identify what flows through the galvanometer

The problem states that when the ammeter reads its full-scale value $I$ , the current through the galvanometer is $0.1\%$ of $I$ :
$I_g = 0.001 \, I = \frac{I}{1000}$
2. Find the current through the shunt

Since the galvanometer and shunt are in parallel, the total current splits between them:
$I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}$
3. Apply the voltage-equality condition

Both the galvanometer and shunt have the same potential difference across them (parallel connection). Using Ohm's law:
$V_g = V_s$ $I_g \cdot G = I_s \cdot S$
Substituting the currents:
$\frac{I}{1000} \cdot G = \frac{999I}{1000} \cdot S$
Simplifying:
$G = 999 \, S$ $S = \frac{G}{999}$
4. Calculate the ammeter's net resistance …
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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 A1 markMCQ
$Q. A galvanometer is converted into ammeter by adding (A) low resistance in parallel (B) high resistance in series (C) low resistance in series (D) high resistance in parallel$
›Reveal solutionSolution

An ammeter = galvanometer + a small shunt resistance in parallel.

An ammeter must (i) read large currents and (ii) have very low resistance so it does not disturb the circuit. A galvanometer is a sensitive, high-resistance device that can carry only a tiny current.

To convert it, a low resistance (shunt) S is connected in parallel with the galvanometer. Most of the current then passes through the shunt and only a small fixed fraction through the coil:

$S = \frac{I_g}{I - I_g}\,G.$

…

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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 F1 markMCQ
$Q. If any ammeter is shunted, then the total resistance of the circuit (A) increases (B) decreases (C) remains same (D) none of these$
›Reveal solutionSolution

A shunt is a small resistance in parallel with the ammeter, so the net resistance decreases.

An ammeter is shunted by connecting a low-value resistance (the shunt) in parallel with it, so that most of the current bypasses the meter coil. Two resistances in parallel give an equivalent resistance smaller than either one:

…

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Q.State the principle of a moving coil galvanometer. Explain its working and obtain the expression for the deflection produced due to the current passed through the coil. Define current sensitivity.

Part (a) — Moving-coil galvanometer

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