JEE MainPhysics

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Current Electricity: JEE Main Physics Question with Solution

A student is provided with a variable voltage source

\mathrm{V}

, a test resistor

\mathrm{R}_{\mathrm{T}}=10 \Omega

, two identical galvanometers

\mathrm{G}_{1}

and

\mathrm{G}_{2}

and two additional resistors,

\mathrm{R}_{1}=10 \mathrm{M} \Omega

and

\mathrm{R}_{2}=0.001 \Omega

. For conducting an experiment to verify ohms law, the most suitable circuit is:

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Four circuit diagrams showing different arrangements of galvanometers G1 and G2 with shunt/series resistors R1 and R2 connected with test resistor RT and variable voltage source V.

Step-by-step solutionView

Correct answer

An ammeter requires a very small resistance in parallel with a galvanometer, while a voltmeter requires a very large resistance in series with a galvanometer; both are then connected in series and parallel with the test resistor, respectively.

Option analysis

Why each option works or fails

A ·

Believing that connecting the small resistor $R_2$ in series with a galvanometer forms an ammeter, or placing the voltmeter in series with the load resistor. To create an ammeter, the low-resistance shunt must be in parallel with the galvanometer. An ammeter must then be connected in series with the test resistor.

B ·

Confusing the shunt resistor of an ammeter with the multiplier resistor of a voltmeter by putting the large resistor $R_1$ in parallel. A voltmeter needs a high resistance in series with the galvanometer, not in parallel. A parallel high resistance does not effectively limit current through a galvanometer.

C ·

None. This circuit correctly configures $G_1$ with $R_2$ in parallel to act as an ammeter in series with $R_T$ , and $G_2$ with $R_1$ in series to act as a voltmeter across $R_T$ . Correctly pairs low resistance in parallel for current measurement and high resistance in series for voltage measurement across the test resistor.

D ·

Reversing the functional roles of the converted meters by placing the voltmeter in series with the circuit and the ammeter in parallel. The ammeter ( $G_1$ with parallel $R_2$ ) must be placed in series with $R_T$ , while the voltmeter ( $G_2$ with series $R_1$ ) must be placed in parallel across $R_T$ .

Reviewed route

Solution

StepWorking

01concept

To verify Ohm's law for $R_T$ , we need an ammeter connected in series with $R_T$ to measure the current $I$ , and a voltmeter connected in parallel across $R_T$ to measure the potential difference $V$ . 1. Ammeter conversion: A galvanometer is converted into an ammeter by connecting a very low resistance (shunt, $R_2 = 0.001\,\Omega$ ) in parallel with it. 2. Voltmeter conversion: A galvanometer is converted into a voltmeter by connecting a very large resistance ( $R_1 = 10\,\text{M}\Omega$ ) in series with it.

02option_verdict

In option (0), the voltmeter is formed by putting $R_1$ in parallel with $G_1$ (which decreases effective resistance instead of increasing it) and the ammeter has $R_2$ in series with $G_2$ (which would unnecessarily add resistance in series).

03option_verdict

In option (1), the ammeter has $R_1$ in parallel and the voltmeter has $R_2$ in series, which swaps the roles of high and low resistors.

04option_verdict

Option (2) correctly puts $R_1 = 10\,\text{M}\Omega$ in series with $G_1$ to make an ideal-like voltmeter connected in parallel with $R_T$ , and puts $R_2 = 0.001\,\Omega$ in parallel with $G_2$ to make an ammeter connected in series with $R_T$ .

05option_verdict

In option (3), the voltmeter branch is incorrectly connected in series with $R_T$ and the ammeter is connected in parallel.

✓discriminator

Look for $R_2 = 0.001\,\Omega$ (small) in parallel with $G_2$ in series with $R_T$ , and $R_1 = 10\,\text{M}\Omega$ (large) in series with $G_1$ across $R_T$ . This uniquely identifies option (2).

Hints that build this answer step by step

How should a galvanometer be modified using an external resistor to function as an ammeter?
Connect a very small resistance ( $R_2 = 0.001\,\Omega$ ) in parallel with the galvanometer.
How should a galvanometer be modified using an external resistor to function as a voltmeter?
Connect a very large resistance ( $R_1 = 10\,\text{M}\Omega$ ) in series with the galvanometer.
How must the constructed ammeter and voltmeter be connected relative to the test resistor $R_T$ to verify Ohm's law?
The ammeter in series with $R_T$ , and the voltmeter in parallel across $R_T$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

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

An ammeter is placed in series, so low resistance ensures it does not alter the circuit current. A voltmeter is placed in parallel, so high resistance ensures it draws negligible current away from the component being measured.

Answer

An ammeter requires a very small resistance in parallel with a galvanometer, while a voltmeter requires a very large resistance in series with a galvanometer; both are then connected in series and parallel with the test resistor, respectively.

Why each option works or fails

A: - Believing that connecting the small resistor $R_2$ in series with a galvanometer forms an ammeter, or placing the voltmeter in series with the load resistor. To create an ammeter, the low-resistance shunt must be in parallel with the galvanometer. An ammeter must then be connected in series with the test resistor.
B: - Confusing the shunt resistor of an ammeter with the multiplier resistor of a voltmeter by putting the large resistor $R_1$ in parallel. A voltmeter needs a high resistance in series with the galvanometer, not in parallel. A parallel high resistance does not effectively limit current through a galvanometer.
C · correct: - None. This circuit correctly configures $G_1$ with $R_2$ in parallel to act as an ammeter in series with $\mathrm{R}_{\mathrm{T}}=10 \Omega$ , and $G_2$ with $R_1$ in series to act as a voltmeter across $R_T$ . Correctly pairs low resistance in parallel for current measurement and high resistance in series for voltage measurement across the test resistor.
D: - Reversing the functional roles of the converted meters by placing the voltmeter in series with the circuit and the ammeter in parallel. The ammeter ( $G_1$ with parallel $\mathrm{R}_{2}=0.001 \Omega$ ) must be placed in series with $R_T$ , while the voltmeter ( $G_2$ with series $R_1$ ) must be placed in parallel across $R_T$ .

Step-by-step solution

concept: To verify Ohm's law for $R_T$ , we need an ammeter connected in series with $R_T$ to measure the current $I$ , and a voltmeter connected in parallel across $R_T$ to measure the potential difference $V$ . 1. Ammeter conversion: A galvanometer is converted into an ammeter by connecting a very low resistance (shunt, $R_2 = 0.001\,\Omega$ ) in parallel with it. 2. Voltmeter conversion: A galvanometer is converted into a voltmeter by connecting a very large resistance ( $R_1 = 10\,\text{M}\Omega$ ) in series with it.
option_verdict: In option (0), the voltmeter is formed by putting $R_1$ in parallel with $G_1$ (which decreases effective resistance instead of increasing it) and the ammeter has $R_2$ in series with $G_2$ (which would unnecessarily add resistance in series).
option_verdict: In option (1), the ammeter has $R_1$ in parallel and the voltmeter has $R_2$ in series, which swaps the roles of high and low resistors.
option_verdict: Option (2) correctly puts $R_1 = 10\,\text{M}\Omega$ in series with $G_1$ to make an ideal-like voltmeter connected in parallel with $R_T$ , and puts $R_2 = 0.001\,\Omega$ in parallel with $G_2$ to make an ammeter connected in series with $R_T$ .
option_verdict: In option (3), the voltmeter branch is incorrectly connected in series with $R_T$ and the ammeter is connected in parallel.
discriminator: Look for $R_2 = 0.001\,\Omega$ (small) in parallel with $G_2$ in series with $R_T$ , and $R_1 = 10\,\text{M}\Omega$ (large) in series with $G_1$ across $R_T$ . This uniquely identifies option (2).

Current Electricity: JEE Main Physics Question with Solution

What feels right?

An ammeter requires a very small resistance in parallel with a galvanometer, while a voltmeter requires a very large resistance in series with a galvanometer; both are then connected in series and parallel with the test resistor, respectively.

Why each option works or fails

Solution

01concept

02option_verdict

03option_verdict

04option_verdict

05option_verdict

✓discriminator

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