JEE Physics 2023 Question with Solution

Given: A variable voltage source $V$, test resistor $R_T = 10 \, \Omega$, two identical galvanometers $G_1$ and $G_2$, and resistors $R_1 = 10M\Omega$ and $R_2 = 0.001\Omega$.

Find: The most suitable circuit for verifying Ohm’s law.

To verify Ohm’s law, the ammeter must be connected in series with the test resistor and the voltmeter must be connected in parallel across it.

A galvanometer is converted into an ammeter by connecting a very small shunt resistance in parallel. Here, $R_2 = 0.001\Omega$ is very small, so it should be used as the shunt.

A galvanometer is converted into a voltmeter by connecting a very large resistance in series. Here, $R_1 = 10M\Omega$ is very large, so it should be used in series with the galvanometer.

Thus, the correct circuit is the one in which:

1. $G_1$ with $R_1$ acts as a voltmeter across $R_T$
2. $G_2$ with $R_2$ acts as an ammeter in series in the main circuit

All these conditions are satisfied by option (4).

Therefore, the correct option is D.

Given: The two added resistors are $R_1 = 10M\Omega$ and $R_2 = 0.001\Omega$.

Find: Which circuit uses the galvanometers correctly.

The required instrument placements are:

• Voltmeter: high resistance, connected in parallel
• Ammeter: low resistance, connected in series

So, for the galvanometers:

• Use $R_1$ in series with a galvanometer to make a voltmeter
• Use $R_2$ in parallel with a galvanometer to make an ammeter

Now compare the circuits. The suitable one is the circuit where the voltmeter combination is across $R_T$ and the ammeter combination is in series with the circuit.

Hence, the correct choice is option (4), that is D.