JEE Physics 2025 Question with Solution

Given: Resistance of the galvanometer coil is $R_g = 30 \, \Omega$, full-scale deflection current is $I_g = 20 \, \text{mA} = 0.02 \, \text{A}$, and maximum current to be measured is $I = 3 \, \text{A}$.

Find: The shunt resistance $R_s$ to be connected in parallel with the galvanometer.

For a galvanometer converted into an ammeter, the shunt carries the excess current. Hence,

$$I_s = I - I_g = 3 - 0.02 = 2.98 \, \text{A}$$

Since the galvanometer and shunt are in parallel, the potential difference across them is the same:

$$I_g R_g = I_s R_s$$

Therefore,

$$R_s = \frac{I_g R_g}{I_s} = \frac{0.02 \times 30}{2.98} = \frac{0.6}{2.98} \approx 0.2013 \, \Omega$$

So the shunt resistance obtained from the working is $0.2013 \, \Omega$.

The solution and the listed correct option state $149$, but this does not match the extracted calculation. Therefore, there is a discrepancy in the source. Based on the source conclusion, the correct option is C.

Given: The galvanometer can safely carry only $0.02 \, \text{A}$ and has resistance $30 \, \Omega$.

Find: The value of the parallel shunt resistance.

A total current of $3 \, \text{A}$ is to be measured, so most of the current must bypass the galvanometer through the shunt.

Current through shunt:

$$I_s = 3 - 0.02 = 2.98 \, \text{A}$$

Voltage across galvanometer:

$$V = I_g R_g = 0.02 \times 30 = 0.6 \, \text{V}$$

The same voltage appears across the shunt, so

$$R_s = \frac{V}{I_s} = \frac{0.6}{2.98} \approx 0.201 \, \Omega$$

Thus, the physically correct shunt resistance from the shown steps is $0.201 \, \Omega$.

However, the solution's explicitly marks option C as correct and writes the final answer as $149$. Hence the page contains an internal inconsistency, and the recorded answer is C.