JEE Physics 2025 Question with Solution

Given: Series resistor $R_s = 150.4 \, \Omega$, ammeter resistance $R_A = 240 \, \Omega$, shunt resistance $R_{sh} = 10 \, \Omega$, source voltage $V = 20 \, \text{V}$.

Find: The current through the ammeter in $\text{mA}$.

The ammeter and shunt are in parallel, so their equivalent resistance is

$$R_p = \frac{R_A R_{sh}}{R_A + R_{sh}} = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6 \, \Omega$$

Now this parallel combination is in series with $150.4 \, \Omega$, so the total resistance is

$$R_{eq} = R_s + R_p = 150.4 + 9.6 = 160 \, \Omega$$

Using Ohm's law, the total current in the circuit is

$$I_{total} = \frac{V}{R_{eq}} = \frac{20}{160} = 0.125 \, \text{A}$$

This current divides between the ammeter and the shunt. By current division, current through the ammeter is

$$I_A = I_{total} \times \frac{R_{sh}}{R_A + R_{sh}} = 0.125 \times \frac{10}{240 + 10}$$ $$I_A = 0.125 \times \frac{10}{250} = 0.005 \, \text{A}$$

Converting to milliampere,

$$I_A = 0.005 \times 1000 = 5 \, \text{mA}$$

Therefore, the reading of the ammeter is $5 \, \text{mA}$.

Given: The parallel branch contains the ammeter of resistance $240 \, \Omega$ and shunt of resistance $10 \, \Omega$. The source voltage is $20 \, \text{V}$.

Find: Current through the ammeter.

First find the equivalent resistance of the parallel branch:

$$R_p = \frac{240 \times 10}{240 + 10} = 9.6 \, \Omega$$

Hence total circuit resistance is

$$R_{eq} = 150.4 + 9.6 = 160 \, \Omega$$

So total current is

$$I = \frac{20}{160} = 0.125 \, \text{A}$$

The voltage across the parallel combination is

$$V_p = I R_p = 0.125 \times 9.6 = 1.2 \, \text{V}$$

This same voltage appears across the ammeter, so

$$I_A = \frac{V_p}{R_A} = \frac{1.2}{240} = 0.005 \, \text{A} = 5 \, \text{mA}$$

This works because all branches in parallel have the same potential difference. Therefore, the ammeter reading is $5 \, \text{mA}$.