JEE Main 2025 — Current Electricity Question with Solution

Step 1: Find the Total Resistance

First, add up the resistor in series ($R_1=150.4\Omega$) with the combined resistance of the ammeter and its shunt ($R_2$).

The ammeter ($240\Omega$) is in parallel with a shunt of ($10\Omega$). The combined resistance of resistors in parallel is found by:

$$R_2=\frac{240\times 10}{240+10}=\frac{2400}{250}=9.6\Omega$$

The total resistance ($R_{eq}$) is:

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

Step 2: Find the Current in the Circuit

Suppose the total current in the circuit is $I$. But to find the ammeter reading, we care about the current through the ammeter coil itself.

The current through the ammeter, $I_1$, is given by how the current splits between the ammeter and the shunt. The fraction going through the ammeter is:

$$I_1=\frac{R_2}{240}\times I$$

But $R_2=9.6\Omega$ (the parallel combination), and $240\Omega$ is the ammeter's resistance:

$$I_1=\frac{9.6}{240}\times I$$

Step 3: Substitute the Circuit Values

The supply voltage is $20V$. The total resistance is $160\Omega$. So the total current is:

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

So: $$I_1=0.125\times \frac{9.6}{240}$$

Calculate this: $$I_1=0.125\times 0.04=0.005A=5\text{mA}$$

Final Answer:

The ammeter reads $5\text{mA}$.