JEE Main 2025 — Current Electricity Question with Solution

Given:

Galvanometer resistance:

$$R_g=30\Omega$$

Full-scale deflection current:

$$I_g=20mA=0.02A$$

Maximum current to be measured:

$$I=3A$$

When using the galvanometer to measure 3 A, the extra current passing through the shunt resistor ($R_s$) is:

$$I_s=I-I_g=3-0.02=2.98A.$$

Since the galvanometer and the shunt resistor are connected in parallel, their voltage drops must be equal. Therefore:

$$I_g{R}_g=I_s{R}_s.$$

Substitute the given values:

$$0.02\times 30=2.98\times R_s\Rightarrow R_s=\frac{0.6}{2.98}\approx 0.2013\Omega .$$

The shunt resistance is represented as:

$$R_s=\frac{30}{X}\Omega .$$

Equate the two expressions:

$$\frac{30}{X}=\mathrm{0.2013.}$$

Solving for $X$:

$$X=\frac{30}{0.2013}\approx 149.$$

Thus, the value of $X$ is approximately $$149.$$