JEE Main 2024 — Current Electricity Question with Solution

To convert a galvanometer into an ammeter, we need to add a shunt resistance in parallel with the galvanometer's coil. The purpose of the shunt resistance is to bypass the majority of the current while allowing only a small fraction of it to pass through the galvanometer, thereby preventing it from being damaged by high currents.

Given parameters:

• Resistance of the galvanometer's coil, $$R_g=200\Omega$$


• Full-scale deflection current of the galvanometer, $$I_g=20\mu \mathrm{A}=20\times 10^{-6}\mathrm{A}$$


• Desired ammeter range, $$I=20\mathrm{mA}=20\times 10^{-3}\mathrm{A}$$

The shunt resistance, $$R_s$$, can be calculated using the formula:

$$\frac{R_s}{R_g+R_s}=\frac{I_g}{I}$$

Solving for $$R_s$$:

$$R_s=R_g\left(\frac{I_g}{I-I_g}\right)$$

Substituting the given values:

$$R_s=200\left(\frac{20\times 10^{-6}}{20\times 10^{-3}-20\times 10^{-6}}\right)$$

Simplifying the expression:

$$R_s=200\left(\frac{20\times 10^{-6}}{19.98\times 10^{-3}}\right)$$

$$R_s\approx 200\left(\frac{20\times 10^{-6}}{20\times 10^{-3}}\right)=200\left(\frac{1}{1000}\right)=0.20\Omega$$

Therefore, the value of the resistance to be added to use the galvanometer as an ammeter of range $$(0-20)\mathrm{mA}$$ is $$0.20\Omega$$. Thus, the correct answer is:

Option C: $$0.20\Omega$$