QJEEM24Ph01a3: JEE Main 2024 Physics - Class 12 JEEM24Ph01a3 Solution: A galvanometer has a coil of resistance with a full scale…

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: $0.20 \Omega$