A galvanometer has a coil of resistance $200 \Omega$ with a full scale deflection at $20 \mu \mathrm{A}$. The value of resistance to be added to use it as an ammeter of range $(0-20) \mathrm{mA}$ is :

<p>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.</p> <p>Given parameters: <ul> <li>Resistance of the galvanometer's coil, $R_g = 200 \Omega$</li><br> <li>Full-scale deflection current of the galvanometer, $I_g = 20 \mu \mathrm{A} = 20 \times 10^{-6} \mathrm{A}$</li><br> <li>Desired ammeter range, $I = 20 \mathrm{mA} = 20 \times 10^{-3} \mathrm{A}$</p></li> </ul> <p>The shunt resistance, $R_s$, can be calculated using the formula:</p> <p> <p>$\frac{R_s}{R_g + R_s} = \frac{I_g}{I}$</p> </p> <p>Solving for $R_s$:</p> <p> <p>$R_s = R_g \left(\frac{I_g}{I - I_g}\right)$</p> </p> <p>Substituting the given values:</p> <p> <p>$$ R_s = 200 \left(\frac{20 \times 10^{-6}}{20 \times 10^{-3} - 20 \times 10^{-6}}\right) $$</p> </p> <p>Simplifying the expression:</p> <p> <p>$R_s = 200 \left(\frac{20 \times 10^{-6}}{19.98 \times 10^{-3}}\right)$</p> </p> <p> <p>$$ R_s \approx 200 \left(\frac{20 \times 10^{-6}}{20 \times 10^{-3}}\right) = 200 \left(\frac{1}{1000}\right) = 0.20 \Omega $$</p> </p> <p>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:</p> <p>Option C: $0.20 \Omega$</p>