A galvanometer has a coil of resistance $50 \Omega$ and shows a full-scale deflection for a current

Question

A galvanometer has a coil of resistance $50 \Omega$ and shows a full-scale deflection for a current of $20 \mathrm{~mA}$. If it is to be operated as a voltmeter of $100 \mathrm{~V}$ range, the external resistance required to be added will be

Professor PC Thomas Classes · Answer

(d) : Given the resistance of the galvanometer, $G = 50 \Omega$, and the current for full-scale deflection, $I_g = 20 \mathrm{~mA} = 0.02 \mathrm{~A}$. The desired voltage range is $V = 100 \mathrm{~V}$. To convert the galvanometer into a voltmeter, a high resistance $R$ is connected in series. The total resistance of the voltmeter is $G + R$. By Ohm's law, $V = I_g(G + R)$. Rearranging for $R$, we get $R = \frac{V}{I_g} - G = \frac{100}{0.02} - 50 = 5000 - 50 = 4950 \Omega$.