A galvanometer of resistance $40 \Omega$ is connected to a $6 \mathrm{~V}$ battery along with a resi

Question

A galvanometer of resistance $40 \Omega$ is connected to a $6 \mathrm{~V}$ battery along with a resistance of $1960 \Omega$ in series. A full-scale deflection of 40 divisions is obtained on the galvanometer. In order to cleanly reduce this deflection to exactly 20 divisions, the total resistance in series should be adjusted to

Professor PC Thomas Classes · Answer

(b) : The initial total resistance of the circuit is $R_{initial} = G + R_1 = 40 + 1960 = 2000 \Omega$. The initial current corresponding to 40 divisions is $I_1 = \frac{V}{R_{initial}} = \frac{6}{2000} = 3 \mathrm{~mA}$. To reduce the deflection to 20 divisions (exactly half), the current must be halved to $I_2 = 1.5 \mathrm{~mA}$. The new total required resistance is $R_{final} = \frac{V}{I_2} = \frac{6}{1.5 	imes 10^{-3}} = 4000 \Omega$. Since the galvanometer itself has a resistance of $40 \Omega$, the new series resistance must be $R_2 = 4000 - 40 = 3960 \Omega$.