Question

The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals ${\theta }_0$ (see figure). On closing $K_2$ also and adjusting $R_{2}to5\Omega$ , the deflection in galvanometer becomes $\frac{{\theta }_0}{5}$ .The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

• A. $12\Omega$
• B. $25\Omega$
• C. $5\Omega$
• D. $22\Omega$

Answer 1

Answer: (D)

$22\Omega$

case I $i_g=\frac{E}{220+R_g}=C{\theta }_0$ ..(i)
caseII $i_g=\left(\frac{E}{220+\frac{5R_g}{5+R_g}}\right)\times \frac{5}{(R_g+5)}=\frac{C{\theta }_0}{5}$...(ii)
$\Rightarrow \frac{5E}{225R_g+1100}=\frac{C{\theta }_0}{5}$ ..(ii)
$\frac{E}{220+R_g}=C\theta$
...(i)
$\Rightarrow \frac{225R_g+1100}{1100+5R_0}=5$
$\Rightarrow 5500+25R_g=225R_g=1100$
$200R_g=4400$
$R_g=22\Omega$
$-4$