Question

In a circuit for finding the resistance of a galvanometer by half deflection method, a $6V$ battery and a high resistance of $11k\Omega$ are used. The figure of merit of the galvanometer is $60muA$ division. In the absence of shunt resistance, the galvanometer produces a deflection of $\theta =9$ divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of $\theta /2$, is closest to :

• A. $550\Omega$
• B. $220\Omega$
• C. $55\Omega$
• D. $110\Omega$

Answer 1

Answer: (D)

Current required by unit deflection is $60\mu A$
For, $\theta =9$ current is $I=9\times 60\mu A$
$\Rightarrow I=540\mu A=540\times 10^{-6}A$
Let $G$ is resistance of galvanometer. Then,
$540\times 10^{-6}=\frac{6}{(11000+G)}$
$[11000+G]90\times 10^{-6}=1$
$99000+9G=10^5$
$9G=100000-99000$
$9G=1000$
$G=\frac{1000}{9}\Omega$
Also in half deflection method,
$G=\frac{RS}{R-S}$
$\Rightarrow \frac{1000}{9}=\frac{11000S}{11000-S}$
$\frac{1}{9}=\frac{11S}{11000-S}$
$\Rightarrow 11000-S=99S$
$100S=11000\Rightarrow S=110\Omega$