Question

In a circuit for finding the resistance of a galvanometer by half deflection method, a $6\, V$ battery and a high resistance of $11 \, k\, \Omega$ are used. The figure of merit of the galvanometer is $60 \, mu A$ 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,
$540 \times 10^{-6} =\frac{6}{(11000+G)}$
$[11000+G] 90 \times 10^{-6}=1$
$99000+9 G =10^{5}$
$9 G =100000-99000$
$9 G =1000$
$G =\frac{1000}{9} \Omega$
Also in half deflection method,
$G =\frac{R S}{R-S}$
$\Rightarrow \frac{1000}{9}=\frac{11000 S}{11000-S}$
$\frac{1}{9} =\frac{11 S}{11000-S}$
$\Rightarrow 11000-S=99\, S$
$100\, S =11000 \Rightarrow S=110\, \Omega$