Question 362 of 380

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 $60\mu A{division}^{-1}$. 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 $\dfrac{\theta }{2}$, is closest to:

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

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Answer: (D) $110\Omega$

We are given a galvanometer connected in the first circuit in the above figure, where $E=6V$, $R=11k\Omega$ and the deflection $\theta =9$ divisions. The galvanometer has a resistance of $G$. The current in the circuit,
$I=\dfrac{E}{R+G}...(1)$
Also, since the figure of merit is $60\mu A/division$
$I=60\times 9\mu A=540\mu A...(2)$
Substituting equation $(2)$ in $(1)$,
$540\times {10}^{-6}=\dfrac{6}{11,000+G} \Rightarrow G=\dfrac{6}{540\times {10}^{-6}}-11,000 \Rightarrow G=111\Omega ...(3)$
In the second circuit, it is given that the deflection is half that of the previous circuit. So, current flowing through the galvanometer half the first circuit. Hence,

$\dfrac{I}{2}=\dfrac{E}{R+\dfrac{GS}{G+S}}\times \dfrac{S}{S+G} \Rightarrow \dfrac{I}{2}=\dfrac{ES}{R\left(S+G\right)+GS}$

$\Rightarrow S=\dfrac{RG\times \dfrac{I}{2}}{E-\dfrac{(R+G)I}{2}}$
Substituting the values from equations $(1),(2)$ and $(3)$,

$S=\dfrac{11\times {10}^{3}\times 111\times \dfrac{540}{2}\times {10}^{-6}}{6-\left(\dfrac{6}{2}\right)} =110\Omega$