In a circuit for finding the resistance of a galvanometer by half… | JEE Main 2018 (16 Apr, Shift 1)

Q172JEE Main 2018 · 16 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

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:

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Answer: (d)

No shunt: $I = 9 \times 60\ \mu\text{A} = 540\ \mu\text{A} = \frac{6}{11000 + G}$
$11000 + G = 11111 \Rightarrow G = 111\ \Omega$
Half deflection method with very large series resistance: $S \approx G$ (exact working gives $S \approx 110\ \Omega$ ).
So the answer is (d) $110\ \Omega$ .

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