A galvanometer having a coil of resistance 30 $Ω$ need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X} Ω,$ where X is

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149

298

596

447

answer is C.

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Detailed Solution

I_g R_g = (I – I_g) r_s
20 × 10^–3 × 30 = (3 – 0.02) × r_s
$\begin{array}{l} r_{5} = (\frac{0.6}{2.98}) = \frac{30}{x} \\ x = (\frac{2.98 \times 30}{0.6}) = 149 \end{array}$

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