A galvanometer of resistance 200 Omega is to | vedclass

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(D) Given:Resistance of galvanometer $G = 200 \Omega$.Current through the galvanometer $i_g = 3 \% 	ext{ of } i = 0.03i$.To convert a galvanometer in

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$6$

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(D) Given:Resistance of galvanometer $G = 200 \Omega$.Current through the galvanometer $i_g = 3 \% 	ext{ of } i = 0.03i$.To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with it.The potential difference across the galvanometer and the shunt resistance is the same:$i_g G = (i - i_g) S$Rearranging for $S$:$S = \frac{i_g G}{i - i_g}$Substitute the given values:$S = \frac{0.03i 	imes 200}{i - 0.03i}$$S = \frac{6i}{0.97i}$$S = \frac{6}{0.97} \approx 6.18 \Omega$Rounding to the nearest integer,the shunt resistance is $6 \Omega$.

$A$ galvanometer of resistance $200 \Omega$ is to be converted into an ammeter. The value of shunt resistance which allows $3 \%$ of the main current through the galvanometer is equal to (nearly) (in $\Omega$)

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