The resistance of a galvanometer is 50,Omega | vedclass

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(C) The resistance of the galvanometer is $G = 50\,\Omega$.Given that the galvanometer requires $2\,\mu A$ for $2$ divisions,the current per division

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$3 	imes 10^{-4}\,\Omega$

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(C) The resistance of the galvanometer is $G = 50\,\Omega$.Given that the galvanometer requires $2\,\mu A$ for $2$ divisions,the current per division is $1\,\mu A/	ext{division}$.Since there are $30$ divisions on one side,the full-scale deflection current $I_g$ is:$I_g = 30 	imes 1\,\mu A = 30 	imes 10^{-6}\,A$.The range of the ammeter is $I = 5\,A$.The shunt resistance $S$ required is given by the formula:$S = \frac{I_g G}{I - I_g}$.Substituting the values:$S = \frac{30 	imes 10^{-6} 	imes 50}{5 - 30 	imes 10^{-6}}$.Since $30 	imes 10^{-6}$ is negligible compared to $5$,we have:$S \approx \frac{30 	imes 10^{-6} 	imes 50}{5} = 30 	imes 10^{-6} 	imes 10 = 300 	imes 10^{-6}\,\Omega = 3 	imes 10^{-4}\,\Omega$.

The resistance of a galvanometer is $50\,\Omega$ and it requires $2\,\mu A$ per two division deflection. The value of the shunt required in order to convert this galvanometer into an ammeter of range $5\,A$ is (The number of divisions on the galvanometer scale on one side is $30$).

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