The resistance of a galvanometer is 25,Omega | vedclass

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Question

(A) Given:Galvanometer resistance,$G = 25\,\Omega$Full-scale deflection current,$i_g = 50\,\mu A = 50 	imes 10^{-6}\,A = 5 	imes 10^{-5}\,A$Required

Vedclass · Accepted Answer

$2.5 	imes 10^{-4}\,\Omega$

Vedclass · Answer

(A) Given:Galvanometer resistance,$G = 25\,\Omega$Full-scale deflection current,$i_g = 50\,\mu A = 50 	imes 10^{-6}\,A = 5 	imes 10^{-5}\,A$Required range of ammeter,$i = 5\,A$The formula for shunt resistance $S$ is given by:$S = \frac{G \cdot i_g}{i - i_g}$Since $i_g$ is very small compared to $i$,we can approximate $i - i_g \approx i$ or use the exact formula:$S = \frac{25 	imes 5 	imes 10^{-5}}{5 - 5 	imes 10^{-5}}$$S = \frac{125 	imes 10^{-5}}{4.99995} \approx \frac{125 	imes 10^{-5}}{5} = 25 	imes 10^{-5} = 2.5 	imes 10^{-4}\,\Omega$

The resistance of a galvanometer is $25\,\Omega$ and it requires $50\,\mu A$ for full-scale deflection. The value of the shunt resistance required to convert it into an ammeter of $5\,A$ range is:

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