The resistance of a galvanometer is 50,Omega | vedclass

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(D) Given: Galvanometer resistance $G = 50\,\Omega$,full-scale deflection current $I_G = 100\,\mu A = 100 	imes 10^{-6}\,A$,and the desired range $I

Vedclass · Accepted Answer

$5 	imes 10^{-4}\,\Omega$

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(D) Given: Galvanometer resistance $G = 50\,\Omega$,full-scale deflection current $I_G = 100\,\mu A = 100 	imes 10^{-6}\,A$,and the desired range $I = 10\,A$.To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.The formula for shunt resistance is $S = \left( \frac{I_G}{I - I_G} ight) G$.Substituting the values: $S = \left( \frac{100 	imes 10^{-6}}{10 - 100 	imes 10^{-6}} ight) 	imes 50$.Since $100 	imes 10^{-6} = 10^{-4}$ is very small compared to $10$,the denominator $10 - 10^{-4} \approx 10$.Thus,$S \approx \left( \frac{10^{-4}}{10} ight) 	imes 50 = 10^{-5} 	imes 50 = 5 	imes 10^{-4}\,\Omega$.

The resistance of a galvanometer is $50\,\Omega$ and the current required to give full-scale deflection is $100\,\mu A$. In order to convert it into an ammeter for reading up to $10\,A$,it is necessary to put a resistance of:

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