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(C) Given: Galvanometer resistance $G = 100 \, \Omega$,full-scale deflection current $I_g = 10^{-5} \, A$,and the desired range $I = 1 \, A$.To conver

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

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(C) Given: Galvanometer resistance $G = 100 \, \Omega$,full-scale deflection current $I_g = 10^{-5} \, A$,and the desired range $I = 1 \, A$.To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel.The formula for shunt resistance is $S = \frac{I_g \cdot G}{I - I_g}$.Substituting the values: $S = \frac{10^{-5} 	imes 100}{1 - 10^{-5}}$.Since $10^{-5}$ is negligible compared to $1$,we have $S \approx \frac{10^{-5} 	imes 10^2}{1} = 10^{-3} \, \Omega$.Thus,the required shunt resistance is $10^{-3} \, \Omega$.

An ammeter of $100 \, \Omega$ resistance gives full-scale deflection for a current of $10^{-5} \, A$. The shunt resistance required to convert it into an ammeter of $1 \, A$ range will be:

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