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(B) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.Given: Galvanometer resistanc

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$50/99\,\Omega$ in parallel

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(B) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.Given: Galvanometer resistance $G = 50\,\Omega$,full-scale current $i_g = 10\,mA = 10 	imes 10^{-3}\,A$,and desired range $i = 1\,A$.The formula for shunt resistance is $S = \frac{i_g 	imes G}{i - i_g}$.Substituting the values: $S = \frac{10 	imes 10^{-3} 	imes 50}{1 - 10 	imes 10^{-3}} = \frac{0.5}{1 - 0.01} = \frac{0.5}{0.99} = \frac{50}{99}\,\Omega$.Therefore,a resistance of $50/99\,\Omega$ must be connected in parallel.

$A$ moving coil galvanometer has a resistance of $50\,\Omega$ and gives full-scale deflection for $10\,mA$. How could it be converted into an ammeter with a full-scale deflection for $1\,A$?

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