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(B) The formula for the shunt resistance $S$ required to convert a galvanometer of resistance $G$ and full-scale deflection current $i_g$ into an amme

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$0.02$

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(B) The formula for the shunt resistance $S$ required to convert a galvanometer of resistance $G$ and full-scale deflection current $i_g$ into an ammeter of range $i$ is given by: $S = \frac{i_g}{i - i_g} G$,which implies $i_g G = (i - i_g) S$.For the first case,with shunt $S_1 = 4r$ and range $i_1 = 0.03 \, A$:$i_g G = (0.03 - i_g) 4r$  --- $(i)$For the second case,with shunt $S_2 = r$ and range $i_2 = 0.06 \, A$:$i_g G = (0.06 - i_g) r$  --- $(ii)$Equating $(i)$ and $(ii)$:$(0.03 - i_g) 4r = (0.06 - i_g) r$$4(0.03 - i_g) = 0.06 - i_g$$0.12 - 4i_g = 0.06 - i_g$$3i_g = 0.06$$i_g = 0.02 \, A$.

When a shunt resistance of $4r$ is connected to a galvanometer,it becomes an ammeter that can measure $0.03 \, A$. When a shunt resistance of $r$ is connected to the same galvanometer,it becomes an ammeter that can measure $0.06 \, A$. What is the current capacity $(i_g)$ of the galvanometer in $A$?

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