A galvanometer having a resistance of $18 \Omega$ is shunted by a wire of resistance $6 \Omega$. If

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A galvanometer having a resistance of $18 \Omega$ is shunted by a wire of resistance $6 \Omega$. If the total current in the external circuit is $4 \mathrm{~A}$, the part of it passing through the galvanometer will be

Professor PC Thomas Classes · Answer

(c) : The galvanometer and the shunt are connected in parallel. Let $I_g$ be the current through the galvanometer and $I_s$ be the current through the shunt. The potential difference across both branches must be equal, so $I_g 	imes G = I_s 	imes S$. We also know the total current is $I = I_g + I_s$. Substituting $I_s = I - I_g$, we get $I_g 	imes G = (I - I_g) 	imes S$, which rearranges to $I_g = I 	imes \frac{S}{G + S}$. Thus, $I_g = 4 	imes \frac{6}{18 + 6} = 4 	imes \frac{6}{24} = 4 	imes \frac{1}{4} = 1 \mathrm{~A}$.