A galvanometer having a resistance of $8 \Omega$ is shunted by a wire of resistance $2 \Omega$. If the total current is 1 A , the part of it passing through the shunt will be

Key Idea : Potential difference across galvanometer should be equal to potential difference across shunt. The shunt and galvanometer are connected as shown in figure. Let total current through the parallel combination is $i$, the current through the galvanometer is $i_g$ and the current through the shunt is $i-i_g$. The potential difference $V_{a b}\left(=V_a-V_b\right)$ is the same for both paths, so $\begin{array}{ll} & i_g G=\left(i-i_g\right) S \\ \text { or } & i_g(G+S)=i S \\ \text { or } & \frac{i_g}{i}=\frac{S}{S+G}\end{array}$ The fraction of current passing through shunt $\begin{aligned} & =\frac{i-i_g}{i}=1-\frac{i_g}{i} \\ & =1-\frac{S}{S+G}=\frac{G}{S+G} \\ & =\frac{8}{2+8}=\frac{8}{10} \\ & =0.8 \mathrm{~A}\end{aligned}$