When the switch S, in the circuit shown, is closed then the valued of current $i$ will be:

Let voltage at $\mathrm{C}=\mathrm{xV}$

From kirchhoff's current law,

$\mathrm{KCL}: \mathrm{i}_{1}+\mathrm{i}_{2}=\mathrm{i}$

$\frac{20-x}{2}+\frac{10-x}{4}=\frac{x-0}{2} \Rightarrow x=10$

$\therefore \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{\mathrm{X}}{\mathrm{R}}=\frac{10}{2}=5 \mathrm{~A}$