As the switch $S$ is closed in the circuit shown in figure, current passed through it is ............. 
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Let $\mathrm{V}$ be the potential at $\mathrm{C}.$

Using Kirchhoff's first law, $ \mathrm{i}_{1}+\mathrm{i}_{2}=\mathrm{i}_{3}$

$\frac{10-V}{4}+\frac{5-V}{2}=\frac{V-0}{2}$

$\Rightarrow \mathrm{V}=4 \mathrm{\,V}$

$\therefore \mathrm{i}_{3}=\frac{\mathrm{V}-0}{2}=\frac{4-0}{2}=2 \mathrm{\,A}$

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