As the switch $S$ is closed in the circuit shown in figure, current passing through it is ................ $\mathrm{A}$
Medium
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By $\mathrm{KCL},$ $\quad \mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}=0$
$\Rightarrow \frac{\mathrm{V}-20}{2}+\frac{\mathrm{V}-5}{4}+\frac{\mathrm{V}-0}{2}=0$
$\Rightarrow \mathrm{V}=9 \mathrm{\,Volt}$
Therefore $\mathrm{I}_{3}=\frac{\mathrm{V}}{2}=\frac{9}{2}=4.5 \mathrm{\,A}$
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