When the switch $S$, in the circuit shown, is closed, then the value of current $i$ will be ................. $A$
JEE MAIN 2019, Medium
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$ \mathrm{i}_{3}+\mathrm{i}_{2} =\mathrm{i}_{1} $
$ \frac{20-\mathrm{v}}{2} +\frac{10-\mathrm{v}}{4}=\frac{\mathrm{v}}{2} $
$ \mathrm{v}=10 \,\mathrm{V} $
$\Rightarrow \quad \mathrm{i}_{1} =\frac{10}{2} $
$=5 \,\mathrm{amp} $
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