As the switch S is closed in the circuit show | vedclass

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(C) Let $V$ be the potential at node $C$.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$i_{1} + i_{2} = i_{3}$$\frac{10 - V}{4} + \frac{5 - V}{

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$2$

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(C) Let $V$ be the potential at node $C$.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$i_{1} + i_{2} = i_{3}$$\frac{10 - V}{4} + \frac{5 - V}{2} = \frac{V - 0}{2}$Multiplying the entire equation by $4$ to clear the denominators:$(10 - V) + 2(5 - V) = 2V$$10 - V + 10 - 2V = 2V$$20 - 3V = 2V$$5V = 20$$V = 4 \, V$Now,the current passing through the switch $S$ is $i_{3}$:$i_{3} = \frac{V - 0}{2} = \frac{4 - 0}{2} = 2 \, A$

As the switch $S$ is closed in the circuit shown in the figure,the current passing through it is ............. (in $, A$)

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