When the switch S, in the circuit shown, is c | vedclass

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(B) Let the potential at node $C$ be $V$. When the switch $S$ is closed, the node $C$ is connected to the ground $(0 \, V)$ through a $2 \, \Omega$ re

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$5$

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(B) Let the potential at node $C$ be $V$. When the switch $S$ is closed, the node $C$ is connected to the ground $(0 \, V)$ through a $2 \, \Omega$ resistor.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$\frac{20 - V}{2} + \frac{10 - V}{4} = \frac{V - 0}{2}$Multiplying the entire equation by $4$ to clear the denominators:$2(20 - V) + (10 - V) = 2V$$40 - 2V + 10 - V = 2V$$50 - 3V = 2V$$5V = 50$$V = 10 \, V$The current $i$ flowing through the $2 \, \Omega$ resistor is given by:$i = \frac{V - 0}{2} = \frac{10}{2} = 5 \, A$.

When the switch $S$, in the circuit shown, is closed, then the value of current $i$ will be ................. $A$.

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