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(A) Let $V$ be the potential of the central junction. Applying Kirchhoff's junction law at this point,the sum of currents entering the junction must e

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$4.5$

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(A) Let $V$ be the potential of the central junction. Applying Kirchhoff's junction law at this point,the sum of currents entering the junction must equal the current leaving it.Assuming the currents $i_1$ and $i_2$ enter the junction from the $20 \ V$ and $5 \ V$ sources respectively,and $i_3$ flows through the switch to the ground $(0 \ V)$:$\frac{20 - V}{2} + \frac{5 - V}{4} = \frac{V - 0}{2}$Multiplying the entire equation by $4$ to clear the denominators:$2(20 - V) + (5 - V) = 2V$$40 - 2V + 5 - V = 2V$$45 - 3V = 2V$$5V = 45$$V = 9 \ V$The current $i_3$ passing through the switch is given by:$i_3 = \frac{V - 0}{2} = \frac{9}{2} = 4.5 \ A$.

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

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