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Question

(A) Let the potential at the junction be $V$. Applying Kirchhoff's Current Law $(KCL)$ at the junction:$\frac{V - 20}{2} + \frac{V - 5}{4} + \frac{V -

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(A) Let the potential at the junction be $V$. Applying Kirchhoff's Current Law $(KCL)$ at the junction:$\frac{V - 20}{2} + \frac{V - 5}{4} + \frac{V - 0}{2} = 0$Multiplying the entire equation by $4$ to clear the denominators:$2(V - 20) + (V - 5) + 2V = 0$$2V - 40 + V - 5 + 2V = 0$$5V - 45 = 0$$5V = 45$$V = 9 \, V$Now,the current $I_3$ passing through the switch $S$ (which is in series with the $2 \, \Omega$ resistor connected to ground) is:$I_3 = \frac{V - 0}{2} = \frac{9}{2} = 4.5 \, A$.

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

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