In the circuit shown in the figure,the curren | vedclass

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(C) Let the potential at junction $P$ be $V$. Applying Kirchhoff's Current Law $(KCL)$ at junction $P$,the sum of currents entering the junction equal

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$4$

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(C) Let the potential at junction $P$ be $V$. Applying Kirchhoff's Current Law $(KCL)$ at junction $P$,the sum of currents entering the junction equals the sum of currents leaving the junction:$I = I_1 + I_2$$\frac{24 - V}{3} = \frac{V - 10}{2} + \frac{V - 9}{1}$Multiplying the entire equation by $6$ to clear the denominators:$2(24 - V) = 3(V - 10) + 6(V - 9)$$48 - 2V = 3V - 30 + 6V - 54$$48 - 2V = 9V - 84$$132 = 11V$$V = 12 \,V$Now,calculate the current $I$ using the potential at $P$:$I = \frac{24 - V}{3} = \frac{24 - 12}{3} = \frac{12}{3} = 4 \,A$

In the circuit shown in the figure,the current $I$ is (in $\,A$)

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