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

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

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$4$

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(C) Let the potential at junction $P$ be $V$. According to Kirchhoff's Current Law $(KCL)$ at junction $P$,the sum of currents entering the junction equals the sum of currents leaving the junction.$\frac{24 - V}{3} = \frac{V - 10}{2} + \frac{V - 9}{1}$Multiply 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$$11V = 132$$V = 12 \,V$Now,calculate the current $I$ flowing through the $3 \,\Omega$ resistor:$I = \frac{24 - V}{3} = \frac{24 - 12}{3} = \frac{12}{3} = 4 \,A$Thus,the current $I$ is $4 \,A$.

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

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