In the circuit shown in the figure,find the c | vedclass

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(A) Let the potential at node $D$ be $0 \, V$. Then the potential at node $B$ is $V_B$. Using Kirchhoff's Current Law $(KCL)$ at node $B$:$\frac{V_B -

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$5$

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(A) Let the potential at node $D$ be $0 \, V$. Then the potential at node $B$ is $V_B$. Using Kirchhoff's Current Law $(KCL)$ at node $B$:$\frac{V_B - 15}{6} + \frac{V_B - 0}{3} + \frac{V_B - 30}{3} = 0$Multiplying by $6$:$(V_B - 15) + 2V_B + 2(V_B - 30) = 0$$V_B - 15 + 2V_B + 2V_B - 60 = 0$$5V_B = 75$$V_B = 15 \, V$The current through branch $BD$ is $I_{BD} = \frac{V_B - V_D}{3} = \frac{15 - 0}{3} = 5 \, A$.

In the circuit shown in the figure,find the current through the branch $BD$ in $A$.

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