In the given circuit diagram,find the current | vedclass

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(B) Let the potential at the left junction be $V_L = 50 \, V$ and at the right junction be $V_R = 0 \, V$. Let the potential at node $C$ be $V_C$ and

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(B) Let the potential at the left junction be $V_L = 50 \, V$ and at the right junction be $V_R = 0 \, V$. Let the potential at node $C$ be $V_C$ and at node $D$ be $V_D$.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$\frac{V_C - 50}{1} + \frac{V_C - V_D}{R_{CD}} + \frac{V_C - 0}{2} = 0$Since $CD$ is a conducting wire,$R_{CD} = 0$,so $V_C = V_D = V$.Applying $KCL$ at the combined node $CD$:$\frac{V - 50}{1} + \frac{V - 50}{3} + \frac{V - 0}{2} + \frac{V - 0}{4} = 0$Multiplying by $12$:$12(V - 50) + 4(V - 50) + 6V + 3V = 0$$16V - 800 + 9V = 0$$25V = 800 \implies V = 32 \, V$Now,calculate the currents:$I_1 = \frac{50 - 32}{1} = 18 \, A$ (flowing towards $C$)$I_3 = \frac{32 - 0}{2} = 16 \, A$ (flowing away from $C$)Current through $CD$ from $C$ to $D$ is $I_{CD} = I_1 - I_3 = 18 - 16 = 2 \, A$.

In the given circuit diagram,find the current passing through the wire $CD$ (in ampere).

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