A constant voltage of 50 V is maintained bet | vedclass

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(B) Let the potential at point $A$ be $V_A = 50 \ V$ and at point $B$ be $V_B = 0 \ V$. Let the potential at node $C$ be $V_C$ and at node $D$ be $V_D

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(B) Let the potential at point $A$ be $V_A = 50 \ V$ and at point $B$ be $V_B = 0 \ V$. Let the potential at node $C$ be $V_C$ and at node $D$ be $V_D$.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$\frac{V_A - V_C}{1} + \frac{V_B - V_C}{2} = I_{CD} \implies (50 - V_C) + \frac{0 - V_C}{2} = I_{CD} \implies 50 - 1.5V_C = I_{CD} \quad (1)$Applying $KCL$ at node $D$:$\frac{V_A - V_D}{3} + \frac{V_B - V_D}{4} = -I_{CD} \implies \frac{50 - V_D}{3} + \frac{0 - V_D}{4} = -I_{CD} \implies \frac{50}{3} - \frac{7V_D}{12} = -I_{CD} \quad (2)$Since the branch $CD$ is a short circuit between nodes $C$ and $D$,$V_C = V_D = V$. The current $I_{CD}$ flows from $C$ to $D$ if $V_C > V_D$,but here they are connected,so we treat $CD$ as a single node. Let the potential at $CD$ be $V$.Applying $KCL$ at the combined node $CD$:$\frac{V - 50}{1} + \frac{V - 0}{2} + \frac{V - 50}{3} + \frac{V - 0}{4} = 0$$V(1 + 0.5 + 0.333 + 0.25) = 50 + 16.667$$V(2.0833) = 66.667 \implies V = 32 \ V$.The current through the $1 \ \Omega$ resistor is $I_1 = \frac{50 - 32}{1} = 18 \ A$.The current through the $2 \ \Omega$ resistor is $I_2 = \frac{32 - 0}{2} = 16 \ A$.By $KCL$ at node $C$,$I_{CD} = I_1 - I_2 = 18 - 16 = 2 \ A$.

$A$ constant voltage of $50 \ V$ is maintained between the points $A$ and $B$ of the circuit shown in the figure. The current through the branch $CD$ of the circuit is (in $A$)

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