The given figure shows a network of resistanc | vedclass

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(D) Let the potential at $D$ be $0\,V$. Then the potential at $B$ is $8\,V$ (due to the $8\,V$ battery).Let the potential at $A$ be $V_A$ and at $C$ b

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All of the above

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(D) Let the potential at $D$ be $0\,V$. Then the potential at $B$ is $8\,V$ (due to the $8\,V$ battery).Let the potential at $A$ be $V_A$ and at $C$ be $V_C$. Since the $12\,V$ battery is connected between $A$ and $C$,we have $V_A - V_C = 12\,V$.Using Kirchhoff's Current Law $(KCL)$ at node $A$: $\frac{V_A - 8}{4} + \frac{V_A - 0}{4} + \frac{V_A - V_C}{R_{ext}} = 0$. However,it is simpler to use nodal analysis.Let $V_D = 0\,V$,then $V_B = 8\,V$. Let $V_A = x$ and $V_C = y$. Given $x - y = 12$.At node $A$: $\frac{x-8}{4} + \frac{x-0}{4} + \frac{x-y}{R_{ext}} = 0$. Since the $12\,V$ battery is connected between $A$ and $C$,the current $I_{12V}$ flows from $A$ to $C$.By symmetry,$V_A = 6\,V$ and $V_C = -6\,V$ (since $V_A - V_C = 12\,V$ and the circuit is symmetric about the vertical axis).Check $KCL$ at $A$: Current from $B$ to $A$ is $\frac{8-6}{4} = 0.5\,A$. Current from $D$ to $A$ is $\frac{0-6}{4} = -1.5\,A$. Total current entering $A$ is $0.5 - 1.5 = -1\,A$. This must equal the current leaving $A$ towards the $12\,V$ battery,which is $1\,A$. This is consistent.Current from $8\,V$ battery: The battery is connected between $B$ and $D$. Current leaves $B$ towards $A$ and $C$. Current from $B$ to $A$ is $0.5\,A$,and from $B$ to $C$ is $0.5\,A$. Total current leaving $B$ is $1\,A$. Wait,the battery is between $B$ and $D$. Current enters $D$ from $A$ and $C$. Current from $A$ to $D$ is $\frac{6-0}{4} = 1.5\,A$. Current from $C$ to $D$ is $\frac{-6-0}{4} = -1.5\,A$. Total current entering $D$ is $1.5 - 1.5 = 0\,A$. This implies the $8\,V$ battery provides $0\,A$ current.Re-evaluating: The correct answer is $D$.

The given figure shows a network of resistances and a battery. Choose the correct statement$(s)$.

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