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(B, C) Let the potential at point $A$ be $V_A = 20\,V$ and at point $C$ be $V_C = 0\,V$.The circuit is symmetric. Let the potential at point $B$ be $V

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$B$ and $C$ both.

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(B, C) Let the potential at point $A$ be $V_A = 20\,V$ and at point $C$ be $V_C = 0\,V$.The circuit is symmetric. Let the potential at point $B$ be $V_B$ and at point $D$ be $V_D$.Applying Kirchhoff's Current Law $(KCL)$ at node $B$:$4(V_A - V_B) + 2(V_D - V_B) + 2(V_C - V_B) = 0$$4(20 - V_B) + 2(V_D - V_B) + 2(0 - V_B) = 0$$80 - 4V_B + 2V_D - 2V_B - 2V_B = 0 \implies 8V_B - 2V_D = 80 \implies 4V_B - V_D = 40$  --- $(1)$Applying $KCL$ at node $D$:$2(V_A - V_D) + 2(V_B - V_D) + 4(V_C - V_D) = 0$$2(20 - V_D) + 2(V_B - V_D) + 4(0 - V_D) = 0$$40 - 2V_D + 2V_B - 2V_D - 4V_D = 0 \implies 8V_D - 2V_B = 40 \implies 4V_D - V_B = 20$ --- $(2)$Solving equations $(1)$ and $(2)$:From $(2)$,$V_B = 4V_D - 20$. Substitute into $(1)$:$4(4V_D - 20) - V_D = 40 \implies 16V_D - 80 - V_D = 40 \implies 15V_D = 120 \implies V_D = 8\,V$.Substituting $V_D = 8\,V$ into $V_B = 4(8) - 20 = 32 - 20 = 12\,V$.Thus,$V_B = 12\,V$ and $V_D = 8\,V$. The correct options are $B$ and $C$.

The figure shows a diagonal symmetric arrangement of capacitors and a battery. The potentials of the points $B$ and $D$ are:

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