The figure shows a diagonal symmetric arrangement of capacitors and a battery. If the potential of C is zero, then (all the capacitors are initially uncharged):

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The cell sets V A ​ − V C ​ = 20 V ; with V C ​ = 0 , V A ​ = + 20 V - (A) correct. Charge conservation at isolated node B: 4 ( V B ​ − V A ​ ) + 2 ( V B ​ − V C ​ ) + 2 ( V B ​ − V D ​ ) = 0 , i.e. 4 ( V A ​ − V B ​ ) + 2 ( V D ​ − V B ​ ) = 2 V B ​ - (B) correct. Similarly at node D: 2 ( V A ​ − V D ​ ) + 2 ( V B ​ − V D ​ ) = 4 V D ​ - (C) correct. Solving both with V A ​ = 20 : V B ​ = 12 V , V D ​ = 8 V , so V A ​ = V B ​ + V D ​ = 20 V - (D) correct.

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