The four identical capacitors in the circuit shown below are initially uncharged. The switch is then thrown first to position $A$, and then to position $B$. After this is done: 

Note: $V_{1,2,3,4}$ are the potential differences across $C_{1,2,3,4}$ and $Q_{1,2,3,4}$ are the final charges stored in $C_{1,2,3,4}$ respectively.

  • A$V_1 = V_0$
  • B$V_1 > V_2 > V_3 >V_4$
  • C$V_1+V_2+V_3=V_4=V_0$
  • D$Q_1 = 3Q_3$
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