Question

Match the following.
Treat all capacitors as identical and initially uncharged

Answer 1

For A: if breakdown voltage of each capacitor is $V$,
$1)V_{ext}=V,2)V_{ext}=V+V=2V,3)V_{ext}=V,4)V_{ext}=V+V+V=3V$
Thus, $A\to 4$
For B: if supplied charge is $Q$ and capacitance of each capacitors is $C$,
$1)C_{eq}=C+C+C=3C,V=\frac{Q}{C_{eq}}=\frac{Q}{3C}$.
$2)C_{eq}=\frac{(C+C)C}{(C+C)+C}=(2/3)C,V=\frac{Q}{C_{eq}}=\frac{3Q}{2C}$.
$3)C_{eq}=\frac{CC}{C+C}+C=(3/2)C,V=\frac{Q}{C_{eq}}=\frac{2Q}{3C}$.
$4)C_{eq}=(1/C+1/C+1/C{)}^{-1}=C/3,V=\frac{Q}{C_{eq}}=\frac{3Q}{C}$.
Thus, $B\to 4$
For C: if applied potential is $V$,
$1)C_{eq}=3C,Q=C_{eq}V=3CV$
$2)C_{eq}=(2/3)C,Q=C_{eq}V=(2/3)CV$
$3)C_{eq}=(3/2)C,Q=C_{eq}V=(3/2)CV$
$4)C_{eq}=C/3,Q=C_{eq}V=(1/3)CV$
thus,$C\to 1$