Q44 · Work done charging two parallel condensers

Question

Two condensers one of capacity $C$ and the other capacity $\dfrac{C}{2}$, are connected to a $V$ volt battery as shown. The work done in charging fully both the condensers is

a) $2CV^2$
b) $\dfrac{1}{4}CV^2$
c) $\dfrac{3}{4}CV^2$
d) $\dfrac{1}{2}CV^2$

The Circuit

Both condensers hang directly across the battery, so each is charged to the same voltage $V$ — a parallel arrangement.

Given

First condenser	$C$
Second condenser	$C/2$
Battery	$V$ (across both)

Asked

Wanted	Total energy stored $W$
Tool	$U=\tfrac12 C V^2$ per capacitor

Concept

The work to fully charge a capacitor equals the energy it stores, $U=\tfrac12 C V^2$. In parallel each capacitor sees the full $V$, so add their energies — or equivalently use the combined capacitance.

$$ W = \tfrac12 C_{eq} V^2, \qquad C_{eq}=C+\tfrac{C}{2}=\tfrac{3C}{2}. $$

Method & Steps

Energy in $C$: $\tfrac12 C V^2.$
Energy in $C/2$: $\tfrac12\!\left(\tfrac{C}{2}\right)V^2=\tfrac14 C V^2.$
Total: $W=\tfrac12 C V^2+\tfrac14 C V^2.$
$W=\tfrac34 C V^2.$

Easy Trick

Parallel ⇒ just add capacitances and use one formula: $C_{eq}=\tfrac{3C}{2}$, so $W=\tfrac12 C_{eq}V^2=\tfrac34 C V^2$. (Energy scales with capacitance at fixed $V$ — the bigger $C$ stores $\tfrac12CV^2$, the smaller stores half that, $\tfrac14CV^2$.)

$W = \dfrac{3}{4}\,C V^2$

Sum of $\tfrac12CV^2$ (in $C$) and $\tfrac14CV^2$ (in $C/2$).

First condenser	\(C\)
Second condenser	\(C/2\)
Battery	\(V\) (across both)

Wanted	Total energy stored \(W\)
Tool	\(U=\tfrac12 C V^2\) per capacitor