Question

In the circuit shown above, the energy stored in $1\mu F$ capacitor is :

• A. $40\mu J$
• B. $64\mu J$
• C. $32\mu J$
• D. $none$

Answer 1

The correct option is C $32\mu J$

The capacitance 4 and (3,5,1) are in parallel so potential across (3,5,1) will be $V=24V$.
The equivalent capacitance of (3,5,1) is $C_{eq}=\frac{3(5+1)}{3+(5+1)}=2\mu F$
and $Q_{eq}=C_{eq}V=2\times 24=48\mu C$
As 3 and (5,1) are in series so charge on 3 and (5,1) is equal to $Q_{eq}$.
Thus potential across (5,1) is $V_{51}=\frac{Q_{eq}}{C_{51}}=\frac{48}{5+1}=8V$
As 5 and 1 are in parallel so potential difference of capacitors will be same i.e $8V$
Now energy stored in 1 is $=\frac{1}{2}{C}_1{V}_{51}^2=\frac{1}{2}\times {10}^{-6}\times 8^2=32\mu J$