Question

In the circuit shown, find the energy stored in $1\mu F$ capacitor.

Answer 1

Charge on $24F=$ $2\times 24=48\mu$ col.

$q_1+q_2=48$

$\frac{q_1}{5}=\frac{q_2}{1}{q}_1=5q_2$

${6q}_2=48$

$q_2=8$

energy stored $=\frac{1}{2}\times \frac{Q^2}{C}=\frac{1}{2}\times \frac{8\times 8}{1}=32\mu J$