In the circuit shown, the energy stored in $1$ $ \mu F$ capacitor is......$\mu J$
Medium
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Charge on $24 F=2 \times 24=48 \mu \mathrm{col}$
$q_{1}+q_{2}=48$
$\frac{q_{1}}{5}=\frac{q_{2}}{1} \quad q_{1}=5 q_{2}$
$6 q_{2}=48$
$q_{2}=8$
energy stored $=\frac{1}{2} \times \frac{Q^{2}}{C}=\frac{1}{2} \times \frac{8 \times 8}{1}=32 \mu J$
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