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Two capacitors of capacities $1 \mu \mathrm{F}$ and $C \mu \mathrm{F}$ are connected in series and the combination is charged to a potential difference of $120 \mathrm{~V}$. If the charge on the combination is $80 \mu \mathrm{C}$, the energy stored in the capacitor of capacity $C$ in $\mu \mathrm{J}$ is
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1600
Capacitances $1 \mu \mathrm{F}$ and $C \mu \mathrm{F}$ are connected in series, then
$C_{\text {eq }}=\frac{C}{1+C}$
Given, $V=120 \mathrm{~V}$ and $q=80 \mu \mathrm{C}$
$\because \quad q=C_{\mathrm{eq}} V$
$80=\frac{C}{C+1} \times 120$
or $\quad C=2 \mu \mathrm{F}$
The energy stored in the capacitor of capacity $C$
$U=\frac{1}{2} \frac{q^2}{C}$
$=\frac{1}{2} \times \frac{\left(80 \times 10^{-6}\right)^2}{2 \times 10^{-6}}$
$=\frac{1}{2} \times \frac{80 \times 10^{-6} \times 80 \times 10^{-6}}{2 \times 10^{-6}}$
$U=1600 \mu \mathrm{J}$
$C_{\text {eq }}=\frac{C}{1+C}$
Given, $V=120 \mathrm{~V}$ and $q=80 \mu \mathrm{C}$
$\because \quad q=C_{\mathrm{eq}} V$
$80=\frac{C}{C+1} \times 120$
or $\quad C=2 \mu \mathrm{F}$
The energy stored in the capacitor of capacity $C$
$U=\frac{1}{2} \frac{q^2}{C}$
$=\frac{1}{2} \times \frac{\left(80 \times 10^{-6}\right)^2}{2 \times 10^{-6}}$
$=\frac{1}{2} \times \frac{80 \times 10^{-6} \times 80 \times 10^{-6}}{2 \times 10^{-6}}$
$U=1600 \mu \mathrm{J}$
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