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The total energy stored in the condenser system shown in the figure will be
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Verified Answer
The correct answer is:
$8 \mu \mathrm{J}$
$6 \mu \mathrm{F}$ and $3 \mu \mathrm{F}$ capacitors are in series
$$
\begin{aligned}
\therefore \quad \frac{1}{\mathrm{C}_{1}} &=\frac{1}{6}+\frac{1}{3} \\
\mathrm{C}_{1} &=2
\end{aligned}
$$
$\mathrm{C}_{1}$ is parallel to $2 \mu \mathrm{F}$ capacitor
$\therefore \quad \mathrm{C}_{\mathrm{eq}}=2+2=4 \mu \mathrm{F}$
Total energy, $\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 4 \times(2)^{2}=8 \mu \mathrm{J}$
$$
\begin{aligned}
\therefore \quad \frac{1}{\mathrm{C}_{1}} &=\frac{1}{6}+\frac{1}{3} \\
\mathrm{C}_{1} &=2
\end{aligned}
$$
$\mathrm{C}_{1}$ is parallel to $2 \mu \mathrm{F}$ capacitor
$\therefore \quad \mathrm{C}_{\mathrm{eq}}=2+2=4 \mu \mathrm{F}$
Total energy, $\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 4 \times(2)^{2}=8 \mu \mathrm{J}$
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