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A $20 \mathrm{~F}$ capacitor is charged to $5 \mathrm{~V}$ and isolated. It is then connected in parallel with an uncharged $30 \mathrm{~F}$ capacitor. The decrease in the energy of the system will be
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$150 \mathrm{~J}$
$\mathrm{C}_{\mathrm{s}}=20 \mathrm{~F}, V=5$ volt,
$F_x=30 \mathrm{~F}$
Decrease in energy,
$\Delta U=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_2-V_2\right)^2$
$\begin{aligned} & =\frac{1}{2} \times \frac{20 \times 30}{20+30}(5-0)^2 \\ & =\frac{300}{50} \times 25=150 \mathrm{~J}\end{aligned}$
$F_x=30 \mathrm{~F}$
Decrease in energy,
$\Delta U=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_2-V_2\right)^2$
$\begin{aligned} & =\frac{1}{2} \times \frac{20 \times 30}{20+30}(5-0)^2 \\ & =\frac{300}{50} \times 25=150 \mathrm{~J}\end{aligned}$
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