$0.2\, F$ capacitor is charged to $600\, V$ by a battery. On removing the battery. It is connected with another parallel plate condenser of $1\, F$. The potential decreases to....$V$
Medium
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A capacitor $=-2 \mu F$ (in series)
Charge $=600 \mathrm{V}$
Parallel capacitor $=1.0 \mu F$ $0.2 \times 600=\frac{V}{0.1}$
or, $V=0.2 \times 0.1 \times 600$
$\quad=\frac{2}{10} \times \frac{1}{10} \times 600$
$2 \times \frac{600}{12}=50 \times 2=100 \mathrm{V}$
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