0.2, F capacitor is charged to 600, V by a battery. On removing battery. It is connected with anothe

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2. Electric Potential and Capacitance

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$0.2\, F$ 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....$V$

A

$100$

B

$120$

C

$300$

D

$600$

Solution

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, $V=0.2 \times 0.1 \times 600$

$\quad=\frac{2}{10} \times \frac{1}{10} \times 600$

$2 \times \frac{600}{12}=50 \times 2=100 \mathrm{V}$

Standard 12

Physics

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