Q181 / 200 · Manipal (MET) 2011 · Physics
A $10 \mu \mathrm{~F}$ capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V . The capacitance of second capacitor is
- A$10 \mu \mathrm{~F}$
- B$20 \mu \mathrm{~F}$
- C$30 \mu \mathrm{~F}$
- Correct answerD$15 \mu \mathrm{~F}$
Solution
$C_1=10 \mu \mathrm{~F}, V=50 \mathrm{~V}$
Charge on first capacitor $\begin{aligned} q & =C_1 V \\ & =10 \times 50=500 \mu \mathrm{C}\end{aligned}$ Common potential $V=\frac{\text { total charge }}{\text { total capacitance }}$ $\therefore \quad 20=\frac{500}{C_1+C_2}$ or $20=\frac{500}{10+C_2}$ or $C_2=15 \mu \mathrm{~F}$
Charge on first capacitor $\begin{aligned} q & =C_1 V \\ & =10 \times 50=500 \mu \mathrm{C}\end{aligned}$ Common potential $V=\frac{\text { total charge }}{\text { total capacitance }}$ $\therefore \quad 20=\frac{500}{C_1+C_2}$ or $20=\frac{500}{10+C_2}$ or $C_2=15 \mu \mathrm{~F}$
Hence, the correct answer is option: 'D'.