Question:
A $10 \mu \mathrm{F}$ capacitor is fully charged to a potential difference of $50 \mathrm{~V}$. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes $20 \mathrm{~V}$. The capacitance of the second capacitor is :
Correct Option: 1
Solution:
(1) Given,
Capacitance of capacitor, $C_{1}=10 \mu \mathrm{F}$
Potential difference before removing the source voltage, $V_{1}=50 \mathrm{~V}$
If $C_{2}$ be the capacitance of uncharged capacitor, then common potential is
$V=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$
$\Rightarrow 20=\frac{10 \times 50+0}{20+C} \Rightarrow C=15 \mu F$