Q. A $10\mu F$ 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 $20\ volt.$ The capacitance of second capacitor is
A) $10\mu F$
B) $20\mu F$
C) $30\mu F$
D) $15\mu F$
Explanation
(d) $V = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}} \Rightarrow 20 = \frac{10 \times 50 + C_{2} \times 0}{10 + C_{2}}$
$\Rightarrow 200 + 20C_{2} = 500 \Rightarrow C_{2} = 15\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.