Question

A capacitor of $20 \, \mu F$ charged upto $500\, V$ is connected in parallel with another capacitor of $10\, \mu F$ which is charged upto $200\,V$. The common potential is

• A. 500 V
• B. 300 V
• C. 400 V
• D. 200 V

Answer 1

Answer: (C)

The common potential on the capacitor is the ratio of net charge to the net capacitance.
Common potential $(V) =\frac{\text { Net charge }}{\text { Net capacitance }} $
$=\frac{q_{1}+q_{2}}{C_{1}+C_{2}} $
$=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$
Herer, $C_{1} =20 \times 10^{-6} F ,$
$V_{1} =500\, V$
$C_{2} =10 \times 10^{-6} F ,$
$V_{2} =200 \,V $
$\therefore V= \frac{20 \times 10^{-6} \times 500+10 \times 10^{-6} \times 200}{30 \times 10^{-6}}$
$= 400\, V$
When a capacitor is charged by battery, then $50 \%$ of energy is liberated or wastage) as heat always.