Question

A capacitor of $10\, \mu F$ charged upto $250\, V$ is connected in parallel with another capacitor of $5\, \mu F$ charged upto $100\, V$. The common potential is :

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

Answer 1

Answer: (A)

The equivalent capacitance of the capacitors joined in parallel is equal to the sum of their individual capacitances.
Let the charge on the capacitor's be $q_{1}$ and $q_{2}$. Then total charge is

$Q=Q_{1}+Q_{2} $
$C V=C_{1} V_{1}+C_{2} V_{2}$
Since, capacitor's are connected in parallel equivalent capacitance is $C=C_{1}+C_{2}$
$\therefore V=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}} $
Given, $C_{1}=10\, \mu F , $
$V_{1}=250 \,V , $
$C_{2}=5 \,\mu F ,$
$V_{2}=100\, V $
$\therefore V=\frac{\left(10 \times 10^{-6} \times 250\right)+\left(5 \times 10^{-6} \times 100\right)}{\left(10 \times 10^{-6}+5 \times 10^{-6}\right)} $
$\Rightarrow V=\frac{3000 \times 10^{-6}}{15 \times 10^{-6}} $
$=200$ volt