Question

A parallel plate capacitor of capacitance $5\, \mu F$ is charged to $120\,V$ and then connected to another uncharged capacitor. If the potential falls to $40\,V, $ and capacitance of the second capacitor is

• A. $5\,\mu F$
• B. $10\,\mu F$
• C. $15\,\mu F$
• D. $20\,\mu F$

Answer 1

Answer: (C)

Change, $Q=C V$
Given, $C=5\, \mu F=5 \times 10^{-6} F , V=120\, V$
$Q=5 \times 10^{-6} \times 120$
Also, $5 \times 10^{-6} \times 120=40 \times C$
$\Rightarrow C=\frac{5 \times 10^{-6} \times 120}{40}$
$=15 \times 10^{-6} F$
$\Rightarrow C=15\, \mu F$