Q. A capacitor of capacitance value $1\mu\ F$ is charged to 30 $V$ and the battery is then disconnected. If it is connected across a $2\mu\ F$ capacitor, the energy lost by the system is
A) 300$\ \mu J$
B) 450$\ \mu J$
C) 225$\ \mu J$
D) 150$\ \mu J$
Explanation
(a) Initial energy $= \frac{1}{2} \times 1 \times 10^{- 6} \times (30)^{2} = 450 \times 10^{- 6}\ J$
Final energy $= \frac{1}{2}\left( C_{1} + C_{2} \right)V_{common}^{2}\ \ \lbrack\because V = \frac{V_{1}C_{1} + V_{2}C_{2}}{C_{1} + C_{2}}\rbrack$
$= \frac{1}{2} \times 3 \times 10^{- 6} \times (10)^{2}$
$= 150 \times 10^{- 6}\ J$
Loss of energy $= (450 - 150) \times 10^{- 6}J = 300 \times 10^{- 6}\ J$
$= 300\mu J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.