Q. A capacitor of capacitance value 1 $\mu$F is charged to 30 V and the battery is then disconnected. If the remaining circuit is connected across a 2$\mu$F capacitor, the remaining circuit 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) Energy $E_{1} = \frac{1}{2}C_{1}V_{1}^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2} \times 1 \times 10^{- 6} \times (30)^{2} = 450 \times 10^{- 6}\ J$
Common potential
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = \ \frac{q_{1} + q_{2}}{C_{1} + C_{2}}$
$= \ \frac{1 \times 30 + 0}{1 + 2} = 10\ volt$
$\ \ \ \ \ \ \ \ \ \ \ \ \ E_{2} = \frac{1}{2}\left( C_{1} + C_{2} \right)V^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}(1 + 2) \times 10^{- 6} \times (10)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.5 \times 100 \times 10^{- 6} = 150 \times 10^{- 6}\ J$
Loss of energy =$E_{2} - E_{1} = 300\mu J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.