Q. A $20F$ capacitor is charged to $5V$ and isolated. It is then connected in parallel with an uncharged $30F$ capacitor. The decrease in the energy of the system will be
A) $25\ J$
B) $200\ J$
C) $125\ J$
D) $150\ J$
Explanation
(d) $\mathrm{\Delta}U = \frac{C_{1}C_{2}}{2\left( C_{1} + C_{2} \right)}\left( V_{1} - V_{2} \right)^{2} = \frac{20 \times 30}{2(20 + 30)}(5 - 0)^{2} = 150J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.