Jee Main 2024 01 02 2024 Shift 1 - Question9
Question 9
Two capacitors are charged as shown. When both the positive terminals and negative terminals of capacitors are connected the energy loss will be
(1) $\frac{1}{2} C V^{2}$
(2) $\frac{3}{4} C V^{2}$
(3) $\frac{1}{4} C V^{2}$
(4) $2 CV^{2}$
Show Answer
Answer: (3)
Solution:
$V _c=\frac{C V+2 C V}{2 C}=\frac{3 V}{2}$
$\therefore \quad$ Energy loss $=\frac{1}{2} C V^{2}+\frac{1}{2} C(2 V)^{2}-\frac{1}{2} 2 C\left(\frac{3 V}{2}\right)^{2}$
$$ =\frac{1}{4} C V^{2} $$
BETA
