Q. A 4$\mu$F capacitor is charged to 400V and then its plates are joined through a resistance. The heat produced in the resistance is
A) 0.16 J
B) 0.32 J
C) 0.64 J
D) 1.28 J
Explanation
(b) The work done in charging the capacitor $(C)$ is stored as potential energy $(U)$ in the vicinity of capacitor, given by.
$U = \frac{1}{2}CV^{2}$
This produces the heat when its plates are joint through a resistance.
Given, $C = 4\mu F = 4 \times 10^{- 6}F,\ V = 400$volt
$\therefore\ \ U = \frac{1}{2} \times 4 \times 10^{- 6} \times (400)^{2}$
$\ \ \ \ \ \ U = 0.32\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.