Q. A capacitor of capacity 10$\mu F\ $is charged to a potential of 400V. When its both plates are connected by a conducting wire , then heat generated will be
A) 80 J
B) 0.8 J
C) $8 \times 10^{- 3}\ J$
D) $8 \times 10^{- 6}\ J$
Explanation
(b) Given, the capacity of capacitor, $C = 10 \times 10^{- 6}\mu F$
Potential $V = 400$volt
Stored energy =Heat generated
$\ \ \ \ \ \ \ \ \ \ \ \ \ U = \frac{1}{2}CV^{2} = \frac{1}{2}\left( 10 \times 10^{- 6} \right) \times (400)^{2} = 0.8\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.