Question

A one microfarad capacitor of a TV is subjected to 4000 V potential difference. The energy stored in capacitor is :

• A. 8J
• B. 16 J
• C. $4\times {10}^{-3}J$
• D. $2\times {10}^{-3}J$

Answer 1

The correct option is A 8J

Given, capacitor $C={10}^{-6}F$ and potential difference $V=4000V$
The energy stored in capacitor $=\frac{1}{2}C{V}^2=\frac{1}{2}\times {10}^{-6}\times (4000{)}^2=8J$