The energy stored in a capacitor of capacitance 5μF is 40J. Calculate the voltage applied across its terminals?
-
4000V
- 200V
- 16V
- 6V
- 4V
The correct answer is: A
Explanation
\(E = \frac{CV^{2}}{2}\)
\(V^{2} = \frac{2E}{C}\)
\(V^{2} = \frac{2\times 40}{5 \times 10^{-6}}\)
\(V^{2} = 16 \times 10^{6}\)
\(V = 4 \times 10^{3} = 4000v\)