The energy stored in a capacitor of capacitance 5μF is 40 J. Calculate the voltage applied across its terminals.
A
6 V
B
200 V
C
16 V
D
4000 V
Verified Correct Answer: D.
Physical Principles & Numerical Solution
E = \(\frac{1}{2}\)QV;
But, Q = CV
∴ E = \(\frac{1}{2}\)CV\(^2\);
40 = \(\frac{1}{2}\) \times 5 \times 10\(^{-6}\) \times V\(^2\)
V\(^2\) = \(\frac{(2 \times 40)}{5 \times 10^{-6}}\)
V\(^2\) = 16 \times 10\(^6\)
∴ V = \(\sqrt{(16 \times 10^6)}\) = 4000 V
Source: WAEC / JAMB / NECO Classroom Archive (Record #76401)