The energy stored in a capacitor of capacitance 5μF is 40 J. Calculate the voltage applied across its terminals.
E = \(\frac{1}{2}\)QV;
But, Q = CV
∴ E = \(\frac{1}{2}\)CV\(^2\);
40 = \(\frac{1}{2}\) × 5 × 10\(^{-6}\) × V\(^2\)
V\(^2\) = \(\frac{(2 \times 40)}{5 \times 10^{-6}}\)
V\(^2\) = 16 × 10\(^6\)
∴ V = \(\sqrt{(16 \times 10^6)}\) = 4000 V
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