Question

A certain capacitor is charged to a potential difference V. If you wish to increase its stored energy by 77%, by what percentage should you increase V

Answer 1

Answer:

33%

Explanation:

Energy stored in a capacitor (U) = $1/2CV^2$

The capacitance is constant. Hence, $\frac{U_2}{U_1} = \frac{V_2^2}{V_1^2}$

V2/V1 = $\sqrt{\frac{U_2}{U_1} }$

If U1 = 1, then U2 is an increase of 77% (0.77), hence U2 = 1.77

V2/V1 = $\sqrt{\frac{1.77}{1} }$

V2/V1 = 1.33

Hence, the v of the capacitor should be increased by 33% in order to increase its stored energy by 77%.

Answer 2

Explanation:

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