Answer:
33%
Explanation:
Energy stored in a capacitor (U) = [tex]1/2CV^2[/tex]
The capacitance is constant. Hence, [tex]\frac{U_2}{U_1} = \frac{V_2^2}{V_1^2}[/tex]
V2/V1 = [tex]\sqrt{\frac{U_2}{U_1} }[/tex]
If U1 = 1, then U2 is an increase of 77% (0.77), hence U2 = 1.77
V2/V1 = [tex]\sqrt{\frac{1.77}{1} }[/tex]
V2/V1 = 1.33
Hence, the v of the capacitor should be increased by 33% in order to increase its stored energy by 77%.
