An ac power supply with frequency 60 Hz is connected to a capacitor C=40uF. The maximum instantaneous current that passes through the circuit is 2.26A. What is the maximum voltage?

Answer 1

C = Q/v = 40*10^-6

Q = 40 * 10^-6 v

i = dQ/dt = 40*10^-6 dv/dt

let v = V sin wt
dV/dt = V w cos wt
so
i = 40*10^-6 V w cos w t
max i = 40*10^-6 V w
w = 2 pi f = 2 pi(60) = 377
so
max i = 40*10^-6 * 377 V
so
V = (2.26/(15080)) 10^6 =1344

Answer 2

134.4