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?

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

134.4

To find the maximum voltage across a capacitor in an AC circuit, we can use the formula:

Vmax = Imax / (2 * π * f * C)

Where:
Vmax is the maximum voltage across the capacitor,
Imax is the maximum instantaneous current,
f is the frequency of the power supply, and
C is the capacitance of the capacitor.

In this case, we are given the values:
Imax = 2.26A
f = 60 Hz
C = 40uF

Now, let's substitute the given values into the formula:

Vmax = 2.26A / (2 * π * 60 Hz * 40uF)

Before we proceed with the calculation, we need to convert the capacitance from microfarads (uF) to farads (F), since the formula requires capacitance in farads.

1uF = 10^-6 F

Therefore, 40uF = 40 * 10^-6 F = 4 * 10^-5 F.

Now we can continue with the calculation:

Vmax = 2.26A / (2 * π * 60 Hz * 4 * 10^-5 F)

Simplifying further:

Vmax = 2.26A / (2 * 3.14159 * 60 Hz * 4 * 10^-5 F)

Vmax ≈ 59.98 volts

Therefore, the maximum voltage across the capacitor in this circuit is approximately 59.98 volts.