A 1.50-mF capacitor is connected to a North American electrical outlet (ΔVrms = 120 V, f = 60.0 Hz). Assuming the energy stored in the capacitor is zero at t = 0, determine the magnitude of the current in the wires at t = 1/171s.

96.10 A

Explanation:

The capacitor is connected to a voltage source defined by the expression,

V (t) = √2 * Vrms*sin2πft

Where V (t) = instantaneous voltage, Vrms = root mean square voltage value, f = frequency, t = time π=pi.

Vrms = 120 V, f = 60.0Hz, t = 1/171s, π = 3.143.

∴ V (t) = √2 * 120 * sin (2*3.143*60*t)

∴V (t) = 169.71 * sin (377.16t) ... (1)

using capacitor's equation,

I (t) = C (dv/dt) ... (2)

Note: the differentiation of sin∅ = cos∅

∴ dv/dt=169.71*377.16*cos (377.16t)

dv/dt = 64109.65 * cos (377.16t)

At time t=1/171s.

∴dv/dt=64109.65 * cos (377.16*1/171)

dv/dt = 64109.65 * cos (2.21)

dv/dt = 64109.65 * 0.999

dv/dt = 64045.54.

I (t) = C (dv/dt)

Where C = I. 5 mF = 0.0015 F

∴ I (t) = 0.0015 * 64045.54

I (t) = 96.07

I (t) ≈ 96.10 A