A 100 μF capacitor in series with a 40Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the voltage maximum?

Given: capacitor C = 100 μf = 100 × 10^-6F

Resistance R = 40Ω

Voltage V = 110 V

Frequency v = 60 Hz

(a) Angular frequency is given by the relation: ω = 2πv

ω = 2π × 60 rad/s

The impedance is given by the relation

Or Z² =

Peak voltage is given by the relation V₀ = V √2

Or V₀ = 110 √ 2 V

The maximum current is given by the following relation:

I₀ = V₀/Z

I₀ =

I₀ =

On calculating, we get

I₀ = 3.24 A

(b) Since in capacitor circuit, the voltage lags behind the current by a phase angle Ф. Therefore, the phase angle is given by the expression

tanФ =

tanФ = 1/ωCR

substituting the values, we get

tanФ = 1/(120π(rad/s) × 10^-4(F) × 40 (Ω))

On calculating, we get

tanФ = 0.6635

or Ф = 33.56°

in radians, Ф = (33.56 π) /180

Time lag is given by the relation

t = Ф/ω

t =

on calculating, we get

t = 1.55 × 10^-3s

on converting in ms, we get time lag t = 1.55 ms.