A 100 μF capacitor in series with a 40 Ω 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?

Capacitance of the capacitor, C = 100 μF = 100 x 10^-6 F = 10⁻⁴ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Ans (a).

Frequency of oscillations, ν= 60 Hz

Angular frequency, ω = 2π ν = 2π x 60 rad/s =120π rad/s

For a RC circuit, we have the relation for impedance as: Z = √(R² +1/ω²c²)

Peak voltage, V₀ = V√2 = 110√2. Maximum current is given as:

I₀= V₀ /Z  =  V₀/√(R² +1/ω²c²)

=110 √2 / √(R² +1/ω²c²)

=110 √2 / √(40² +1/(120π)²(10⁻⁴)²)

=3.24A

Ans (b).

In a capacitor circuit, the voltage lags behind the current by a phase angle of φ. This angle is given by the relation:

tan φ =(1/ωC)/R = 1/ωCR

=1/(120 π10⁻⁴x 40)

= 0.6635

Therefore, φ =tan⁻¹ (0.6635) = 33.56º

= 33.56π/180 rad

Therefore ,Time lag = φ/ω = 33.56 π/(180 x 120 π) = 1.55 x 10⁻³ s

=1.55 ms

Hence, the time lag between maximum current and maximum voltage is 1.55 ms.