Question

A series RCL circuit contains a 65.2-Ω resistor, a 2.26-μF capacitor, and a 2.08-mH inductor. When the frequency is 2400 Hz, what is the power factor of the circuit?

Answer 1

The power factor of the circuit is 0.99

Step-by-step explanation:

Given;

resistance of the resistor, R = 65.2 ohms

capacitance of the capacitor, C = 2.26 μF = 2.26 x 10⁻⁶ F

inductance, L = 2.08 mH = 2.08 x 10⁻³ H

frequency of the AC, f = 2400 Hz

The angular frequency is given by;

ω = 2πf

ω = 2π(2400) = 15081.6 rad/s

The inductive reactance is given by;

XL = ωL

XL = (15081.6 x 2.08 x 10⁻³)

XL = 31.37 ohms

The capacitive reactance is given by;

`X_c = (1)/(\omega C) \\\\X_c = (1)/((15081.6)(2.26*10^(-6) ))\\\\X_c = 29.34 \ ohms`

The phase difference is given by;

`tan\phi = (X_l - X_c)/(R)\\\\ tan\phi =(31.37-29.34)/(65.2) \\\\tan\phi = 0.0311 \\\\\phi = tan^(-1) (0.0311)\\\\\phi = 1.78^0`

The power factor is given by;

CosФ = Cos(1.78) = 0.99

Therefore, the power factor of the circuit is 0.99