Answer:
The power factor of the circuit is 0.99
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;
[tex]X_c = \frac{1}{\omega C} \\\\X_c = \frac{1}{(15081.6)(2.26*10^{-6} )}\\\\X_c = 29.34 \ ohms[/tex]
The phase difference is given by;
[tex]tan\phi = \frac{X_l - X_c}{R}\\\\ tan\phi =\frac{31.37-29.34}{65.2} \\\\tan\phi = 0.0311 \\\\\phi = tan^{-1} (0.0311)\\\\\phi = 1.78^0[/tex]
The power factor is given by;
CosФ = Cos(1.78) = 0.99
Therefore, the power factor of the circuit is 0.99
