Question

An L-R-C series circuit L = 0.123 H , R = 244 Ω , and C = 7.27 μF carries an rms current of 0.452 A with a frequency of 410 Hz .a) What is the phase angle? b) What is the power factor for this circuit? c) What is the impedance of the circuit? d) What is the rms voltage of the source? e) What is the average rate at which electrical energy is converted to thermal energy in the resistor? f) What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor?, g) In the inductor?

Answer 1

Answer:

A) phase angle = 0.82 rad

B) power factor = 0.68

C) Z = 244 + 264j

D) V(rms) = 162.5 V

E) Pr = 50 W

F) Qc = 11 VA

G) Ql = 65 VA

Explanation:

The impedance is:

R + X*j

The reactance is:

X = Xl - Xc

The inductive reactance is:

Xl = w * L = 2π * f * L = 2π * 410 * 0.123 = 317

The capacitive reactance is:

$Xc = \frac{1}{w * C} = \frac{1}{2 \pi * f * C} = \frac{1}{2 \pi * 410 * 7.27e-6} = 53.4$

So

X = 317 - 53.4 = 264

Z = 244 + 264j

$V(rms) = |Z| * I(rms) = \sqrt{244^2 + 264^2} * 0.452 = 162.5 V$

The phase angle is in the first quadrant, it is between 0 and π/2

$\phi = atan(\frac{X}{R}) =  = atan(\frac{264}{244}) = 0.82$

The power factor is the cosine of the phase angle:

$cos(\phi) = cos(0.82) = 0.68$

The power consumed by the resistor will be the resistance multiplied by the current(rms) squared

Pr = 244 * 0.452^2 = 50 W

Similar for the capacitor and inductor but with reactance instead of resistance

Qc = 53.4 * 0.452^2 = 11 VA

Ql = 317 * 0.452^2 = 65 VA