Question

A series ac circuit contains a 250-Ω resistor, a 15-mH inductor, a 3.5-μF capacitor, and an ac power source of voltage amplitude 45 V operating at an angular frequency of 360 rad/s. (a) What is the power factor of this circuit? (b) Find the average power delivered to the entire circuit. (c) What is the average power delivered to the resistor, to the capacitor, and to the inductor?

Answer 1

Answer:

Explanation:

power factor  in a series AC circuit  (cosΘ) =$\frac{R}{Z}$

R = resistance

Z = impedance of the circuit

Z²= R² + (XL - XC)²

XL = wL

from the question w is the angular frequency = 360 rad/s

360× 15× 10⁻³

= 5.4Ω

XC = $\frac{1}{wc}$

= $\frac{1}{360 * 3.5 * 10-6}$

= 793.65Ω

Z²= 250² + (5.4 - 793.65)²

= 62500 + (-788.25)²

62500+621338.06

Z²= 683838.0625

Z =√ 683838.0625

Z= 826. 95Ω

(a) power factor = R / Z

= 250 / 826.95

= 0.30

b) average power delivered to the entire circuit.

V = I × Z

I = $\frac{V}{Z} = \frac{45}{826.95}$

I =0.054A

P = I× V

= 0.054× 45

= 2.43 W

c) average power delivered to the resistor

P = I²× R

= 0.054²× 250

= 0.729 W

average power delivered to the capacitor

P = I² ×XC

=0.054²× 793.65

= 2.31 W

average power delivered to the inductor

P = I²× XL

= 0.054²× 5.4

= 0.02W