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?

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Answer:

Explanation:

power factor  in a series AC circuit  (cosΘ) =[tex]\frac{R}{Z}[/tex]

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 = [tex]\frac{1}{wc}[/tex]

= [tex]\frac{1}{360 * 3.5 * 10-6}[/tex]

= 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 = [tex]\frac{V}{Z} = \frac{45}{826.95}[/tex]

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

The power factor of the circuit is 0.30

The average power delivered to the entire circuit is 2.43 W.

The average power delivered to the resistor, capacitor and to the inductor are 2.43w, 2.31w, and 0.02W respectively.

The formula for

power factor in a series AC circuit is (cosΘ) =R² + (XL - XC)²

Where,

R is the resistance

Z is impedance of the circuit

w is the angular frequency = 360 rad/s

Z²= R² + (XL - XC)²

XL = wL

w = 360 rad/s

360× 15× 10⁻³

= 5.4Ω

XC = 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

The power factor = R / Z

= 250 / 826.95

= 0.30

To get the average power delivered to the entire circuit.

The formula is

V = I × Z

v potential difference or voltage

I = current

I =0.054A

but

P = I× V

where p is power

I is current=0.054A

v is voltage which is 45v

P= 0.054× 45

= 2.43 W

To get the average power delivered to the resistor

P = I²× R

where R is resistance.

= 0.054²× 250

= 0.729 W

To get the average power delivered to the capacitor.

P = I² ×XC

=0.054²× 793.65

= 2.31 W

To get average power delivered to the inductor

P = I²× XL

where xl is inductance.=5.4

= 0.054²× 5.4

= 0.02W

What is a circuit?

An electric circuit is comprises of electronic components like esistors, transistors, capacitors, inductors and diodes which are connected together by conductive wires through which electric current flow.

Therefore, The power factor of the circuit is 0.30

The average power delivered to the entire circuit is 2.43 W.

The average power delivered to the resistor, capacitor and to the resistor are 2.43w, 2.31w, and 0.02W respectively.

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