Math, asked by abhishekchowdary6302, 2 months ago

Question:
In a series RL circuit excited by an AC source of 10 V, 50 Hz and the current flowing through the circuit is 2A.
Resistance(R) of the circuit is 20. Compute the power factor angle, active power and reactive power consumed by
the circuit​

Answers

Answered by amitnrw
2

Given : In a series RL circuit excited by an AC source of 10 V, 50 Hz and the current flowing through the circuit is 2A.

Resistance(R) of the circuit is 2Ω .  

To Find :

the power factor angle,

active power

reactive power

Solution:

V = 10 V

f = 50 Hz

I = 2A

R - 2  Ω

L = ?

XL  = ωl = 2πfL    = 100πL   ( f = 50 Hz)

Z =  V/I

=> Z = 10/2  = 5 Ω

Z = √(R² + (XL)²

=> 5 = √(2² + (XL)²

=> 25 = 4 + (XL)²

XL = √21 =  4.583 Ω

=> 21 =   (100πL)²

=>  L =  14.58  mH

Active power = P = I²R    = 2² * 2  = 8 W

Reactive power = Q = I²XL   = 2²  * 4.583 = 18.33 W

Apparent power = S = I²Z  = 2² * 5  = 20W

Power factor =  Active power  / Apparent power   = 8/20

= 2/5

= 0.4 = cos ( power factor angle )

Power factor Angle = 66.42°

or Simpler method

V = 10  and I = 2

Hence Apparent power  = VI  = 10 * 2 = 20W

Active Power  =  2² * 2  = 8 W

Reactive power = √(20)² - 8²  = 18.33 W

Power factor =  Active power  / Apparent power   = 8/20

= 2/5

= 0.4 = cos ( power factor angle )

Power factor Angle = 66.42°

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