Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN: 9781337900348
Author: Stephen L. Herman
Publisher: Cengage Learning
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Chapter 24, Problem 1RQ

An AC circuit contains a 24 Ω resistor, a 15.9-mH inductor, and a 13.3 μF capacitor connected in parallel. The circuit is connected to a 240-V, 400-Hz power supply. Find the following values.

X L =  ____________ Ω X C =  ____________ Ω I R =  ____________A I L =  ____________A I C =  ____________A P =  ____________W VARs L =  ________ VARs C =  ________ I T =  ____________A VA =  __________ PF =  ___________% θ =  ____________

Expert Solution & Answer
Check Mark
To determine

The missing values.

Answer to Problem 1RQ

XL=39.96 ΩXC=29.91 ΩIR=10 AIL=6 AIC=8.02 AP=2400 WVARSL=1440VARSC=1924.8IT=10.20 AVA=2448PF=98.03 %θ=11.36°

Explanation of Solution

Given data:

  R=24 Ω  L=15.9 mH  C= 13.3 μFET=240 V  f=400Hz

Since the circuit is a parallel RLC,

ET=ER=EL=EC=240 V

The reactance of the inductor is given by,

XL=2πfL = 2π×400×15.9×103= 39.96 Ω

The reactance of the capacitor is given by,

XC=12πfC=12π×400×13.3×106=29.91 Ω      

The current flowing in the resistor is,

IR=ERR=24024=10 A

The current flowing in the inductor is,

IL=ELXL=24039.96=6 A

The current flowing in the capacitor is,

IC=ECXC=24029.91=8.02 A

The true power in the circuit is,

P=ERIR=240×10=2400 W

The reactive power in the inductor is,

VARSL=ELIL=240×6=1440 

The reactive power in the capacitor is,

VARSC=ECIC=240×8.02=1924.8 

The total current in the parallel R-L-C circuit is given,

IT=IR2+(ILIC)2IT=102+(68.02)2IT=102+(2.02)2IT=10.20 A

The apparent power in the circuit is,

VA=ETIT=240×10.206=2448 

The power factor in the circuit is,

PF=PVA×100=24002448×100=98.03%

The power factor angle is,

cosθ=PFcosθ=98.03100cosθ=0.9803     θ=cos10.9803     θ=11.36°

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