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 3PP

The circuit in Figure 24-2 is connected to a 60-Hz line. The apparent power in the circuit is 48.106 VA. The resistor has a resistance of 12 Ω . The inductor has an inductive reactance of 60 Ω , and the capacitor has a capacitive reactance of 45 Ω .

E T  _____ I T  _____ Z _____ VA 48 .106  PF _____ E R  _____ I R  _____ R 12  Ω P _____ θ _____ E L _ _ _ _ _ I L  _____ X L 60   Ω VARs L  _____ L_____ E _ _ _ _ _ I C _ _ _ _ _ X C  45  Ω   VARs C  _____ _ _ _ _ _

Expert Solution & Answer
Check Mark
To determine

The missing values in the table.

Answer to Problem 3PP

ET = 23.99 V ER = 23.99 V EL = 23.99 V EC = 23.99 V
IT = 2.004 A IR = 1.999 A IL =0.399 A IC =0.533 A
Z = 11.973 Ω R = 12 Ω XL = 60 Ω XC = 45 Ω
VA = 48.106 P = 47.96 W VARSL = 9.575 VARSC = 12.791
PF = 99.725 % ∠θ = 4.25° L = 0.159 H C = 58.94 µF

Explanation of Solution

Given data :

VA=48.106   f=60 Hz   R = 12 Ω XL= 60 Ω XC= 45 Ω

The value of the inductor is given by,

L=XL2πf=602π×60 = 0.159 H

The value of the capacitor is given by,

C=12πfXC=12π×60×45=58.94μF

The impedance of the parallel circuit is given as,

Z=1(1R)2+(1XL1XC)2Z=1(112)2+(160145)2Z=1(112)2+(152700)2Z=1(112)2+(1180)2Z=11.973 Ω

To determine the value of total voltage across the parallel R-L-C circuit,

VA=ETITVA=ET(ETZ)VA=ET2ZRearrange the terms to get an expression for ET,ET=VA×ZET=48.106×11.973ET=23.999 V

Since the circuit is parallel R-L-C circuit,

ET=ER=EL=EC=23.999 V

The total current flowing through the parallel RLC circuit will be,

IT=ETZ=23.99911.973=2.004 A

The current flowing through the resistor in the parallel RLC circuit will be,

IR=ERR=23.99912=1.999 A

The current flowing through the inductor in the parallel RLC circuit will be,

IL=ELXL=23.99960=0.399 A

The current flowing through the capacitor in the parallel RLC circuit will be,

IC=ECXC=23.99945=0.533 A

The true power in the parallel circuit will be,

P=ERIR=23.999×1.999=47.974 W

The reactive power of the inductor is given by,

VARSL=ELIL=23.999×0.399=9.575

The reactive power of the capacitor is given by,

VARSC=ECIC=23.999×0.533=12.791 

The Power factor of the circuit is calculated as,

PF=PVA×100PF=47.97448.106×100PF=99.725 %

Power factor angle θ will be,

cosθ=PFcosθ=99.725100cosθ=0.99725     θ=cos10.99725     θ=4.25°

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