Assume that the circuit in Figure 18-1 has a power factor of 78%, an apparent power of 374.817 VA, and a frequency of 400 Hz. The inductor has an inductance of 0.0382 H.
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The missing values in the table.
Answer to Problem 4PP
ET = 150.062 V | ER = 150.062 V | EL = 150.062 V |
IT = 2.498 A | IR = 1.948 A | IL = 1.563 A |
Z = 60.073 Ω | R = 77.024 Ω | XL = 96.007 Ω |
VA = 374.817 | P = 292.357 W | VARSL = 234.55 |
PF = 78% | L = 0.0382 H |
Explanation of Solution
Given data:
The inductive reactance XL is calculated as,
Using the Power factor (PF) and Apparent Power(VA), we can calculate True Power (P) as,
The reactive power VARSL is calculated as,
The relation between reactive power, voltage and inductive reactance is given as,
Since the circuit is parallel R-L,
The current through the inductor is,
The relation between true power P, voltage ER and resistance R is given as,
The current through the resistor is,
The current through the inductor is,
The total current in the circuit
The impedance Z is calculated as,
Power factor angle θ will be,
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Chapter 18 Solutions
Delmar's Standard Textbook Of Electricity
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