Q6:- For the system in Fig. 5: a) Find Pr. Qr. ST, Vr and the power factor Fp. b) Draw the powertriangle.er lep oV 100 VZ0$301800 W= 30 A20°1600 VARInductive load4 0U8

Answered: Q6:- For the system in Fig. 5: a) Find…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Q6:- For the system in Fig. 5: a) Find Pr. Qr. ST, Vr and the power factor Fp. b) Draw the power
triangle.
er lep o
V 100 VZ0
$30
1800 W
= 30 A20°
1600 VAR
Inductive load
4 0
U8

Expert Solution

Step 1

Given:

Step 2

The given circuit is redrawn as shown in the below figure:

In the above figure, the current I₁ is assumed to be flowing through the resistor of $8 Ω$ and I₃ is the current flowing through the impedance of $3 - j 4$ .

The current through the 3-j4 impedance can be calculated as shown below:

$I_{3} = \frac{100 ∠ 0 °}{3 - j 4} I_{3} = \frac{100 ∠ 0 °}{5 ∠ - 53.13 °} I_{3} = 20 ∠ + 53.13 ° A I_{3} = 12 + j 16 A$

The power through the 3-j4 impedance can be calculated as shown below:

$P_{3} = I^{2} R P_{3} = {I_{3}}^{2} \times (3) P_{3} = 20^{2} \times 3 P_{3} = 1200 W Q_{3} = I^{2} X_{C} Q_{3} = {I_{3}}^{2} X_{C} Q_{3} = 20^{2} \times 4 Q_{3} = 1600 V A R [Since, capacitor is present therefore it is leading]$

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