Answered: Suppose that a network can be divided…

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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Related questions

Question

Suppose that a network can be divided into two subnets connected by only two conductors, as shown below. Two points on these conductors are labeled a and b. The power factor seen to the right of a and b is defined as cos (ϕ) if ϕ is the angle formed by the phasor V and I. (Given the parity of the cosine, it does not matter if it is the angle by which V is advanced with relative to I or for the one that is delayed.) (Image 1)

Now consider the following circuit, in which six points have been marked. (Image 2)

a) Determine the power factor seen to the right of a3 and b3
analyze your result. Does it make physical sense? What is happening to the currents in the inductor and in the capacitor?

b) Determine the power factor seen to the right first of a2 and b2 and then of a1 and b1. Re-analyze your results.

c) Calculate the capacitance that must be connected between a1 and b1 so that the power factor seen to the right of these points is 0.95 in lag. Assume the source frequency is 60 Hz.

100
ai
a2
a3
102
240 V
rms /45°
E-j10n gj10N
Sj102
bị
b2

Expert Solution

Step 1

According to question:

a) The equivalent impedance of -j10ohm and +j10ohm is calculated as

$∵ Z_{e q} = \frac{j 10 \times - j 10}{j 10 - j 10} \Rightarrow Z_{e q} = \frac{j 10 \times - j 10}{0} \Rightarrow Z_{e q} = \infty \Rightarrow Z_{e q} = open circuit .$

So the resultant current in the combination of -j10ohm and +j10ohm is 0A as the combination is acting as a open circuit. As the current is 0A due to open circuit so power factor cannot be defined for this case as power factor is defined as the cosine of angle difference between voltage and current PHASORS. The given combination of -j10ohm and +j10ohm does not make any sense as it acts as an open circuit.

The inductor current and capacitor current is calculated as

$∵ I = \frac{240 ∠ 45^{0}}{10 + (10 + j 10)} \Rightarrow I = 10.733 ∠ 18 . 434^{0} A ∴ V_{a_{2} b_{2}} = V_{a_{3} b_{3}} = I (10 + j 10) \Rightarrow V_{a_{2} b_{2}} = V_{a_{3} b_{3}} = 10.733 ∠ 18 . 434^{0} \times (10 + j 10) \Rightarrow V_{a_{2} b_{2}} = V_{a_{3} b_{3}} = 151.787 ∠ 63 . 434^{0} V ∵ I_{C} = \frac{V_{a_{3} b_{3}}}{- j 10} \Rightarrow I_{C} = \frac{151.787 ∠ 63 . 434^{0}}{- j 10} \Rightarrow I_{C} = 15.178 ∠ 153 . 434^{0} A ∵ I_{L} = \frac{V_{a_{3} b_{3}}}{j 10} \Rightarrow I_{L} = \frac{151.787 ∠ 63 . 434^{0}}{10 ∠ 90^{0}} \Rightarrow I_{L} = 15.178 ∠ - 26 . 566^{0} A$