Answered: A load on a 120Vac 60Hz power source…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

A load on a 120Vac 60Hz power source consists of a 10 Ohm resistor in series with a 27 mH inductance. What value of capacitance should be put in parallel with the load to correct the power factor?

a) 133 uF

b) 270 uF

c) 2.2 uF

d) 470 uF

Please show the calculations. Thanks

Expert Solution

Step 1

Given: Voltage across the load, V=120 V

Frequency, f=60 Hz

Load resistance, R_L=10 $Ω$

Load inductance, L=27 mH

Capacitance is connected in parallel to improve power factor as shown in the figure below:

Step 2

Now, inductive reactance is given by:

$\begin{array}{rcl} X_{L} & = & 2 πfL \\ = & 2 π \times 60 \times 27 \times 10^{- 3} \\ = & 10.178 Ω \end{array}$

So, the load impedance will be:

$\begin{array}{rcl} Z & = & \sqrt{R^{2} + {X_{L}}^{2}} \\ = & \sqrt{10^{2} + 10 . 178^{2}} \\ = & 14.269 Ω \end{array}$

The load power factor will be:

$\begin{array}{rcl} \cos ϕ_{1} & = & \frac{R}{Z} \\ = & \frac{10}{14.269} \\ = & 0.7 l a g \end{array}$

The power is lagging because the load is a series RL load.

Active power, P will be:

$P = \frac{V^{2}}{R} = \frac{{(120)}^{2}}{10} = 1440 w$

Reactive power, Q₁ will be:

$\begin{array}{rcl} Q_{1} & = & P \tan ϕ_{1} \\ = & 1440 \tan (\cos^{- 1} 0.7) \\ = & 1469.09 V A R \end{array}$

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