Basic Calculation For AC ElectricalCircuit

BASIC
CALCULATION FOR
AC ELECTRICAL
CIRCUIT

JUNAINAH BINTI ABD. KADIR
FAIZAL MOHAMAD TWON TAWI

JABATAN KEJURUTERAAN ELEKTRIK
POLITEKNIK SEBERANG PERAI

Basic Calculation for AC Electrical Circuit i

BASIC
CALCULATION FOR

AC ELECTRICAL
CIRCUIT

JUNAINAH ABD. KADIR
FAIZAL MOHAMAD TWON TAWI

2021
JABATAN KEJURUTERAAN ELEKTRIK

POLITEKNIK SEBERANG PERAI

ii Basic Calculation for AC Electrical Circuit

ALL RIGHTS RESERVED

No part of this publication may be translated or reproduced in any
retrieval system, or transmitted in any form or by any means,
electronic, mechanical, recording, or otherwise, without prior
permission in writing from Politeknik Seberang Perai.

PUBLISHED BY

Politeknik Seberang Perai
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[email protected] www.psp.edu.my
politeknikseberangperai politeknikseberangperai

Perpustakaan Negara Malaysia Cataloguing-in-Publication Data
Junainah Abd. Kadir
BASIC CALCULATION FOR AC ELECTRICAL CIRCUIT / JUNAINAH ABD. KADIR,
FAIZAL MOHAMAD TWON TAWI.
Mode of access: Internet
eISBN 978-967-2774-01-3
1. Electric circuits--Alternating current.
2. Electric circuits--Problems, exercises, etc.
3. Government publications--Malaysia.
4. Electronic books.
I. Faizal Mohamad Twon Tawi, 1982-.
II. Title.
621.31913076

Basic Calculation for AC Electrical Circuit iii

ACKNOWLEDGEMENT

Alhamdulillah, all thanks to the All Mighty for allowing us to complete
this book of Basic Calculation for AC Electrical Circuit.
Peace and blessings be upon our beloved Prophet Muhammad PBUH
who was never tired in spreading his love and guidance for us to be on
the foundation of truth.
We also would like to thank our colleagues in Electrical Engineering
Department from Politeknik Seberang Perai that helped us to complete
this book.

iv Basic Calculation for AC Electrical Circuit

PREFACE

Generally, studying of AC circuit subject is continuing of DC circuit
subject for electrical engineering students.
After study the DC circuit subject, they must continue the subject AC to
complete their knowledge of basic electrical circuit calculation.
Therefore, this book is written after the Basic Calculation for DC
Electrical Circuit was published.
It aims to help faculty members and students to make a revision on the
basis calculation of an AC circuit.
Hopefully, this book can help students improve their understanding of
the basic calculation in the subject of AC electrical circuits.

Basic Calculation for AC Electrical Circuit v

TABLE OF CONTENT

Introduction ...............................................................................................1

Chapter 1: AC WAVEFORM ...................................................................... 2
Express Note .............................................................................................. 2
Example ....................................................................................................... 4
Practice Problem ........................................................................................ 6

Chapter 2: AC SINUSOIDAL GENERAL EQUATION ............................. 12
Express Note ............................................................................................ 12
Example ..................................................................................................... 13
Practice Problem ...................................................................................... 15

Chapter 3: COMPLEX NUMBERS IN RECTANGULAR AND POLAR
FORM ......................................................................................... 21

Express Note ............................................................................................ 21
Example ..................................................................................................... 22
Practice Problem ...................................................................................... 24
Chapter 4: INDUCTOR (L) CIRCUIT ........................................................ 26
Express Note ............................................................................................ 26
Example ..................................................................................................... 27
Practice Problem ...................................................................................... 27
Chapter 5: CAPACITOR(C) CIRCUIT ...................................................... 30
Express Note ............................................................................................ 30
Example ..................................................................................................... 31
Practice Problem ...................................................................................... 32
Chapter 6: COMBINATION RLC SERIES CIRCUIT ................................ 34
Express Note ............................................................................................ 34
Example ..................................................................................................... 35
Practice Problem ...................................................................................... 37
Answers of Practice Problem ................................................................. 41

Bibliography ............................................................................................. vi



Basic Calculation for AC Electrical Circuit 1

Introduction

In the electrical circuit, the current that changes its direction constantly
according to time is called Alternating Current (AC). Changes in polarity
of the AC voltage generator makes the circuit change of current flow.

The basic waves found in AC waveform are like sine waves, cosine
waves, rectangular waves, triangular waves and turquoise waves.

Basic calculation of AC circuit include AC waveform, AC sinusoidal
general equation, and complex numbers in rectangular, complex
numbers in polar form, inductor circuit, and capacitor circuit and
combination series circuit.

Each chapter will consist of express note, example practice problem.
Answer of practice problem will be shown on the page before
bibliography.

This will make it easier for the students to review all answers students
made for each chapter.

2 Basic Calculation for AC Electrical Circuit

Chapter 1: AC WAVEFORM

Express Note

Sine wave is one of the shapes for AC waveforms. Figure 1 shows the sine
wave of AC waveform.

Figure 1: AC waveform

Period (T) – The time required for any AC wave to complete one cycle. Unit
in second (s)

Frequency (f) – The number of complete cycles which is repeated per second
generated by AC waveform. Unit in hertz (Hz).

= 1


Peak / Amplitude/Maximum Value (Vp/Ip) – The maximum current or voltage
that can be achieved by an AC sinewave from zero.

Peak to Peak Value (Vpp/Ipp) – The voltage or current peak to peak value is
twice its peak value.

= 2

= 2

=
2

Basic Calculation for AC Electrical Circuit 3
Root Mean Square Value (Vrms/Irms) – The effective voltage or current that
can generate the same amount power as a direct current (DC) that flows
through the resistor.

= 0.707
= 0.707
Average Value (Vave/Iave) = Average value represents the quotient of the
area under AC wave form with respect to time.
= 0.637 ×
= 0.637
All term in AC waveform are shown in Figure 2.

Figure 2: Labels of AC waveform

4 Basic Calculation for AC Electrical Circuit

Peak factor is ratio of peak value and rms value.

= =
0.707

RMS Value is ratio of peak value and peak factor.

=


Form factor is ratio of rms value and average value.

= = 0.707
0.637

Example

Question

Based on Figure 3, calculate the value of;

a. Period.
b. Frequency.
c. Amplitude/ peak voltage.
d. Root mean square voltage.
e. Average voltage.
f. Form factor.
g. Peak factor.

V

20V

S
10 20

Figure 3 : AC sinusoidal waveform

Basic Calculation for AC Electrical Circuit 5

Answer

a. Period.

= 20

b. Frequency.

= 1 = 1 = 50
20

c. Amplitude/ peak voltage.

= 20

d. Root mean square voltage.
= 0.707 × = 0.707 × 20 = 14.14

e. Average voltage.
= 0.637 × = 0.637 × 20 = 12.74

f. Form factor.

= = 14.14 = 1.11
12.74

g. Peak factor.

= = 20 = 1.41
14.14

6 Basic Calculation for AC Electrical Circuit

Practice Problem S
60
Question 1
Based on Figure 4, calculate the value of;

a. Period.
b. Frequency.
c. Amplitude/ peak voltage.
d. Root mean square voltage.
e. Average voltage.
f. Form factor.
g. Peak factor.

V

10V

30

Figure 4: AC sinusoidal waveform

Answer = 60
a. Period.

b. Frequency.

= 1 = 1 = 16.67
60

Basic Calculation for AC Electrical Circuit 7

c. Amplitude/ peak voltage.
= 10

d. Root mean square voltage.
= 0.707 × = 0.707 × 10 = 7.07

e. Average voltage.
= 0.637 × = 0.637 × 10 = 6.37

f. Form factor.

= = 7.07 = 1.11
6.37

g. Peak factor.

= = 10 = 1.41
7.07

8 Basic Calculation for AC Electrical Circuit s
50
Question 2
Based on Figure 5, calculate the value of;

a. Period.
b. Frequency.
c. Amplitude/ peak current.
d. Root mean square current.
e. Average current.
f. Form factor.
g. Peak factor.

I

30A

10 30

Figure 5: AC sinusoidal waveform

Answer

a. Period.

= 50

b. Frequency.

= 1 = 1 = 20
50

Basic Calculation for AC Electrical Circuit 9

c. Amplitude/ peak current.
= 30

d. Root mean square current.
= 0.707 × = 0.707 × 30 = 21.21

e. Average current.
= 0.637 × = 0.637 × 30 = 19.11

f. Form factor.

= = 21.21 = 1.11
19.11

g. Peak factor.

= = 30 = 1.41
21.21

10 Basic Calculation for AC Electrical Circuit

Question 3

Based on Figure 6, calculate the value of;

a. Period.
b. Frequency.
c. Amplitude/ peak current.
d. Root mean square current.
e. Average current.
f. Form factor.
g. Peak factor.

V

50V

-60 0 ms
60

Figure 6: AC sinusoidal waveform

Answer

a. Period.

= 120

b. Frequency.

= 1 = 1 = 8.33
120

Basic Calculation for AC Electrical Circuit 11

c. Amplitude/ peak voltage.
= 50

d. Root mean square voltage.
= 0.707 × = 0.707 × 50 = 35.35

e. Average voltage.
= 0.637 × = 0.637 × 50 = 31.85

f. Form factor

= = 35.35 = 1.11
31.85

g. Peak factor

= = 50 = 1.41
35.35

12 Basic Calculation for AC Electrical Circuit

Chapter 2: AC SINUSOIDAL
GENERAL EQUATION

Express Note

The general sinusoidal equation of AC voltage and AC current are:

Voltage:

= ( ± )( )

Where;

= ℎ

= ℎ @ @
= ℎ

= ℎ ℎ

=

Current:

= ( ± )( )

Where;

= ℎ

= ℎ @ @
= ℎ

= ℎ ℎ

=

Basic Calculation for AC Electrical Circuit 13

Example

Question
An Alternating voltage is given by V = 120 sin 314.2t V. Find;
a) The amplitude.
b) The root mean square voltage.
c) The average.
d) The frequency.
e) The period.
f) The instantaneous value of voltage when t = 4ms.
g) The form factor.
h) The peak factor.

Answer
Compare the given equation with the general sinusoidal equation

= 120 314.2 ↔ = ( ± )

a. The amplitude

= = 120

b. The root mean square voltage
= 0.707 × 120 = 84.84

c. The average
= 0.637 × 120 = 76.44

d. The frequency

To find the frequency we can use this formula
= 2

From the equation, =314.2

Therefore;

14 Basic Calculation for AC Electrical Circuit

= 314.2 = 50
2
e. The period

To find the period (T), we can use this formula

= 1 = 1 = 0.02
50

f. The instantaneous value of voltage when t = 4ms.

= 120 314.2
= 120 314.2(4 )

= 120 1.256
= 114.1

f) The form factor

= = 84.84 = 1.11
76.44

g) The peak factor

= = 120 = 1.41
84.84

Basic Calculation for AC Electrical Circuit 15

Practice Problem

Question 1

Calculate the amplitude, root mean square current, average, frequency,
instantaneous value of current when t = 2m, form factor, peak factor of the
AC voltage equation.

= 18 (50 + 100) .

Answer

Given:

= 18 (50 + 100)

Compare the equation given by general equation:
= 18 (50 + 100) ↔ = sin( + )

The amplitude

= = 18

The root mean square voltage (rms)

= 0.707 × 18 = 12.73

The average = 0.637 × 18 = 11.47
The frequency

To find the frequency we can use this formula

From the equation, =50 = 2

Therefore;

16 Basic Calculation for AC Electrical Circuit

= 50 = 7.96
2

The period

To find the period (T), we can use this formula

= 1 = 1 = 0.13
7.96

The instantaneous value of voltage when t = 2ms.

= 18 (50 + 100) .
= 18 (50(2 ) + 100) .

= 18 (0.1 + 100) . Change to radian
= 18 sin(0.1 + 0.175) .

= 18 sin(0.275) .
= 4.888

The form factor

= = 12.73 = 1.11
11.47

The peak factor

= = 18 = 1.41
12.73

Basic Calculation for AC Electrical Circuit 17

Question 2

The current in an AC circuit at any time t seconds is given by =
100 (100 + 0.36) amperes. Calculate:

a. The amplitude.
b. The root mean square current.
c. The average.
d. The frequency.
e. The period.
f. The instantaneous value of current when t = 9ms.
g. The form factor.
h. The peak factor.

Answer

Given

= 100 sin(100 + 0.36)

Compare the equation given by general equation
= 100 sin(100 + 0.36) ↔ = sin( + )

a. The amplitude = = 100

b. The root mean square current
= 0.707 × 100 = 70.70

c. The average
= 0.637 × 100 = 63.7

d. The frequency
To find the frequency we can use this formula

= 2

From the equation, =100

Therefore; = 100 = 50
e. The period 2

18 Basic Calculation for AC Electrical Circuit

To find the period (T), we can use this formula

= 1 = 1 = 0.02
50

f. The instantaneous value of current when t = 9ms.

= 100 sin(100 (9 ) + 0.36)

= 100 sin(2.83 + 0.36)

= 100 3.19
= −4.58

g. The form factor

= = 70.7 = 1.11
63.7

h. The peak factor

= = 100 = 1.41
70.7

Basic Calculation for AC Electrical Circuit 19

Question 3
For the AC equation of = 160 (157.1 + 600) .

a. Determine the value of peak, rms, average, frequency, period
and different phase in radian term.

b. Find the value of V when t=1ms.
c. Calculate the form and peak factor.

Answer

a. The value of peak, rms, average, frequency, period and different
phase in radian term.

Peak value

= 160

RMS value

= 0.707 × 160 = 113.12
Average value

= 0.637 × 160 = 101.92

Frequency

= 157.1 = 25
2

Period

= 1 = 1 = 0.04
25

Different phase in radian,

= π × 60 = 1.047 rad
180

b. At t=1ms

= 160 sin(157.1(1 ) + 1.047) Change to radian

= 160 (1.2041)

= 149.36

20 Basic Calculation for AC Electrical Circuit

c. The form factor and the peak factor

= = 113.12 = 1.11
101.92

= = 160 = 1.41
113.12

Basic Calculation for AC Electrical Circuit 21

Chapter 3: COMPLEX NUMBERS IN
RECTANGULAR AND POLAR FORM

Express Note

There are two types of complex number, which are polar form and rectangular
form.
Before knowing these two forms, we need to know first about the impedance
triangle.

Figure 7: Impedance Triangle

Based on the impedance triangle we can know the rectangular form is
expressed as:

= ± (Ω)
Z = the impedance
R = the resistance
X = the reactance
Based on the impedance triangle also we can know the polar form is:

= | |∠ 0(Ω)
0= the phase angle
|Z| = the magnitude of impedance
How to convert polar form to rectangular form

22 Basic Calculation for AC Electrical Circuit

=
=
= ± (Ω) →

Convert rectangular form to polar form

| | = �( 2 + 2)
= −1 � �

= | |∠ 0(Ω) →

Add Operation

( 1 + 1) + ( 2 + 2) = ( 1 + 2) + ( 1 + 2)

Subtract Operation

( 1 + 1) − ( 2 + 2) = ( 1 + 2) − ( 1 + 2)

Multiple Operation

| 1|∠ 10 × | 2|∠ 20 = (| 1| × | 2|)∠( 10 + 20)

Division Operation

| 1|∠ 10 = (| 1| ÷ | 2|)∠( 10 − 20)
| 2|∠ 20

Example

Question 1
Convert = 8∠450in rectangular form

Answer

= 8 450 = 5.66
= 8 450 = 5.66

So, = 5.66 + 5.66

Basic Calculation for AC Electrical Circuit 23

Question 2
Express = 2 + 5 in polar form

Answer

= �22 + 52 = √29 = 5.3

= −1 �52� = 68.20

So,
= 5.3∠68.20

Question

Solve the following complex numbers
. (8 + 2) + (5 + 3)

. (8 + 2) − (5 + 3)

. (8 + 2) × (5 + 3)

. (8 + 2) ÷ (5 + 3)

. (8 + 2) + (5 + 3)
(8 + 2) × (5 + 3)

Answer

a. (8 + 2) + (5 + 3) = (8 + 5) + (2 + 3) = 13 + 5

b. (8 + 2) − (5 + 3) = (8 − 5) + ( 2 − 3) = 3 −

c. (8 + 2) × (5 + 3) = (8.25∠14.040) × (5.83∠30.960) = 48.10∠450

d. (8 + 2 ) ÷ (5 + 3) = (8.25∠14.040) ÷ (5.83∠30.960) = 1.42∠−16.920

e. (8+ 2)+(5+ 3) = 13+ 5 = 13.93∠21.04 = 0.29∠ − 23.960
(8+ 2)×(5+ 3) 48.10∠450 48.10∠450

24 Basic Calculation for AC Electrical Circuit

Practice Problem

Question 1
Convert = 20∠530in rectangular form

Answer

= 20 530 = 12.04
= 20 530 = 15.97

So,
= 12.04 + 15.97

Question 2
Express = 12 + 15 in polar form

Answer

= �122 + 152 = √369 = 19.21

= −1 �1125� = 51.340
So,

= 19.21∠51.340

Question 3

Solve the following complex numbers
. (7 + 4) + (6 + 3)
. (7 + 4) − (6 + 3)
. (7 + 4) × (6 + 3)
. (7 + 4) ÷ (6 + 3)

Basic Calculation for AC Electrical Circuit 25

. (7 + 4) + (6 + 3)
(7 + 4) × (6 + 3)

Answer

. (7 + 4) + (6 + 3) = (7 + 6) + (4 + 3) = 13 + 7

. (7 + 4) − (6 + 3) = (7 − 6) + ( 4 − 3) = 1 +

. (7 + 4) × (6 + 3) = (8.06∠29.740) × (6.71∠26.570) = 54.08∠56.310

. (7 + 4) ÷ (6 + 3) = (8.06∠29.740) ÷ (6.71∠26.570) = 1.20∠3.170

. (7 + 4) + (6 + 3) = 13 + 7 = 14.76∠28.300 = 0.27∠ − 280
(7 + 4) × (6 + 3) 54.08∠53.310 54.08∠53.310

26 Basic Calculation for AC Electrical Circuit

Chapter 4: INDUCTOR (L) CIRCUIT

Express Note

In Figure 8, an inductor connected with AC voltage supply.

Figure 8: AC Inductor Circuit

When AC voltage is supplied across the inductor, the current flows in the

inductor will establish the magnetic flux changes. This change of magnetic
flux induces across the coil in the direction to oppose the current

conductions.

Inductive reactance, XL can be calculated as:
= 2

Where
= ( )

= ( )

Ohms ‘Law

=

Basic Calculation for AC Electrical Circuit 27

Example

Question

An inductor of 0.2mH inductance is connected to a 235V, 50Hz voltage
source. Calculate:

a. Inductive reactance.
b. The current flows through the inductor.

Answer

Given = 0.2 , = 50 and voltage source, = = 235

a. Inductive reactance.
= 2 = 2 (50)(0.2) = 62.83Ω

b. The current flows through the inductor.

= = 235 = 3.74
62.83

Practice Problem

Question 1

Based on Figure 9, calculate:

a. Inductive reactance.
b. The current flows through the inductor.

Figure 9: AC Inductor circuit

28 Basic Calculation for AC Electrical Circuit

Answer
Given = 0.8 , = 60 and voltage source, = = 100

a. Inductive reactance.
= 2 = 2 (60)(0.8) = 301.59Ω

b. The current flows through the inductor.

= = 100 = 331.58
301.59

Question 2

An inductor of 40mH inductance is connected to a 240V, 50Hz
voltage source. Calculate:

a. Inductive reactance.
b. The current flows through the inductor.

Answer

Given = 40 , = 50 and voltage source, = = 240

a. Inductive reactance.
= 2 = 2 (50)(40 ) = 12.57Ω

b. The current flows through the inductor.

= = 240 = 19.09
12.57

Basic Calculation for AC Electrical Circuit 29

Question 3

A 240V, 60Hz voltage supply is connected to an inductor and the
current flow to the circuit is 3A. Calculate the value of inductor.

Answer

Given: voltage source, = = 240 and = 3 .

Inductive reactance.

= = 3 = 12.5 Ω
240

The value of inductor

= 2

=
2

12.5 =
2 (60)

33.2µ =

30 Basic Calculation for AC Electrical Circuit

Chapter 5: CAPACITOR(C)
CIRCUIT

Express Note

In Figure10, a capacitor connected with AC voltage supply.

Figure 10: RC circuit

When a capacitor circuit supplied by an AC voltage supply, the electrons will
travel between the two metal plates in the capacitor. The current flowing in
the circuit will be opposed by the capacitor and the capacitor's function is
named capacitive reactance, XC.

The equation of capacitive reactance is

= 1
2

= ( )

= ( )

Ohms ‘Law

=

Basic Calculation for AC Electrical Circuit 31

Example

Question
Figure 11 shows a circuit consisting of 20μF capacitor with 240
voltage source and 60Hz frequency. Determine:
a. Capacitive reactance.
b. The current flows through the capacitor.

Figure 11: Capacitor circuit

Answer

Given = 20 , = 60 and voltage source, = = 240

a. Capacitive reactance.

= 1 = 1 = 132.63Ω
2 2 (60)(20 )

b. The current flows through the capacitor.

= = 240 = 1.81
132.63

32 Basic Calculation for AC Electrical Circuit

Practice Problem

Question 1
Figure 12 shows a circuit consisting of 2mF capacitor with 100V
voltage source and 50Hz frequency. Determine:
a. Capacitive reactance.
b. The current flows through the capacitor.

Figure 12: Capacitor circuit

Answer

Given = 2 , = 50 and voltage source, = = 100

a. Capacitive reactance.

= 1 = 1 = 1.59Ω
2 2 (50)(2 )

b. The current flows through the capacitor.

= = 100 = 62.89
1.59

Question 2

A capacitor of 100μF capacitance is connected to a 240V, 50Hz
voltage source. Calculate:

a. Capacitive reactance.

Basic Calculation for AC Electrical Circuit 33

b. The current flows through the capacitor.

Answer

Given = 100 , = 50 and voltage source, = = 240

a. Capacitive reactance.

= 1 = 1 = 31.83Ω
2 2 (50)(100 )

b. The current flows through the inductor.

= = 240 = 7.54
31.83

Question 3

A 240V, 60Hz voltage supply is connected to an capacitor and the
current flow to the circuit is 2A. Calculate the value of capacitor.

Answer

Given: voltage source, = = 240 and = 2 .

Capacitive reactance.

= = 2 = 8.33 Ω
240

The value of capacitor

= 1
2

= 1
2

= 1
(8.33 )2 (60)

= 0.318

34 Basic Calculation for AC Electrical Circuit

Chapter 6: COMBINATION RLC
SERIES CIRCUIT

Express Note

Figure 13 shows a resistor, an inductor and a capacitor connected in series
with an AC voltage source.

Figure 13: RLC series circuit.

In AC circuit, the total opposition to the flow of current in a sin wave is known
as impedance, Z.
Impedance consist of real (R) or imaginary part (X), measured in ohm (Ω).
The component of real part is resistor and the component of imaginary part
may represent inductor or capacitor or both of inductor & capacitor.
The general equation in polar form as:

= + ( − )(Ω)
The general equation in polar form as:

= | |∠ ± (Ω)
The equation of total impedance in series circuit is:

= 1 + 2 + ⋯

Basic Calculation for AC Electrical Circuit 35

Ohm’s law is used:

=


Summary of impedances of passive elements is shown in Table 1

Table 1: Impedance of passive elements

Elements Impedances Formula
Resistor (R)
Inductor (L) R
Capacitor (C) = 2
1
= 2

Example

Question

Based on the circuit in Figure 14, calculate:

a. The inductive reactance, XL.
b. The capacitive reactance, XC..
c. The total impedance, ZTotal .
a. The current in the circuit.
d. The voltage across the inductor.

Figure 14: RLC series circuit

36 Basic Calculation for AC Electrical Circuit

Answer

a. The inductive reactance, XL.
= 2 = 2 (60)(0.2) = 75.40Ω

b. The capacitive reactance, XC

= 1 = 1 = 1.33Ω
2 2 (60)(2 )

c. The total impedance, ZTotal
= + ( − )(Ω)

= 8 + (75.4 − 1.33)(Ω)

= 8 + 74.07Ω
= 74.5∠83.840Ω

d. The current in the circuit

= = 100∠00 = 100∠00 = 1.34∠ − 83.840
8 + 74.07 74.5∠83.840

e. The voltage across the inductor.
= × = (1.34∠ − 83.840)(75.4∠900) = 101.04∠6.160

Basic Calculation for AC Electrical Circuit 37

Practice Problem

Question 1
Based on the circuit in Figure 15, calculate:
a. The inductive reactance, XL.
b. The capacitive reactance, XC.
c. The total impedance, ZTotal.
d. The current in the circuit.
e. The voltage across the inductor.

Figure 15: RLC series circuit

Answer

a. The inductive reactance, XL
= 2 = 2 (50)(0.6) = 188.50Ω

b. The capacitive reactance, XC

= 1 = 1 = 0.80Ω
2 2 (50)(4 )

c. The total impedance, ZTotal
= + ( − )(Ω)

38 Basic Calculation for AC Electrical Circuit

= 10 + (188.5 − 0.80)(Ω)

= 10 + 187.7Ω
= 187.97∠86.950Ω

d. The current in the circuit

= = 120∠00 = 120∠00 = 0.64∠ − 86.950
10 + 187.7 187.97∠86.950

e. The voltage across the inductor.
= × = (0.64∠ − 86.950)(188.5∠900) = 121.64∠3.050

Question 2

An RLC series circuit consists of R = 20Ω, C = 200µF, and L =
100mH with a voltage source of 120V, 60Hz. Determine:

a. The inductive reactance, XL.
b. The capacitive reactance, XC.
c. The total impedance, ZTotal.
d. The current in the circuit.
e. The voltage across the capacitor.

Answer

a. The inductive reactance, XL
= 2 = 2 (60)(100 ) = 37.70Ω

b. The capacitive reactance, XC

= 1 = 1 = 13.26Ω
2 2 (60)(200 )

c. The total impedance, ZTotal
= + ( − )(Ω)

= 10 + (37.7 − 13.26)(Ω)

= 10 + 24.44Ω

Basic Calculation for AC Electrical Circuit 39

= 26.41∠67.750Ω

d. The current in the circuit

= = 120∠00 = 4.54∠ − 67.750
26.41∠67.750

e. The voltage across the capacitor.
= × = (4.54∠ − 67.750)(13.26∠−900) = 60.20∠−157.750

Question 3
Voltage source = 120∠00 is supplied to three impedances that
connected in series circuit which are:
1 = 6∠450, 2 = 5∠800 3 = 2∠ − 450,

Calculate:

a. The total impedance.
b. The current in the circuit.
c. The voltage across the 1.

Answer

The total impedance

a. Convert the impedance from polar to rectangular form first
1 = 6∠450 ↔ 4.24 + 4.24
2 = 5∠800 ↔ 0.87 + 4.92

3 = 2∠ − 450 ↔ 1.41 − 1.41
= 1 + 2 + 3

= 4.24 + 4.24 + 0.87 + 4.92 + 1.41 − 1.41 = 6.52 + 7.75Ω

b. The current in the circuit

40 Basic Calculation for AC Electrical Circuit

= = 120∠00 = 120∠00 = 11.84∠ − 49.910
6.52 + 7.75 10.13∠49.930

c. The voltage across the 1
1 = × 1 = (11.84∠ − 49.910)(6∠450) = 71.04∠−4.910 V

Basic Calculation for AC Electrical Circuit 41

Answers of Practice Problem

Chapter 1: AC WAVEFORM
Question 1

a. = 60
b. = 16.67
c. = 10 ,
d. = 7.07
e. = 6.37
f. = 1.11
g. = 1.41
Question 2
a. = 40
b. = 25
c. = 30
d. = 21.21
e. = 19.11
f. = 1.11
g. = 1.41
Question 3

a. = 120
b. = 8.33
c. . = 50
d. = 35.35
e. = 31.85
f. = 1.11
g. = 1.41

42 Basic Calculation for AC Electrical Circuit

Chapter 2: AC SINUSOIDAL GENERAL EQUATION

Question 1
a. = 18
b. = 12.73
c. = 11.47
d. = 50
e. = 0.02
f. = 4.88V
g. = 1.11
h. = 1.41

Question 2
a. = = 100
b. = 0.707 × 100 = 70.70
c. = 0.637 × 100 = 63.7
d. = 50
e. = 0.02
f. = −4.84
g. = 1.11
h. = 1.41

Question 3
a. = 160 , = 101.92 , = 25 , = 0.04 , = 1.047 rad
b. = 149.36
c. 1.11, = 1.41

Basic Calculation for AC Electrical Circuit 43

Chapter 3: COMPLEX NUMBERS IN RECTANGULAR AND POLAR FORM

Question 1
= 12.04 + 15.97

Question 2
= 19.21∠51.340

Question 3
a. 13 + 7
b. 1 +
c. 54.08∠56.310
d. 1.20∠3.170
e. 0.27∠ − 280