basic calculation for dc electrical circuitrepository.psp.edu.my/jspui/bitstream/123456789... ·...

i

BASIC CALCULATION FOR DC ELECTRICAL CIRCUIT

PREPARING STUDENTS FOR AC ELECTRICAL CIRCUIT PROBLEM SOLVING

FAIZAL BIN MOHAMAD TWON TAWI

FAIZAL BIN MOHAMAD TWON TAWI NOR HANIDA BINTI AHMAD

NURUL HIDAYAH BT. AHMAD SHAIRAZI

ii

ISBN 978-967-0783-31-4 First Print 2017 © POLITEKNIK SEBERANG PERAI All rights reserved. Not part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, whether print, electronic, mechanical, photocopying, recording or otherwise than by written permission of the author and publisher.

Perpustakaan Negara Malaysia Cataloguing-in-Publication Data

Faizal Bin Mohamad Twon Tawi, 1982- BASICS CALCULATION FOR DC ELECTRICAL CIRCUIT: PREPARING STUDENTS FOR AC

ELECTRICAL CIRCUIT PROBLEM SOLVING / FAIZAL BIN MOHAMAD TWON TAWI, NOR HANIDA BINTI AHMAD, NURUL HIDAYAH BT. AHMAD SHAIRAZI.

Bibliography: page 38

ISBN 978-967-0783-31-4 1. Electric circuit analysis. 2. Electronic circuits. . Nor Hanida Ahmad, 1981-. . Nurul

Hidayah Ahmad Shairazi, 1980 -. . Title. 621.3192

Review and edited by:-

Unit Penyelidikan & Penerbitan Perpustakaan Politeknik Seberang Perai. Publisher:- Perpustakaan, Politeknik Seberang Perai, Jalan Permatang Pauh, 13500 Permatang Pauh, Pulau Pinang. Tel: 04-5383322 Fax: 04-5389266 Email: [email protected]. Graphic Designer:- Faizal Bin Mohamad Twon Tawi

iii

Acknowledgments

Alhamdulillah, all thanks to the All mighty for allowing us to complete this book of Basic

Calculation for 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.

Finally, we would like to thank the staff of Unit Penyelidikan & Penerbitan Perpustakaan

Politeknik Seberang Perai (UPPen) for their helpful efforts during this project. The

cheerful professionalism of the staffs were appreciated.

iv

Preface

After several years of teaching the subject of electrical circuit, lecturers found that many

students were still weak in terms of basic DC circuit.

Whereas, almost all the basic knowledge of the DC circuit will be used when studying the

subject of AC circuits.

This is because the concept of calculation in DC circuit and AC circuits do not have any

change. There are still the same.

Consequently, the idea arose to collect all the important basic DC circuit.

It aims to help faculty members and students to make a revision on the basis calculation

of a DC circuit.

Hopefully, this book can help students improve their understanding of the basic DC circuit

and thus facilitating their studies in the subject of AC electrical circuits.

Contents

Acknowledgments .......................................................................................................... iii

Preface ............................................................................................................................. iv

Introduction ...................................................................................................................... 2

Chapter 1: Ohm's Law ..................................................................................................... 3

Express Note .......................................................................................................................... 3

Example .................................................................................................................................. 3

Practice Problem ................................................................................................................... 4

Chapter 2: Kirchhoff’s Voltage Law (KVL) ..................................................................... 6

Express Note .......................................................................................................................... 6

Example .................................................................................................................................. 7

Practice Problem ................................................................................................................... 8

Chapter 3: Kirchhoff’s Current Law (KCL) ................................................................... 10

Express Note ........................................................................................................................ 10

Example ................................................................................................................................ 11

Practice Problem ................................................................................................................. 12

Chapter 4: Series Circuit ............................................................................................... 16

Express Note ........................................................................................................................ 16

Example ................................................................................................................................ 18


Chapter 5: Parallel Circuit ............................................................................................. 22

Express Note ........................................................................................................................ 22

Example ................................................................................................................................ 24


Chapter 6: Series Parallel Circuit .................................................................................. 28

Express Note ........................................................................................................................ 28

Example ................................................................................................................................ 31


Solutions of Practice Problem ...................................................................................... 37

Bibliography ................................................................................................................... 38

2

Introduction

In the electrical circuit, the current flowing in one direction is called Direct Current (DC). A direct current

electrical circuit consisting of a dc supply source, a switch, a conductor and a load.

Basic components that are commonly found in the direct current circuit is as such as resistors,

capacitors, inductors, direct current supply source and etc.

To analyse the direct current electric circuit, some of the most basic principles of calculation must be

understood and mastered by students.

This is because, the basic knowledge of direct current electrical circuits will be used in learning basic

Alternating Current (AC) electrical circuits.

Some basic calculations that need to be understood by students are as follows:

1. Ohm’s Law

2. Kirchhoff’s Voltage Law

3. Kirchhoff’s Current Law

4. Series Circuit

5. Parallel Circuit

6. Series Parallel Circuit

When students have mastered all the six basic calculations, it will make it easier for students to learn

Alternating Current (AC) electrical circuits.

3

Chapter 1: Ohm's Law

Express Note

The voltage across a resistor is directly proportional to the current flowing through the resistor

as shown in Figure 1.

Figure 1 Current flow in a resistor

The mathematical form of Ohm's Law is

𝑉 = 𝐼𝑅

Example

1. An electric iron draws 3A at 240V. Find its resistance.

Solution

𝑉 = 𝐼𝑅

𝑉

𝐼= 𝑅

240

3= 𝑅

80Ω = 𝑅

2. Calculate the value of current that flows in the 60Ω resistor if the voltage drop across it is 6V.

Solution

𝑉 = 𝐼𝑅

𝑉

𝑅= 𝐼

6

60Ω= 𝐼

0.1𝐴 = 𝐼

4

Practice Problem

1. Find the value of resistor in Figure 2.

Figure 2 Resistive circuit

Solution

𝑉 = 𝐼𝑅1

𝑅1 =𝑉

𝐼=

40

8= 5Ω

2. In the circuit shown in Figure 3, calculate the current I.


Solution

𝑉 = 𝐼𝑅𝑡𝑜𝑡𝑎𝑙

𝐼 =𝑉

𝑅𝑡𝑜𝑡𝑎𝑙=

40

𝑅1 + 𝑅2 + 𝑅3=

40

20 + 10 + 10= 1𝐴

5

3. In the circuit shown in Figure 4, calculate the voltage across R1.


Solution

Find I first;

𝑉 = 𝐼𝑅𝑡𝑜𝑡𝑎𝑙

𝐼 =𝑉


40

𝑅1 + 𝑅2 + 𝑅3=

40

20 + 10 + 8= 1.05𝐴

Then, calculate

𝑉 = 𝐼𝑅1 = 1.05 × 20 = 21𝑉

6

Chapter 2: Kirchhoff’s Voltage Law (KVL)

Express Note The sum of all voltages or potential differences in an electrical circuit loop is 0. (I.e. see figure 5).


Σ𝑉𝑐𝑙𝑜𝑠𝑒𝑑 𝑙𝑜𝑜𝑝 = 0

E − 𝑉1 − 𝑉2 = 0

Or, in any closed loop, the algebraic sum of the electromotive force (e.m.f.) applied is equal to the

algebraic sum of the voltage drops in the elements.

𝐸𝑒𝑚𝑓 = ΣV𝑑𝑟𝑜𝑝

E = 𝑉1 + 𝑉2

7

Example

Based on Figure 6, determine the voltage drop, V3 at resistance, R3 by using Kirchhoff’s Voltage Law

(KVL).


Solution

Apply Kirchhoff’s Voltage Law (KVL) Theorem


10 + 20 + 𝑉3 − 40 = 0

𝑉3 = 40 − 30 = 10𝑉

Alternatively,

𝐸𝑒𝑚𝑓 = ΣV𝑑𝑟𝑜𝑝

40 = 20 + 10 + 𝑉3

𝑉3 = 40 − 30 = 10𝑉

8

Practice Problem

1. Find V2 in the circuit of Figure 7.


Solution

Apply KVL Theorem


8 + 𝑉2 + 20 − 45 = 0

𝑉2 = 45 − 28 = 17𝑉

2. Find V3 in the circuit of Figure 8.


Solution

Apply KVL Theorem


18 − 4 − 8 − 𝑉3 = 0, 𝑉3 = 18 − 12 = 6𝑉

9

4. Find the value of voltage drop Vx for a circuit in Figure 9 using KVL Theorem.


Solution

Apply KVL Theorem


15 + 𝑉𝑋 + 8 − 20 = 0

𝑉𝑋 = 20 − 8 − 15 = −3𝑉

10

Chapter 3: Kirchhoff’s Current Law (KCL)

Express Note

The algebraic sum of incoming currents in any electrical circuit to a point and the outgoing currents

from that point is Zero as shown in Figure 10.

Figure 10 Current flow in a point

Σ𝐼𝐴𝑡 𝑛𝑜𝑑𝑒 𝑎 = 0

𝐼𝑎 − 𝐼𝑏 − 𝐼𝑐 + 𝐼𝑑 − 𝐼𝑒 − 𝐼𝑓 = 0

Alternatively, the entering currents to a point are equal to the leaving currents of that point as shown in

Figure 11.

Figure 11 Current flow in a point

𝐼𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 = 𝐼𝑙𝑒𝑎𝑣𝑖𝑛𝑔

𝐼1 + 𝐼𝑑 = 𝐼𝑏 + 𝐼𝑐 + 𝐼𝑒 + 𝐼𝑓

11

Example

Use Kirchhoff’s Current Law (KCL) to find Current Supply, Is in Figure 12.

Figure 12 Current flow in parallel resistive circuit

Solution

Apply Kirchhoff’s Current Law (KCL)

Σ𝐼𝐴𝑡 𝑛𝑜𝑑𝑒 1 = 0

𝐼𝑆 − 𝐼1 − 𝐼2 = 0

𝐼𝑆 − 8 − 4 = 0

𝐼𝑆 = 8 + 4 = 12𝐴

Alternatively,

𝐼𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 = 𝐼𝑙𝑒𝑎𝑣𝑖𝑛𝑔

𝐼𝑆 = 𝐼1 + 𝐼2

𝐼𝑆 = 8 + 4

𝐼𝑆 = 8 + 4 = 12𝐴

12

Practice Problem

1. Determine I3 in Figure 13 using Kirchhoff’s Current Law (KCL).


Solution

Apply KCL Theorem


3 + 4 − 1.5 − 𝐼3 = 0

𝐼3 = 7 − 1.5 = 5.5𝐴

13

2. Find the current I1, I2 and I3 in Figure 14 by using Kirchhoff’s Current Law (KCL).


Solution

To calculate I1, we need to find Rtotal.

1


1

𝑅1+

1

(𝑅2 + 𝑅3)

1


1

2+

1

(1 + 2)=

1

2+

1

3

𝑅𝑡𝑜𝑡𝑎𝑙 = 1.2Ω

Then, we use Ohm’s Law to find I1

𝑉 = 𝐼𝑅

𝐼1 =𝑉𝑆


8

1.2= 6.67𝐴

To find, I2 we use Ohm’s Law

𝐼2 =𝑉𝑆

𝑅1=

8

2= 4𝐴

Apply Kirchhoff’s Current Law (KCL) to find I3


𝐼1 − 𝐼2 − 𝐼3 = 0

6.67 − 4 − 𝐼3 = 0

𝐼3 = 2.67𝐴

14

3. Refer to Figure 15, find each current, I1, I2, I3, I4, I5 and I6 using Kirchhoff’s Current Law (KCL).


Solution

To find I1, we can apply KCL at node a


6 + 2 − 𝐼1 = 0

𝐼1 = 8𝐴

To find I2, we can apply KCL at node b

Σ𝐼𝐴𝑡 𝑛𝑜𝑑𝑒 𝑏 = 0

𝐼1 − 𝐼2 − 1 = 0

8 − 𝐼2 − 1 = 0

𝐼2 = 7𝐴

To find I5, we can apply KCL at node c

Σ𝐼𝐴𝑡 𝑛𝑜𝑑𝑒 𝑐 = 0

1 − 2 + 𝐼5 = 0

𝐼5 = 1𝐴

To find I3, we can apply KCL at node f

Σ𝐼𝐴𝑡 𝑛𝑜𝑑𝑒 𝑓 = 0

3 + 𝐼3 − 15 = 0

15

𝐼3 = 12𝐴

To find I4, we can apply KCL at node e

Σ𝐼𝐴𝑡 𝑛𝑜𝑑𝑒 𝑒 = 0

𝐼2 + 𝐼4 − 𝐼3 = 0

7 + 𝐼4 − 9 = 0

𝐼4 = 2𝐴

To find I5, we can apply KCL at node d

Σ𝐼𝐴𝑡 𝑛𝑜𝑑𝑒 𝑒 = 0

𝐼6 − 𝐼4 − 2 = 0

𝐼6 − 2 − 2 = 0

𝐼6 = 4𝐴

16

Chapter 4: Series Circuit

Express Note

Series circuit is a circuit that provides only ONE path for current flow between two points.

Resistive Circuit

Figure 16 Series resistive circuit

In series resistive circuit:

a) The total resistance in the circuit is EQUAL to the sum of individual resistance in the circuit

𝑅𝑇 = 𝑅1 + 𝑅2 + 𝑅3 + ⋯ + 𝑅𝑁

b) The current flow in each load is SAME and EQUAL to the total current

𝐼𝑇 = 𝐼𝑅1 = 𝐼𝑅2 = 𝐼𝑅3 = 𝐼𝑅𝑁

c) The total voltage drop in each load is EQUAL to the voltage supply

𝑉𝑇 = 𝑉𝑅1 + 𝑉𝑅2 + 𝑉𝑅3 + ⋯ + 𝑉𝑅𝑁

Where;

𝑉𝑅𝑁 = 𝐼𝑅𝑁 × 𝑅𝑁 With;

𝑁 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟

d) The total power is EQUAL to the sum of power in each load

𝑃𝑇 = 𝑃𝑅1 + 𝑃𝑅2 + 𝑃𝑅3 + ⋯ + 𝑃𝑅𝑁

Where

𝑃𝑅𝑁 = 𝐼𝑅𝑁 × 𝑉𝑅𝑁

17

Capacitive Circuit

Figure 17 Series capacitive circuit

In series capacitive circuit:

a) The total capacitance is less than the smallest capacitance in the circuit

𝐶𝑇 =1

1𝐶1

+1𝐶2

+1𝐶3

b) Voltage across each capacitor is inversely proportional to the individual capacitance value

𝑉𝐶𝑁 = (𝐶𝑇

𝐶𝑁) 𝑉𝑇

With;


Inductive Circuit

Figure 18 Series inductive circuit

In series inductive circuit:

a) The total inductance in the circuit is EQUAL to the sum of individual inductance in the circuit

𝐿𝑇 = 𝐿1 + 𝐿2 + 𝐿3 + ⋯ + 𝐿𝑁

With;


18

Example

1. Refer to Figure 19, calculate:

a) The total resistance in the circuit

b) The total current in the circuit

c) The current flows in each resistor

d) The voltage drop in each resistor


Solution:

a) 𝑅𝑇 = 𝑅1 + 𝑅2 + 𝑅3 + ⋯ + 𝑅𝑁

𝑅𝑇 = 100 + 470 + 820

𝑹𝑻 = 𝟏. 𝟑𝟗𝑲Ω

b) 𝐼𝑇 =𝑉𝑇

𝑅𝑇=

10

1.39𝐾= 𝟕. 𝟏𝟗𝒎𝑨

c) 𝐼𝑇 = 𝐼𝑅1 = 𝐼𝑅2 = 𝐼𝑅3 = 𝐼𝑅𝑁

Therefore, 𝑰𝑹𝟏 = 𝑰𝑹𝟐 = 𝑰𝑹𝟑 = 𝟕. 𝟏𝟗𝒎𝑨

d) 𝑉𝑅1 = 𝐼𝑅1 × 𝑅1

𝑉𝑅1 = (7.19𝑚) × 100 = 𝟕𝟏𝟗𝒎𝑽

𝑉𝑅2 = 𝐼𝑅2 × 𝑅2

𝑉𝑅2 = (7.19𝑚) × 470 = 𝟑. 𝟑𝟖𝑽

𝑉𝑅3 = 𝐼𝑅3 × 𝑅3

𝑉𝑅3 = (7.19𝑚) × 820 = 𝟓. 𝟗𝑽

19

2. Refer to Figure 20, calculate the total capacitance in the circuit.


Solution:

𝐶𝑇 =1

1𝐶1

+1𝐶2

+1𝐶3

𝐶𝑇 =1

1100𝜇

+1

220𝜇+

1470𝜇

= 𝟓𝟗. 𝟗𝟖𝝁𝑭

3. Refer to Figure 21 , determine the total inductance in the circuit.


Solution:

𝐿𝑇 = 𝐿1 + 𝐿2 + 𝐿3 + ⋯ + 𝐿𝑁

𝐿𝑇 = 100𝑚 + 220𝑚 + 470𝑚 = 𝟕𝟗𝟎𝒎𝑯

20

Practice Problem







Solution:

a) 𝑅𝑇 = 𝑅1 + 𝑅2

𝑅𝑇 = 2.2𝐾 + 5.6𝐾

𝑹𝑻 = 𝟕. 𝟖𝑲Ω

b) 𝐼𝑇 =𝑉𝑇

𝑅𝑇=

20

7.8𝑘= 𝟐. 𝟓𝟔𝒎𝑨

c) 𝐼𝑇 = 𝐼𝑅1 = 𝐼𝑅2

Therefore, 𝑰𝑹𝟏 = 𝑰𝑹𝟐 = 𝟐. 𝟓𝟔𝒎𝑨

d) 𝑉𝑅1 = 𝐼𝑅1 × 𝑅1

𝑉𝑅1 = (2.56𝑚) × 2.2𝐾 = 𝟓. 𝟔𝟑𝑽

𝑉𝑅2 = 𝐼𝑅2 × 𝑅2

𝑉𝑅2 = (2.56𝑚) × 5.6𝐾 = 𝟏𝟒. 𝟑𝟒𝑽

21



Solution:

𝐶𝑇 =1

1𝐶1

+1𝐶2

+1𝐶3

𝐶𝑇 =1

15𝜇

+1

10𝜇 +1

15𝜇

= 𝟐. 𝟕𝟑𝝁𝑭

3. Refer to Figure 24 , determine the total inductance in the circuit.


Solution:

𝐿𝑇 = 𝐿1 + 𝐿2 + 𝐿3

𝐿𝑇 = 5 + 6 + 3 = 𝟏𝟒𝑯

22

Chapter 5: Parallel Circuit

Express Note

Parallel circuit is a circuit that provides more than ONE path for current flow between two points.

Resistive Circuit

Figure 25 Parallel resistive circuit

In parallel resistive circuit:

a) The total resistance in the circuit is less than the smallest resistance in the circuit

𝑅𝑇 =1

1𝑅1

+1

𝑅2+

1𝑅3

+ ⋯ +1

𝑅𝑁

b) The current flow in each load IS EQUAL to the total current

𝐼𝑇 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 + ⋯ + 𝐼𝑅𝑁

Where;

𝐼𝑅𝑁 =𝑉𝑇

𝑅𝑁

c) The total voltage drop in each load is EQUAL to the voltage supply

𝑉𝑇 = 𝑉𝑅1 = 𝑉𝑅2 = 𝑉𝑅3 = 𝑉𝑅𝑁

Where;

𝑉𝑅𝑁 = 𝐼𝑅𝑁 × 𝑅𝑁

With;


23

Capacitive circuit

Figure 26 Parallel capacitive circuit

The total capacitance in the circuit is EQUAL to the sum of individual capacitance in the circuit

𝐶𝑇 = 𝐶1 + 𝐶2 + 𝐶3 + ⋯ + 𝐶𝑁

Inductive circuit

Figure 27 Parallel inductive circuit

The total inductance in the circuit is less than the smallest inductance in the circuit

𝐿𝑇 =1

1𝐿1

+1𝐿2

+1𝐿3

+ ⋯ +1

𝐿𝑁

24

Example







Solution


𝑅𝑇 =1

1𝑅1

+1

𝑅2+

1𝑅3

𝑅𝑇 =1

130 +

150

+1

100

𝑹𝑻 = 𝟏𝟓. 𝟕𝟗Ω

b) The total current

𝐼𝑇 =𝑉𝑇

𝑅𝑇=

10

15.79= 633.3𝑚𝐴


𝐼𝑅1 =𝑉𝑇

𝑅1=

10

30= 333. 𝑚𝐴

𝐼𝑅2 =𝑉𝑇

𝑅2=

10

50= 200𝑚𝐴

𝐼𝑅3 =𝑉𝑇

𝑅3=

10

100= 100𝑚𝐴

d) The total voltage drop in each load is EQUAL to the voltage supply

𝑉𝑇 = 𝑉𝑅1 = 𝑉𝑅2 = 𝑉𝑅3 = 10𝑉

25



Solution

The total capacitance in the circuit

𝐶𝑇 = 𝐶1 + 𝐶2 + 𝐶3 = 5𝜇 + 10𝜇 + 15𝜇 = 𝟑𝟎𝝁𝑭

3. Refer to Figure 30, calculate the total inductance in the circuit


Solution

The total inductance in the circuit

𝐿𝑇 =1

1𝐿1

+1𝐿2

+1𝐿3

𝐿𝑇 =1

120 +

150

+1

100

𝑳𝑻 = 𝟏𝟐. 𝟓𝑯

26

Practice Problem







Solution

e) The total resistance in the circuit

𝑅𝑇 =1

1𝑅1

+1

𝑅2+

1𝑅3

𝑅𝑇 =1

110𝐾 +

15𝐾

+1

4𝐾

𝑹𝑻 = 𝟏. 𝟖𝟐𝑲Ω

f) The total current

𝐼𝑇 =𝑉𝑇

𝑅𝑇=

25

1.82𝐾= 𝟏𝟑. 𝟕𝟓𝒎𝑨

g) The current flows in each resistor

𝐼𝑅1 =𝑉𝑇

𝑅1=

25

10𝐾= 𝟐. 𝟓𝒎𝑨

𝐼𝑅2 =𝑉𝑇

𝑅2=

25

5𝐾= 𝟓𝒎𝑨

𝐼𝑅3 =𝑉𝑇

𝑅3=

25

4𝐾= 𝟔. 𝟐𝟓𝒎𝑨

h) The total voltage drop in each load is EQUAL to the voltage supply

𝑉𝑇 = 𝑉𝑅1 = 𝑉𝑅2 = 𝑉𝑅3 = 25𝑉

27



Solution

The total capacitance in the circuit

𝐶𝑇 = 𝐶1 + 𝐶2 + 𝐶3 = 10 + 25 + 45 = 𝟖𝟎𝑭

3. Refer to Figure 33, calculate the total inductance in the circuit.


Solution

The total inductance in the circuit

𝐿𝑇 =1

1𝐿1

+1𝐿2

+1𝐿3

𝐿𝑇 =1

15𝑚

+1

8𝑚 +1

10𝑚

𝑳𝑻 = 𝟐. 𝟑𝟓𝒎𝑯

28

Chapter 6: Series Parallel Circuit

Express Note

A combination circuit, combines both series and parallel connections. The series connected have the same current and for parallel connected have the same voltage.

Resistive Circuit

Figure 34 Series parallel resistive circuit

Analysis of the circuit


At junction Z1:

𝑅𝑎 = 𝑅1𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝑅2 = 𝑅1 + 𝑅2

At junction Z2:

𝑅𝑏 = 𝑅3𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝑅4 = 𝑅3 + 𝑅4

Total resistance of the circuit

Ra Rb

29

𝑅𝑇𝑜𝑡𝑎𝑙 = 𝑅𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑤𝑖𝑡ℎ 𝑅𝑏 = 𝑅𝑎//𝑅𝑏 =𝑅𝑎 × 𝑅𝑏

𝑅𝑎 + 𝑅𝑏

Capacitive Circuit

Figure 36 Series parallel capacitive circuit



At junction Z1:

𝐶𝑎 = 𝐶1𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐶2 = 𝐶1// 𝐶2 =𝐶1 × 𝐶2

𝐶1 + 𝐶2

At junction Z2:

𝐶𝑏 = 𝐶3𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐶4 = 𝐶3// 𝐶4 =𝐶3 × 𝐶4

𝐶3 + 𝐶4

Total capacitance of the circuit

𝐶𝑇𝑜𝑡𝑎𝑙 = 𝐶𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑤𝑖𝑡ℎ 𝐶𝑏 = 𝐶𝑎 + 𝐶𝑏

Ca Cb

30

Inductive Circuit

Figure 38 Series parallel inductive circuit



At junction Z1:

𝐿𝑎 = 𝐿1𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐿2 = 𝐿1 + 𝐿2

At junction Z2:

𝐿𝑏 = 𝐿3𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐿4 = 𝐿3 + 𝐿4


𝐿𝑇𝑜𝑡𝑎𝑙 = 𝐿𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑤𝑖𝑡ℎ 𝐿𝑏 = 𝐿𝑎//𝐿𝑏 =𝐿𝑎 × 𝐿𝑏

𝐿𝑎 + 𝐿𝑏

La Lb

31

Example

1. Calculate the total resistance of the circuit in Figure 40.


Solution


At junction Z1:

𝑅𝑎 = 𝑅1𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝑅2 = 𝑅1 + 𝑅2 = 4 + 5 = 9Ω

At junction Z2:

𝑅𝑏 = 𝑅3𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝑅4 = 𝑅3 + 𝑅4 = 10 + 2 = 12Ω



𝑅𝑎 + 𝑅𝑏=

9 × 12

9 + 12= 5.14Ω

Ra Rb

32

2. Calculate the total capacitance of the circuit in Figure 42.


Solution


At junction Z1:


𝐶1 + 𝐶2=

22𝜇 × 30𝜇

22𝜇 + 30𝜇= 12.69𝜇𝐹

At junction Z2:


𝐶3 + 𝐶4=

100𝜇 × 47𝜇

100𝜇 + 47𝜇= 31.97𝜇𝐹


𝐶𝑇𝑜𝑡𝑎𝑙 = 𝐶𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑤𝑖𝑡ℎ 𝐶𝑏 = 𝐶𝑎 + 𝐶𝑏 = 12.69𝜇 + 31.97𝜇 = 44.66𝜇𝐹

Ca Cb

33

3. Calculate the total inductance of the circuit in Figure 44.


Solution


At junction Z1:

𝐿𝑎 = 𝐿1𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐿2 = 𝐿1 + 𝐿2 = 10𝑚𝐻 + 20𝑚𝐻 = 30𝑚𝐻

At junction Z2:

𝐿𝑏 = 𝐿3𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐿4 = 𝐿3 + 𝐿4 = 15𝑚𝐻 + 37𝑚𝐻 = 52𝑚𝐻


𝐿𝑇𝑜𝑡𝑎𝑙 = 𝐿𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑤𝑖𝑡ℎ 𝐿𝑏 = 𝐿𝑎//𝐿𝑏 =𝐿𝑎 × 𝐿𝑏

𝐿𝑎 + 𝐿𝑏=

30𝑚 × 52𝑚𝐻

30𝑚 + 52𝑚𝐻= 19.02𝑚𝐻

La Lb

34

Practice Problem

1. Calculate the total resistance of the circuit in figure 46.


Solution


At junction Ra:

𝑅𝑎 = 𝑅1𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝑅2 = 𝑅1 + 𝑅2 = 4 + 5 = 9Ω

At junction Rb:

𝑅𝑏 = 𝑅3𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝑅4 = 𝑅3 + 𝑅4 + 𝑅5 = 3 + 10 + 2 = 15Ω



𝑅𝑎 + 𝑅𝑏=

9 × 15

9 + 15= 5.625Ω

Ra

Rb

35

2. Calculate the total capacitance of the circuit in Figure 48.


Solution


At junction Z1:


𝐶1 + 𝐶2=

4𝜇 × 3𝜇

4𝜇 + 3𝜇= 1.71𝜇𝐹

At junction Z2:


𝐶3 + 𝐶4=

10𝜇 × 2𝜇

10𝜇 + 2𝜇= 1.67𝜇𝐹

𝐶𝑎 + 𝐶𝑏 = 1.71𝜇𝐹 + 1.67𝜇𝐹 = 3.38𝜇𝐹


𝐶𝑇𝑜𝑡𝑎𝑙 = 𝐶𝑠 𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ (𝐶𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑤𝑖𝑡ℎ𝐶𝑏) = 𝐶𝑠// (𝐶𝑎 + 𝐶𝑏)

=10𝜇 × 3.38𝜇

10𝜇 + 3.38𝜇= 2.53𝜇𝐹

Ca Cb

36

3. Calculate the total inductance of the circuit in Figure 50.


Solution


At junction Z1:

𝐿𝑎 = 𝐿1𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐿2 = 𝐿1 + 𝐿2 = 37𝑚𝐻 + 20𝑚𝐻 = 57𝑚𝐻

At junction Z2:

𝐿𝑏 = 𝐿3𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝐿4 = 𝐿3 + 𝐿4 = 15𝑚𝐻 + 10𝑚𝐻 = 25𝑚𝐻


𝐿𝑇𝑜𝑡𝑎𝑙 = 𝐿𝑠 + (𝐿𝑎//𝐿𝑏) = 𝐿𝑠 +𝐿𝑎 × 𝐿𝑏

𝐿𝑎 + 𝐿𝑏= 20𝑚𝐻 +

57𝑚 × 25𝑚𝐻

57𝑚 + 25𝑚𝐻= 37.38𝑚𝐻

La Lb

37

Solutions of Practice Problem

Chapter 1: Ohm's Law

1. 𝑅1 = 5Ω

2. 𝐼 = 1𝐴

3. 𝑉 = 21𝑉

Chapter 2: Kirchhoff’s Voltage Law (KVL)

1. 𝑉2 = 17𝑉

2. 𝑉3 = 6𝑉

3. 𝑉𝑥 = −3𝑉

Chapter 3: Kirchhoff’s Current Law (KCL)

1. 𝐼3 = 5.5𝐴

2. 𝐼1 = 6.67𝐴, 𝐼2 = 4𝐴 & 𝐼3 = 2.67𝐴

3. 𝐼1 = 8𝐴, 𝐼2 = 7𝐴, 𝐼3 = 12𝐴, 𝐼4 = 2𝐴, 𝐼5 = 1𝐴 & 𝐼6 = 4𝐴

Chapter 4: Series Circuit

1. 𝑎)𝑅𝑇 = 7.8𝑘 Ω, 𝑏)𝐼𝑇 = 2.56𝑚𝐴, 𝑐) 𝐼𝑅1 = 𝐼𝑅2 = 2.56𝑚𝐴 𝑑) VR1 = 5.63V, VR2 = 14.34V

2. CT = 2.73uF

3. LT = 14H

Chapter 5: Parallel Circuit

1. 𝑎)𝑅𝑇 = 1.82𝑘Ω, 𝑏) 𝐼𝑇 = 13.75𝑚𝐴, 𝑐) 𝐼𝑅1 = 2.5𝑚𝐴, 𝐼𝑅2 = 5𝑚𝐴, 𝐼𝑅3 = 6.25𝑚𝐴, 𝑑) 𝑉𝑇 =

𝑉𝑅1 = 𝑉𝑅2 = 𝑉𝑅3 = 25𝑉

2. 𝐶𝑇 = 80𝐹

3. LT = 2.35mH

Chapter 6: Series Parallel Circuit

1. 𝑅𝑇𝑜𝑡𝑎𝑙 = 5.625Ω

2. 𝐶𝑇𝑜𝑡𝑎𝑙 = 2.53𝜇𝐹

3. 𝐿𝑇𝑜𝑡𝑎𝑙 = 37.38𝑚𝐻

38

Bibliography

Alexander, C., & Sadiku, M. (2016). Fundamentals of Electric Circuits (6th ed.). Mc Graw Hill Education.

Allan H Robbins, & Miller, W. C. (2013). Circuit Analysis - Theory and Practice (3rd ed.). Cengage Learning.

Bird, C. M. eld J., Bolton, M. A. L. W., Sueker, A. L. R. S. K., Moura, T. W. M. T. L., & Warne, I. D. W. K. A. B. D. (2008). Electrical Engineering Know It All. Newnes.

Bird, J. (2007). Electrical Circuit Theory and Technology. Routledge. doi:10.4324/9780080549798

Chapman, S. J. (2002). Electric Machinery and Power System Fundamentals. Mc Graw Hill.

Schultz, M. E. (2010). Grob ’ s Basic Electronics (11th ed.). Maney Publishing Education.

basic calculation for dc electrical circuitrepository.psp.edu.my/jspui/bitstream/123456789... ·...

Documents