BASIC CALCULATION FOR DC ELECTRICAL CIRCUIT

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 i

ISBN 978-967-0783-3-4 First Print 207 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, 982- 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 978-967-0783-3-4. Electric circuit analysis. 2. Electronic circuits.. Nor Hanida Ahmad, 98-.. Nurul Hidayah Ahmad Shairazi, 980 -.. Title. 62.392 Review and edited by:- Unit Penyelidikan & Penerbitan Perpustakaan Politeknik Seberang Perai. Publisher:- Perpustakaan, Politeknik Seberang Perai, Jalan Permatang Pauh, 3500 Permatang Pauh, Pulau Pinang. Tel: 04-5383322 Fax: 04-5389266 Email: webmaster@psp.edu.my. Graphic Designer:- Faizal Bin Mohamad Twon Tawi ii

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. iii

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. iv

Contents Acknowledgments... iii Preface... iv Introduction... 2 Chapter : 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)... 0 Express Note... 0 Example... Practice Problem... 2 Chapter 4: Series Circuit... 6 Express Note... 6 Example... 8 Practice Problem... 20 Chapter 5: Parallel Circuit... 22 Express Note... 22 Example... 24 Practice Problem... 26 Chapter 6: Series Parallel Circuit... 28 Express Note... 28 Example... 3 Practice Problem... 34 s of Practice Problem... 37 Bibliography... 38

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:. 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. 2

Chapter : Ohm's Law Express Note The voltage across a resistor is directly proportional to the current flowing through the resistor as shown in Figure. The mathematical form of Ohm's Law is Figure Current flow in a resistor V = IR Example. An electric iron draws 3A at 240V. Find its resistance. 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. V = IR V R = I 6 60Ω = I 0.A = I 3

Practice Problem. Find the value of resistor in Figure 2. Figure 2 Resistive circuit V = IR R = V I = 40 8 = 5Ω 2. In the circuit shown in Figure 3, calculate the current I. Figure 3 Resistive circuit I = V R total = V = IR total 40 R + R 2 + R 3 = 40 20 + 0 + 0 = A 4

3. In the circuit shown in Figure 4, calculate the voltage across R. Find I first; Figure 4 Resistive circuit Then, calculate I = V R total = V = IR total 40 R + R 2 + R 3 = V = IR =.05 20 = 2V 40 20 + 0 + 8 =.05A 5

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). Figure 5 Resistive circuit ΣV closed loop = 0 E V V 2 = 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. E emf = ΣV drop E = V + V 2 6

Example Based on Figure 6, determine the voltage drop, V 3 at resistance, R 3 by using Kirchhoff s Voltage Law (KVL). Figure 6 Resistive circuit Apply Kirchhoff s Voltage Law (KVL) Theorem ΣV closed loop = 0 0 + 20 + V 3 40 = 0 V 3 = 40 30 = 0V Alternatively, E emf = ΣV drop 40 = 20 + 0 + V 3 V 3 = 40 30 = 0V 7

Practice Problem. Find V 2 in the circuit of Figure 7. Figure 7 Resistive circuit Apply KVL Theorem ΣV closed loop = 0 8 + V 2 + 20 45 = 0 V 2 = 45 28 = 7V 2. Find V 3 in the circuit of Figure 8. Figure 8 Resistive circuit Apply KVL Theorem ΣV closed loop = 0 8 4 8 V 3 = 0, V 3 = 8 2 = 6V 8

4. Find the value of voltage drop V x for a circuit in Figure 9 using KVL Theorem. Figure 9 Resistive circuit Apply KVL Theorem ΣV closed loop = 0 5 + V X + 8 20 = 0 V X = 20 8 5 = 3V 9

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 0. Figure 0 Current flow in a point ΣI At node a = 0 I a I b I c + I d I e I f = 0 Alternatively, the entering currents to a point are equal to the leaving currents of that point as shown in Figure. Figure Current flow in a point I entering = I leaving I + I d = I b + I c + I e + I f 0

Example Use Kirchhoff s Current Law (KCL) to find Current Supply, Is in Figure 2. Figure 2 Current flow in parallel resistive circuit Apply Kirchhoff s Current Law (KCL) ΣI At node = 0 I S I I 2 = 0 I S 8 4 = 0 I S = 8 + 4 = 2A Alternatively, I entering = I leaving I S = I + I 2 I S = 8 + 4 I S = 8 + 4 = 2A

Practice Problem. Determine I 3 in Figure 3 using Kirchhoff s Current Law (KCL). Apply KCL Theorem Figure 3 Resistive circuit ΣI At node a = 0 3 + 4.5 I 3 = 0 I 3 = 7.5 = 5.5A 2

2. Find the current I, I 2 and I 3 in Figure 4 by using Kirchhoff s Current Law (KCL). To calculate I, we need to find R total. Figure 4 Resistive circuit = + R total R (R 2 + R 3 ) = R total 2 + ( + 2) = 2 + 3 R total =.2Ω Then, we use Ohm s Law to find I To find, I 2 we use Ohm s Law I = V = IR V S = 8 R total.2 = 6.67A I 2 = V S R = 8 2 = 4A Apply Kirchhoff s Current Law (KCL) to find I 3 ΣI At node a = 0 I I 2 I 3 = 0 6.67 4 I 3 = 0 I 3 = 2.67A 3

3. Refer to Figure 5, find each current, I, I 2, I 3, I 4, I 5 and I 6 using Kirchhoff s Current Law (KCL). Figure 5 Resistive circuit To find I, we can apply KCL at node a ΣI At node a = 0 6 + 2 I = 0 I = 8A To find I 2, we can apply KCL at node b ΣI At node b = 0 I I 2 = 0 8 I 2 = 0 I 2 = 7A To find I 5, we can apply KCL at node c ΣI At node c = 0 2 + I 5 = 0 I 5 = A To find I 3, we can apply KCL at node f ΣI At node f = 0 3 + I 3 5 = 0 4

I 3 = 2A To find I 4, we can apply KCL at node e ΣI At node e = 0 I 2 + I 4 I 3 = 0 7 + I 4 9 = 0 I 4 = 2A To find I5, we can apply KCL at node d ΣI At node e = 0 I 6 I 4 2 = 0 I 6 2 2 = 0 I 6 = 4A 5

Chapter 4: Series Circuit Express Note Series circuit is a circuit that provides only ONE path for current flow between two points. Resistive Circuit In series resistive circuit: Figure 6 Series resistive circuit a) The total resistance in the circuit is EQUAL to the sum of individual resistance in the circuit R T = R + R 2 + R 3 + + R N b) The current flow in each load is SAME and EQUAL to the total current I T = I R = I R2 = I R3 = I RN c) The total voltage drop in each load is EQUAL to the voltage supply V T = V R + V R2 + V R3 + + V RN Where; With; V RN = I RN R N N = number of resistor d) The total power is EQUAL to the sum of power in each load P T = P R + P R2 + P R3 + + P RN Where P RN = I RN V RN 6

Capacitive Circuit Figure 7 Series capacitive circuit In series capacitive circuit: a) The total capacitance is less than the smallest capacitance in the circuit C T = + + C C 2 C 3 b) Voltage across each capacitor is inversely proportional to the individual capacitance value With; Inductive Circuit V CN = ( C T C N ) V T N = number of resistor In series inductive circuit: Figure 8 Series inductive circuit a) The total inductance in the circuit is EQUAL to the sum of individual inductance in the circuit L T = L + L 2 + L 3 + + L N With; N = number of resistor 7

Example. Refer to Figure 9, 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 Figure 9 Series resistive circuit : a) R T = R + R 2 + R 3 + + R N R T = 00 + 470 + 820 R T =. 39KΩ b) I T = V T = 0 = 7. 9mA R T.39K c) I T = I R = I R2 = I R3 = I RN Therefore, I R = I R2 = I R3 = 7. 9mA d) V R = I R R V R = (7.9m) 00 = 79mV V R2 = I R2 R 2 V R2 = (7.9m) 470 = 3. 38V V R3 = I R3 R 3 V R3 = (7.9m) 820 = 5. 9V 8

2. Refer to Figure 20, calculate the total capacitance in the circuit. Figure 20 Series capacitive circuit : C T = + + C C 2 C 3 C T = 00μ + 220μ + = 59. 98μF 470μ 3. Refer to Figure 2, determine the total inductance in the circuit. : Figure 2 Series inductive circuit L T = L + L 2 + L 3 + + L N L T = 00m + 220m + 470m = 790mH 9

Practice Problem. Refer to Figure 22, 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 : Figure 22 Series resistive circuit a) R T = R + R 2 R T = 2.2K + 5.6K R T = 7. 8KΩ b) I T = V T = 20 = 2. 56mA R T 7.8k c) I T = I R = I R2 Therefore, I R = I R2 = 2. 56mA d) V R = I R R V R = (2.56m) 2.2K = 5. 63V V R2 = I R2 R 2 V R2 = (2.56m) 5.6K = 4. 34V 20

2. Refer to Figure 23, calculate the total capacitance in the circuit. Figure 23 Series capacitive circuit : C T = C + C + 2 C 3 C T = 5μ + 0μ + = 2. 73μF 5μ 3. Refer to Figure 24, determine the total inductance in the circuit. Figure 24 Series inductive circuit : L T = L + L 2 + L 3 L T = 5 + 6 + 3 = 4H 2

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 In parallel resistive circuit: Figure 25 Parallel resistive circuit a) The total resistance in the circuit is less than the smallest resistance in the circuit R T = R + R + 2 R + + 3 R N b) The current flow in each load IS EQUAL to the total current I T = I R + I R2 + I R3 + + I RN Where; I RN = V T R N c) The total voltage drop in each load is EQUAL to the voltage supply V T = V R = V R2 = V R3 = V RN Where; With; V RN = I RN R N N = number of resistor 22

Capacitive circuit Figure 26 Parallel capacitive circuit The total capacitance in the circuit is EQUAL to the sum of individual capacitance in the circuit Inductive circuit C T = C + C 2 + C 3 + + C N Figure 27 Parallel inductive circuit The total inductance in the circuit is less than the smallest inductance in the circuit L T = L + L + 2 L + + 3 L N 23

Example. Refer to Figure 28, 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 Figure 28 Parallel resistive circuit a) The total resistance in the circuit R T = R + R + 2 R 3 R T = 30 + 50 + 00 R T = 5. 79Ω b) The total current I T = V T R T = 0 5.79 = 633.3mA c) The current flows in each resistor I R = V T = 0 = 333. ma R 30 I R2 = V T R 2 = 0 50 = 200mA I R3 = V T R 3 = 0 00 = 00mA d) The total voltage drop in each load is EQUAL to the voltage supply V T = V R = V R2 = V R3 = 0V 24

2. Refer to Figure 29, calculate the total capacitance in the circuit. Figure 29 Parallel capacitive circuit The total capacitance in the circuit C T = C + C 2 + C 3 = 5μ + 0μ + 5μ = 30μF 3. Refer to Figure 30, calculate the total inductance in the circuit Figure 30 Parallel inductive circuit The total inductance in the circuit L T = + + L L 2 L 3 L T = 20 + 50 + 00 L T = 2. 5H 25

Practice Problem. Refer to Figure 3, 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 Figure 3 Parallel resistive circuit e) The total resistance in the circuit R T = R + R + 2 R 3 R T = 0K + 5K + 4K R T =. 82KΩ f) The total current I T = V T = 25 = 3. 75mA R T.82K g) The current flows in each resistor I R = V T = 25 = 2. 5mA R 0K I R2 = V T R 2 = 25 5K = 5mA I R3 = V T = 25 = 6. 25mA R 3 4K h) The total voltage drop in each load is EQUAL to the voltage supply V T = V R = V R2 = V R3 = 25V 26

2. Refer to Figure 32, calculate the total capacitance in the circuit. Figure 32 Parallel capacitive circuit The total capacitance in the circuit C T = C + C 2 + C 3 = 0 + 25 + 45 = 80F 3. Refer to Figure 33, calculate the total inductance in the circuit. Figure 33 Parallel inductive circuit The total inductance in the circuit L T = L + L + 2 L 3 L T = 5m + 8m + 0m L T = 2. 35mH 27

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 Analysis of the circuit Figure 34 Series parallel resistive circuit Ra Rb Figure 35 Series parallel resistive circuit At junction Z : R a = R series with R 2 = R + R 2 At junction Z 2: R b = R 3 series with R 4 = R 3 + R 4 Total resistance of the circuit 28

R Total = R a parallel with R b = R a //R b = R a R b R a + R b Capacitive Circuit Figure 36 Series parallel capacitive circuit Analysis of the circuit Ca Cb Figure 37 Series parallel capacitive circuit At junction Z : C a = C series with C 2 = C // C 2 = C C 2 C + C 2 At junction Z 2: Total capacitance of the circuit C b = C 3 series with C 4 = C 3 // C 4 = C 3 C 4 C 3 + C 4 C Total = C a parallel with C b = C a + C b 29

Inductive Circuit Figure 38 Series parallel inductive circuit Analysis of the circuit La Lb Figure 39 Series parallel inductive circuit At junction Z : L a = L series with L 2 = L + L 2 At junction Z 2: L b = L 3 series with L 4 = L 3 + L 4 Total resistance of the circuit L Total = L a parallel with L b = L a //L b = L a L b L a + L b 30

Example. Calculate the total resistance of the circuit in Figure 40. Figure 40 Series parallel resistive circuit Ra Rb Figure 4 Series parallel resistive circuit At junction Z : R a = R series with R 2 = R + R 2 = 4 + 5 = 9Ω At junction Z 2: Total resistance of the circuit R b = R 3 series with R 4 = R 3 + R 4 = 0 + 2 = 2Ω R Total = R a parallel with R b = R a //R b = R a R b R a + R b = 9 2 9 + 2 = 5.4Ω 3

2. Calculate the total capacitance of the circuit in Figure 42. Figure 42 Series parallel capacitive circuit Ca Cb Figure 43 Series parallel capacitive circuit At junction Z : C a = C series with C 2 = C // C 2 = C C 2 22μ 30μ = C + C 2 22μ + 30μ = 2.69μF At junction Z 2: Total capacitance of the circuit C b = C 3 series with C 4 = C 3 // C 4 = C 3 C 4 00μ 47μ = C 3 + C 4 00μ + 47μ = 3.97μF C Total = C a parallel with C b = C a + C b = 2.69μ + 3.97μ = 44.66μF 32

3. Calculate the total inductance of the circuit in Figure 44. Figure 44 Series parallel inductive circuit La Lb Figure 45 Series parallel inductive circuit At junction Z : L a = L series with L 2 = L + L 2 = 0mH + 20mH = 30mH At junction Z 2: Total resistance of the circuit L b = L 3 series with L 4 = L 3 + L 4 = 5mH + 37mH = 52mH L Total = L a parallel with L b = L a //L b = L a L b 30m 52mH = L a + L b 30m + 52mH = 9.02mH 33

Practice Problem. Calculate the total resistance of the circuit in figure 46. Figure 46 Series parallel resistive circuit Ra Rb Figure 47 Series parallel resistive circuit At junction R a: R a = R series with R 2 = R + R 2 = 4 + 5 = 9Ω At junction R b: R b = R 3 series with R 4 = R 3 + R 4 + R 5 = 3 + 0 + 2 = 5Ω Total resistance of the circuit R Total = R a parallel with R b = R a //R b = R a R b R a + R b = 9 5 9 + 5 = 5.625Ω 34

2. Calculate the total capacitance of the circuit in Figure 48. Figure 48 Series parallel capacitive circuit Ca Cb Figure 49 Series parallel capacitive circuit At junction Z : C a = C series with C 2 = C // C 2 = C C 2 4μ 3μ = C + C 2 4μ + 3μ =.7μF At junction Z 2: C b = C 3 series with C 4 = C 3 // C 4 = C 3 C 4 0μ 2μ = C 3 + C 4 0μ + 2μ =.67μF Total capacitance of the circuit C a + C b =.7μF +.67μF = 3.38μF C Total = C s series with (C a parallel withc b ) = C s // (C a + C b ) = 0μ 3.38μ 0μ + 3.38μ = 2.53μF 35

3. Calculate the total inductance of the circuit in Figure 50. Figure 50 Series parallel inductive circuit La Lb Figure 5 Series parallel inductive circuit At junction Z : L a = L series with L 2 = L + L 2 = 37mH + 20mH = 57mH At junction Z 2: L b = L 3 series with L 4 = L 3 + L 4 = 5mH + 0mH = 25mH Total resistance of the circuit L Total = L s + (L a //L b ) = L s + L a L b 57m 25mH = 20mH + L a + L b 57m + 25mH = 37.38mH 36

s of Practice Problem Chapter : Ohm's Law. R = 5Ω 2. I = A 3. V = 2V Chapter 2: Kirchhoff s Voltage Law (KVL). V 2 = 7V 2. V 3 = 6V 3. V x = 3V Chapter 3: Kirchhoff s Current Law (KCL). I 3 = 5.5A 2. I = 6.67A, I 2 = 4A & I 3 = 2.67A 3. I = 8A, I 2 = 7A, I 3 = 2A, I 4 = 2A, I 5 = A & I 6 = 4A Chapter 4: Series Circuit. a)r T = 7.8k Ω, b)i T = 2.56mA, c) I R = I R2 = 2.56mA d) V R = 5.63V, V R2 = 4.34V 2. C T = 2.73uF 3. L T = 4H Chapter 5: Parallel Circuit. a)r T =.82kΩ, b) I T = 3.75mA, c) I R = 2.5mA, I R2 = 5mA, I R3 = 6.25mA, d) V T = V R = V R2 = V R3 = 25V 2. C T = 80F 3. L T = 2.35mH Chapter 6: Series Parallel Circuit. R Total = 5.625Ω 2. C Total = 2.53μF 3. L Total = 37.38mH 37

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