Applied Electricity (EE 151) Slides PDF

APPLIED ELECTRICITY (EE 151) Taught by: Prof. Emmanuel Asuming Frimpong Address: Department of Electrical Engineering KNUST, Kumasi, Ghana E-mail: [email protected] Phone: 0501349207/0246665284 Office: Room 318...

APPLIED ELECTRICITY (EE 151) Taught by: Prof. Emmanuel Asuming Frimpong Address: Department of Electrical Engineering KNUST, Kumasi, Ghana E-mail: [email protected] Phone: 0501349207/0246665284 Office: Room 318 Semester 1, 2022/2023 1  TEACHING ASSISTANT Sharon Amoah Mensah (NSP) 0247994889 ANDOH BUABENG COLLINS (ABC) 2  AIM To equip students with the tools required to analyze electric and magnetic circuits 3  METHODOLOGY ❖Classroom lectures 4  MATERIAL REQUIRED ❖ PowerPoint presentation ❖ Recommended books for further reading ✓ E. Hughes, Electrical and Electronic Technology ✓ P. Y. Okyere and E. A. Frimpong, Fundamentals of electric and magnetic circuits. ✓ S. N. Singh, Basic Electrical Engineering 5  COURSE OUTLINE Unit 1: Circuits and Network Theorems – Kirchhoff's laws, Thevenin’s Theorem, Norton's Theorem, Superposition Theorem, Reciprocity Theorem and Delta- Star Transformation. Unit 2: Alternating current circuits – Determination of Average and RMS values, Harmonics, Phasors, impedance, current and power in ac circuits 6  COURSE OUTLINE Unit 3: Three-phase circuits – Connection of three- phase windings, three phase loads, power in three-phase circuits, solving three-phase circuit problems Unit 4: Magnetic circuits – Components and terminologies, solving magnetic circuit problems 7  LESSON PLAN Week Date Topic 1 Chapter 1 – Course overview and DC circuit fundamentals 2 Chapter 1 – Course overview and DC circuit fundamentals 3 Chapter 1 – DC circuit fundamentals(cont.) and Kirchhoff’s laws 4 Chapter 1 – Thevenin’s theorem and Norton’s theorem 5 Chapter 1 – Superposition theorem, and Reciprocity theorem Chapter 2 – Average value 8  LESSON PLAN Week Date Topic 6 Chapter 2 – RMS Harmonics, Phasors, and Impedance 7 Chapter 2 – Power in AC circuits and Application of complex numbers to AC circuits 8 Three-phase circuits 9 Three-phase circuits/Mag. cct 10 Magnetic circuits 11 Mid-semester examination 12 Mid-semester break 9  LESSON PLAN Week Date Topic 13 Revision 14 Revision 15 End-of-semester examination ANDOH BUABENG COLLINS (ABC 10  CLASSROOM NORMS 1. No phone usage 2. No eating 3. No noise-making 4. No lateness 5. No intimidation 6. No sleeping ❖Students who fail to abide by these norms will be asked to leave the classroom. 11  ASSESSMENT 1. Quizzes 30% 2. Mid-semester examination 3. End of semester examination 70% 12  EXTRAS ❖Daily admonishment and encouragement ❖Radom call-ups to ask and answer questions 13 Questions or Comments 14  THE WORD Ecclesiastes 9:10 Whatsoever thy hand finds to do, do it with thy might; for there is no work, nor device, nor knowledge, nor wisdom, in the grave, where thou goes.  UNIT 1: CIRCUIT AND NETWORK THEOREMS ❖ Definitionof a circuit An interconnection of elements forming a closed path along which current can flow. R + V(t) L - C ❖Elements of an electric circuit Active elements: Energy producing elements eg. Batteries, Generators, Solar cells Passive elements: Energy using elements eg. Resistors, inductors, capacitors 16  CIRCUIT TERMINOLOGIES ❖ Node (Junction) –A point where currents split or come together [ points c, d, e and f] ❖Path – Any connection where current flows [eg bc, be, fa] b c e g a d f h 17  CIRCUIT TERMINOLOGIES ❖Branch – A connection (path) between two nodes [eg. cd, cbad, df] ❖Loop/Mesh – a closed path of a circuit [ eg. cghdc] b c e g a d f h 18  CIRCUIT TERMINOLOGIES ❖ Short-circuit– A branch of theoretically zero resistance. It diverts to itself all currents that would have flown in adjacent branches (branches hooked to the same node) except branches with sources. a b R1 c R3 d R1 a R2 R4 V R2 V R3 h g e f b 19  CIRCUIT TERMINOLOGIES ❖ Short-circuit cont. Self assessment Which of the resistors in the circuit below have been short- circuited? R 1 ANS: R2 and R3 V R2 R3 20  CIRCUIT TERMINOLOGIES ❖ Open circuit – A branch of theoretically infinite resistance. It prevents current from flowing in its branch. R2 R3 R5 V R1 R4 21 RESISTORS IN PARALLEL 22 RESISTORS IN SERIES 23  RESISTORS IN SERIES Resistors are in series when the same current flows through them. There is NO JUNCTION between them. R1 R2 R1 R2 V V R3 Fig. 1 Fig. 2 In Fig. 1 : R1 and R2 are in series In Fig. 2: None of the resistors are in series 24  RESISTORS IN SERIES ❖Self assessment 1 Which of the following resistors are in series? R2 R5 R6 R3 R1 R4 ANS: R1&R2, R3 & R4 and R5 & R6 25  RESISTORS IN SERIES ❖ Total (effective) resistance of series resistors The total resistance RT for resistors R1, R2, R3, ….., RN which are in series if given by: RT = R1 + R2 +... + RN 26  THE WORD Matthew 6:33 But seek ye first the kingdom of God, and his righteousness; and all these things shall be added unto you.  RESISTORS IN PARALLEL Resistors are said to be in parallel when the voltage across them is the same. R1//R2 R1//R2 R1//R2 R1//R3 R3//R2 28  RESISTORS IN PARALLEL Colloquially, TWO resistors are in parallel if it is possible to traverse them without passing through another element. R1 V R2 29  RESISTORS IN PARALLEL Self assessment Which of the resistors in the circuit below are in parallel? 1.6Ω 3Ω 6Ω 4Ω ANS: 4 // 6 30  RESISTORS IN PARALLEL Total resistance When resistors R1 and R2 are in parallel, the total resistance RT is given by: 1 1 1 R1 R2 = +  RT = RT R1 R2 R1 + R2 31  EFFECTIVE RESISTANCE OF A CIRCUIT Effective circuit resistance is found by identifying and putting together series and or parallel resistors Eg 1. Find the total resistance of the circuit below. 23 RT = (2 // 3) + 1 = 11 ❖Solution +1 =  2+3 5 32  EFFECTIVE RESISTANCE OF A CIRCUIT Eg. 2. Find the total resistance of the circuit below. ❖Solution 2 2 RT = (2 // 2) + 1 = + 1 = 2 2+2 33  EFFECTIVE RESISTANCE OF A CIRCUIT Self Assessment 1 Find the total resistance of the circuit below. 2Ω 3Ω 6Ω 4Ω V Answer  4 6  22 66 RT = (4 // 6) + 2// 3 =  + 2 // 3 = // 3 =  4 + 6  34 5 37  EFFECTIVE RESISTANCE OF A CIRCUIT Self Assessment 2 Find the total resistance of the circuit below. Answer 1  RT = (1 // 1) + 1 // 1 // 1 =  + 1 // 1 // 1 = // =  3 1 3 2  2 2 8 35  CURRENT DIVISION RULE The current division rule is applied to share current between parallel branches. Consider the circuits below I I1 I2 V R1 R2 R1 R2 I V R1 + R2 R1 R2 1 R2 I1 = = =I  =I R1 R1 R1 + R2 R1 R1 + R2 36  CURRENT DIVISION RULE Similarly, I I1 I2 V R1 R2 R1 I2 = I R1 + R2 37  CURRENT DIVISION RULE Comparing currents I I1 I2 V R1 R2 R2 R1 I1 = I I2 = I R1 + R2 R1 + R2 38  CURRENT DIVISION RULE Consider the figure below, I I1 I2 V R1 R2 R2 R1 I1 = I I2 = − I R1 + R2 R1 + R2 39  CURRENT DIVISION RULE Example 1 Find the values of I1 and I2 in the circuit below. 10A I1 I2 V 3Ω 2Ω Solution R2 2 I1 = I=  10 = 4 A R1 + R2 2+3 R1 3 I2 = − I =−  10 = −6 A R1 + R2 2+3 40  CURRENT DIVISION RULE Example 1 Find the values of I1 and I2 in the circuit below. 10A I1 I2 V 3Ω 2Ω Solution R2 2 I1 = I =  10 = 4 A R1 + R2 2+3 R1 3 I2 = − I =−  10 = −6 A R1 + R2 2+3 41  CURRENT DIVISION RULE Example 2 Find the value of I1 in the circuit below. 10A I1 V 3Ω 2Ω 2Ω Solution 10A I1 1 I1 =  10 = 2.5 A V 3Ω 1Ω 1+ 3 42  VOLTAGE DROP ❖Any time voltage drives current through a resistor, some of the voltage drops across the resistor. ❖The magnitude of the drop is the product of the resistance and current I V1 V2 R1 R2 V V2 = V − V1 43  VOLTAGE DROP Example Find the values of I and R in the circuit below. I 4V R 3 10V Solution Voltage across 3Ω resistor = 10 – 4 = 6V Current in 3Ω resistor = I = 6/3 = 2A Resistance R = 4V/I = 4/2 = 2Ω 44  REVISION EXERCISE Find the value of I in the circuit below. I 3 3.5 4 1 2 5 7 25V Solution I I 3 6  4 1 2 1 4 5 25V 25V 45  REVISION EXERCISE 12  13 1 25V 25 V 25 RT =  IT = = = 13 A 13 RT 25 13 46  REVISION EXERCISE I 6  4 1 5 25V 6 I =− 5  13 = −3 A 6 +4 5 47  THE WORD TEACH US WISDOM Psalm 90: 12 So teach us to number our days, That we may gain a heart of wisdom.  KIRCHHOFF'S CURRENT LAW(KCL) ❖ The Law The sum of currents entering a node equals the sum of currents leaving the node. i5 i6 i1 i4 i3 i2 Sum of currents entering i1 + i3 + i5 Sum of currents Leaving i2 + i4 + i6 Applying KCL 49  KIRCHHOFF'S CURRENT LAW(KCL) ❖ Example Find the value of i in the figure below. i 2 3 4 Solution 7 i+2+3= 4+7 i + 5 = 11 i =6 50  KIRCHHOFF'S CURRENT LAW(KCL) ❖ Self assessment Find the value of i in the figure below. i 2 3 4 7 ANS i=2 51  KIRCHHOFF'S VOLTAGE LAW(KVL) ❖ The law The algebraic sum of the voltages in a loop (closed path) equals zero. Alternatively, in a loop, the algebraic sum of voltage sources equals the algebraic sum of voltage drops. R1 R2 R4 a I1 b I2 c I4 d I3 I5 R3 R5 V1 V2 h g f e Loop abgha V1=I1R1+I3R3 Loop adeha V1-V2=I1R1+I2R2-I4R4 52  KIRCHHOFF'S VOLTAGE LAW(KVL) R1 R2 R4 a I1 b I2 c I4 d I3 I5 R3 R5 V1 V2 h g f e Loop cbgfc 0 = -I2R2 + I3R3 +I5R5 Loop acfha V1=I1R1 + I2R2 – I5R5 53  KIRCHHOFF'S VOLTAGE LAW(KVL) Example 1 Find the current in all parts of the circuit below. 2Ω 4Ω b c e I1 I2 32V 8Ω 20V I3 a d f ❖Applying KVL to loop bcdab 32 – 2I1 – 8I3 = 0 32 = 2I1 + 8I3 (1) ❖Applying KVL to loop ecdfe 20 – 4I2 – 8I3 = 0 20 = 4I2 + 8I3 (2) 54  KIRCHHOFF'S VOLTAGE LAW(KVL) 2Ω 4Ω b c e I1 I2 32V 8Ω 20V I3 a d f ❖Applying KCL to node c: I 3 = I1 + I 2 (3) Solving the equations simultaneously yields I1 = 4A, I2 = -1A and I3 = 3A 55  KIRCHHOFF'S VOLTAGE LAW(KVL) Example 2 Find the currents in all parts of the circuit below. 2Ω c 5Ω e 2Ω g b I1 I1-I2 I2 I2-I3 I3 7V 3Ω 2Ω 10 V I2 a d f h Solution 2Ω Apply KVL to loop cefdc 5I2 + 2 ( I2 – I3 ) + 2I2 - 3 ( I1 – I2 ) = 0 56  KIRCHHOFF'S VOLTAGE LAW(KVL) 0 = -3I1 + 12I2 - 2I3 (1) 2Ω c 5Ω e 2Ω g b I1 I1-I2 I2 I2-I3 I3 7V 3Ω 2Ω 10 V I2 a d f h 2Ω Apply KVL to loop abcda: 7 = 2I1 + 3(I1 – I2) 7 = 5I1 – 3I2 (2) Apply KVL to loop ghfeg: 10 = -2 ( I2 – I3 ) + 2I3 10 = -2I2 + 4I3 (3) 57  KIRCHHOFF'S VOLTAGE LAW(KVL) Solving the three equations: 0 = -3I1 + 12I2 - 2I3 (1) 7 = 5I1 – 3I2 (2) 10 = -2I2 + 4I3 (3) Simultaneously, I1 = 2.0A, I2 = 1.0A and I3 = 3.0A 58  Thevenin’s Theorem Theorem: Any linear circuit connected between two terminal can be replaced by a Thevenin’s voltage(VTH) in series with a Thevenin's resistance (RTH). VTH is the open-circuit voltage across the two terminals RTH is the resistance seen from the two terminals when all sources have been deactivated 12 Ω RTH A B A 6Ω VTH 42 V 35 V B 59  Thevenin’s Theorem To find the current through a resistor in a circuit, the following steps are taken: 1. Remove the resistor from the circuit and mark the two terminals. 12 Ω 3 A B 42 V 35 V 6Ω 60  Thevenin’s Theorem 2. Find the open-circuit voltage (VTH) across the two terminals by applying KVL. Treat VTH as a source. 12 Ω A B VTH 42 V 35 V 6Ω 61  Thevenin’s Theorem 3. Recall the circuit created in step 1 and deactivate all sources. Short-circuit voltage sources and Open-circuit current sources. 12 Ω A B 42 V 35 V 6Ω 62  THEVENIN’S THEOREM 4. Find the total resistance of the circuit resulting from step 3 as seen from the two terminals 12 Ω A B 6Ω 63  THEVENIN’S THEOREM 5. Reproduce the Thevenin’s equivalent circuit and connect the resistor whose current is to be found. RTH A VTH 3 B 64  THEVENIN’S THEOREM 6. Calculate the current in the circuit in step 5. This is the current being sought. I RTH A VTH VTH 3 i= RTH + R B 65  THEVENIN’S THEOREM Example 1 Using Thevenin’s theorem, determine the current in the 3-Ω resistor of the circuit below. 12 Ω 3Ω 42 V 35 V 6Ω Solution 12 Ω A B Steps 1 & 2 VTH 6Ω 42 V 35 V I 66  THEVENIN’S THEOREM 12 Ω A B a b c VTH 6Ω 42 V 35 V I f e d Applying KVL to loop dcbed: 35 + VTH = 6I (1) Applying KVL to loop fabef: 42 = (12 + 6) I 7 I= A 67 3  THEVENIN’S THEOREM 7 Substituting for I in equation 1: 35 + VTH = 6( ) 3 Steps 3 & 4 VTH = −21V 12 Ω RTH A B 12  6 RTH = 12 // 6 = = 4 6Ω 12 + 6 68  THEVENIN’S THEOREM Steps 5 & 6 RTH = 4Ω A I3 VTH= -21V 3Ω B VTH 21 I3 = =− = −3 A RTH + 3 4+3 69  THEVENIN’S THEOREM Example 2 Find the current in the 10-Ω resistor of the circuit below using Thevenin’s theorem. 5Ω 10 Ω 12 Ω 4V 15 Ω 8Ω 6V Solution 5Ω VTH 12 Ω Steps 1 & 2 A B 4V I1 15 Ω I2 8Ω 6V 70  THEVENIN’S THEOREM 5Ω VTH a b c 12 Ω d A B 4V I1 15 Ω I2 8Ω 6V h g f e Applying KVL to loop cbgfc: VTH = 15I1 - 8I2 (1) 1 Applying KVL to loop abgha: 4 = (5+15)I1 I1 = A 5 Applying KVL to loop dcfed: 6 = (12 + 8) I 2 3 I2 = A 71 10  THEVENIN’S THEOREM Substituting for I1 and I2 in equation 1 yields:  1  3  3 VTH = 15  − 8  = V  5   10  5 5Ω RTH 12 Ω Steps 3 & 4 A B 15 Ω 8Ω 171 RTH = (5 // 15) + (12 // 8) =  20 72  THEVENIN’S THEOREM Steps 5 & 6 RTH A I VTH 10 Ω B VTH 3 3  20 I = = = = 0.032 A RTH + 10  171  5  371 5 + 10   20  73  NORTON’S Theorem Theorem: Any linear circuit connected between two terminals can be replaced by a Norton’s current(IN) in parallel with a Norton's resistance (RN). IN is the short-circuit current between the two terminals RN is the resistance seen from the two terminals when all sources have been deactivated (RN = RTH) 12 Ω A B A 6Ω IN RN 42 V 35 V B 74  NORTON’S Theorem To find the current through a resistor in a circuit, the following steps are taken: 1. Remove the resistor from the circuit and mark the two terminals. 12 Ω 3 A B 42 V 35 V 6Ω 75  NORTON’S Theorem 2. Find the short-circuit current (IN) through the two terminals by applying KVL. 12 Ω IN A B 42 V 35 V 6Ω 76  NORTON’S Theorem 3. Recall the circuit created before step 2 and deactivate all sources. Short-circuit voltage sources and Open-circuit current sources. 12 Ω A B 42 V 35 V 6Ω 77  NORTON’S THEOREM 4. Find the total resistance of the circuit resulting from step 3 as seen from the two terminals 12 Ω A B 6Ω 78  NORTON’S THEOREM 5. Reproduce the Norton’s equivalent circuit and connect the resistor whose current is to be found. A IN RN 3 B 79  NORTON’S THEOREM 6. Calculate the current in the circuit in step 5. This is the current being sought for. I A IN RN 3 RN i=  IN RN + 3 B 80  NORTON’S THEOREM Example 1 Using Norton’s theorem, determine the current in the 3-Ω resistor of the circuit below. 12 Ω 3Ω 42 V 35 V 6Ω Solution I+IN 12 Ω A IN B Steps 1 & 2 I 6 Ω 42 V 35 V 81  NORTON’S THEOREM a I+IN 12 Ω b IN A B c I 6 Ω 42 V 35 V f e d Applying KVL to loop abefa: 42 = 12(I+IN) + 6I 42 = 18I + 12IN (1) Applying KVL to loop cbedc: 35 = 6I 35 I= A 6 82  NORTON’S THEOREM  35  Substituting for I in equation 1: 42 = 18  + 12 I N  6 Steps 3 & 4 12 Ω RN A B 12  6 RN = 12 // 6 = = 4 12 + 6 6Ω 83  NORTON’S THEOREM Steps 5 & 6 I3 A − 21 A 4 3 4 B 4 − 21 I3 =  = −3 A 4+3 4 84  NORTON’S THEOREM Example 2 Determine the current in the load resistor RL A R1 R2 R3 R4 RL E1 E2 E3 E4 B Solution A I1 I2 I3 I4 R1 R2 R3 R4 IN E1 E2 E3 E4 B 85  NORTON’S THEOREM A I1 I2 I3 I4 R1 R2 R3 R4 IN E1 E2 E3 E4 B Solution Applying KCL I N = I1 + I 2 + I 3 + I 4 E1 E2 E3 E4 = + + + R1 R2 R3 R4 86  NORTON’S THEOREM Finding RN A R1 R2 R3 R4 RN B 1 1 1 1 1 = + + + R N R1 R2 R3 R4 87  NORTON’S THEOREM Finding IL A IL IN RN RL B RN IL =  IN RN + RL 88  SUPERPOSITION THEOREM The Theorem The current through(or the voltage across) any element in a multiple-source linear circuit can be found by taking the algebraic sum of the current through(or the voltage across) that element due to each individual source acting alone. 89  SUPERPOSITION THEOREM The Theorem I 12Ω 6Ω 3Ω 42V 35V IA IB 12Ω 12Ω 6Ω 3Ω 6Ω 3Ω 42V + 35V 90  SUPERPOSITION THEOREM Example 1 Use superposition theorem to find the current supplied by the 35V battery of the circuit below. I 12Ω 6Ω 3Ω 42V 35V Solution With the 42V battery acting alone, IA RT = (3 // 6) + 12 = 14 12Ω 42 6Ω 3Ω IT = = 3A 14 6 42V IA = 3 = 2A 6+3 91  SUPERPOSITION THEOREM With the 35V battery acting alone, IB 12Ω 6Ω 3Ω 35V RT = (12 // 6) + 3 = 7 35 IT = = 5A 7 I B = IT = 5 A 92  SUPERPOSITION THEOREM With both batteries acting, I IA IB 12Ω 12Ω 12Ω 6Ω 3Ω 6Ω 3Ω 42V 35V = 6Ω 3Ω + 35V 42V I = IB − I A = 5 − 2 = 3A 93  RECIPROCITY THEOREM The Theorem An ideal ammeter and ideal voltage source when inserted in two different branches of a linear network can be interchanged without changing the reading of the ammeter R1 R2 R1 R2 + + V(t) R3 A = A R3 V(t) - - 94  RECIPROCITY THEOREM Similarly, An ideal voltmeter and ideal current source when connected across two different branches of a network can be interchanged without changing the reading of the voltmeter. R3 R3 + + R1 R2 + R1 R2 + i R4 V = V R4 - i - - - 95  RECIPROCITY THEOREM Example 1 Jointly use superposition and reciprocity theorems to find the current supplied by the 35V battery of the circuit below. I 12Ω 6Ω 3Ω 42V 35V Solution With the 42V battery acting alone, RT = (3 // 6) + 12 = 14 IA 42 12Ω IT = = 3A 6Ω 3Ω 14 6 IA = 3 = 2A 42V 6+3 96  RECIPROCITY THEOREM The second circuit to be solved is: IB IA 12Ω 6Ω 3Ω 12Ω is comparable 6Ω 3Ω 35V to 42V Applying the reciprocity theorem IA IA 12Ω 12Ω 6Ω 3Ω 6Ω 3Ω 42V = 42V A B 97  RECIPROCITY THEOREM I B IA 12Ω 6Ω 3Ω 12Ω 6Ω 3Ω is comparable 35V to circuit B 42V Applying proportion, If 42V = IA =2A, Then 35V =(35*2)/42=5/3A 5 A 3 B Applying KVL, I 12Ω 5 6Ω 3Ω 35 = 3I + 12  B 3 35V  I = 5A B 98 RECIPROCITY THEOREM I = IB − I A = 5 − 2 = 3A 99  DELTA-STAR TRANSFORMATION ❖The transformation is employed in situations where neither series nor parallel arrangements can be identified. ❖An arrangement of three(3) resistors where the resistors are connected to each other is a delta A arrangement. A A R2 R3 R2 R3 B R1 C B R1 C 100  DELTA-STAR TRANSFORMATION ❖An arrangement of three(3) resistors where all resistors have a common point of connection through one terminal is a star(wye) arrangement. Ra Ra Rc Rb Rc Rb 101  DELTA-STAR TRANSFORMATION ❖A delta arrangement can be changed to star and vice versa using the following relations: A R2 R3 Ra = R1 + R2 + R3 Ra R3 R1 R2 R3 Rb = R1 + R2 + R3 Rc Rb R1 R2 Rc = B R1 C R1 + R2 + R3 102  DELTA-STAR TRANSFORMATION A Rb Rc R1 = Rb + Rc + Ra Ra R2 R3 Ra Rc Rb R2 = Ra + Rc + Rc Rb B R1 C Ra Rb R3 = Ra + Rb + Rc When all values are the same, delta values are 3 times star values 103  DELTA-STAR TRANSFORMATION Example 1 Determine the voltage V0 across the 6Ω resistor of the circuit below 6Ω 2Ω 2Ω a 10 V 2Ω 6Ω Vo b Solution 6 Ra = Rb = Rc = 2  3 = 6 Ra 2 2 2 Rb 6 10 R c 104  DELTA-STAR TRANSFORMATION 6Ω 6Ω 10V 6Ω 6Ω 6Ω 3Ω a 3 V0 =  10V = 5V 3Ω 3+3 10V 6Ω Vo b 105  ALTERNATING CURRENT CIRCUITS ❖Alternating current (AC) circuits are circuits with currents and voltages which are time-varying ❖Examples of AC waveforms are: ❑ Sine wave ❑Square wave ❑Triangular wave 106  TERMINOLOGIES IN AC CIRCUITS ❖Amplitude (peak): The maximum deviation of the function from its center position ❖Cycle: A repeating portion of a function (wave). ❖Period (T): The duration of a cycle 1 f = ❖Frequency(f): The inverse of period. T v(V) E A B C t(s) -E 107  AVERAGE VALUE ❖Average value: The average value of a periodic function is its dc value. If i = f (t ) 1 area[ f (t )]  T Then I av = f (t )dt = T 0 T 108  AVERAGE VALUE The following steps are followed when finding average values of waveforms: 1. Identify a cycle of the wave 2. Note the period 3. Find the area of the cycle 4. Divide the area by the period 109  ROOT MEAN SQUARE VALUE The Root Mean Square (RMS) or Effective value of an alternating quantity is the value of a direct current which when flowing through a given resistance for a given time produces the same heat as produced by the alternating current when flowing through the same resistance. The RMS value of an alternating current i = f (tis:) area[( f (t )) ] 1   2 1 T =  0  f (t ) dt  = 2 2 I rms T  T 110  ROOT MEAN SQUARE VALUE The following steps are taken when finding the RMS value of a waveform : 1. Identify a cycle of the waveform 2. Note the period 3. Square the cycle 4. Find the area under the squared cycle 5. Divide the area by the period 6. Take the square root of the result 111  ROOT MEAN SQUARE VALUE Example 1 Find the average and rms values of the waveform below. v E 2 4 6 8 s -E Solution Average Value ❑Cycle spans 0 to 4 = (2  E ) − (2  E ) ❑Period= 4s ❑Area of cycle 112  ROOT MEAN SQUARE VALUE =0 Area 0 Vavg = = = 0V period 4 113  ROOT MEAN SQUARE VALUE v RMS value E ❑Cycle spans 0 to 4 2 4 6 8 ❑Period = 4s s ❑Squared cycle E2 -E 2 4 ❑Area covered by squared cycle = 4  E = 4E2 2 2 4E ❑Division of area by period = =E 2 4 114  ROOT MEAN SQUARE VALUE ❑Taking square root Vrms = E =E 2 115  ROOT MEAN SQUARE VALUE Example 2 Calculate the effective value of the voltage below. sine Vm θ π 2π Solution ❑Cycle spans 0 to 2π ❑Area of sine part Vm (1 − cos 2 ) 2   As =  Vm sin d =  d 2 2 Vm 0 2 0 2 =  2 116  ROOT MEAN SQUARE VALUE sine Area of triangular part Vm θ π 2π   2  Vm  2  Vm   2 Vm  2 3 At =     d =  2   = 0     3  0 3 V  V  5 2 2 2 Total area: = + m = Vm  m 5 2 2 3 6 Vm  5 Mean =6 = V 2 2 m 12 117  ROOT MEAN SQUARE VALUE 5 2 5 RMS value = Vm = Vm 12 12 Note: The area of a squared right-angled triangular wave is 2 bh = 3 118  SINUSOIDAL VOLTAGES AND CURRENT Voltages and currents of commercial ac generators have the following expressions: v = Vm sin tor v = Vm sin 2ft Vmis the peak voltage f is the frequency in Hz  Issecond. the angular frequency in radian per It specifies how many oscillations occur in a unit time interval 119  SINUSOIDAL VOLTAGES AND CURRENT i = I m sin t or i = I m sin 2ft I m is the peak current f is the frequency in Hz  Issecond. the angular frequency in radian per It specifies how many oscillations occur in a unit time interval 120  RMS VALUE OF SINUSOIDAL QUANTITIES The RMS value of a sinusoidal voltage v = Vm sin 2ft is given by: 1 1 T 2 2  2 V =   Vm sin 2ftdt  T 0  1  1 TV  2 2 =  0 (1 − cos 4ft )dt  m T 2 1   m  2 V T 2 Vm =  =  2 T 2 121  RMS VALUE OF SINUSOIDAL QUANTITIES Similarly, The RMS value of a sinusoidal current i = I m sin 2ft is given by: Im I = 2 122  RMS VALUE OF SINUSOIDAL QUANTITIES Example 1 Find the rms values of the following quantities: (a) i = 10 2 sin 100t (b) v = 20 sin 100t Solution 10 2 (a) I = = 10 2 20 (b) V = = 14.14 2 123  HARMONICS Non-sinusoidal periodic voltages and currents can be expressed as the sum of sine waves in which the lowest frequency is f and all other frequencies are integral multiples of f. For example, a square wave v(t ) of amplitude E can be expressed as: 4E  1 1  v(t ) = sin 2ft + sin 6ft + sin 10ft + − − −   3 5  124  HARMONICS ❖Any quantity which contains multiple frequencies is a harmonic quantity. ❖The frequency of which others have been expressed as multiples of is the fundamental frequency. ❖An odd multiple of the fundamental is an odd harmonic. ❖An even multiple of the fundamental is an even harmonic. 125  RMS VALUE OF A HARMONIC QUANTITY The effective value of a harmonic quantity is obtained by: ❖First obtaining the square of the rms value of each term ❖Adding the obtained squared rms values ❖Taking the square root of the sum v(t ) = ao + a1 sin (t + 1 ) + a2 sin (2t +  2 ) + a3 sin (3t + 3 ) +........ 2 2 2  a1   a 2   a3  V = ao +   +   +   +.......... 2  2  2  2 126  RMS VALUE OF A HARMONIC QUANTITY Example Find the RMS value of the current i(t ) = 2 + 5 sin wt + 3 2 sin (3wt + 30 0 ) Solution 2 5  3 2 2  I = 2 +  +  2  2  2  = 5.05 127  PHASORS ❖Phasors are used to represent sinusoidal quantities to avoid drawing the sine waves. ❖A phasor is a straight line whose length is proportional to the rms voltage or current it represents. ❖To show the phase angle or phase displacement between voltages and currents, the phasors bear an arrow. 128  PHASORS ❖Two phasors are said to be in phase when they point in the same direction. The phase angle between them is then zero. V I I V 129  PHASORS ❖Two phasors are said to be out of phase when they point in different directions. ❖The phase angle between them is the angle through which one of them has to be rotated to make it point in the same direction as the other. V φ I 130  PHASOR DIAGRAMS ❖It is used to show at a glance the magnitude and phase relations among the various quantities within a network. This is often helpful in the analysis of the network. ❖Example A 50 Hz source having rms voltage of 240 V delivers a rms current of 10 A to a circuit. The current lags the voltage by 30°. (a) Draw the phasor diagram for the circuit. (b) Express the voltage and current as functions of time. 131  PHASOR DIAGRAMS ❖Solution (a) Take V as the reference V = 240V 30o I = 10A (b) v(t ) = 240 2 sin 100t i(t ) = 10 2 sin(100t − 30 )  132/244  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES ❖The sum of sinusoidal quantities is obtained by taking the vector sum of their phasors. ❖The difference of sinusoidal quantities is obtained by first reversing the subtracted quantity and adding it as a vector to the other phasors. ❖A sinusoidal quantity is reversed by adding 1800 to its angle ❖Only sinusoidal quantities of the SAME FREQUENCY can be added or subtracted. 133  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES ❖Example 1 Let v1(t) = 60 sin ωt and v2(t) = 80 sin (ωt – 90°). Determine (a) v1 + v2 and (b) v1 – v2 ❖Solution Φ V1 ≡ 60 (a) Phasor diagram V1 + V2 V2 ≡ 80 V1 +V2 = 602 + 802 = 100 −1  80   = tan   = 53 0  60  134  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES  v1 + v2 = 100 sin( t − 53 ) 0 (b) v1 − v2 = v1 + (− v2 ) = 60 sin t + 80 sin (t − 90 + 180 0 0 ) = 60 sin t + 80 sin (t + 90 ) 0 Phasor diagram -V2 ≡ 80 V1 ≡ 60 135  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES -V2 ≡ 80 V1 – V2 Φ V1 ≡ 60 V1 −V2 = 60 + 80 = 100 2 2 −1  80   = tan   = 53 0  60   v1 − v2 = 100 sin t + 53 ( 0 ) 136  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES ❖Example 2 Four conductors meet at a junction. The following relationship exists between them. i = i + i + i 4 1 2 3 Find the value of i 4given that i1 = 5 sint i 2 = 8 sin(t + 3 ) + 5 sin 3t i3 = 15 sin(t − 4 ) + 8 sin( 3t + 3 ) 137  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES ❖Solution (a) There are two different frequencies. They must therefore be added separately. Addition of  part. 8 X-component = 8 cos 60 + 5 + 15 cos 45 o 0 60o = 19.607 45 o 5 Y-component = 8 sin 60 + 0 − 15 sin 45 o 0 = −3.678 138  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES Amplitude = (19.607 ) + (− 3.678 ) 2 2 = 19.949  − 3.678  Angle = tan −1  = −10.62 0  19.607  Therefore,  part of i 4 is 19.949 sin(t − 10.62 ) 0 139  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES Addition of 3part. X-component 8 = 8 cos 60 + 50 =9 60o 5 Y-component = 8 sin 60 + 00 = 6.928 Amplitude = 9 + 6.928 2 2 = 11.358 140  ADDITION AND SUBTRACTION OF SINUSOIDAL QUANTITIES Angle −1  6.928  0 = tan   = 37.59  9  3 part of i4is Therefore, 11.358 sin(3t + 37.59 ) 0 Hence ( i4 = 19.949 sin t − 10.62 0)+ 11.358 sin(3t + 37.59 ) 0 141  IMPEDANCE (Z) ❖The opposition to current flow in ac circuits owing to the presence of combinations of resistive, inductive and capacitive elements. ❖Opposition due to inductance (L) is called inductive reactance(XL). ❖Opposition due to capacitance is called capacitive reactance(XC). 142  IMPEDANCE (Z) ❖Phase relationship between the current and voltage in a resistor i v i= + R R v Let v = Vm sint - v Vm sint i= = = I m sint R R It is noted that the voltage across and the current through a resistor are in phase. 143  IMPEDANCE (Z) ❖Phase relationship between the current and voltage iin an inductor + di v=L VL = I L X L dt L v Let i = I m sint - v = L  I m cos t = LI m sin(t + 90 ) 0 = Vm sin(t + 90 ) 0 It is noted that the current through an inductor lags the voltage by 900. X L = L 144  IMPEDANCE (Z) ❖Phase relationship between the current and voltage in a i capacitor i=c dv VC = I C X C + v C dt - Let v = Vm sint i = C  Vm cos t = CVm sin(t + 90 )0 = I m (sint + 90 ) 0 It is noted that the current through a capacitor leads the voltage by 900. Xc = 1 c 145  IMPEDANCE (Z) ❖Series circuit containing R and L VR VL I VL V R XL ϕ VR I V V = VR + VL Z = R + XL 2 2 2 V = VR + VL 2 2 2 Z= R + XL 2 2 ( IZ ) = ( IR ) + ( IX L ) 2 2 2 146  IMPEDANCE (Z) ❖Phase angle between current and voltage in a series RL circuit  XL  VL V  = tan  −1   R  ϕ VR I ❖The current in a series RL circuit lags the voltage but not by 900 147  IMPEDANCE (Z) ❖Series circuit containing R and C VR I VR VC I R ϕ XC VC V V V = VR + VC Z = R + XC 2 2 2 V = VR + VC 2 2 2 Z= R + XC 2 2 ( IZ ) = ( IR) + ( IX C ) 2 2 2 148  IMPEDANCE (Z) ❖Phase angle between current and voltage in a series RL circuit  XC  VR I  = tan  −1  ϕ  R  VC V ❖The current in a series RC circuit leads the voltage but not by 900 149  IMPEDANCE (Z) ❖Series circuit containing R, L and C VR VL VC VL I R XL XC V VR I V VC V = VR + VL + V c V = VR + (VL − VC ) 2 2 2 150  IMPEDANCE (Z) V = VR + (VL − VC ) 2 2 2 = ( IR ) + ( IX L − IX c ) 2 2 V= ( IR ) + ( IX L − IX c ) 2 2 IZ = I R + ( X L − X c ) 2 2  Z = R + (X L − Xc ) 2 2 151  IMPEDANCE (Z) ❖Phase angle between current and voltage in a series RLC XL  X L − XC  circuit  = tan  −1  Z  R  ϕ R XC ❖Current in a series RLC circuit may lead or lag the voltage depending on the relative values of XL and XC 152  IMPEDANCE (Z) ❖Example 1 A coil has R=12Ω and L=0.1H. It is connected across a 100V, 50Hz supply. Calculate (a) the reactance and impedance of the coil (b) the current and (c) the phase difference or angle between the current and the applied I voltage. R XL X L Z ϕ ❖Solution R I 2fL = 2  50  0.1 = 31.416 (a) X L = V Z = R + X L = 12 + 31.416 2 2 2 2 = 33.630 153  IMPEDANCE (Z) V 100 (b) I= = = 2.974 A Z 33.630 X L V ϕ R I  XL  − 1  31.416  (c)  = tan  −1  = tan    R   12  = 69.09 0 154  IMPEDANCE (Z) ❖Example 2 A metal filament lamp, rated at 750W, 100V is to be connected in series with a capacitor across a 230V, 50Hz supply. Calculate (a) the capacitance required and (b) the phase angle between the current and supply voltage. R I VC I ϕ XC = R XC ❖Solution 2 IC XC V (a) 2 V = VR + VC 2 V VC = V − VR = 230 − 100 2 2 2 2 = 207.123V 155  IMPEDANCE (Z) PR 750 (a) I R = I C = I = = = 7.5 A VR 100 VC 207.123  XC = = = 27.616 IC 7.5 1 1 Hence, c= = 2fX C 2  50  27.616 = 115 F 156  IMPEDANCE (Z)  XC  − 1  VC  (b)  = tan   = tan   −1  R   VR  − 1  207.123  = tan    100  = 64.23 0 157  POWER IN AC CIRCUITS ❖There are three kinds of power in ac circuits 1. Apparent Power (S) which is measured in Volt- amperes (VA) 2. Active Power (P) which is measured in Watts (W). Active Power is also called Actual Power, Useful Power, True Power, Real Power or simply, Power 3. Reactive Power (Q) which is measured in Volt- amperes reactive (VAR) 158  POWER IN AC CIRCUITS ❖There following relationships exist between S, P and Q Q S = VI S S = P +Q 2 2 2 ϕ P = S cos  P P= VI cos Q = S sin  Q = VI sin  cos  is called power factor (pf) P pf = S 159  POWER IN AC CIRCUITS ❖Power factor may be said to be lagging or leading. ❖Power factor is lagging when current lags voltage ❖Power factor is leading when current leads voltage 160  POWER IN AC CIRCUITS ❖Relationships between the three passive elements , P and Q. Q XL S Z ϕ R ϕ P XC 1. Resistors consume only P 2. Inductors consume only Q 3. Capacitors do not consume P and Q. They rather supply Q or reduce the consumption of Q. 161  POWER IN AC CIRCUITS ❖Example 1 A single-phase motor connected to a 400-V, 50-Hz supply is developing 10 kW with efficiency of 84 per cent and a power factor of 0.7 lagging. Calculate (a) the input kVA (b) the active and reactive components of the current and (c) the reactive kVA. ❖Solution Pout 10 (a) P = = = 11.905kW in  0.84 162  POWER IN AC CIRCUITS Pin 11.905 S= = = 17.007 kVA pf 0.7 S 17.007  10 3 (b) S = VI  I = = V 400 = 42.518 A I r I I a = I cos = 42.518  0.7 = 29.766 A Ir = I − Ia 2 2 ϕ I = 30.361 A a 163  POWER IN AC CIRCUITS (b) Q = VI sin  = VI r = 400  30.361 = 12.144kVAR 164  POWER IN AC CIRCUITS ❖Example 2 An emf whose instantaneous value is given by 283sin(314t + π/4)V is applied to an inductive circuit and the current in the circuit is 5.66sin(314t – π/6)A. Determine (a) the frequency of the emf (b) the R and L (c) the power absorbed. ❖Solution (a) 314 2f = 314  f = = 50 Hz (b) 2 V 283 5.66 Z= =  = 50 I 2 2 165  POWER IN AC CIRCUITS   10 Z (b)  = − − =  = 75 XL 0 4 6 24 R = Z cos  = 50 cos75 0 ϕ = 12.941 R X L = Z sin = 50 sin75 = 48.296 0 XL 48.296 X L = 2fL  L = = 2f 2  50 = 0.154 H 166  POWER IN AC CIRCUITS 283 5.66 (c) P = VI cos  =  cos75 0 2 2 = 207.286W 167  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS ❖The ability to make a vector quantity appear as a scalar quantity using complex numbers is utilized in the analysis of ac circuits. ❖All the mathematical manipulations in complex algebra hold when employing complex numbers in analyzing ac circuits. ❖The operator ‘i’ is replaced with ‘j’ in other to avoid confusing it with current. 168  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS ❖The three passive elements are represented as follows: 1. R as a complex number is R 2. X L as a complex number is jX L 3. X C as a complex number is − jX C XL 4. Series impedance Z Z = R + j( X L − X C ) ϕ R XC 169  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS ❖Example 1 Express in rectangular and polar notations, the impedance of each of the following circuits at a frequency of 50 Hz: (a) a resistance of 20 Ω (b) a resistance of 20 Ω in series with an inductance of 0.1 H (c) a resistance of 50 Ω in series with a capacitance of 40 μF. ❖Solution (a) (b) Z = 20+ j 0 = 200 0 X L = 2fL = 2  50  0.1 = 31.416  Z = 20 + j 31.416 = 37.24257.52 0 170  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS 1 1 (c) XC = = 2fC 2  50  40  10 −6 = 79.577   Z = 50 − j79.577 = 93.981 − 57.86 0 171  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS ❖Example 2 A circuit is arranged as indicated in the figure below, the values being as shown. Calculate the value of the current in each branch and its phase relative to the supply voltage. A 150 μF 20 Ω ❖Solution C Z A = 20 + j 0 B 0.1 H 5Ω 200 V, 50 Hz Z B = R + jX L = 5 + j 31.4 ZC = − jX C = − j 21.2  Z AB = Z A // Z B = 15.84 29.48 0 172  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS A 150 μF = 13.78 + j7.8 20 Ω C ZT = ZC + Z AB B 0.1 H 5Ω 200 V, 50 Hz = − j 21.2 + (13.78 + j7.8 ) = 19.22 − 44.2 0 Choosing the voltage a the reference phasor, V 2000 0 I C = IT = = Z T 19.22 − 44.2 0 = 10.4 44.2 0 173  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS Current leads supply voltage by 44.20 VAB = IZ AB = (10.4 44.2 )  (15.84 0 A  29. 48 0 ) = 164.873.68 150 μF 0 20 Ω VAB C IA = B 0.1 H 5Ω ZA 200 V, 50 Hz 164.873.68 0 = Current leads supply 20 = 8.2473.68 0 voltage by 73.680 174  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS VAB 164.873.68 0 IB = = 0 = 5.18 − 7.27 0 Z B 31.7980.95 Current lags supply voltage by 7.270 A 150 μF 20 Ω C B 0.1 H 5Ω 200 V, 50 Hz 175  USING COMPLEX NUMBERS TO SOLVE AC CIRCUIT PROBLEMS S = VI * = P + jQ ❖Q is positive when the current lags the voltage ❖Q is negative when the current leads the voltage 176  CALCULATION OF COMPLEX POWER ❖Example 1 The potential difference across and the current in a circuit are represented by 100 + j200 v and 10 + j5 a respectively. Calculate the power and reactive voltamperes (or vars). ❖Solution S = VI = (100 + j 200 )(10 + j 5 ) * * = (100 + j 200 )(10 − j 5 ) = 2000 + j 1500 P = 2000W Q = 1500VAR 177  CALCULATION OF COMPLEX POWER ❖Example 2 A small installation consists of the following loads connected in parallel across a single-phase 240V, 50Hz supply: (a) a fan motor taking an input of 1.5kVA at 0.75pf lag, (b) a 1000W radiator operating at unity power factor (c) a number of fluorescent lamps taking a total load of 1.2kVA at 0.95pf lagging Find the total current, kW, kVA and power factor of the load. 178  CALCULATION OF COMPLEX POWER ❖Solution The problem is solved by obtaining the active and reactive component of each load and then summing all reactive components and all active components Load (kVA) cosϕ P(kW) = sinϕ Q(kVAR) = Scosϕ Ssinϕ ( a ) 1.5 0.75 1.125 0.66 0.99 ( b ) 1.0 1.0 1.0 0 0 ( c ) 1.2 0.95 1.14 0.312 0.374 TOTAL 3.265 1.364 179  CALCULATION OF COMPLEX POWER  Total kW = 3.265 Total kVA = P + Q 2 2 = 3.265 + 1.364 2 2 = 3.54 S 3.54  10 3 I= = = 14.8 A V 240 P 3.265 pf = = = 0.923 lagging S 3.54 180  CALCULATION OF COMPLEX POWER ❖NOTE In the problem above, if one load had a leading power factor, then its Q would have a negative sign and thus would have subtracted from the others. 181  THREE-PHASE CIRCUITS ❖A single-phase generator produces a single sinusoidal voltage. ❖A 3-phase generator on the other hand produces three equal voltages which are out of phase with one another by 120o. ❖The three voltages are generated in three separate windings arranged in a special way in the machine. 182  THREE-PHASE CIRCUITS ❖A 3-phase system is a power supply system consisting of three voltages which are 120o out of phase with one another. ❖Three-phase systems have the following advantages over single-phase systems ❑Three-phase motors, generators and transformers are simpler, cheaper and more efficient ❑Three-phase transmission lines can deliver more power for a given weight and cost 183  THREE-PHASE CIRCUITS ❑The voltage regulation of three-phase transmission lines is inherently better ❑A 1-phase supply can be obtained from a 3- phase one 184 Winding arrangement of a three-phase generator 185 A three-phase transmission line 186 11kV Distribution Feeder 187 A distribution line 188  THREE-PHASE CIRCUITS ❖The two main connections of three-phase windings 1. A star arrangement where all winding have a common point a line L1 Neutral point Va Vb Vc L2 c b L3 ❑Letters a, b and c , colours red (R), yellow(Y) and blue (B) or numbers 1, 2 and 3 are used to name the windings 189  THREE-PHASE CIRCUITS 2. A delta arrangement where all winding are connected to each other line a L1 Va Vb c L2 b Vc ❑Letters a, b and c , colours red (R), yellow(Y) and blue (B) or numbers 1, 2 and 3 are used to name the windings 190  THREE-PHASE CIRCUITS ❖The phasor diagram for the three-voltages (in star or delta) is indicated below. V3,Vc or VB Vc = Vm sin(t + 120 o ) V1,Va or VR Va = Vm sint V2,Vb or VY Vb = Vm sin(t − 120 o ) 191  THREE-PHASE CIRCUITS ❖Line and phase voltages ❑The voltage from one line to another is called a line-to- line voltage or simple a line voltage ❑The voltage across each winding is a phase voltage ❑On a phasor diagram, a line voltage is drawn from the end of one phase to another in the anti clockwise a direction Van = V0 Vab N Vbn = V − 120 b Vcn = V  + 120 192 c  THREE-PHASE CIRCUITS ❖Relationship between line and phase voltages for a star connection Vab = Van − Vbn = V0 − V − 120 o 0 = V (1 − 1 − 120 ) 0 = V (1 − cos(− 120 ) − j sin(− 120 )) 3 3 =V + j  = 3V 30 0  2 2  Hence, for a star connection, the line voltage is 3 times the phase voltage. VL = 3V p 193  THREE-PHASE CIRCUITS ❖Relationship between Line and phase voltages for a delta connection VL = V p Va Vb VL Vc 194  THREE-PHASE CIRCUITS ❖Relationship between Line and phase current for a star connection IL L a 1 IL = I p Ia Ib Ic L2 b c L3 195  THREE-PHASE CIRCUITS ❖Relationship between Line and phase currents for a delta ILa connection I La = I a − I b a I b I = I0 − I − 120 o 0 = I (1 − 1 − 120 ) 0 Ic = I (1 − cos(− 120 ) − j sin(− 120 )) 3 3 = I + j  = 3I 30 0  2 2  Hence, for a delta connection, the line current is 3 times the phase current I L = 3I p 196  THREE-PHASE CIRCUITS ❖Analysis of three-phase balanced circuits ❑A balanced three-phase circuit is that in which identical loads are connected in each phase. ❑The currents that flow in a balanced three-phase system are equal in magnitude and also 120° out of phase. ❑A balanced three-phase circuit is analysed by considering just one phase ❑When finding total power, the per phase power is multiplied by three ❑1-phase power factor in the same as 3-phase 197  THREE-PHASE CIRCUITS ❖Example 1 Three identical resistors are connected in star across a 3- phase, 415-V supply. If each resistor has a resistance of 50 ohms, calculate (a) the voltage across each resistor (b) the current in each resistor (c) the total power supplied to the load ❖Solution V 415 (a) V = p L = = 240 3 3 V p 240 (b) Ip = = = 4.8 A R 50 198  THREE-PHASE CIRCUITS (c) Pp = V p I p = 240  4.8 = 1152W  PT = 3  Pp = 3  1152 = 3456W 199  THREE-PHASE CIRCUITS ❖Example 2 Three identical impedances are connected in delta across a 3-phase, 415-V supply. If the line current is 10 A, calculate (a) the current in each impedance (b) the value of each impedance. ❖Solution ❖(a) I = I 10 p L = = 5.78 A 3 3 V p 415 ❖(b) Zp = = = 71.80  I p 5.78 200  THREE-PHASE CIRCUITS ❖Example 3 A 3-phase, 450-V system supplies a balanced delta- connected load of 12 kW at 0.8 power factor lagging. Calculate (a) the phase currents (b) the line currents and (c) the effective impedance per phase. ❖Solution ❖(a) P = V I cos p p 12 p Pp  10 3  Ip = = 3 = 11.1 A V p cos 450  0.8 201  THREE-PHASE CIRCUITS (b) IL = 3I p = 3  11.1 = 19.2 A Vp (c) Zp = Ip 4500 o = = 40.5  36.9 0 11.1 − cos (0.8 ) −1 202  THREE-PHASE CIRCUITS ❖Power in three-phase circuits ❑The total apparent power of a balanced three-phase circuit (star or delta) is given by: S= 3VL I L ❑The total active power of a balanced three-phase circuit (star or delta) is given by: P= 3VL I L cos ❑The total reactive power of a balanced three-phase circuit (star or delta) is given by: Q= 3VL I L sin 203  THREE-PHASE CIRCUITS ❖Analysis of parallel balanced three-phase circuit problems ❑When the parallel loads are not of the same kind(say one is delta and the other is star), then they must be changed either all to star or all to delta. ❑When all are in star, the circuit is analyzed by taking one phase of each and applying the star characteristics. ❑When all are in delta, the circuit is analyzed by taking one phase of each and applying the delta characteristics. 204  THREE-PHASE CIRCUITS ❑Parallel circuit are better analysed by reverting to complex numbers. ❖ Example 1 A Y-connected impedance Z1 = 20.0 + j37.75 Ω per phase is connected in parallel with ∆-connected impedance Z2 = 30.0 – j159.3 Ω per phase. For an impressed 3-phase voltage of 398 V line to line, compute the line current, power factor and the power taken by the parallel combination. ❖Solution The circuit is solved by making all star. 205  THREE-PHASE CIRCUITS Y-connected impedance Z1 = 20.0 + j37.75 Ω ∆-connected impedance Z2 = 30.0 – j159.3 Ω 30.0 − j 159.3 Changing the delta to star Z 2e = 3 398 = 10 I.0 − j 53.1 Vp = = 230V p 3 Z1 Z2e Vp Vp Vp Ip = + Z1 Z 2e = 3.37  − 9.9 0 206  THREE-PHASE CIRCUITS I L = I p = 3.37 A pf = cos  = cos 9.9 = 0.985 lagging 0 P = 3VL I L cos = 3  398  3.37  0.985 = 2.29kW 207  THREE-PHASE CIRCUITS ❖Example 2 A manufacturing plant draws 415 kVA from a 2400 V, 3- phase line. If the plant power factor is 0.875 lagging, calculate (a) the impedance of the plant per phase (b) the phase angle between the phase voltage and phase current. ❖Solution The load is considered to be star. (a) VL 2400 Vp = = = 1390V 3 3 208  THREE-PHASE CIRCUITS S S= 3VL I L  I L = 3VL 415000  I p = IL = = 100 A 3  2400 V p 1390 Zp = = = 13.9  Ip 100  = cos (0.875 ) = 29 −1 0 209  THREE-PHASE CIRCUITS ❖Example 3 A star-connected motor is connected to a 4000 V, 3-phase, 50 Hz line. The motor produces an output power of 2681 kW at efficiency of 93 % and a power factor of 0.9 lagging. Calculate (a) the active power absorbed by the motor, (b) the reactive power absorbed by the motor, (c) the apparent power supplied by the transmission line and (d) the motor line current. ❖Solution Pout 2681 (a) Pin = = = 2883kW  0.93 210  THREE-PHASE CIRCUITS (a) P = Pout 2681 = = 2883kW in  0.93 Pin 2883 (b) S = = = 3203kVA pf 0.9 Q= S −P2 2 = 3203 − 2883 = 1395 kVAR 2 2 (c) S = 3203kVA 211  THREE-PHASE CIRCUITS S 3203  103 (d) I = = = 462 A L 3VL 3  4000 212  THREE-PHASE CIRCUITS (a)Three-phase four-wire feeding unbalanced star-connected loads ❑Three-phase unbalanced load does not have the same impedance in all the three phases. ❑The neutral points of the source and the load are the same, so the voltage across each phase of the load is the phase voltage of the source ❑Three-phase four wire is usually used for low-voltage power distribution ❑The neutral carries current if the load is unbalanced I N = I pa + I pb + I pc 213  THREE-PHASE CIRCUITS ❑Each phase is analyzed separately when the load is unbalanced ❖Example 1 A three-phase four-wire system has the following data: Supply voltage is 415 V, Z1 = 8 + j0 Ω, Z2 = 0 - j8 Ω and Z3 = 0 + j8 Ω. Determine the load currents and the current in the neutral. ❖Solution VL 415 Vp = = = 240V 3 3 214  THREE-PHASE CIRCUITS V p 240 0 0 I p1 = = = 30 0 0 Z1 8 + j0 V p 240  − 120 0 I p2 = = = 30  − 30 0 Z2 0 − j8 V p 240120 0 I p3 = = = 30 30 0 Z3 0 + j8 I N = I p1 + I p 2 + I p 3 I N = 300 + 30 − 30 + 30 30 0 0 0 215  THREE-PHASE CIRCUITS I N = 82 A 216  THREE-PHASE CIRCUITS ❖Example 2 In a three-phase four-wire system, the line voltage is 400V and resistive loads of 10 kW, 8 kW and 5 kW are connected between the three lines and neutral. Calculate (a) the current in each line and (b) the current in the neutral conductor. ❖Solution VL 400 (a) Vp = = = 231V 3 3 217  THREE-PHASE CIRCUITS Let Ia, Ib and Ic be the currents drawn by the 10kW, 8kW and 5kW loads respectively. Pa 10  10 3 Ia = = = 43.3 A V p cos 231  1 Pb 8  10 3 Ib = = = 34.6 A Vp 231 Pc 5  10 3 Ic = = = 21.65 A Vp 231 218  THREE-PHASE CIRCUITS I N = Ia + Ib + Ic = 43.30 + 34.6  − 120 + 21.65120 0 0 0 = 18.9 A 219  MAGNETIC CIRCUITS ❖A magnetic circuit is a closed path followed by any group of lines of magnetic flux. ❖Magnetic circuits are created with coils and ferromagnetic (iron, cobalt, nickel, etc) or permanent magnetic materials. 220  MAGNETIC CIRCUITS ❖Terminologies in magnetic circuits ❑Flux(ϕ) It is a measure of the amount of magnetic field passing through a given surface. The SI unit of flux is Weber(Wb). ❑Flux density (B) It is the flux per unit area. It is the flux divided by the cross-sectional area. The SI unit is Weber per square meter (Wb/m2) or Tesla (T)  B= A 221  MAGNETIC CIRCUITS ❖Terminologies in magnetic circuits ❑Magnetomotive force (F) It is the source which sets up the flux flowing around a magnetic circuit. The unit is Amperes (A) or Ampere- turns(AT). It is the product of current in coil and number of turns of coil F = NI 222  MAGNETIC CIRCUITS ❑Reluctance(S or  ) It is likened to resistance. It is the opposition offered by a material to the flow of magnetic flux. L S= o r A L is the length of the magnetic path in meters (m) A is the cross sectional area in square meter (m2) The unit of reluctance is ampere-turns per weber (AT/Wb) F = S 223  MAGNETIC CIRCUITS ❖Terminologies in magnetic circuits ❑Magnetic field intensity (H) It is the mmf per unit length. The unit is Amperes per meter (A/m)or Amper-Turns per meter (AT/m) F H= L 224  MAGNETIC CIRCUITS ❖B-H curves They are curves which show the relation between magnetic flux density and magnetic field intensity for various materials B ❑The B-H curve of vacuum is as shown B = o H o is a constant called magnetic constant or permeability of free space H o = 4  10 H / m−7 225  MAGNETIC CIRCUITS ❑B-H curve of no