ELECTRICAL CIRCUITS DET20033 SURIA SINDHI SHARIATI BT HASSAN SAIFENAH BT SAIPUDIN
ELECTRICAL CIRCUITS SURIA SINDHI SHARIATI BINTI HASSAN SAIFENAH BINTI SAIPUDIN POLITEKNIK KOTA BHARU
Department of Electrical Engineering Politeknik Kota Bharu, KM24, Kok Lanas, 16450 Ketereh, Kelantan ELECTRICAL CIRCUITS First Edition 2023 ©2023 Suria Sindhi Shariati Binti Hassan , Saifenah Binti Saipudin All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without permission of the publisher. ELECTRICAL CIRCUITS/ Suria Sindhi Shariati Binti Hassan , Saifenah Binti Saipudin
i PREFACE Assalamualaikum and greetings to all. The first version of Electrical Circuits (DET20033) is intended as a e-Book for Electrical Circuits (DET20033) course taken by semester 2 students in Electrical Engineering Department, Politeknik Kota Bharu. It’s also can be used as a reference book for all undergraduate students in electrical and electronics engineering. Electrical Circuits course is designed to provide students with the knowledge related to AC of electrical circuits. It emphasized on the principles of an alternating current AC waveform and sinusoidal steady state circuit analysis. This course also covers the applications of three phase system and operation of various types of transformers. This course also includes the fundamental of electrical circuits and exposes students to the understanding of basic circuit operations and calculations for different types of circuits. Hopefully this e-Book will help the polytechnics students to develop the better understanding and confident to answer any electrical circuits questions. Suria Sindhi Shariati Bt Hassan Saifenah binti Saipudin Department of Electrical Engineering Politeknik Kota Bharu, KM24, Kok Lanas, 16450 Ketereh, Kelantan
ii ABOUT AUTHOR Suria Sindhi Shariati Binti Hassan was born in 1986 in Kota Bharu, Kelantan. Obtained primary education at Sek Keb Sri Chempaka, Kota Bharu and secondary education at Sek Men Keb Tanjung Mas and Sek Men Teknik Kuala Selangor. After that, the author continues her studies at Matrikulasi Kolej Mara Kulim and then continued her studies for the Bachelor of Electrical Engineering at Universiti Tun Hussein Onn Malaysia (UTHM). Thereafter, she continued her studies in the Master of Technical and Vocational Education at the same place (UTHM). The author started her career as a lecturer at Politeknik Kuala Terengganu in 2012 until 2021 and now she works at Department of Electrical Engineering Politeknik Kota Bharu. Saifenah Binti Saifudin was born in 1975 in Pasir Mas, Kelantan. Obtained primary education at Sek Keb Sultan Ibrahim 2, Pasir Mas and secondary education at Sek Men Keb Sultan Ibrahim 2, Pasir Mas. After that, the author continues her studies at UTM KL and then continued her studies for the bachelor of Electrical Engineering at Universiti Tun Hussein Onn Malaysia (UTHM). Thereafter, she continued her studies in the Master of Technical and Vocational Education at the same place (UTHM). The author started her career as a lecturer at Politeknik Kuching in 2001 until the end of 2004, next PLIMAS 2004-Mei 2008, PTSB Jun 2008-Mei 2013, PSAS Jun 2013-April 2021 and now she works at Department of Electrical Engineering Politeknik Kota Bharu.
iii Table of Contents NO TOPIC PAGE 1.0 ALTERNATING VOLTAGE AND CURRENT 1-11 1.1Introduction to Alternating Current 1.2 Generation of an Alternating Current 1.3Sinusoidal Voltage and Current Values of a Sine Wave 1.4 Sinusoidal Wave for an Angular Measurement 1.5Phasor to Represent a Sine Wave Tutorial 1 2.0 Sinusoidal Steady State Circuit Analysis 12-20 2.1AC Basic Circuits 2.2 Circuit with Inductive and Capacitive Load 2.3Power in AC Circuit Tutorial 2 3.0 RESONANCE 21-28 3.1Resonance In RLC Series Circuit
iv 3.2 Resonance In RLC Parallel Circuit Tutorial 3 NO TOPIC PAGE 4.0 TRANSFORMER 29-37 4.1Several Types Of Transformer 4.2Non-ideal transformer 4.3construction and operation of a transformer. 4.4Transformer Increases and Decreases Voltage 4.5Effect Of A Resistive Load Across The Secondary Winding Tutorial 4 5.0 THREE PHASE SYSTEM 38-43 5.1Introduction To Three Phase System 5.2 Three Phase System Configuration Tutorial 5
1 CHAPTER 1 ALTERNATING VOLTAGE AND CURRENT
2 1.1 Remember An Alternating Current Distinguish Between Alternating Urrent (Ac) And Direct Current (Dc) DC AC The flow of electrical charge is only in one direction The movement of electrical charge periodically reverses directions. The output voltage will remain essentially constant over time AC source of electrical power charges constantly in amplitude & regularly changes polarity Why AC is used in preference to DC DC CRITERIA AC When a large amount of electrical energy is required, it is much difficult to generate DC (Expensive) COST When a large amount of electrical energy is required, it is much economical & easier to generate & transmit AC (Cheaper) Difficult to convert voltage. DC à AC : complex, expensive & less efficient. CONVERT VOLTAGE Easy to change AC voltage to a higher @ low voltage using transformer. Easy to convert to DC, so can be used to operate various types of DC circuits @ equipment DC does get used in some local commercial applications USAGE AC is the form in which electrical power is delivered to business & residences. AC may also converted into electromagnetic waves (radio waves) which can radiate @ travel through space. (wireless). Use extensively in electronic to carry information from 1 point to another. Source of alternating current • Alternating current generator • Generating plant • Wind power station • Thermal power generator • Hydroelectric power plant
3 1.2 Generating an alternating current Faraday’s and Lenz’s Law - Faraday’s Law: Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be induced in the coil. - Lenz’s Law: There is an induced current in a closed conducting loop if and only if the magnetic flux through the loop is changing. The direction of the induced current is such that the induced magnetic field always opposes the change in the flux. AC waveforms produced by a simple alternating current generator (one loop in 2-pole magnet) Generators convert rotational energy to electrical energy. When a conductor is in a magnetic field and either the field or the conductor moves, an emf (voltage) is induced in the conductor. This effect is called electromagnetic induction. A loop of wire rotating in a magnetic field produces a voltage which constantly changes in amplitude and direction. The waveform produced is called a sine wave and is a graphical picture of alternating current (ac). One complete revolution (360°) of the conductor produces one cycle of ac. The cycle is composed of two alternations: a positive alternation and a negative alternation. One cycle of ac in one second is equal to 1 hertz (1 Hz).
4 Equation of A Sinusoidal Waveform e = Em sin ( ωt ± θ ) where, Em = amplitude or maximum voltage ω = angular velocity (rad/s) ; ω= 2πf t = time (s) θ = phase angle (lagging or leading)
5 1.3 Sinusoidal voltage and current values of sine wave Frequency, Period, Peak Value and Amplitude PeRiod, T: Time required for completing one full cycle (Unit: s)T =1/f FReQuency, f: Number of cycles that is completed one second (Unit: Hz) f = 1/T PeaK VaLue @ AMPLiTude, VP@IP: Value measured from the baseline of an AC waveform to its maximum or peak level (Unit : Volt (V) @ Ampere (A)) PeaK TO PeaK VaLue, VPP@IPP: Value measured from the maximum positive level to the maximum negativelevel (Unit : Volt (V) @ Ampere (A)) Vpp = 2Vp Varioss voltage and current values of sine wave INSTANTANEOUS VALUE AT ANY POINT: - The instantaneous values of a sine wave voltage @ current are different at any different point along the curve, having +ve and –ve value. - Represent as: v @ I RMS VALUE: - The rms ( root mean square ) value @ effective value of a sinusoidal voltage is equal to the dc voltage that produces the same amount of heat in a resistance as does the sinusoidal voltage. - V rms = 0.707 . Vp - I rms = 0.707 . Ip NOTE: 0.707 = 1_ √2
6 AVERAGE VALUE: - By definition, the average value is as 0.637 times the peak value - The average value is the total area under the half cycle curve divided by the distance in radians of the curve along the horizontal axis. - Vavg = 0.637 Vp @ 2/π Vp - Iavg = 0.637 Ip @ 2/π Ip Form Factor: Ratio between rms value and the average value Form Factor= (rms value)/(average value)=1.11 Peak Factor: Ratio between rms value and the peak value Form Factor= (rms value)/(peak value)=1.414 FORMULA UNIT Frequency f = 1 T Hz Period T = 1 f Sec Amplitude Vp @ Ip Volt @ A Peak to Peak value 2 x Vp @ 2 x Ip Volt @ A RMS value 0.707 x Vp @ 0.707 x Ip Volt @ A Average value 0.637 x Vp @ 0.637 x Ip Volt @ A Form Factor RMS value__ = 1.11 Average value - Peak Factor Peak value = 1.414 RMS value -
7 1.4 Sinusoidal wave for an Angular Measurement Sine wave in terms of Angles • As angle A increases, the values of the trigonometric functions of A undergo a periodic cycle from 0, to a maximum of 1, down to a minimum of -1, and back to 0. • There are several ways to express the measure of the angle A. One way is in degrees, where 360 degrees defines a complete circle • Another way to measure angles is in a unit called the radian, where 2π radians defines a complete circle. Phase Angle of Sine wave in degree and Radians. - Because there are 2π radians in one complete revolution and 360o in a revolution, the conversion between radians and degrees is easy to write.
8 1.5 Phasor to Represent a Sine Wave Phasor related to Sine wave formula - The sine wave can be represented as the projection of a vector rotating at a constant rate. This rotating vector is called a phasor. - The phasor represented by the arrow is rotating in an anticlockwise direction about the centre origin point, describing the sine wave as it rotates. - Phasors allow AC calculations to use basic trigonometry. The sine function in trigonometry is the ratio of the opposite side of a right triangle to the adjacent side.
9 TUTORIAL 1 1. List two method to generate an alternating current. 2. State two advantages of alternating current compared to direct current. 3. List two (2) laws related in generating of alternating current. 4. Calculate the rms voltage if peak to peak value voltage is 14V. 5. Calculate the rms voltage if average voltage is 24V. 6. An alternating voltage is given by = ( − . )v. Calculate: i. Value of peak to peak ii. Value of RMS iii. Time period and frequency iv. The value of the voltage when t=8ms.
10 7. Calculate the following value for an AC voltage circuit given by V(t) = . i. The frequency and time period ii. The value of current at t = 4ms 8. An alternating current is given by = ( − . ). Calculate: i. The amplitude value ii. The peak to peak value iii. The phase angle in degree 9. The current in an AC circuit at any time t seconds is given by = ( + . ). Calculate time period, frequency and the value of current when t = 0s. 10. An alternating current is given by = ( +. ). Calculate the period time, frequency, RMS value and instantaneous value when t = 0s.
11 11. The voltage in AC circuit is given by V= . . Calculate the value of frequency, period time and the value of the voltage at t = 2ms. 12. Based on the figure, find the value of frequency, period time, peak voltage and write the equation for the figure. 13. Calculate the the value of period time , frequency and the value of voltage when t= 4ms for an AC voltage circuit given by v(t) = 100 sin 314t V .
12 CHAPTER 2 SINUSOIDAL STEADY STATE CIRCUIT ANALYSIS
13 2.1 AC Basic Circuit Purely Resistive Purely Inductive Purely Capacitive Circuit Diagram Current & Voltage Waveform - Current and voltage are in the same phase - current lagging voltage I - Voltage lagging Current Current & Voltage Phasor Diagram Current & Voltage Relationship - Current and voltage are in the same phase - Current lagging voltage by 90° - Current leading voltage by 90° Load Resistor, R Inductor, L = Capacitor,C = PHASE RELATIONSHIP CAPACITOR : INDUCTOR : LEAD (+ve) LAGG (-ve)
14 IMPEDANCES, Z - Impedance is the total opposition a device or circuits offer to the flow of periodic current. - Impedance is a complex quantity: R = Real part of Z = Resistance X = Imaginary part of Z = Reactance 2.2 Circuit With Inductive and Capacitive Load RL, RC and RLC Series Circuit RL Series Circuit RC Series Circuit RLC Series Circuit Circuit Diagram Impedance Triangle Current = Voltage Drop VR = I x R VL = I x XL VR = I x R Vc = I x Xc VR = I x R VL = I x XL Vc = I x Xc Voltage Relationship Element Impedance Polar form Rectangular Form R ZR = R R<0° R + j0 L ZL= jωL XL<90° 0 + jXL C ZC = - XC<-90° 0 - jXC Z = R ± jX
15 RL, RC and RLC Parallel Circuit RL Parallel Circuit RC Parallel Circuit RLC Parallel Circuit Circuit Diagram Voltage Drop Voltage drop at component is equal to voltage supply Current = = = = = = = 2.3 Series and Parallel R-L-C circuit - When the circuit contain the series and parallel component position, it’s called combination circuit - Example circuit for series and Parallel R-L-C circuit 2.4 Power in AC Circuit TRUE POWER/ ACTUAL POWER (P) - True power is the actual of power being used or dissipated in a circuit. - Its unit measured in watts(W) P = I2R = IV cos ɵ
16 REACTIVE POWER (Q) - Reactive Power is a function of a circuit's reactance (X). (power stored and returned to generator by circuit’s inductive and capacitive components) - Reactive power is caused by inductor load or capacitor load.The unit is measured in Volt Ampere Reactive (VAR). APPARENT POWER (S) - Apparent power is a combination of true power and reactive power.Apparent power is measured in the unit of Volt Ampere (VA). POWER FACTOR AND POWER FACTOR CORRECTION - The cos θ function is defined as power factor, pf, used to describe how muchof its apparent power is actually real power. - It can be categorized into unity, lagging and leading power factor which isdepend on the type of load.
17 TUTORIAL 2 1. By referring to the figure 1, calculate the total impedance, ZT for the series circuit witch has a frequency of 30KHz Figure 1 2. By referring to the figure 2, calculate the total impedance, ZT for the series circuit witch has a frequency of 20KHz Figure 2 3. Calculate the total impedance of a circuit and its circuit current when RLC series circuit containing a resistance of 12Ω, an inductance of 0.15H and a capacitor of 100µF are connected in series across a 100V, 50Hz supply. 4. A coil of inductance 80 mH and resistance 60Ω is connected in series to a 100V, 50 Hz supply. Calculate the circuit impedance and the current taken from the supply.
18 5. Based on Figure 3 given, calculate impedance total, Z and total of current, I. Figure 3 6. A 5Ω resistance, a 120mH inductance and 100µF are connected in series to a 240V, 50Hz voltage supply. Calculate the current flowing through the circuit and the voltage across capacitor and inductor. 7. A parallel circuit consist of a resistor, inductor and capacitor. Refer to the Figure 4, calculate the value of the total current, IT flowing in the circuit Figure 4
19 8. Referring to Figure 5, a coil of inductance 0.12H and resistance 3kΩ is connected in parallel with a 0.02μF capacitor across a 240V, 50 Hz supply. Find the value of the total current I, flows in the circuit. Figure 5 9. A 100Ω resistor, a 0.5H coil , a 10µF capacitor and 1H coil are all connected in series to 100V, 50Hz supply as shown in Figure C1. Calculate the total impedance (Zt) of the circuit , total current drawn from the supply (Is) , the voltage across the inductance L2 (VL2) , power factor (Cos Ø) and the real power (P) dissipation. Figure 6
20 10. A circuit consist of a resistor connected in series with a capacitor and true power 100W at a power factor of 0.5 and voltage supply 240V, 50Hz. Calculate current flowing in the circuit, the phase angle, the resistance value, the total impedance, the capacitance value, the potential difference at each component and draw the voltage phasor diagram. 11. A 240V, 50Hz supply is applied to a series capacitive circuit. The current flowing is 2A and the power dissipated is 150W. Calculate the values of the resistance and capacitance. 12. A coil of resistance 10Ω and inductance 100mH in series with capacitor 100uF, is connected across 200V, 50Hz power supply as in Figure 7. Calculate : Figure 7 i) Circuit impedance, Z ii) Circuit current, I iii) Phase angle iv) Voltage across inductor v) Voltage across capacitor
21 CHAPTER 3 RESONANCE
22 3.1 Characteristic of Resonance Circuit Resonance is a condition in RLC circuit in which the capacitive and inductive reactance are equal in magnitude, thereby resulting in a purely resistive impedance. Characteristic of resonance circuit Circuit reactance equal zero because capacitive and inductive are equal in magnitude The impedances is purely resistive The voltage Vs and current I are in phase The frequency response of a circuit is maximum Power factor equal to one XL = XC
23 3.2 Resonance in RLC Series Circuits. Current will be maximum & offering minimum impedance. Effect of changing frequency in R-L-C Parallel Circuits A series RLC circuit’s reactance changes as you change the voltage source’s frequency. Its total impedance also changes. At low frequencies, Xc > XL and the circuit is primarily capacitive. At high frequencies, XL > Xc and the circuit is primarily inductive.
24 3.3 Resonance in RLC Parallel Circuits. Current will be minimum & offering maximum impedance. Effect of changing frequency in R-L-C Parallel Circuits Reactance change as you change the voltage source’s frequency. At low frequencies, XL < Xc and the circuit is primarily inductive. At high frequencies, Xc< XL and the circuit is primarily capacitive.
25 3.4 Summary of Characteristic of Resonance RLC Circuit CHARACTERISTIC SERIES CIRCUIT PARALLEL CIRCUIT Resonant frequency, fr = √ or = √ Quality factor,Q = = . = √ = = = √ Bandwidth, BW = , − , (), () Half power frequency, fL & fH = − & = +
26 TUTORIAL 3 1. With the aid of an appropriate diagram, explain the effects of changing frequency to RLC series circuit. 2. Express the resonant frequency equation for RLC series circuits. 3. A coil of inductance 100mH is connected in series with a capacitance of 2µF and a resistance of 10Ω across a 240V, variable frequency supply. Calculate the resonant frequency, the current at resonance, voltages across inductor and capacitor at resonance and Q-factor of the circuit. 4. A series of resonance circuit consist of 100Ω resistor, a capacitor of 16µF and an inductor of 80mH is connected to a 240Vac supply. Calculate resonance frequency, current during resonance, bandwidth, lower and upper cut-off frequency. Sketch and label the resonance graph for current versus frequency with necessary value.
27 5. Calculate the total impedance of a circuit and its circuit current when a RLC series circuit containing a resistance of 12Ω, an inductance of 0.15H and capacitor of 100µF are connected in series across a 100V, 50 Hz supply. 6. A circuit which consists of a 12Ω resistor, 45mH inductor and 100µF capacitor is connected in series across a 220V AC supply. Calculate the upper and lower cut-off frequency. Then sketch and label the resonance graph current versus frequency with the obtained value. 7. A series resonance circuit consists of 50Ω resistor, a capacitor of 8µF and inductor of 40mH. When the circuit is connected to a 10V AC supply, calculate the current and the lower cut-off frequency and upper cut-off frequency during resonance. Then sketch and label the resonance graph current versus frequency with the obtained value.
28 8. A circuit which consists of 25Ω resistor, 2.5mH and a 550pF capacitor is connected in series across 2.5V AC supply. Calculate the upper and lower cut-foo frequency. Then sketch the resonance graph current versus frequency with the obtained value. 9. A coil of 100Ω resistor and 60mH inductance connected in series with capacitance of 0.6µF across a 240V with the variable frequency supply. Calculate the upper and lower cut-off frequency during resonance. Illustrate in detail the corresponding current waveform for all frequencies.
29 CHAPTER 4 TRANSFORMER
30 TRANSFORMER 4.1 Introduction A transformer is a device which uses the phenomenon of mutual induction to change the values of alternating voltages and currents. In fact, one of the main advantages of a.c. transmission and distribution is the ease with which an alternating voltage can be increased or decreasedby transformers. Transformers range in size from the miniature units used in electronic applications to the large power transformers used in power stations. The principle of operation is the same for each. A transformer consisting of two electrical circuits linked by a common ferromagnetic core. One coil is termed the primary winding which is connected to the supply of electricity, and the other the secondary winding, which may be connected to a load.
31 • The transformer is the static device which works on the principle of electromagnetic induction. • In electromagnetic induction, the transfer of energy from one circuit to another takes places by the help of the mutual induction. the flux induced in the primary winding is linked with the secondary winding. • The main flux is induced in the primary winding of the transformer. • This flux passes through the low reluctance path of the magnetic coreand linked with the secondary winding of the transformer.
32 4.2 Describe center-tapped transformer The center tap (CT) transformer is equivalent to two secondary windings with half the voltage across each. Center tap windings are used for rectifier supplies and impedance-matching transformers. 4.2.1 Describe multiple-winding transformers. Multiple-winding transformers have more than one winding on a common core. They are used to operate on, or provide, different operating voltages. 4.2.2 Describe autotransformers. In an autotransformer, one winding serves as oth the primary and the secondary. The windings is tapped at the proper points to achieve the desired turns ratio for stepping up or down the voltage.
33 4.3Non-ideal transformer Non ideal transformer is an transformer which has i.) has copper losses (no winding resistance) ii) has iron loss in core iii) has leakage flux iv) An ideal transformer gives output power unequal to the input power.(output power = input power) v) The efficiency of an idea transformer is less than 100%. Efficiency is the ratio of output power to input power.It is usually stated percentage. Ideally, efficiency of transformer is 100% 4.4Describe power rating transformer. A transformer is usually described in ratings (volts) and sometimes in apparent power (kVA) in which the transformer can function without overheating. Referring to figure below, a transformer’s rating can be obtained by either using VpIp or VsIs where Ip is the primary current and Is is the secondary current on full load.
34 4.5 Construction and operation of a transformer 4.5.1 Describe Parts Of A Basic Transformer. A transformer consists of two windings connected by a magnetic core. One winding is connected to a power supply and the other to a load. A circuit diagram symbol for a transformer is shown in figure below: A transformer construction 4.5.2 Turn Ratio Turn ratio (η) is defined as the ratio of the number of turns in the primary winding (Np) to the number of turns in the secondarywinding (Ns). Direction Of Windings Affect Voltage Polarities
35 4.6 Transformer Increases and Decreases Voltage 4.6.1 Step Up Transformer • A transformer in which the secondary voltage (Vs) is greater than the primary voltage (Vp) is calleda step-up transformer. • The number of turns in the secondary winding (Ns) is larger than the number of turns in theprimary winding (Np) • The secondary current (Is) is smaller than the primary current (Ip).The turn ratio of a transformer is less than 1 4.6.2 Step Down Transformer • A transformer in which the secondary voltage (Vs) is smaller than the primaryvoltage (Vp) is called a step-up transformer. • The number of turns in the secondary winding (Ns) is smaller than the numberof turnsin the primary winding (Np) • The secondary current (Is) is larger than the primary current (Ip).The turn ratio of a transformer is more than 1
36 4.6 Effect Of A Resistive Load Across The Secondary Winding 4.6.1 Current Delivered By The Secondary . • Since Np/Ns equal the turns ratio,η,the relationship voltage in a transformer is • For an ideal transformer : • Resistance value can be determined by : 4.6.2 Power In A Transformer The power delivered by the primary is: The power delivered by the secondary is:
37 TUTORIAL 4 1. Given a transformer with a voltage ratio of 10:1, if the input voltage is 220V, what is the output voltage produced by the transformer? (22V) 2. A transformer has a turns ratio of 5:2. If the input current is 4A, what is the output current obtained? (1.6A) 3. If a transformer has an efficiency of 95% and an input power of 1000VA, what is the output power obtained? (950VA) 4. A transformer has a primary winding of 100 turns and a secondary winding of 2000 turns. If the input voltage is 240V, what is the output voltage produced by the transformer? (4800V)
38 CHAPTER 5 THREE PHASE SYSTEM
39 5.1 Introduction To Three Phase System 5.1.1 Basic Principle Of A Three Phase System • Three phase supply is generated when THREE coils are placed 120° apart and the whole rotated in auniform magnetic field. • The result is three independence supplies of equal voltages which are each displaced by 120°. • The convention adopted to identify each of the phase voltages is R-red, Yyellow, and B-blue. • Most consumers are fed by means of a single-phase a.c. supply. • Two wires are used, one called the live conductor (usually coloured red) and the other is called the neutral.conductor (usually coloured black). • The neutral is usually connected via protective gear to earth, the earth wire being coloured green.The standard voltage for a single-phase a.c.supply is 240 V.
40 5.2 Three Phase E.M.F Generation The loops are being rotated anti-clockwise and each loop is producing exactlythe same emf with the same amplitude and frequency but the loop Y Lags loop Rby 1200 and the loop B lagsloop Y by 1200. This is the same for the associated loop Y 1, B 1, and R 1 . At any moment the e.m.f generated in the three loops are as follows : 2 ways to generate 3 phase: Rotate the coil in the constant magnetic field. Rotate the magnetic field around the static coils. 5.3 Balance Load In A Three Phase System Load can be connected to a 3-phase supply using certain connections. For simplicity, consider abalanced load for balanced threephase system. For a balanced system: • each load has the same magnitude of impedance (Z1 =Z2=Z3); • each load impedance has the same phase angle (f1= f2= f3); • each phase voltage (and current) is equal in magnitude; • each phase voltage is displaced by 120° from each other. It can be proved that at every instant, for balanced loads the algebraic sum of the currents flowing in the three conductors is zero.
41 5.4 Three Phase System Configuration Advantages And Application Of Three Phase System . • It is more economical as it requires less conductor material compared to a single-phase system. • A three-phase machine gives more output compared to a single-phase machine of the same size. • Three-phase motors are very robust, relatively cheap, generally smaller, have self-starting properties, provide a steadier output and require little maintenance compared with single- phase motors. • Domestic power and industrial or commercial power can be supplied from the same source.Has higher efficiency and minimum losses. • Voltage regulation is better in a three-phase system.
42 5.2.1 Compare the delta and star quintities. Star Connection Delta Connection Real Power Reactive Power Apparent Power
43 TUTORIAL 5 1. In a balanced three-phase system, the line voltage is 400V. What is the phase voltage in a star (Y) connection? (230.94V). 2. A three-phase motor is connected in a delta (Δ) configuration. The line current is measured as 20A. What is the phase current? (20A). 3. In a three-phase system, the line-to-line voltage is 480V, and the line current is 50A. If the power factor is 0.8, calculate the total power. (55.3 kVA) 4. A three-phase system has a power factor of 0.9 and an apparent power of 100 kVA. What is the real power in the system? (90 kW)