ZURIDA KHATIM FASIHAH ARSHAD ZAHILAWATI AWANG KECHIK CIRCUIT Electrical AC DC & SINUSOIDAL STEADY STATE PART1
ZURIDA KHATIM FASIHAH ARSHAD ZAHILAWATI AWANG KECHIK Electrical circuit QUICK STUDY PART1 FORPOLYTECHNIC'SSTUDENT
Copyright @2022 Electrical Engineering Department Politeknik Sultan Abdul Halim Mu’adzam Shah 06000 Jitra, Kedah ZURIDA BINTI KHATIM FASIHAH BINTI ARSHAD ZAHILAWATI BINTI AWANG KECHIK e ISBN 978-967-0055-26-8 [email protected] [email protected] [email protected] All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any former by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the publisher, except in the case of brief quotations embodied in critical reviews and certain other noncommercial uses permitted by copyright law. For permission requests, write to the publisher at the address below
Table of Contents General outcome How to generate an AC voltage Phase Difference Faraday and Len'z Law ALTERNATING VOLTAGE & 0 CURRENT 1 SINUSOIDAL STEADY STATE ANALYSIS 02 General outcome Purely Resistive Purely Capacitive Purely Inductive Impedance R-L Series R-C Series R-L-C Series R-L Parallel R-C Parallel R-L-C Parallel Understand power in AC circuit 1 2 9 12 15 16 17 18 22 23 25 27 31 34 34 40
PREFACE The e-book has been produced to provide basic knowledge about electrical circuits. This e-book is specifically for DET20033 students and discusses AC, DC, and sinusoidal steady-state analysis. Alhamdulillah, praise be to Allah SWT, with His grace and mercy, the eBook "Quick Study: Electrical Circuit For Polytechnic Student's Part 1" is finally complete. We would like to express our appreciation to eLearning Politeknik Sultan Abdul Halim Mu’adzam Shah for providing us with the opportunities and inspiration to produce this eBook. Our experience in the teaching and learning process of the previous subject has been an advantage that guided us along the way. The editor would like to give the highest regards and appreciation to all who have been involved, either directly or indirectly, in making this eBook project work. Any positive feedback from lecturers and students is most welcome and appreciated. It is hoped that this eBook will be one of the tiny steps that we have taken to start the long journey towards excellence. ZURIDA KHATIM FASIHAH ARSHAD ZAHILAWATI AWANG KECHIK
Remember an alternating current 1.1 Understand the generation of an alternating current Apply the sinusoidal voltage and current values of a sine wave Apply a sinusoidal wave for an angular measurement ALTERNATING VOLTAGE & CURRENT GENERAL OUTCOME At the end of this topic, you should be able to: 1.2 1.3 1.4
HOW TO GENERATE AN AC VOLTAGE ? HOW TO GENERATE AN AC VOLTAGE ? Figure (a) Figure (b) Figure (c) To generate ac voltage, coils and magnets are needed The coil needs to be rotated between the 2 magnet poles A coil that cuts through a magnetic field will produce a current then the bulb will light up Figure (a) -If the coil does not cut the magnetic field no current is produced then the bulb does not light up Figure (b) -The coil is rotated again through a magnetic field will produce a current then the bulb will light up again Figure (c) 1 2 3 4 5
The resulting voltage value changes according to the angle of coil rotation, refer to the diagram
The resulting waveform is called a sine wave
You should know on Vrms Vave Vp Vpp Time Period What you should know about sinewave Sinewave
How to expression the sinewave equation ? for voltage e = Em sin wt for current i = Im sin wt Solution
Example 1 Calculate the peak to peak voltage , rms and average value for a given waveform. i. The peak-to-peak voltage is Vp-p=40V ii. The rms voltage is Vrms=0.707x Vp = 0.707x 20 =14.14 V iii.The average value for the sinusoidal voltage is Vave=0.6366x Vp =0.6366x20 =12.73 V Solution
Example 2 Calculate Vp, Vpp, Vrms, Vavg, frequency and peak factor. ii. Vpp = 2 x Vp =2 x 30 =60V iii. V rms = 0.707 x Vp =0.707 x 30 = 21.21 V iv. Vave = 0.6366 x Vp = 0.6366 x 30 = 19.11V v. f = 1/T = 1/8ms = 125 Hz i. Vp = 30V vi. Peak factor = Vp/Vrms = 30/21.21 = 1.414 Solution
e = Em sin ωt i = im sin ωt e = Em sin ωt i = im sin (ωt + θ) e = Em sin ωt i = im sin (ωt – θ) Phase difference refers to the angular displacement between different waveforms of the same frequency. Phase Difference
Example of awave that lags the reference Example of awave that leads the reference
Sources of AC Signal generator Function generator AC generator
FARADAY’S AND LEN’Z LAW INVOLVED IN GENERATING AC VOLTAGE
Len’z Law
2.1 Understand the ACbasic circuit Apply the circuit with inductive & capacitive load Apply the series and parallel R-L-C circuit Understand power in AC circuit Apply the understanding of power cosumption in AC circuit SINUSOIDAL STEADY STATE ANALYSIS GENERAL OUTCOME At the end of this topic, you should be able to: 2.2 2.3 2.4 2.5
Purely resistive AC circuit the currentIR and applied voltage VR are in phase (sefasa). PURELY RESISTIVE CIRCUIT R VR IR IR VR Circuit diagram Phasor diagram Current and voltage waveform
Purely capacitive AC circuit the currentIc LEADS the applied voltage VL by 90◦. XC varies with frequency frequency as the opposition to the flow of alternating currentis called the capacitive reactance, XC. Purely capacitive AC circuit Purely capacitive AC circuit PURELY CAPACITIVE CIRCUIT Circuit diagram Phasor diagram Current and voltage waveform
XL is proportionalto frequency as Purely inductive circuit the currentIL LAGS the applied voltage VL by 90◦. the opposition to the flow of alternating currentis called the inductive reactance, XL. Purely inductive circuit Purely inductive circuit PURELY INDUCTIVE CIRCUIT Circuit diagram Phasor diagram Current and voltage waveform
C I V I L C L I I V V lead lead = = So, V lag I So, I lag V Where : C : Capacitor L : Inductor I : Current V : Voltage Capacitor : Current lead Voltage Inductor : Voltage lead Current
The voltage across a 10 Ω resistor is given by the expression; Find the expression for the current i through the resistor and sketch the curves for v and i. EXAMPLE 2.1 SOLUTION 2.1
The voltage across a 10 Ω resistor is given by the expression; Find the expression for the currentithrough the resistor and sketch the curves for v and i. EXAMPLE 2.2 SOLUTION 2.2
IMPEDANCE ??? IMPEDANCE ?? MEASURE Complex Ratio FORMULA the opposition that an electrical circuit present the passage of current when a voltage is applied. of the voltage to the current in an alternating current (AC) circuit |Z|∠θ or Z = R + jX
R-L SERIES CIRCUIT Series R-L circuit, voltage leads applied current by 0o to 90ᵒ (IT as reference)
EXAMPLE 2.3 Find the i) total impedance, ZT ii) total current, IT iii) voltage at inductor, VL SOLUTION 2.3 :
R-C SERIES CIRCUIT In the series capacitor circuits, the voltage lags current by -90ᵒ ( IT as reference) The phase angles of resistive and capacitive impedance are always 0o and -90ᵒ, respectively, regardless of the given phase angles for voltage or current
EXAMPLE 2.4 SOLUTION 2.4 : Find total current, IT
R-L-C SERIES CIRCUIT In an a.c. series circuit containing resistance R, inductance L and capacitance C, the applied voltage V is the phasor sum of VR, VL and VC. VL and VC are anti-phase by 180◦, and there are three phasor diagrams possible — each depending on the relative values of VL and VC. R-L-C Series circuit
Lagging Leading "mengekor" "mendahului" remember ...
A coil of resistance 5ohm and inductance 120mH in series with a 100μF capacitor, is connected to a 300V, 50 Hz supply. Calculate : (a) the current flowing (b) the phase difference between the supply voltage and current (c) the voltage across the coil (inductor) (d) the voltage across the capacitor. EXAMPLE 2.5 SOLUTION 2.5 :
SOLUTION 2.5 cont.. :
R–L PARALLEL CIRCUIT In the two branch parallel circuit containing resistance R and inductance L shown in Fig. below, IR is in-phase with the supply voltage V and the current flowing in the inductor, IL, lags V by 90◦. The supply current I is the phasor sum of IR and IL and thus the current I lags the applied voltage V.
R–L PARALLEL CIRCUIT Example A 20ohm resistor is connected in parallel with an inductance of 2.387mH across a 60V, 1kHz supply. Calculate: (a) the current in each branch (b) the supply current (c) the circuit phase angle (d) the circuit impedance (e) the power consumed.
Solution
R–C PARALLEL CIRCUIT The supply current I is the phasor sum of IR and IC and thus the current I leads the applied voltage V. In the two branch parallel circuit containing resistance R and capacitance C shown in Fig. below, IR is in-phase with the supply voltage V and the current flowing in the capacitor, IC, leads V by 90◦.
R–C PARALLEL CIRCUIT Example A 30μF capacitor is connected in parallel with an 80 resistor across a 240V, 50 Hz supply. Calculate : (a) the circuit impedance, (b) the current in each branch, (c) the supply current (d) the circuit phase angle, (e) the power dissipated, (f) the apparent power
Solution
R–L-C PARALLEL CIRCUIT In the three branch parallel circuit containing resistance R and Inductance L, capacitance C shown in Fig. below. IR is in-phase with the supply voltage V and the current flowing in the capacitor, IC, leads V by 90◦ and the current flowing in the inductor, IL, lags V by 90◦.
R–L-C PARALLEL CIRCUIT Example Three branches, possessing a resistance of 50Ω, an inductance of 0.15H and a capacitance of 100uF respectively, are connected in parallel across a 100V, 50Hz supply. Calculate: a) the current in each branch b) the supply current c) the phase angle between the supply current and the supply voltage
Solution a) the current in each branch XL = 2πfL = 47.12Ω XC = 1 = 31.83Ω 2πfC IR¬ = V = 100 = 2A R 50 IL= V = 100 = 2.12A XL 47.12 IC = V = 100 = 3.14A XC 31.83 b) the supply current IT = √ (IC - IL)2 + IR2 IT = √ (3.14-2.12)2 + 22 IT = 2.25A c) the phase angle between the supply current and the supply voltage tan Ɵ = IC - IL = 27.02 leading IR
Power is distributed into the resistance and reactance in AC circuit. The power delivered to a load at any instant is defined by the product of the applied voltage and the resulting current: Since v and i are sinusoidal quantities: Positive power means that power has been distributed from supply into circuit. Negative power means that the power has been distributed from circuit into supply. There are three type of power in AC circuit: i) Average/active/real/true power, P ii) Apparent Power, S iii) Reactive Power, Q Understand power in AC circuits
Average/active/real/true power, P Apparent Power, S S = Vrms Irms S = Irms² Z S = Vrms ² Z The power delivered to the load is simply determined by P=VI, with no concern for the components of the load. Therefore P=VI is called apparent power, S (unit volt- ampere,VA) Where, Or θ : phase angle between Vrms and Irms The average power (real power) is the power delivered to the load and dissipated by the load The unit of average power is Watt (W) Or Where, θ : phase angle between Vrms and Irms P = Irms² R P = Vrms ² R P = Vrms Irms cos θ Apparent power, S
Reactive power, Q Power Factor pf = cos θ = 0 = 1 (unity) pf = cos θ ≠ 0 (has value between 0 dan 1, lagging due to current lags voltage) pf = cos θ ≠ 0 (has value between 0 dan 1, lagging due to current leads voltage) Purely inductive load: Purely inductive load: The cos θ function is defined as power factor, pf used to describe how much of its apparent power is actually real power. Purely resistive load: In general, the reactive power associated with any circuit is defined to be Or For the resistor, Q = 0 VAR ( volt-ampere reactive, VAR) Where, θ : phase angle between Vrms and Irms Q = Irms ²X P = Vrms ² X Q= Vrms Irms sin θ
POWER TRIANGLE For capacitive load, the phasor power S, is defined by S = P + QC With P = P <0ᵒ QL = P <90ᵒ QC = P <-90ᵒ For an inductive load, the phasor power S, as it is often called, is defined by S = P + QL The three quantities average power P, apparent power S and reactive power Q can be related in the vector domain by S S = P + Q Where P = IV cos θ Q = IV sin θ S P QL θ P QC θ
POWER FACTOR Describe how much of its apparent power is actually real power If pf<1, the current that must be supply will become larger For utility company, this will increase the cost. To improve the power factor, capacitor will be install to make pf close to unity Pf=COS θ POWER FACTOR CORRECTION Basically, the load supplied consists of resistance and inductance. This will lead to the appearance of an active and reactive power. A capacitance is placed in parallel to load to reduced excessive current drawn from the supply due to low power factor The reason to do this correction is to avoid any damage at the supply since each supply has its own current/voltage rating
Tutorial Tutorial A circuit having a resistance of 12Ω and an inductance of 100mH is connected in series to a sinusoidal supply voltage of 100V, 50Hz. Calculate: a) The reactance and impedance of the coil b) The current c) The phase different between the current and supply voltage Answer: a) 31.41 ,33.62 b)2.97A c) 69˚ A circuit having a resistance of 12Ω and a capacitance of 10uF is connected in series to a sinusoidal supply voltage of 200V, 50Hz. Calculate: a) The reactance and impedance of the coil b) The current c) The phase different between the current and supply voltage Answer: a) 318.31, 322.21 b)0.627A c) 87.84˚ Exercise 1 Exercise 2