Lecture notes for Basic Electrical and Electronics engineering

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GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING A Course Material on GE 6252 BASIC ELECTRICAL AND ELECTRONICS ENGINEERING By Mrs. R.HEMALATHA Mrs.K.UMARANI Mr.S.VIJAY Mr.R.GUNASEKARAN ASSISTANT PROFESSOR DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING SASURIE COLLEGE OF ENGINEERING VIJAYAMANGALAM – 638 056 SCE Page 1 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING CONTENTS S.NO TOPIC PAGE NO. UNIT I ELECTRICAL CIRCUITS & MEASUREMENTS 1.1 Basic definitions 7 1.2 DC circuits 8 1.3 Ohm’s Law 10 1.4 AC Circuits 11 1.5 Kirchoff’s Laws 12 1.6 Steady State Solution of DC Circuits 13 1.7 Simple problems using ohm’s law 14 1.8 Introduction to AC Circuits 18 1.9 Waveforms and RMS Value 18 1.10 Power and Power factor 18 1.11 Single Phase and Three Phase Balanced Circuits 19 1.12 Operating Principles of Moving Coil Ammeters and Voltmeters 36 Operating Principles of Moving Iron Instruments Ammeters and 1.13 40 Voltmeters 1.14 Dynamometer type Watt meters 43 1.15 Dynamometer type Energy meters 47 UNIT II ELECTRICAL MECHANICS 2.1 Construction, Principle of Operation of DC Generators 50 2.2 Basic Equations and Applications of DC Generators 56 2.3 Construction, Principle of Operation of DC Motor 65 SCE Page 3 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING 2.4 Basic Equations and Applications of DC Motor 69 2.5 Construction, Principle of Operation of Single Phase Transformer 71 2.6 Basic Equations and Applications of Single Phase Transformer 74 Construction, Principle of Operation of Single phase induction 2.7 82 Motor 2.8 Types of Single phase induction Motor 85 UNIT III SEMICONDUCTOR DEVICES AND APPLICATIONS 3.1 Characteristics of PN Junction Diode 98 3.2 Zener Effect 99 3.3 Zener Diode and its Characteristics 99 3.4 Half wave Rectifiers 101 3.5 Full wave Rectifiers 103 3.6 Voltage Regulation 104 3.7 Bipolar Junction Transistor 104 3.8 CB Configurations and Characteristics 108 3.9 CE Configurations and Characteristics 111 3.10 CC Configurations and Characteristics 115 3.11 Elementary Treatment of Small Signal Amplifier 118 UNIT IV DIGITAL ELECTRONICS 4.1 Binary Number System 119 4.2 Logic Gates 130 4.3 Boolean algebra 136 4.4 Half and Full Adders 138 SCE Page 4 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING UNIT I ELECTRICAL CIRCUITS & MEASUREMENTS 12 Ohm’s Law – Kirchoff’s Laws – Steady State Solution of DC Circuits – Introduction to AC Circuits – Waveforms and RMS Value – Power and Power factor – Single Phase and Three Phase Balanced Circuits. Operating Principles of Moving Coil and Moving Iron Instruments (Ammeters and Voltmeters), Dynamometer type Watt meters and Energy meters. UNIT II ELECTRICAL MECHANICS 12 Construction, Principle of Operation, Basic Equations and Applications of DC Generators, DC Motors, Single Phase Transformer, single phase induction Motor. UNIT III SEMICONDUCTOR DEVICES AND APPLICATIONS 12 Characteristics of PN Junction Diode – Zener Effect – Zener Diode and its Characteristics – Half wave and Full wave Rectifiers – Voltage Regulation. Bipolar Junction Transistor – CB, CE, CC Configurations and Characteristics – Elementary Treatment of Small Signal Amplifier. UNIT IV DIGITAL ELECTRONICS 12 Binary Number System – Logic Gates – Boolean Algebra – Half and Full Adders – Flip- Flops – Registers and Counters – A/D and D/A Conversion (single concepts) UNIT V FUNDAMENTALS OF COMMUNICATION ENGINEERING 12 Types of Signals: Analog and Digital Signals – Modulation and Demodulation: Principles of Amplitude and Frequency Modulations. Communication Systems: Radio, TV, Fax, Microwave, Satellite and Optical Fibre (Block Diagram Approach only). TOTAL: 60 PERIODS TEXT BOOKS: 1. V.N. Mittle “Basic Electrical Engineering”,Tata McGraw Hill Edition, New Delhi, 1990. 2. R.S. Sedha, “Applied Electronics” S. Chand & Co., 2006. REFERENCES: 1. Muthusubramanian R, Salivahanan S and Muraleedharan K A, “Basic Electrical, Electronics and Computer Engineering”,Tata McGraw Hill, Second Edition, (2006). 2. Nagsarkar T K and Sukhija M S, “Basics of Electrical Engineering”, Oxford press (2005). 3. Mehta V K, “Principles of Electronics”, S.Chand & Company Ltd, (1994). 4. Mahmood Nahvi and Joseph A. Edminister, “Electric Circuits”, Schaum’ Outline Series, McGraw Hill, (2002). 5. Premkumar N, “Basic Electrical Engineering”, Anuradha Publishers, (2003). SCE Page 6 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING UNIT – I ELECTRIC CIRCUITS & MEASUREMENTS Prerequisites Solid, Liquid and gas particles called molecules. These molecules are made up of atoms which can be further spilt into electrons, protons and neutrons. The electrons revolve around the nucleus. The electrons presents in the outer most orbits experience a very weak force of attraction for the obvious reason that according to coulomb’s law, the force between two charges varies inversely with the square of the distance. These electrons are known as free electrons. The movement of electrons are known as electric current Introduction 1.1 Basic Definitions Electric current: The continuous flow of electrons constitutes electric current. It is denoted by ‘I’ and is measured in amperes. ‘I’ is also given by I = coulomb / sec Electric Potential: The electric potential at any point in an electric field is defined as the work done in brining an unit positive charge (Q) from infinity to that point against the electric field ‘V’ is given by V = Resistance: It is the property of a conductor by which it opposes the flow of current. It is denoted by R and its unit is ohms (Ω) Laws of resistance: The resistance of a conductor (i). Varies directly with its length (l) (ii).Varies inversely with its cross sectional area (A) (iii). Depends on the nature of the material (iv). Depends on the temperature R α L And R α 1/A R α L / A R = ρ L / A Where ρ is called specific resistance Specific resistance: It is defined as the resistance offered by unit cube of the material between its opposite faces. It is denoted by ρ and its unit is ohm – meter ρ = RA / L SCE Page 7 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING Temperature effect on resistance: In the case of pure metals the resistance increases with increases in temperature. In case of alloys the increase in resistance with increases in temperature is relatively small and irregular. The resistance of electrolytes and insulators decreases with increases in temperature Temperature co-efficient of resistance It is defined as the change in resistance per ohm per degree change in temperature from 0°C. If a material has resistance of R , R , and R at temperature of 0°C, t °C and t °C respectively, then 0 1 2 1 2 R = R (1 + α t ) 1 0 0 1 R = R (1+ α t ) 2 0 0 2 = R2 = R1 R2 = R1 R = R (1+α (t -t )) 2 1 0 2 1 αt = 1.2. DC Circuits: Prerequisites: A DC circuit (Direct Current circuit) is an electrical circuit that consists of any combination of constant voltage sources, constant current sources, and resistors. In this case, the circuit voltages and currents are constant, i.e., independent of time. More technically, a DC circuit has no memory. That is, a particular circuit voltage or current does not depend on the past value of any circuit voltage or current. This implies that the system of equations that represent a DC circuit do not involve integrals or derivatives. Introduction: In electronics, it is common to refer to a circuit that is powered by a DC voltage source such as a battery or the output of a DC power supply as a DC circuit even though what is meant is that the circuit is DC powered. If a capacitor and/or inductor is added to a DC circuit, the resulting circuit is not, SCE Page 8 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING strictly speaking, a DC circuit. However, most such circuits have a DC solution. This solution gives the circuit voltages and currents when the circuit is in DC steady state. More technically, such a circuit is represented by a system of differential equations. The solution to these equations usually contains a time varying or transient part as well as constant or steady state part. It is this steady state part that is the DC solution. There are some circuits that do not have a DC solution. Two simple examples are a constant current source connected to a capacitor and a constant voltage source connected to an inductor. Electro-magnetic force(E.M.F): Electromotive Force is, the voltage produced by an electric battery or generator in an electrical circuit or, more precisely, the energy supplied by a source of electric power in driving a unit charge around the circuit. The unit is the volt. A difference in charge between two points in a material can be created by an external energy source such as a battery. This causes electrons to move so that there is an excess of electrons at one point and a deficiency of electrons at a second point. This difference in charge is stored as electrical potential energy known as emf. It is the emf that causes a current to flow through a circuit. Voltage: Voltage is electric potential energy per unit charge, measured in joules per coulomb. It is often referred to as "electric potential", which then must be distinguished from electric potential energy by noting that the "potential" is a "per-unit-charge" quantity. Like mechanical potential energy, the zero of potential can be chosen at any point, so the difference in voltage is the quantity which is physically meaningful. The difference in voltage measured when moving from point A to point B is equal to the work which would have to be done, per unit charge, against the electric field to move the charge from A to B. Potential Difference: A quantity related to the amount of energy needed to move an object from one place to another against various types of forces. The term is most often used as an abbreviation of "electrical potential difference", but it also occurs in many other branches of physics. Only changes in potential or potential energy (not the absolute values) can be measured. Electrical potential difference is the voltage between two points, or the voltage drop transversely over an impedance (from one extremity to another). It is related to the energy needed to move a unit of electrical charge from one point to the other against the electrostatic field that is present. The unit of electrical potential difference is the volt (joule per coulomb). Gravitational potential difference between two points on Earth is related to the energy needed to move a unit mass from one point to the other against the Earth's gravitational field. The unit of gravitational potential differences is joules per kilogram. Electromagnetism: When current passes through a conductor, magnetic field will be generated around the SCE Page 9 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING conductor and the conductor become a magnet. This phenomenon is called electromagnetism. Since the magnet is produced electric current, it is called the electromagnet. An electromagnet is a type of magnet in which the magnetic field is produced by a flow of electric current. The magnetic field disappears when the current ceases. In short, when current flow through a conductor, magnetic field will be generated. When the current ceases, the magnetic field disappear. Applications of Electromagnetism: Electromagnetism has numerous applications in today's world of science and physics. The very basic application of electromagnetism is in the use of motors. The motor has a switch that continuously switches the polarity of the outside of motor. An electromagnet does the same thing. We can change the direction by simply reversing the current. The inside of the motor has an electromagnet, but the current is controlled in such a way that the outside magnet repels it. Another very useful application of electromagnetism is the "CAT scan machine." This machine is usually used in hospitals to diagnose a disease. As we know that current is present in our body and the stronger the current, the strong is the magnetic field. This scanning technology is able to pick up the magnetic fields, and it can be easily identified where there is a great amount of electrical activity inside the body The work of the human brain is based on electromagnetism. Electrical impulses cause the operations inside the brain and it has some magnetic field. When two magnetic fields cross each other inside the brain, interference occurs which is not healthy for the brain. Ohm’s Law: Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference or voltage across the two points, and inversely proportional to the resistance between them. The mathematical equation that describes this relationship is: where I is the current through the resistance in units of amperes, V is the potential difference measured across the resistance in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm's law states that the R in this relation is constant, independent of the current. AC Circuits: Prerequisites: An alternating current (AC) is an electrical current, where the magnitude of the SCE Page 10 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING current varies in a cyclical form, as opposed to direct current, where the polarity of the current stays constant. The usual waveform of an AC circuit is generally that of a sine wave, as this results in the most efficient transmission of energy. However in certain applications different waveforms are used, such as triangular or square waves Introduction: Used generically, AC refers to the form in which electricity is delivered to businesses and residences. However, audio and radio signals carried on electrical wire are also examples of alternating current. In these applications, an important goal is often the recovery of information encoded (or modulated) onto the AC signal. Kirchhoff’s law: Kirchhoff's Current Law: First law (Current law or Point law): Statement: The sum of the currents flowing towards any junction in an electric circuit equal to the sum of currents flowing away from the junction. Kirchhoff's Current law can be stated in words as the sum of all currents flowing into a node is zero. Or conversely, the sum of all currents leaving a node must be zero. As the image below demonstrates, the sum of currents I , I , and I , must b c d equal the total current in I . Current flows through wires much like water flows a through pipes. If you have a definite amount of water entering a closed pipe system, the amount of water that enters the system must equal the amount of water that exists the system. The number of branching pipes does not change the net volume of water (or current in our case) in the system. SCE Page 11 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING Kirchhoff's Voltage Law: Second law (voltage law or Mesh law): Statement: In any closed circuit or mesh, the algebraic sum of all the electromotive forces and the voltage drops is equal to zero. Kirchhoff's voltage law can be stated in words as the sum of all voltage drops and rises in a closed loop equals zero. As the image below demonstrates, loop 1 and loop 2 are both closed loops within the circuit. The sum of all voltage drops and rises around loop 1 equals zero, and the sum of all voltage drops and rises in loop 2 must also equal zero. A closed loop can be defined as any path in which the originating point in the loop is also the ending point for the loop. No matter how the loop is defined or drawn, the sum of the voltages in the loop must be zero SCE Page 12 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING Steady State Solution of DC Circuits: Resistance in series connection: The resistors R , R , R are connected in series across the supply voltage “V”. The total current 1 2 3 flowing through the circuit is denoted as “I”. The voltage across the resistor R , R and R is V , 1 2 3 1 V , and V respectively. 2 3 V = IR (as per ohms law) 1 1 V = IR 2 2 V = IR 3 3 V = V +V +V 1 2 3 = IR +IR +IR 1 2 3 = (R +R +R ) I 1 2 3 IR = (R +R +R ) I 1 2 3 R = R +R +R 1 2 3 Resistance in parallel connection: The resistors R , R , R are connected in parallel across the supply voltage “V”. The total 1 2 3 current flowing through the circuit is denoted as “I”. The current flowing through the resistor R , R and R is I , I , and I respectively. 1 2 3 1 2 3 SCE Page 13 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING I = V / R (as per ohms law) I = V / R 1 1 1 I = V / R 2 2 2 I = V / R 3 3 3 V = V = V = V 1 2 3 From the above diagram I = I +I +I 1 2 3 = V / R + V / R + V / R 1 1 2 2 3 3 = V / R + V/R +V/R 1 2 3 I = V (1/R +1/R +1/R ) 1 2 3 V / R = V (1/R +1/R +1/R ) 1 2 3 1/R = 1/R +1/R +1/R 1 2 3 Problems based on ohm’s law Problem 1: A current of 0.5 A is flowing through the resistance of 10Ω.Find the potential difference between its ends. Given data: Current I= 0.5A. Resistance R=1Ω To find Potential difference V = ? Formula used: V = IR Solution: V = 0.5 × 10 = 5V. Result : The potential difference between its ends = 5 V Problem :2 A supply voltage of 220V is applied to a 100 Ω resistor. Find the current flowing through it. Given data Voltage V = 220V Resistance R = 100Ω To find: SCE Page 14 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING Current I = ? Formula used: Current I = V / R Solution: Current I = 220/100 = 2.2 A Result: The current flowing through the resistor = 2.2 A Problem : 3 Calculate the resistance of the conductor if a current of 2A flows through it when the potential difference across its ends is 6V. Given data Current I = 2A Voltage V = 6V To find: Resistance R = ? Formula used: Resistance R = V / I Solution: Resistance R = 6 / 2 = 3 Ω Result: The value of resistance R = 3Ω Problem: 4 Calculate the current and resistance of a 100 W, 200V electric bulb. Given data: Power P = 100W Voltage V = 200V To find: Current I =? Resistance R =? Formula used: Power P = V I Current I = P / V Resistance R = V / I SCE Page 15 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING Solution: Current I = P / V = 100 / 200 = 0.5 A Resistance R = V / I = 200 / 0.2 = 400 Ω Result: The value of the current I = 0.5 A The value of the Resistance R = 400 Ω Problem: 5 A circuit is made of 0.4 Ω wire, a 150Ω bulb and a 120Ω rheostat connected in series. Determine the total resistance of the circuit. Given data: Resistance of the wire = 0.4Ω Resistance of bulb = 150 Ω Resistance of rheostat = 120Ω To find: The total resistance of the circuit R =? T Formula used: The total resistance of the circuit R = R +R +R T 1 2 3 Solution: Total resistance ,R = 0.4 + 150 +120 = 270.4Ω Result: The total resistance of the circuit R = 270.4 Ω T Problem 6: Three resistances of values 2Ω, 3Ω and 5Ω are connected in series across 20 V, D.C supply .Calculate (a) equivalent resistance of the circuit (b) the total current of the circuit (c) the voltage drop across each resistor and (d) the power dissipated in each resistor. Given data: R = 2Ω 1 R = 3Ω 2 R = 5Ω 3 V = 20V SCE Page 16 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING To find: R =? T I =? T V , V , V =? 1 2 3 P , P , P =? 1 2 3 Formula used: R = R +R +R (series connection) T 1 2 3 I = V / R T T T V = R I 1 1 1 V = R I 2 2 2 V = R I 3 3 3 P =V I 1 1 1 P =V I 2 2 2 P =V I 3 3 3 Solution: R = R +R +R T 1 2 3 = 2+3+5 R = 10Ω T I = V / R T T T = 20 / 10 I = 2 A T In series connection I = I = I = I = 2A 1 2 3 T V = I R 1 1 1 = 22 V = 4 V 1 V = I R 2 2 2 = 23 V = 6 V 2 V = I R 3 3 3 = 52 V = 10V 3 P = V I 1 1 1 = 42 P = 8W 1 P = V I 2 2 2 = 62 SCE Page 17 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING P = 12W 2 P = V I 3 3 3 = 102 P = 20W 3 Result: (a). Equivalent resistance of the circuit R = 10Ω T (b). The total current of the circuit I = 2A T (c). Voltage drop across each resistor V = 4 V, V = 6 V, V = 10V 1 2 3 (d). The power dissipated in each resistor P = 8W, P = 12W, P = 20W 1 2 3 AC Instantaneous and RMS: Instantaneous Value: The Instantaneous value of an alternating voltage or current is the value of voltage or current at one particular instant. The value may be zero if the particular instant is the time in the cycle at which the polarity of the voltage is changing. It may also be the same as the peak value, if the selected instant is the time in the cycle at which the voltage or current stops increasing and starts decreasing. There are actually an infinite number of instantaneous values between zero and the peak value. RMS Value: The average value of an AC waveform is NOT the same value as that for a DC waveforms average value. This is because the AC waveform is constantly changing with time and the 2 heating effect given by the formula ( P = I .R ), will also be changing producing a positive power consumption. The equivalent average value for an alternating current system that provides the same power to the load as a DC equivalent circuit is called the "effective value". This 2 effective power in an alternating current system is therefore equal to: ( I .R. Average). As power is proportional to current squared, the effective current, I will be equal to √ I Ave. 2 Therefore, the effective current in an AC system is called the Root Mean Squared or R.M.S. Pure Resistive circuit: SCE Page 18 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING Resistors are “passive” devices that are they do not produce or consume any electrical energy, but convert electrical energy into heat. In DC circuits the linear ratio of voltage to current in a resistor is called its resistance. However, in AC circuits this ratio of voltage to current depends upon the frequency and phase difference or phase angle ( φ ) of the supply. So when using resistors in AC circuits the term Impedance, symbol Z is the generally used and we can say that DC resistance = AC impedance, R = Z. It is important to note, that when used in AC circuits, a resistor will always have the same resistive value no matter what the supply frequency from DC to very high frequencies, unlike capacitor and inductors. For resistors in AC circuits the direction of the current flowing through them has no effect on the behaviour of the resistor so will rise and fall as the voltage rises and falls. The current and voltage reach maximum, fall through zero and reach minimum at exactly the same time. i.e, they rise and fall simultaneously and are said to be “in-phase” as shown below. We can see that at any point along the horizontal axis that the instantaneous voltage and current are in-phase because the current and the voltage reach their maximum values at the same time, o that is their phase angle θ is 0 . Then these instantaneous values of voltage and current can be compared to give the ohmic value of the resistance simply by using ohms law. Consider below the circuit consisting of an AC source and a resistor. The instantaneous voltage across the resistor, V is equal to the supply voltage, V and is given R t as: V = V sinωt R max The instantaneous current flowing in the resistor will therefore be: I = V / R R R = V sinωt / R max = I sinωt max In purely resistive series AC circuits, all the voltage drops across the resistors can be added together to find the total circuit voltage as all the voltages are in-phase with each other. Likewise, in a purely resistive parallel AC circuit, all the individual branch currents can be added together to find the total circuit current because all the branch currents are in-phase with each other. SCE Page 19 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING Since for resistors in AC circuits the phase angle φ between the voltage and the current is zero, o then the power factor of the circuit is given as cos 0 = 1.0. The power in the circuit at any instant in time can be found by multiplying the voltage and current at that instant. Then the power (P), consumed by the circuit is given as P = Vrms Ι cos Φ in watt’s. But since cos Φ = 1 in a purely resistive circuit, the power consumed is simply given as, P = Vrms Ι the same as for Ohm’s Law. This then gives us the “Power” waveform and which is shown below as a series of positive pulses because when the voltage and current are both in their positive half of the cycle the resultant power is positive. When the voltage and current are both negative, the product of the two negative values gives a positive power pulse. Then the power dissipated in a purely resistive load fed from an AC rms supply is the same as that for a resistor connected to a DC supply and is given as: P = V rms I rms = I 2 rms R = V 2 rms / R Pure Inductive circuits: This simple circuit above consists of a pure inductance of L Henries ( H ), connected across a sinusoidal voltage given by the expression: V(t) = V sin ωt. When the switch is closed this max sinusoidal voltage will cause a current to flow and rise from zero to its maximum value. This rise SCE Page 20 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING or change in the current will induce a magnetic field within the coil which in turn will oppose or restrict this change in the current. But before the current has had time to reach its maximum value as it would in a DC circuit, the voltage changes polarity causing the current to change direction. This change in the other direction once again being delayed by the self-induced back emf in the coil, and in a circuit o containing a pure inductance only, the current is delayed by 90 . The applied voltage reaches its maximum positive value a quarter ( 1/4ƒ ) of a cycle earlier than the current reaches its maximum positive value, in other words, a voltage applied to a purely o inductive circuit “LEADS” the current by a quarter of a cycle or 90 as shown below. The instantaneous voltage across the resistor, V is equal to the supply voltage, V and is given R t as: V = V sin (ωt + 90) L max I = V / X L L X = 2πfL L Pure Capacitive circuits: When the switch is closed in the circuit above, a high current will start to flow into the capacitor as there is no charge on the plates at t = 0. The sinusoidal supply voltage, V is increasing in a positive direction at its maximum rate as it crosses the zero reference axis at an instant in time o given as 0 . Since the rate of change of the potential difference across the plates is now at its maximum value, the flow of current into the capacitor will also be at its maximum rate as the maximum amount of electrons are moving from one plate to the other. SCE Page 21 of 226 DEPARTMENT OF EEE GE 6252 BASIC ELECTRICAL AND ELECTRONICSENGINEERING o As the sinusoidal supply voltage reaches its 90 point on the waveform it begins to slow down and for a very brief instant in time the potential difference across the plates is neither increasing nor decreasing therefore the current decreases to zero as there is no rate of voltage change. At o this 90 point the potential difference across the capacitor is at its maximum ( V ), no current max flows into the capacitor as the capacitor is now fully charged and its plates saturated with electrons. At the end of this instant in time the supply voltage begins to decrease in a negative direction o down towards the zero reference line at 180 . Although the supply voltage is still positive in nature the capacitor starts to discharge some of its excess electrons on its plates in an effort to maintain a constant voltage. These results in the capacitor current flowing in the opposite or negative direction. o When the supply voltage waveform crosses the zero reference axis point at instant 180 , the rate of change or slope of the sinusoidal supply voltage is at its maximum but in a negative direction, consequently the current flowing into the capacitor is also at its maximum rate at that instant. o Also at this 180 point the potential difference across the plates is zero as the amount of charge is equally distributed between the two plates. o o Then during this first half cycle 0 to 180 , the applied voltage reaches its maximum positive value a quarter (1/4ƒ) of a cycle after the current reaches its maximum positive value, in other words, a voltage applied to a purely capacitive circuit “LAGS” the current by a quarter of a cycle o or 90 as shown below. I = I sin (ωt + 90) C max I = V / X L C X = 1 / 2πfC C RL Series circuit: SCE Page 22 of 226 DEPARTMENT OF EEE

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