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This ppt presentation is about basic electrical engineering subject for btech students

- 1. (19ES1EE07) ELEMENTS OF ELECTRICAL AND ELECTRONICS ENGINEERING COURSE OBJECTIVES: Student shall be able to • Study and understand the performance of basic electric circuits • Understand the performance of electrical machines • Know the utilization of electrical energy in day to day to affairs • Understand the operation of diode and transistor COURSE OUTCOMES: After the completion of the course, students will be able to CO-1: Analyse the performance of electrical circuits CO-2: Test, analyse and find the applications of different electric machines CO-3: Know the use of electric power for domestic and industrial purposes CO-4: Understand the principles of semiconductor devices and their applications UNIT – I: Fundamentals of Electrical Circuits: Basic R-L-C parameters, Ohm’s Law, kirchhoff’s Laws, Series-parallel connections, Star/Delta Transformation, Generation of A.C, Average, RMS values and Form Factor of Sinusoidal Voltages, AC through RL,RC and RLC, concept of impedance, power, power factor, simple problems UNIT – II: D.C Machines: D.C Generator, Basic Construction, Operation, emf Equation, types, Open Circuit Characteristics, simple problems. D.C Motor-principle-back emf -Torque equation, Speed control, Swinburne’s test. UNIT – III: Introduction to 3-phase circuits, relation between line and phase quantities in balanced star and delta networks. A.C Machines: Single phase transformer: principle-emf equation-types-OC and SC tests- Voltage Regulation -Efficiency-Simple problems.
- 2. UNIT – IV: Three phase induction motor: Working principle – slip- torque equation- Torque slip characteristics, Principle of Alternator. Electrical Power Systems and Utilization: Hydro Power Plant: Lay out -Efficiency Calculation, Illumination: Definitions-Laws of Illumination- working of Incandescent Fluorescent lamps. UNIT – V: Electronics Devices: Semiconductor materials, Review of P-N junction, Diode Characteristics, Basic Operation of Half-wave and Full wave Rectifiers, Zener Diode as Voltage Regulator, BJT, biasing, Characteristics, applications. UNIT – VI: Digital Circuits and Transducers: Logic gates, Combinational Logic circuits, Basic operation of SR-JK-T and D Flip-Flops, Transducers -Overview - Passive Sensors - Working 24 of Strain Gauge, Pressure Gauge, Dial Gauge - Piezoelectric Accelerometer Model- Galvanometer. TEXT BOOKS: 1. Principles of Electrical and Electronics Engineering by V. K. Mehta, S. Chand & Co. 2. Fundamentals of Electrical Engineering by Ashafaq Hussain, 2nd Edition, Dhanpat Rai & Co. 3. Electronic Devices and Circuits by S. Salivahanan, N. Suresh Kumar, 3rd Edition, McGraw- Hill Education. REFERENCES: 1. Electrical Technology by Edward Hughes, ELBS Longman Publisher 2. Basic Electrical Engineering by D. P. Kothari & I. J. Nagrath, TMH Publications, 2nd Edition. 3. Utilization of Electric Power and Electric Traction by G. C. Garg, Khanna Publishers.
- 4. Electric Current (1) • Electric current i = dq/dt. The unit of ampere can be derived as 1 A = 1C/s. A direct current (dc) is a current that remains constant with time. An alternating current (ac) is a current that varies sinusoidal with time. The direction of current flow:
- 5. Electric Voltage • Voltage (or potential difference) is the energy required to move a unit charge through an element, measured in volts (V). Mathematically, (volt) – w is energy in joules (J) and q is charge in coulomb (C). Electric voltage, vab, is always across the circuit element or between two points in a circuit. vab > 0 means the potential of a is higher than potential of b. vab < 0 means the potential of a is lower than potential of b. Power and Energy • The law of conservation of energy ∑ p = 0 • Energy is the capacity to do work, measured in joules (J). • Mathematical expression
- 7. Definition of Electrical Resistance • Resistance (also known as ohmic resistance or electrical resistance) is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). • When a voltage is applied across a substance there will be an electric current through it. The applied voltage across the substance is directly proportional to the current through it. The constant of proportionality is resistance. Hence resistance is defined as the ratio of the applied voltage to the current through the substance. Where v is supply voltage I is current
- 13. Materials used in the field of Electrical Engineering are called the Electrical Engineering materials. Based on properties and area of applications, Electrical Engineering materials can be classified as Conductors, Semiconductors, Insulators Magnetic material 1.Conductors: Conductors are the materials which have very high conductivity. The number of free electrons are very high in a conductor at room temperature, which is the basic reason of high conductivity of conductors. Examples: Silver, Copper, Gold, Aluminium etc. 2.Semiconductors: Semiconductors are materials which have the conductivity between conductors and insulators. Semiconductors are the elements of group-III, group-IV and group-IV elements. Semiconducting materials have covalent bond. At normal temperature the conductivity of semiconductors is very low. With increase in temperature the conductivity of semiconductors increases exponentially. Example: Germanium, Silicon, Gallium Arsenic etc.
- 14. 3.Insulating Materials The conductivity of insulating materials is very low. These material are having a very high resistivity which makes them very suitable to insulate the current carrying parts from earthed metallic structure. In insulating materials the electrons are tightly bounded with nucleus. Due to which they cannot be freed for movement in materials. Due to which the resistivity of insulating materials is very high. Example:- Plastics, Ceramics, PVC etc. 4. Magnetic Materials These materials play an important role for existence of various electrical machines. The magnetic materials having high permeability are used for building the core to from the low reluctance path for magnetic flux. Magnetic materials can be further divided in following categories Ferromagnetic materials: Example: Iron, Cobalt, Nickel. Paramagnetic material : Example: Aluminum, Platinum, oxygen, Air etc. Diamagnetic materials : Mostly all metals i.e. silver, copper, gold, hydrogen etc. Anti ferromagnetic materials : Example: Cr, MNO, FeO, CoO, NiO, Mn etc. Ferrites : Example: Fe3O4, BaO.6Fe2O3 etc.
- 15. Classification of circuit element:- (i) Unilateral and Bilateral element:- If the element property and characteristic does not change with direction of current, then the element is called bilateral element; otherwise unilateral element. (ii) Linear and non linear element:- If the element satisfy homogeneity and additivity property then element is called linear element, otherwise non linear. (iii) Active Elements: Active Elements: When the element is capable of delivering the energy, it is called active element. Example: Voltage source, Current source, Transistor, Diode, Op-amp etc (iv) Passive Elements: When the element is not capable of delivering the energy, it is called passive element. Example: Resistance, capacitor, inductor etc. Identify weather the element is: i. Linear or non linear ii. Active or passive iii. Bilateral or unilateral
- 17. Inductor:- Inductance is the property of the inductor which opposes the sudden change in current. Concept:- When a time varying current is flowing through the coil, then magnetic flux is induced and it is given by CAPACITANCE: Capacitance is the property of capacitor which opposes the sudden change in voltage. Resistance: Resistance is a value that measures how much the component “resist” the passage of electrical current, the value is measured in ohms (Ω). One way to calculate resistance: R= ρL/A where ρ is the resistivity, a material’s property. Another form to calculate the resistance is applying Ohm’s law. V = iR, This constant of proportionality is
- 27. Convert the following Delta Resistive Network into an equivalent Star Network. Convert the following Star Resistive Network into an equivalent Delta Network.
- 28. Generation of AC waveforms https://www.youtube.com/watch?v=JQ8h50HpbsY The sine wave is a common type of alternating current (AC) and alternating voltage The voltage which changes polarity at regular interval of time is known as the alternating voltage. The one complete cycle of an alternating quantity consists two half cycles. And the direction of a half cycle changes after every particular interval of time. The machine which generates the alternating voltage is known as the alternator. Consider the stationary coil places inside the uniform magnetic field. The load is connected across the coil with the help of brushes and the slip rings. When the coil rotates in the anticlockwise direction at constant angular velocity ω the electromotive force induces in the coil. The cross-sectional view of the coil at the different position is shown in the figure below.
- 29. The magnitude of the emf induced in the coil depends on the rate of the flux cut by the conductor. The figure below shows that the no current induces in the coil when they are parallel to the magnetic line of forces. i.e., at the position (1), (3) and (5). And the total flux cut by the conductor becomes zero. The magnitude of the induces emf becomes maximum when the conductor becomes perpendicular to the magnetic line of force. The conductor cuts the maximum flux at this position. The direction of the emf induces in the conductor is determined by Fleming’s right-hand rule. When the coil is at position (2) the emf induces in the outward direction of the conductor whereas at position (4) the direction of the inducing emf becomes inward. In other words, the direction of emf induces in the conductor at position (2) and (4) becomes opposite to each other.
- 30. Amplitude The maximum positive or negative value attained by an alternating quantity in one complete cycle is called Amplitude or peak value or maximum value. The maximum value of voltage and current is represented by Em or Vm and Im respectively. Alternation One-half cycle is termed as alternation. An alternation span is of 180 degrees electrical. Cycle When one set of positive and negative values completes by an alternating quantity or it goes through 360 degrees electrical, it is said to have one complete Cycle. Instantaneous Value The value of voltage or current at any instant of time is called an instantaneous value. It is denoted by (i or e). Wave Form The shape obtained by plotting the instantaneous values of an alternating quantity such as voltage and current along the y-axis and the time (t) or angle (θ=wt) along the x-axis is called a waveform.
- 31. Frequency ( f ) is the number of cycles that a sine wave completes in one second . The more cycles completed in one second, the higher the frequency . Frequency is measured in hertz (Hz) • Relationship between frequency ( f ) and period (T) is: f = 1/T Peak :The peak value of a sine wave is the value of voltage or current at the positive or negative maximum with respect to zero • Peak values are represented as: Vp and I p Peak-to-peak :The peak-to-peak value of a sine wave is the voltage or current from the positive peak to the negative peak • The peak-to-peak value is twice the actual voltage value • Not Often Used • The peak-to-peak values are represented as: Vpp and Ipp where: Vpp = 2Vp and Ipp = 2Ip Time period: The time required for a sine wave to complete one full cycle is called the period (T) • A cycle consists of one complete positive, and one complete negative alternation T= 1/ f
- 32. The angular measure measurement of Sine Waves can be done in Degrees or Radians . A degree is an angular measurement corresponding to 1/360 of a circle or a complete revolution
- 33. Average Value Definition: The average of all the instantaneous values of an alternating voltage and currents over one complete cycle is called Average Value. If we consider symmetrical waves like sinusoidal current or voltage waveform, the positive half cycle will be exactly equal to the negative half cycle. Therefore, the average value over a complete cycle will be zero. The work is done by both, positive and negative cycle and hence the average value is determined without considering the signs. So, the only positive half cycle is considered to determine the average value of alternating quantities of sinusoidal waves. Let us take an example to understand it. Divide the positive half cycle into (n) number of equal parts as shown in the above figure Let i1, i2, i3…….. in be the mid ordinates The Average value of current Iav = mean of the mid ordinates
- 34. Average Voltage Equation of sinusoidal wave form Where: 0 and π are the limits of integration since we are determining the average value of voltage over one half a cycle. Then the area below the curve is finally given as Area = 2VP. Since we now know the area under the positive (or negative) half cycle, we can easily determine the average value of the positive (or negative) region of a sinusoidal waveform by integrating the sinusoidal quantity over half a cycle and dividing by half the period. For example, if the instantaneous voltage of a sinusoid is given as: v = Vp.sinθ and the period of a sinusoid is given as: 2π, then:
- 35. R.M.S Value Definition: That steady current which, when flows through a resistor of known resistance for a given period of time than as a result the same quantity of heat is produced by the alternating current when flows through the same resistor for the same period of time is called R.M.S or effective value of the alternating current. In other words, the R.M.S value is defined as the square root of means of squares of instantaneous values. Let I be the alternating current flowing through a resistor R for time t seconds, which produces the same amount of heat as produced by the direct current (Ieff). The base of one alteration is divided into n equal parts so that each interval is of t/n seconds as shown in the figure below. Let i1, i2, i3,………..in be the mid ordinates Then the heat produced in Then, heat produced in first interval = i1 2Rt/n joules Second interval = i2 2Rt/n joules Third interval = i3 2Rt/n joules nth interval = in 2Rt/n joules Total heat produced in time t = Rt[(i12 + i2 2 + i3 2 + ……+ in 2)/n] …..(1) Since Ieff is considered as the effective value of this current. Then heat produced by this current in time t = Ieff 2Rt …….(2)
- 36. By definition, equations (1) and (2) are equal. Therefore, Ieff 2Rt = Rt[(i1 2 + i2 2 + i3 2 + ……+ in 2)/n] Or Ieff 2 = [(i1 2 + i2 2 + i3 2 + ……+ in 2)/n] Or Ieff = [(i1 2 + i2 2 + i3 2 + ……+ in 2)/n]1/2 Or Ieff = IRMS = square root of the mean of the squares of the instantaneous current. Now, by using the integral calculus the RMS or effective value of the alternating current over a time period can be calculated as follows: Total heat produced in time t= Rt[(i12 + i2 2 + i3 2 + ……+ in 2)/n] …..(1) t = Ieff 2Rt …….(2) https://www.youtube.com/watch?v=E7RvsWITCjs
- 37. RMS Current Equation of sinusoidal wave form Let i = Im sinωt be the alternating current flowing through a resistance of R ohms for time t seconds and produces the same heat as produced by Ieff (a direct current).
- 38. Form Factor The ratio of the root mean square value to the average value of an alternating quantity (current or voltage) is called Form Factor. Mathematically, it is expressed as: Ir.m.s and Er.m.s are the roots mean square values of the current and the voltage respectively, and Iav and Eav are the average values of the alternating current and the voltage respectively. For the current varying sinusoidally, the Form Factor is given as: The value of Form Factor is 1.11 Peak Factor Definition: Peak Factor is defined as the ratio of maximum value to the R.M.S value of an alternating quantity. Mathematically it is expressed as: Where, Im and Em are the maximum value of the current and the voltage respectively, and Ir.m.s and Er.m.s are the roots mean square value of the alternating current and the voltage respectively. For the current varying sinusoidally, the peak factor is given as:
- 39. https://www.youtube.com/watch?v=HaFrY0qQ-NU Pure Resistive AC Circuit: The circuit containing only a pure resistance of R ohms in the AC circuit is known as Pure Resistive AC Circuit. The presence of inductance and capacitance does not exist in a purely resistive circuit. The alternating current and voltage both move forward as well as backwards in both the direction of the circuit. In an AC circuit, the ratio of voltage to current depends upon the supply frequency, phase angle, and phase difference. In an AC resistive circuit, the value of resistance of the resistor will be same irrespective of the supply frequency. Let the alternating voltage applied across the circuit be given by the equation Then the instantaneous value of current flowing through the resistor shown in the figure below will be:
- 40. From equation (1) and (3), it is clear that there is no phase difference between the applied voltage and the current flowing through a purely resistive circuit, i.e. phase angle between voltage and current is zero. Hence, in an AC circuit containing pure resistance, the current is in phase with the voltage as shown in the waveform figure below. Instantaneous power, p= vi Where, •P – average power •Vr.m.s – root mean square value of supply voltage •Ir.m.s – root mean square value of the current Hence, the power in a purely resistive circuit is given by:p =VI
- 41. Pure inductive Circuit The circuit which contains only inductance (L) and not any other quantities like resistance and capacitance in the circuit is called a Pure inductive circuit. In this type of circuit, the current lags behind the voltage by an angle of 90 degrees. Let the alternating voltage applied to the circuit is given by the equation: As a result, an alternating current i flows through the inductance which induces an emf in it. The equation is shown below:
- 43. Pure Capacitor Circuit The circuit containing only a pure capacitor of capacitance C farads is known as a Pure Capacitor Circuit. The capacitors stores electrical power in the electric field, their effect is known as the capacitance. It is also called the condenser.
- 44. Instantaneous power is given by p = vi Hence, from the above equation, it is clear that the average power in the capacitive circuit is zero.
- 46. RL,RC, and RLC Circuits A circuit that contains a pure resistance R ohms connected in series with a coil having a pure inductance of L (Henry) is known as RL Series Circuit. When an AC supply voltage V is applied, the current, I flows in the circuit.So, IR and IL will be the current flowing in the resistor and inductor respectively, but the amount of current flowing through both the elements will be same as they are connected in series with each other. The circuit diagram of RL Series Circuit is shown below: Where, VR – voltage across the resistor R VL – voltage across the inductor L V – Total voltage of the circuit Phasor Diagram of the RL Series Circuit :Current I is taken as a reference. The Voltage drop across the resistance VR = IR is drawn in phase with the current I. The voltage drop across the inductive reactance VL =IXL is drawn ahead of the current I. As the current lags voltage by an angle of 90 degrees in the pure Inductive circuit. The vector sum of the two voltages drops VR and VL is equal to the applied voltage V. Now,In right-angle triangle OAB VR = IR and VL = IXL where XL = 2πfL
- 47. Where, Z is the total opposition offered to the flow of alternating current by an RL Series circuit and is called impedance of the circuit. It is measured in ohms (Ω). Phase Angle: In RL Series circuit the current lags the voltage by 90 degrees angle known as phase angle. It is given by the equation:
- 48. The various points on the power curve are obtained by the product of voltage and current. If you analyze the curve carefully, it is seen that the power is negative between angle 0 and ϕ and between 180 degrees and (180 + ϕ) and during the rest of the cycle the power is positive. The current lags the voltage and thus they are not in phase with each other. Waveform and Power Curve of the RL Series Circuit
- 49. RC Series Circuit A circuit that contains pure resistance R ohms connected in series with a pure capacitor of capacitance C farads is known as RC Series Circuit. A sinusoidal voltage is applied and current I flows through the resistance (R) and the capacitance (C) of the circuit. Where, VR – voltage across the resistance R VC – voltage across capacitor C V – total voltage across the RC Series circuit Phasor Diagram of RC Series Circuit The phasor diagram of the RC series circuit is shown below: VR = IR and VC = IXC Where XC = I/2πfC In right triangle OAB,
- 50. Where, Z is the total opposition offered to the flow of alternating current by an RC series circuit and is called impedance of the circuit. It is measured in ohms (Ω). Phase angle From the phasor diagram shown above, it is clear that the current in the circuit leads the applied voltage by an angle ϕ and this angle is called the
- 51. Waveform and Power Curve of the RC Series Circuit
- 52. RLC Series Circuit When a pure resistance of R ohms, a pure inductance of L Henry and a pure capacitance of C farads are connected together in series combination with each other then RLC Series Circuit is formed. As all the three elements are connected in series so, the current flowing through each element of the circuit will be the same as the total current I flowing in the circuit. In the RLC Series circuit XL = 2πfL and XC = 1/2πfC When the AC voltage is applied through the RLC Series circuit the resulting current I flows through the circuit, and thus the voltage across each element will be: VR = IR that is the voltage across the resistance R and is in phase with the current I. VL = IXL that is the voltage across the inductance L and it leads the current I by an angle of 90 degrees. VC = IXC that is the voltage across capacitor C and it lags the current I by an angle of 90 degrees. The phasor diagram of the RLC series circuit when the circuit is acting as an inductive circuit that means (VL>VC) is shown below and if (VL< VC) the circuit will behave as a capacitive circuit.
- 53. It is the total opposition offered to the flow of current by an RLC Circuit and is known as Impedance of the circuit. Phase Angle From the phasor diagram, the value of phase angle will be
- 54. Impedance Triangle of RLC Series Circuit When the quantities of the phasor diagram are divided by the common factor I then the right angle triangle is obtained known as impedance triangle. The impedance triangle of the RL series circuit, when (XL > XC) is shown below: If the inductive reactance is greater than the capacitive reactance than the circuit reactance is inductive giving a lagging phase angle.
- 55. Impedance triangle is shown below when the circuit acts as an RC series circuit (XL< XC) When the capacitive reactance is greater than the inductive reactance the overall circuit reactance acts as capacitive and the phase angle will be leading. Applications of RLC Series Circuit •It acts as a variable tuned circuit •It acts as a low pass, high pass, bandpass, bandstop filters depending upon the type of frequency. •The circuit also works as an oscillator •Voltage multiplier and pulse discharge circuit
- 56. EX1: 200 V, 50 Hz, inductive circuit takes a current of 10A, lagging 30 degree. Find (i) the resistance (ii) reactance (iii) inductance of the coil. EX2: A Capacitor of capacitance 79.5μF is connected in series with a non inductive resistance of 30 across a 100V, 50Hz supply. Find (i) impedance (ii) current (iii) phase angle
- 57. EX3: A 230 V, 50 Hz ac supply is applied to a coil of 0.06 H inductance and 2.5 resistance connected in series with a 6.8 μF capacitor. Calculate (i) Impedance (ii) Current (iii) Phase angle between current and voltage (iv) power factor EX4: A resistance R, an inductance L=0.01 H and a capacitance C are connected in series. When an alternating voltage v=400sin( 3000t-20º)is applied to the series combination, the current flowing is 10 2 sin(3000t-65º). Find the values of R and C.
- 59. A wound coil that has an inductance of 180mH and a resistance of 35Ω is connected to a 100V 50Hz supply. Calculate: a) the impedance of the coil, b) the current, c) the power factor, and d) the apparent power consumed. Also draw the resulting power triangle for the above coil. Data given: R = 35Ω, L = 180mH, V = 100V and ƒ = 50Hz. (a) Impedance (Z) of the coil:
- 60. UNIT – II: D.C Machines: D.C Generator, Basic Construction, Operation, emf Equation, types, Open Circuit Characteristics, simple problems. D.C Motor-principle-back emf-Torque equation, Speed control, Swinburne’s test. These can be divided into: generators – which convert mechanical energy into electrical energy motors – which convert electrical energy into mechanical energy Both types operate through the interaction between a magnetic field and a set of windings A Simple AC Generator • We noted earlier that Faraday’s law dictates that if a coil of N turns experiences a change in magnetic flux, then the induced voltage V is given by • If a coil of area A rotates with respect to a field B, and if at a particular time it is at an angle to the field, then the flux linking the coil is BAcos, and the rate of change of flux is given by t Φ N E d d cos cos d d d sin d t t BA dt dΦ
- 61. Basic ac generator operation
- 62. One revolution of the wire loop generates one cycle of the sinusoidal voltage
- 63. A Simple DC Generator The alternating signal from the earlier AC generator could be converted to DC using a rectifier A more efficient approach is to replace the two slip rings with a single split slip ring called a commutator. this is arranged so that connections to the coil are reversed as the voltage from the coil changes polarity hence the voltage across the brushes is of a single polarity adding additional coils produces a more constant output. A dc generator is an electrical machine which converts mechanical energy into direct current electricity. This energy conversion is based on the principle of production of dynamically induced emf. Use of a commutator
- 64. A simple generator with two coils Working Principle of DC Generator We can see that in the first half of the revolution current always flows along ABLMCD, i.e., brush no 1 in contact with segment a. In the next half revolution, in the figure, the direction of the induced current in the coil is reversed.
- 65. But at the same time the position of the segments a and b are also reversed which results that brush no 1 comes in touch with the segment b. Hence, the current in the load resistance again flows from L to M. The waveform of the current through the load circuit is as shown in the figure. This current is unidirectional. The above content is the basic working principle of DC generator, explained by single loop generator model. The positions of the brushes of DC generator are so that the change over of the segments a and b from one brush to other takes place when the plane of rotating coil is at a right angle to the plane of the lines of force. It is to become in that position, the induced EMF in the coil is zero.
- 66. A four-pole DC generator
- 67. Basic Construction, Operation Theoretically, a DC generator can be used as a DC motor without any constructional changes and vice versa is also possible. Thus, a DC generator or a DC motor can be broadly termed as a DC machine. These basic constructional details are also valid for the construction of a DC motor. Hence, let's call this point as construction of a DC machine instead of just 'construction of a dc generator'. The above figure shows constructional details of a simple 4-pole DC machine. A DC machine consists of two basic parts; stator and rotor. Basic constructional parts of a DC machine are described below.
- 68. 1.Yoke: The outer frame of a dc machine is called as yoke. It is made up of cast iron or steel. It not only provides mechanical strength to the whole assembly but also carries the magnetic flux produced by the field winding. 2.Poles and pole shoes: Poles are joined to the yoke with the help of bolts or welding. They carry field winding and pole shoes are fastened to them. Pole shoes serve two purposes; (i) they support field coils and (ii) spread out the flux in air gap uniformly. 3.Field winding: They are usually made of copper. Field coils are former wound and placed on each pole and are connected in series. They are wound in such a way that, when energized, they form alternate North and South poles. 4. Armature core: Armature core is the rotor of a dc machine. It is cylindrical in shape with slots to carry armature winding. The armature is built up of thin laminated circular steel disks for reducing eddy current losses. It may be provided with air ducts for the axial air flow for cooling purposes. Armature is keyed to the shaft. 5.Armature winding: It is usually a former wound copper coil which rests in armature slots. The armature conductors are insulated from each other and also from the armature core. Armature winding can be wound by one of the two methods; lap winding or wave winding. Double layer lap or wave windings are generally used. A double layer winding means that each armature slot will carry two different coils. 6.Commutator and brushes: The function of a commutator, in a dc generator, is to collect the current generated in armature conductors. Whereas, in case of a dc motor, commutator helps in providing current to the armature conductors. A commutator consists of a set of copper segments which are insulated from each other. The number of segments is equal to the number of armature coils. Each segment is connected to an armature coil and the commutator is keyed to the shaft. 7.Brushes:are usually made from carbon or graphite. They rest on commutator segments and slide on the segments when the commutator rotates keeping the physical contact to collect or supply the current.
- 69. E.M.F Equation of DC Generator We know that the working principle of dc generator, that when conductors begin to cut the magnetic lines of force and therefore, the e.m.f. induces in the conductors according to 'Faraday's Law of Electromagnetic Induction'. The value of induced e.m.f. depends upon the lengths of the conductor, the magnetic field strength, and the speed at which the coil rotates. Let us see the equation for induced e.m.f. Number of revolutions per minute,= N / 60
- 70. Time taken for one revolution,dt = 60 / N sec E.m.f generated per conductor, Therefore, the total emf E generated between the terminals if given as, E = Average e.m.f generated per conductor * Number of conductor in each parallel path Where,A = P for lap winding. A = 2 for wave winding. EX-1 A four pole generator, having wave wound armature winding has 51 slots, each slot containing 20 conductor. What will be the voltage generated in the machine when driven at 1500 rpm assuming the flux per pole to be 7mWb?
- 71. EX-2 : An 8 pole Dc generator has 500 armature conductors, and a useful flux of 0.05Wb per pole. what will be the emf generated if it is lap-connected and runs at 1200 rpm? What must be the speed at which it is to be driven produce the same emf if it is wave-wound?
- 72. Types of DC Machines There are 2 main types of DC machines first one is a generator and the second one is a dc motor, dc motor uses dc current and provides mechanical power and dc generator generates dc voltage. As we discuss when current-carrying conductor when placed in field emf induced force apply on it. There are two main categories by which the dc machine gets excited first one is self-excited and the second one is separately excited. In self-excited machines field current produced by the machines itself while in separately excited machines the field current is provided by the separated source. Due to these two excitations methods, the dc machines are further divided into different types that are described here. 1 .Separately excited DC machine Application: laboratories for testing as they have a wide range of voltage output. Used as a supply source of DC motors. 2.self excited I. Shunt DC machine II. Series DC machine III. Compound DC machine
- 73. Separately Excited DC Machine If there is no internal voltage induced at the field windings and there is no field current so external power source is connected at the field winding to produce current and voltage these winding is known as separate windings. In the given figure, you see the detailed circuit of separately excited machines.
- 74. Shunt DC Machine In this type of dc machine the field winding (it is located at the stator of a machine) linked with the armature winding (wound rotor) in parallel. Due to a parallel connection, the voltage across the field windings is equalled to the supply voltage in the case of a motor and equals generated voltage in the case of a generator. So these windings have large no of turns. In a given diagram, you can see the circuit of shunt dc machines. Applications generator: lighting purposes. Used to charge the battery. Providing excitation to the alternators Applications motor: Lathe Machines, Centrifugal Pumps, Fans, Blowers, Conveyors, Lifts, Weaving Machine, Spinning machines, etc. Rsh = Shunt winding resistance Ish = Current flowing through the shunt field Ra = Armature resistance Ia = Armature current IL = Load current V = Terminal voltage Eg = Generated EMF
- 75. Series Wound DC Machine In these dc machines, the field windings are connected with the armature windings in series connection schemes. As this winding is in series so the armature current also passes through it that has high value for less power losses the turns of the field windings are less in this machine. In the given figure, you can see the circuit diagram of these machines. Applications generator: DC locomotives, booster in distribution Applications motor: The traction system, cranes, air compressors, Vaccum Cleaner, Sewing machine, etc. Here:Rsc = Series winding resistance Isc = Current flowing through the series field Ra = Armature resistance Ia = Armature current IL = Load current V = Terminal voltage Eg = Generated EMF
- 76. Compound DC Machine This generator/motor consists of series and shunts windings in its circuitry. 2 windings are placed at every pole of the machine. The number of turns in the series is less due to the large value of armature current flowing through it and shunt windings have large no of turns. There are 2 methods by which these two windings are connected in these machines if the field windings are in parallel with the armature windings only then the machines are known as the short shunt compound. In the given figure, you can see these arrangements. If the field windings is in parallel with the armature and series windings the machines are known as the long shunt compound machines. In the given figure, you can see resultant circuitry. Applications motor: Conveyors, Elevators, Rolling Mills, Heavy Planners, Applications generator: lighting and heavy power supply, arc welding purpose, networks, offices, hotels, homes, schools, etc.
- 77. If the brush contact drop is included, the terminal voltage equation is written as: Terminal voltage is given as: The shunt field current is given as: Series field current is given as: Short Shunt Compound Wound Long Shunt Compound Wound If the brush contact drop is included, the terminal voltage equation is written as: Terminal voltage is given as: Series field current is given as: The shunt field current is given as:
- 78. EX-1: A long shunt compound DC generator delivers a load current of 50A at 500V and has armature, series field and shunt field resistances of 0.05Ω, 0.03Ω and 250Ω respectively. Calculate the generated voltage and the armature current. Allow 1V per brush for contact drop. EX-2: A 230V, 50kW short-shunt compound generator has an armature resistance of Ra = 0.06Ω, series winding resistance of 0.04Ω and shunt field winding of resistance 120Ω. Calculate the induced armature voltage,E (in Volts) at rated load and terminal voltage. The total brush contact drop is 2V?
- 79. EX-3: A compound DC generator has an armature resistance of 0.04Ω, shunt field resistance of 48Ω and series field resistance of 0.02Ω. The field excitation is adjusted to get a rated terminal voltage of 220V. Find out the internal generated voltage, E(in Volts), when the motor is supplying 4.4kW at rated voltage for long shunt connection? EX-4 :A short shunt compound DC generator delivers a load current of 30A at 220V and has armature, series field and shunt field resistances of 0.05Ω, 0.3Ω and 200Ω respectively. Calculate the induced emf and the armature current. Allow 1V per brush for contact drop.
- 80. Open Circuit Characteristic (O.C.C.) (E0/If) The above figure shows a typical no-load saturation curve or open circuit characteristics for all types of DC generators. Open circuit characteristic is also known as magnetic characteristic or no-load saturation characteristic. This characteristic shows the relation between generated emf at no load (E0) and the field current (If) at a given fixed speed. The O.C.C. curve is just the magnetization curve and it is practically similar for all type of generators. The data for O.C.C. curve is obtained by operating the generator at no load and keeping a constant speed. Field current is gradually increased and the corresponding terminal voltage is recorded. The connection arrangement to obtain O.C.C. curve is as shown in the figure below. For shunt or series excited generators, the field winding is disconnected from the machine and connected across an external supply
- 81. Now, from the emf equation of dc generator, we know that Eg = kɸ. Hence, the generated emf should be directly proportional to field flux (and hence, also directly proportional to the field current). However, even when the field current is zero, some amount of emf is generated (represented by OA in the figure below). This initially induced emf is due to the fact that there exists some residual magnetism in the field poles. Due to the residual magnetism, a small initial emf is induced in the armature. This initially induced emf aids the existing residual flux, and hence, increasing the overall field flux. This consequently increases the induced emf. Thus, O.C.C. follows a straight line. However, as the flux density increases, the poles get saturated and the ɸ becomes practically constant. Thus, even we increase the If further, ɸ remains constant and hence, Eg also remains constant. Hence, the O.C.C. curve looks like the B-H characteristic.
- 82. D.C Motor-principle-back emf WORKING PRINCIPLE OF A DC MOTOR When a current-carrying conductor is placed in a magnetic field, it experiences a torque and has a tendency to move. or In other words, when a magnetic field and an electric field interact, a mechanical force is produced. The DC motor or direct current motor works on that principle. This is known as motoring action. The DC motor is the device which converts the direct current into the mechanical work. It works on the principle of Lorentz Law, which states that “the current carrying conductor placed in a magnetic and electric field experience a force”. And that force is called the Lorentz force. The Fleming left-hand rule gives the direction of the force. Fleming Left Hand Rule If the thumb, middle finger and the index finger of the left hand are displaced from each other by an angle of 90°, the middle finger represents the direction of the magnetic field. The index finger represents the direction of the current, and the thumb shows the direction of forces acting on the conductor. The formula calculates the magnitude of the force,
- 83. where, P=number of poles of dc motor ,Φ= flux per pole , Z=total number of armature conductors N=armature speed ,A=number of parallel paths in armature winding As all other parameters are constant, therefore, Eb ∝ N As per Lenz's law, "the induced emf always opposes the cause of its production”. Here, the cause of generation of back emf is the rotation of armature. Rotation of armature is due to armature torque. Torque is due to armature current and armature current is due to supply dc voltage V. Therefore, the ultimate cause of production of Eb is the supply voltage V. Therefore, back emf is always directed opposite to supply voltage V. When a dc voltage V is applied across the motor terminals, the armature starts rotating due to the torque developed in it. As the armature rotates, armature conductors cut the pole magnetic field, therefore, as per law of electromagnetic induction, an emf called back emf is induced in them. The back emf (also called counter emf) is given by BACK EMF
- 84. Torque equation The turning or twisting force about an axis is called torque. Consider a wheel of radius R meters acted upon by a circumferential force F new tons as shown in the figure.
- 86. Types of DC Motor A Direct Current Motor, DC is named according to the connection of the field winding with the armature. Mainly there are two type of DC Motors. One is Separately Excited DC Motor and other is Self-excited DC Motor. The self-excited motors are further classified as Shunt wound or shunt motor, Series wound or series motor and Compound wound or compound motor. Shunt Wound Motor This is the most common types of DC Motor. Here the field winding is connected in parallel with the armature as shown in the figure below:
- 87. Series Wound Motor In the series motor, the field winding is connected in series with the armature winding. The connection diagram is shown below:
- 88. SPEED CONTROL OF DC MOTR According to the speed equation of a dc motor N ∞ Eb/φ, N = V- Ia Ra/ φ Thus speed can be controlled by-1.Flux control method: By Changing the flux by controlling the current through the field winding. 2.Armature control method: By Changing the armature resistance which in turn changes the voltage applied across the armature, The speed is directly proportional to the voltage applied across the armature . The speed of a d.c. motor is controlled (i) By varying the flux per pole (f) known as flux control method. (ii) By varying the Ra and is known as armature control method. (iii) By varying the applied voltage V and is known as voltage control method.
- 89. i) Field control method: 1. In this field control method the variable is flux (f) 2. The rheostat is placed in series to the field winding, as the field resistance increases the field current decreases and this weakens the flux 3. The weakening of the flux increases the speed since speed is inversely proportional to the flux. 4. Thus using the field control, above base speeds can be controlled. 5. This method is also known as constant power method or variable torque method. Advantages 1. This is an easy and convenient method. 2. It is an inexpensive method since very little power is wasted in the shunt field rheostat due to relatively small value of If 3. The speed control exercised by this method is independent of load on the machine. Disadvantages 1. Only speeds higher than the normal speed can be obtained. 2. There is a limit to the maximum speed obtainable by this method. It is because if the
- 90. ii) Armature control method 1. In this armature resistance control method the variable is Ra 2. The rheostat is placed in series to the armature winding, as the Ra increases the IaRa drop increases and this decreases the speed. 3. The decreasing of the back emf decreases the speed since speed is directly proportional to Eb. 4. Thus using the Ra control method, below base speeds can be controlled. 5. This method is also known as constant torque method or variable power method. Disadvantages 1. A large amount of power is wasted in the controller resistance since it carries full armature current Ia. 2. The speed varies widely with load since the speed depends upon the voltage drop in the controller resistance and hence on the armature current demanded by the load. 3. The output and efficiency of the motor are reduced. 4. This method results in poor speed regulation.
- 91. iii) Voltage control method by Ward-Leonard system 1. This method is used to get the wide range of speed control 10:1. 2. As the speed of the motor is directly proportional to the applied voltage to the armature, thus by applying the variable voltage the speed is controlled. 3. The armature of the shunt motor M (whose speed is to be controlled) is connected directly to a d.c. generator G driven by a constant-speed a.c. motor A. 4. The field of the shunt motor is supplied from a constant-voltage exciter E. 5. The field of the generator G is also supplied from the exciter E. 6. The voltage of the generator G can be varied by means of its field regulator. 7. By reversing the field current of generator G by controller FC, the voltage applied to the motor may be reversed. Advantages 1. The speed of the motor can be adjusted through a wide range without resistance losses which results in high efficiency. 2. The motor can be brought to a standstill quickly, simply by rapidly reducing the voltage of generator G. 3. The disadvantage of the method is that a special motor-generator set is required for each motor and the losses in this set are high if the motor is operating under light loads for long periods.
- 92. SWINBURNE’S TEST. This test is a no load test and hence cannot be performed on series motor. The circuit connection is shown in Figure The machine is run on no load at rated speed which is adjusted by the shunt field resistance. Advantages 1. Economical, because no load input power is sufficient to perform the test 2. Efficiency can be pre-determined 3. As it is a no load test, it cannot be done on a dc series motor Disadvantages 1. Change in iron loss from no load to full load is not taken into account. (Because of Armature reaction, flux is distorted which increases iron losses). 2. Stray load loss cannot be determined by this test and hence efficiency is over estimated. 3. Temperature rise of the machine cannot be determined. 4. The test does not indicate whether commutation would be satisfactory when the machine is loaded.
- 93. It is a simple indirect test which is applicable where flux' is constant like shunt machine. The machine is run as motor at no load at its rated speed with the help of shunt field resistance. The supply voltage, no load input current and field current are measured by ammeters. Then the no load armature current Ia0 = Io-Ish amps. Where Io = No load input current Ish = Shunt field current No load input power = V * Io watts Constant losses (WC) = input - Ar. cu. losses = V Io – (Iao)2Ra Where Ra = armature resistance
- 94. Efficiency of Machine as a motor: Let motor input Current = I amps Terminal voltage = V volts Input Power = VI watts Output power =Input - Losses =V I - (WC + (Ia)2Ra) watts Where Ia = I - Ish Amps % Efficiency = (output/input) 100 Efficiency of machine as a generator: Let output current supplied by the generator = I Amps Terminal voltage = V volts Output of the generator = V I watts Input = Output + Losses = V I + (Wc + [Ia]2Ra) watts, where Ia = I + Ish % Efficiency of generator = (output/input) 100
- 95. Efficiency of Machine as a motor: Let ‘I’ be the line current at which efficiency is to be predetermined. Input Power (Pi) = VI watts Output power (Po) =Input – Total Losses Total losses = PL = Wc + Ia 2 Ra Where Ia = I - Ish Amps Po =V I - (WC + (Ia)2Ra) watts % Efficiency = (output/input) 100 = (Po / Pi) x 100 Efficiency of machine as a generator: Let ‘I’ be the line current at which efficiency is to be predetermined Output of the generator (Po) = V I watts Total losses (PL) = Wc + Ia 2 Ra Where Ia = I + Ish Amps Input (Pi)= Output + Losses(PL) = V I + (Wc + [Ia]2Ra) watts, where Ia = I + Ish % Efficiency of generator = (output/input) 100 = (Po / Pi) x 100
- 96. UNIT – III: Introduction to 3-phase circuits, relation between line and phase quantities in balanced star and delta networks. A.C Machines: Single phase transformer: principle-emf equation- types-OC and SC tests- Voltage Regulation -Efficiency-Simple problems.
- 99. WHY WE STUDY 3 PHASE SYSTEM ? •ALL electric power system in the world used 3-phase system to GENERATE, TRANSMIT and DISTRIBUTE •One phase, two phase, or three phase i can be taken from three phase system rather than generated independently. • Instantaneous power is constant (not pulsating).– thus smoother rotation of electrical machines , High power motors prefer a steady torque • More economical than single phase – less wire for the same power transfer ,The amount of wire required for a three phase system is less than required for an equivalent single phase system.
- 104. relation between line and phase quantities in balanced star and delta networks. Star Connection in a 3 Phase System •In the Star Connection, the similar ends (either start or finish) of the three windings are connected to a common point called star or neutral point. The three-line conductors run from the remaining three free terminals called line conductors. •The wires are carried to the external circuit, giving three-phase, three-wire star Connected systems. However, sometimes a fourth wire is carried from the star point to the external circuit, called neutral wire, forming three-phase, four-wire star connected systems. •Considering the above figure, the finish terminals a2, b2, and c2 of the three windings are connected to form a star or neutral point. The three conductors named as R, Y and B run from the remaining three free terminals as shown in the above figure. •The current flowing through each phase is called Phase current Iph, and the current flowing through each line conductor is called Line Current IL. Similarly, the voltage across each phase is called Phase Voltage Eph, and the voltage across two line conductors is known as the Line Voltage EL.
- 105. As the system is balanced, a balanced system means that in all the three phases, i.e., R, Y and B, the equal amount of current flows through them. Therefore, the three voltages ENR, ENY and ENB are equal in magnitude but displaced from one another by 120° electrical. The Phasor Diagram of Star Connection is shown below:
- 106. Hence, in a 3 Phase system of star connections, the line current is equal to phase current.
- 107. Example: A balanced star connected load of (8+j6)Ω per phase is connected to a balanced 3-phase 400 V supply. Find the line current, power factor, power and total volt-amperes.
- 108. Delta Connection In a 3 Phase System In Delta (Δ) or Mesh connection, the finished terminal of one winding is connected to start terminal of the other phase and so on which gives a closed circuit. The three-line conductors are run from the three junctions of the mesh called Line Conductors. •To obtain the delta connections, a2 is connected with b1, b2 is connected with c1 and c2 is connected with a1 as shown in the above figure. •The three conductors R, Y and B are running from the three junctions known as Line Conductors. •The current flowing through each phase is called Phase Current (Iph), and the current flowing through each line conductor is called Line Current (IL). •The voltage across each phase is called Phase Voltage (Eph), and the voltage across two line conductors is called Line Voltage (EL).
- 109. Relation Between Phase Voltage and Line Voltage in Delta Connection
- 110. The phasor diagram is shown below: As in the balanced system the three- phase current I12, I23 and I31 are equal in magnitude but are displaced from one another by 120° electrical.
- 112. Definition of Transformer Electrical transformer is a static device which transforms electrical power from one circuit to another without any direct electrical connection and without changing frequency of power but maybe in different voltage levels with the help of mutual induction. Working Principle of Transformer The working principle of transformer is very simple. Mutual induction between two or more windings is responsible for transformation action in an electrical transformer Basic Theory of Transformer The alternating current through the winding produces a continually changing and alternating flux that surrounds the winding. If any other winding is brought nearer to the previous one, obviously some portion of this flux will link with the second. As this flux is continually changing in its amplitude and direction, there must be a changing flux linkage in the second winding or coil. According to Faraday's law of electromagnetic induction, there must be an EMF induced in the second. If the circuit of the later winding is closed, there must be a current flowing through it. This is the most basic thing on which the working principle of transformer stands. The winding which takes electrical power from the source, is known as the primary winding. The winding which gives the desired output voltage due to mutual induction is commonly known as the secondary winding.
- 113. The rate of change of flux linkage depends upon the amount of linked flux with the second winding. So, almost all flux of primary winding should link to the secondary winding. This is effectively and efficiently done by placing one low reluctance path common to both of the winding. The form mentioned above of a transformer is theoretically possible but not practically, because in open air very tiny portion of the flux of the first winding will link with second; so the current that flows through the closed circuit of later, will be so small in amount that it will be difficult to measure.
- 114. Where: VP – is the Primary Voltage VS – is the Secondary Voltage NP – is the Number of Primary Windings NS – is the Number of Secondary Windings Φ (phi) – is the Flux Linkage Principle of Single Phase Transformer The single-phase transformer works on the principle of Faraday’s Law of Electromagnetic Induction. Typically, mutual induction between primary and secondary windings is responsible for the transformer operation in an electrical transformer.
- 115. Working of Single Phase Transformer A transformer is a static device that transfers electric power in one circuit to another circuit of the same frequency. It consists of primary and secondary windings. This transformer operates on the principle of mutual inductance. When the primary of a transformer is connected to an AC supply, the current flows in the coil and the magnetic field build-up. This condition is known as mutual inductance and the flow of current is as per the Faraday’s Law of electromagnetic induction. As the current increases from zero to its maximum value, the magnetic field strengthens and is given by dɸ/dt. This electromagnet forms the magnetic lines of force and expands outward from the coil forming a path of magnetic flux. The turns of both windings get linked by this magnetic flux. The strength of a magnetic field generated in the core depends on the number of turns in the winding and the amount of current. The magnetic flux and current are directly proportional to each other. As the magnetic lines of flux flow around the core, it passes through the secondary winding, inducing voltage across it. The Faraday’s Law is used to determine the voltage induced across the secondary coil and it is given by: N. dɸ/dt
- 116. Core-type Transformer In this type of construction, only half of the windings are wound cylindrically around each leg of a transformer to enhance magnetic coupling as shown in the figure below. This type of construction ensures that magnetic lines of force flow across both the windings simultaneously. The main disadvantage of the core- type transformer is the leakage flux that occurs due to the flow of a small proportion of magnetic lines of force outside the core. Based on Voltage Levels Commonly used transformer type, depending upon voltage they are classified as: Step-up Transformer: They are used between the power generator and the power grid. The secondary output voltage is higher than the input voltage. Step down Transformer: These transformers are used to convert high voltage primary supply to low voltage secondary output.
- 117. Shell-type Transformer In this type of transformer construction, the primary and secondary windings are positioned cylindrically on the center limb resulting in twice the cross-sectional area than the outer limbs. There are two closed magnetic paths in this type of construction and the outer limb has the magnetic flux ɸ/2 flowing. Shell type transformer overcomes leakage flux, reduces core losses and increases efficiency. Applications To step-down long-distance signals to support both residential and light-commercial electronic devices In television sets for voltage regulation To step-up power in home inverters To supply power to non-urban areas To isolate two circuits electrically as primary and secondary are placed far from each other
- 118. As the magnetic flux varies sinusoidally, Φ = Φmax sinωt, then the basic relationship between induced emf, ( E ) in a coil winding of N turns is given by: emf = turns x rate of change Where: ƒ – is the flux frequency in Hertz, = ω/2π Ν – is the number of coil windings. Φ – is the amount of flux in webers This is known as the Transformer EMF Equation. For the primary winding emf, N will be the number of primary turns, ( NP ) and for the secondary winding emf, N will be the number of secondary turns, ( NS ). Transformer EMF Equation
- 119. EMF Equation of a Transformer When a sinusoidal voltage is applied to the primary winding of a transformer, alternating flux ϕm sets up in the iron core of the transformer. This sinusoidal flux links with both primary and secondary winding. The function of flux is a sine function. The rate of change of flux with respect to time is derived mathematically. Let ϕm be the maximum value of flux in Weber f be the supply frequency in Hz N1 is the number of turns in the primary winding N2 is the number of turns in the secondary winding Φ is the flux per turn in Weber As shown in the above figure that the flux changes from + ϕm to – ϕm in half a cycle of 1/2f seconds. By Faraday’s Law,Let E1 be the emf induced in the primary winding Where Ψ = N1ϕ Since ϕ is due to AC supply ϕ = ϕm Sinwt
- 120. So the induced emf lags flux by 90 degrees. Maximum valve of emf But w = 2πf Root mean square RMS value is Putting the value of E1max in equation (6) we get Putting the value of π = 3.14 in the equation (7) we will get the value of E1 as Similarly Now, equating the equation (8) and (9) we get The above equation is called the turn ratio where K is known as the transformation ratio.
- 121. 1. A transformer has 600 turns of the primary winding and 20 turns of the secondary winding. Determine the secondary voltage if the secondary circuit is open and the primary voltage is 140 V. 2. A transformer has a primary coil with 1600 loops and a secondary coil with 1000 loops. If the current in the primary coil is 6 Ampere, then what is the current in the secondary coil.
- 128. Copper losses : This loss occur in both the primary and secondary windings due to their ohmic resistance. If I1, I2 are the primary and secondary currents and R1, R2 are the primary and secondary resistances, respectively.
- 129. Core or iron losses: When AC supply is given to the primary winding of a transformer an alternating flux is set up in the core, therefore, hysteresis and eddy current losses occur in the magnetic core. Hysteresis and : When the magnetic material is subjected to reversal of magnetic flux, it causes a continuous reversal of molecular magnets. This loss can be minimized by using silicon steel material for the construction of core. Eddy current losses :The emf in the core and circulates eddy currents. This power is dissipated in the form of heat . This loss can be minimized by making the core of thin laminations.
- 130. Voltage regulation: When a transformer is loaded, with a constant supply voltage, the terminal voltage changes due to voltage drop in the internal parameters of the transformer i.e., primary and secondary resistances and inductive reactance. The voltage drop at the terminals also depends upon the load and its power factor. The change in terminal voltage from no-load to full-load at constant supply voltage with respect to no-load voltage is known as voltage regulation of the transformer. Let, E2 = Secondary terminal voltage at no-load. V2 = Secondary terminal voltage at full-load. Then, voltage regulation =𝐸2− 𝑉2 /𝐸2 (𝑝𝑒𝑟 𝑢𝑛𝑖𝑡) In the form of percentage, % Reg = 𝐸2− 𝑉2 /𝐸2 × 100
- 131. Explain OC and SC tests on a single phase transformer Purpose of conducting OC and SC tests is to find i) Equivalent circuit parameters ii) Efficiency iii) Regulation Open Circuit Test: 1. The OC test is performed on LV side at rated voltage and HV side is kept opened. 2. As the test is conducted on LV side the meters selected will be at low range values like smaller voltmeter, smaller ammeter and low pf wattmeter 3.As the no-load current is quite small about 2 to 5% of the rated current, the ammeter required here will be smaller range even after on LV side which are designed for higher current values. 4. The voltmeter, ammeter and the wattmeter readings V0, I0and W0 respectively are noted by applying rated voltage on LV side. 5.The wattmeter will record the core loss because of no load input power.
- 133. Short Circuit Test This test is carried out to determine the (i) Copper losses at full load (or at any desired load). These losses are required for the calculations of efficiency of the transformer. (ii) Equivalent impedance (Zes or Zep), resistance (Res or Rep) and leakage reactance (Xes or Xep) of the transformer. This test is usually carried out on the high-voltage side of the transformer i.e., a wattmeter W, voltmeter V and an ammeter A are connected in high-voltage winding . The other winding is then short circuited by a thick strip. A low voltage at normal frequency is applied to the high voltage winding with the help of on autotransformer so that full-load current flows in both the windings.
- 134. Short Circuit Test: 1. The SC test is performed on HV side at rated current and LV side is kept Shorted. 2. As the test is conducted on HV side the meters selected will be at low range values like smaller voltmeter, smaller ammeter and unity pf wattmeter 3. As the voltage required to circulate the short circuit rated current is very small about 10 to 15% of the rated HV voltage, so the voltmeter required here will be smaller range even the test is conducted on HV side. 4. The voltmeter, ammeter and the wattmeter readings Vsc, Isc and Wsc respectively are noted by passing rated current on HV side. 5.The wattmeter will record the copper loss corresponding to the Isc.
- 137. Variation of voltage regulation and efficiency with respect to load and load power factors Why Transformers Are Rated In KVA? From the above transformer tests, it can be seen that Cu loss of a transformer depends on current, and iron loss depends on voltage. Thus, total transformer loss depends on volt- ampere (VA). It does not depend on the phase angle between voltage and current, i.e. transformer loss is independent of load power factor. This is the reason that transformers are rated in kVA.
- 138. Ex1:The core loss of the single phase 2200/220‐V, 50‐Hz transformer is measured using the low‐ voltage side as the primary. The following data are obtained: Voltmeter reading = 240 V Exciting current = 1.0 A Wattmeter reading = 24 W. Calculate: a) The core loss, in watts. b) The power factor and phase angle. c) the Core loss current. d) The lagging (magnetizing) component of the current.