Unit 1 (part A).pdf

Fundamentals of Electrical Engineering (ELE_1072) Department of Electrical and Electronics L00- Introduction Course Code: [ELE 1071] 1 ...

Fundamentals of Electrical Engineering (ELE_1072) Department of Electrical and Electronics L00- Introduction Course Code: [ELE 1071] 1  Objective of the Lecture Vision, Mission & Goal Syllabus Course Plan Course Outcomes(COs) Assessment Plan Lesson Plan Text and References books Overview of Power Scenario in India Voltage, Current, Resistance, Ohm’s Law, Power and Energy Electrical Engineering and Network Elements Problems on Resistance Series and Parallel, Current and voltage division Source transformation Star delta Transformation  Vision Excellence in Emerging Technical Education through Research, Innovation, and Teamwork Mission Educate students professionally to face social challenges by providing a healthy learning environment grounded well in emerging technologies, value-based education , creativity, and nurturing teamwork. Goal Our goal is to be a world class technical institution fostering innovation, leadership and entrepreneurial spirit.  COURSE STRUCTURE TOTAL UNITS – 5 1. DC circuit analysis 2. AC circuit analysis 3. Magnetic circuits 4. Power systems 5. Power electronics TOTAL LECTURE HOURS – 36 (3 LECTURE PER WEEK) Course Code: [ELE 1071] 4 Course outline DC circuit AC circuit Magnetic Power Power analysis analysis circuits systems electronics Circuit elements Generation Laws of Generation- Power Network AC through R, L magnetism transmission- semiconductor reduction and C Magnetic circuits distribution devices Mesh current Power & power Electromagnetic Utilization of Power converter analysis factor induction electric classification Node voltage Resonance Coupled circuits power Power stages in a analysis Three-phase star Transformers Costing of laptop/ desktop Network loads electricity Switched-mode DC and AC theorems Three-phase delta motors Power connection power supply Transients in DC loads for critical loads UPS Special machines circuits Analysis of Electrical safety Electrical vehicles balanced and unbalanced three phase loads 5  Course outcomes Course Outcome Analyse DC Circuit CO1 Analyze single-phase and three-phase AC CO2 circuits Analyze magnetic circuits and explain working CO3 of various types of electrical machines Describe electrical power system components CO4 Illustrate applications of power electronics CO5 6 Course outcomes At the end of this course, Contact Bloom’s CO Marks the student should be able to Hours Level (BL) CO1 Analyze DC circuit 9 25 2, 3, 4 CO2 Analyze single-phase and three-phase AC circuit 10 35 2, 3, 4 Analyze magnetic circuits and explain working of CO3 6 14 2, 3, 4 various types of electrical machines CO4 Describe electrical power system components 5 12 2 CO5 Illustrate applications of power electronics 6 14 2 Total 36 100 7  Assessment Continuous evaluation (Quizzes/Numericals/ Assignments) In-Semester Mid semester exam Assessment End-Semester Written exam Note: Detailed course plan & assessment plan will be shared with you in due course of time 8 Assessment Brief Overview Component Marks Details In-Semester (Midterm) Exam – Approximately 30 Marks Continuous Assessment on weekly basis through Internal 50 Learner Management System (LMS – Lighthouse) – Approximately 20 Marks Quiz / Assignment (If required) External 50 End Semester Examination 9  ASSESSMENT RUBRICS In-Sem assessment – 50 M ( >18 M pass) End-Sem assessment – 50 M ( >18 M pass) Exam Topic covered Weightage Marks Time Mid-Term Test Unit 1,2,3. Objective and Descriptive 30 -- min Quiz unit 1,2 20 MCQs × ½ = 10 10 30 Mins Assignment Unit 1,2,3,4,5 Questions (Open Book) 10 End Sem Exam Unit 1,2,3,4,5 All question Compulsory 50 3 hours ATTENDANCE > 75 % 10  TEXT BOOKS 1. William H. Hayt, Jack E. Kemmerly & Steven M. Durbin, Engineering Circuit Analysis (8th ed), McGraw-Hill, 2012. 2. Nagrath I.J. and D. P. Kothari; Basic Electrical Engineering (4Th Ed); Tata McGraw Hill (2019). 3. Nagasarkar T. K. & Sukhija M. S., Basic Electrical Engineering (3rd ed), OUP; 2017 4. Daniel W. Hart; Power Electronics; McGraw Hill (2011). 5. Debapriya Das; Fundamentals of Electrical Engineering; NPTEL Course – IIT Kharagpur (2018). Reference Books 1. Hughes E., Electrical and Electronic Technology (9e), Pearson Education, 2008 2. D. C. Kulshreshtha, Basic Electrical Engineering, McGraw Hill, 2012. 3. S. N. Singh, Basic Electrical Engineering, PHI, 2011. 11  KEY ACADEMIC TIMELINE You Need A Scientific Calculator Click Here for App Link Physical calc. Click Here for App Link Click Here for App Link 13 Scientific Calculator ▪ Should be non – programmable ▪ Should be non - graphical ▪ Suggestions: o Casio fx-991ES plus (2nd edition) o Casio fx-991MS (2nd edition) o Casio fx-991ES plus o Casio fx-991MS ▪ Android/Apple OS based apps available ▪ Windows app (trial for 3 months) available from Casio Casio fx-991ES plus (2nd edition) 14 Quick Recap Electrostatic Electric Charges Moving Charges Electromagnetic Alternating Potential & Electric Current and Fields & Magnetism Induction Current Capacitance 15 Quiz Time 16 Quiz 1 The electrical installations at our home are connected in _______ A) Series B) Parallel Ans B) Parallel 17  Quiz 2 In the circuit shown, the equivalent resistance of the network shown is _______ A) R/5 Ω B) 5R Ω C) 6R/5 Ω D) 2R Ω Ans B) 5R Ω 18  Quiz 3 The equivalent resistance of the network shown is _______ A) 16/3 Ω B) 10/3 Ω C) 15/3 Ω D) 15 Ω Ans A) 16/3 Ω 19 Reference Source of pictures in Slides 14, 15 & 16: http://ncertbooks.prashanthellina.com/class_12.Physics.PhysicsPartI/Chapter%2 03.pdf 20 Electricity 21 Classification Of Materials Based On Valance Electrons Based on electrical conductivity materials are classified into Conductors Semi-conductors Insulators Conductors When the no. of valance electrons of an atom are less than four, then the material is called as a metal or a conductor No. of valance electrons < 4 Eg : Copper, Silver & aluminum  Insulators When the no. of valance electrons of an atom is more than 4, then the material is usually a non- metal and an insulator No. of valance electrons > 4 Eg : Organ, Neon & Mica. Semi-Conductors When the no. of valance electrons of an atom are 4, then the material is called as a semi- conductor. It has both the metal and non-metal properties No. of valance electrons = 4 Eg : Germanium ,Silicon and carbon.  Types Of Charge Negative charge: Excess of electrons gives negative charge to a material Positive charge Deficit of electrons gives positive charge to a material  History of Electricity Evolution Of 300 years of research history and development, Electricity turned from scientific interest into daily necessity. 1. Ancient Egyptians- Electric Ray fishes 2. Greek Philosopher THALES (600 BCE) discovered static electricity (AMBER) 3. English Scientist William Gilbert (1600) studied multiple experiments on static electricity and worked on Electro-Magnetism and invented Electroscope. Course Code: [ELE 1071] Dept. of Electrical & Electronics Engg., MIT - Manipal 27 4. American scientist Benjamin Franklin (1752) studied on electricity through lightening and discovered battery. Course Code: [ELE 1071] Dept. of Electrical & Electronics Engg., MIT - Manipal 28 5. Italian physicist Luigi Galvin (1771) studied on animal electricity: He discovered that the muscles of dead frogs' legs twitched when struck by an electrical spark. Course Code: [ELE 1071] Dept. of Electrical & Electronics Engg., MIT - Manipal 29 6. Voltas (1775) studied on metal electricity: He discovered battery (Zinc copper and salt water) and produced DC Electricity. Volta proved that electricity could be generated chemically and debunked the prevalent theory that electricity was generated solely by living beings. Course Code: [ELE 1071] Dept. of Electrical & Electronics Engg., MIT - Manipal 30 7. Orrested (1820) observed that electric current carrying conductor produced magnetic field around it. 8. Micheal Faraday (1821) invented Electromagnetism, electric motor dc motor and dynamo. 9. George Ohm (1872) analysed electrical circuits. 10. James Clark Maxewell studied on Electricity + Magnetism + Light. 11. American Scientist Thomas Alva Edison (1879): Invented BULB : Electricity production company using DC generators : Electricity would transmit till 2-3 km : So in every 2-3 km house production house has to be installed. Course Code: [ELE 1071] Dept. of Electrical & Electronics Engg., MIT - Manipal 31  12. Nikola Tesla (1887) invented AC current and AC motor. 13. Albert Einstein (1905) published on photo electric effect. Electronics began, with the invention of first working transistor in 1947. A new era of electricity began i.e. Era of electronics Course Code: [ELE 1071] Dept. of Electrical & Electronics Engg., MIT - Manipal 32  Applications of Electricity 1. Lighting, heating, cooling and other domestic electrical appliances used in home. 2. Street lighting, flood lighting of sporting arena, office building lighting, powering PCs etc. 3. Irrigating vast agricultural lands using pumps and operating cold storages for various agricultural products. 4. Running motors, furnaces of various kinds, in industries. 5. Running locomotives (electric trains) of railways. The point is, without electricity, modern day life will simply come to a stop. In fact, the advancement of a country is measured by the index per capita consumption of electricity – the more it is , more advanced the country is.  Electric Power System Branch of Electrical Engineering deals with the Technology of Generation, Transmission and Distribution  Power System Components Generation subsystem Transmission subsystem Sub-transmission subsystem Distribution subsystem Protection & Control subsystem Power System Structure Typical Voltage Levels in a Power System GENERATION 3 Phase, 50 Hz 11 kV to 22 kV TO OTHER TO OTHER POOL MEMBER POOL MEMBER POOL EHV AC: 765 kV, 400 kV TRANSMISSION HV AC: 220 kV HVDC:  NETWORK 500 kV Very large Customers 132 kV, 110 kV, SUB TRANSMISSION 66 kV, 33 kV NETWORK Large Customers Primary 11 kV DISTRIBUTION 415 V NETWORK Se condary Me dium Cu Customers  Transmission & Distribution Voltage Levels Transmission Networks – EHV AC or HVDC Operates @ 765 kV/ 400 kV/ 220 kV AC or  500 kV DC AC Sub-Transmission Networks Operates @ 132 kV/ 110 kV/ 66 kV/ 33 KV AC Distribution Network Primary side: 11 kV Secondary side: 415 V, 4 wire  Transmission of power The huge amount of power generated in a power station (hundreds of MW) is to be transported over a long distance (hundreds of kilometers) to load centers to cater power to consumers with the help of transmission line and transmission towers as shown in figure. Figure : Transmission tower.  Power In this section we briefly outline the basics of the three most widely found generating stations : - Thermal plants - Hydel plants - Nuclear plants in our country and elsewhere. Other sources of power generation currently used are : - Wind Power - Solar Power  Thermal plant Figure : Basic components of a thermal generating unit.  Thermal power is the "largest" source of power in India. 115 Thermal Projects are operated at several sites by the National Thermal Power Corporation Limited (NTPC) of India  Hydel plant Figure : Basic components of a hydel generating unit. 47 Hydro Projects are operated at several sites by the National Hydroelectric Power Corporation (NHPC) of India  Nuclear plant Figure : Nuclear power generation. 20 Nuclear Power Reactors are operated at seven sites by the Nuclear Power Corporation of India  Wind Power India has the world's fifth largest wind power industry Solar Power  Indian Power Scenario Total approximate generation capacity and contribution by thermal, hydel and nuclear generation in our country are given below. Method of generation in MW % contribution Thermal 77 340 69.4 Hydel 29 800 26.74 Nuclear 2 720 3.85 Total generation 1 11 440 -  Electric Potential Generally, all atoms are electrically neutral. But due to some external Force, some atoms may gain or loose electrons & become excited atoms The excited atoms will always try to neutralize their charges.  Thus, the excited atoms will create an electric force to move the electrons from one place to other in a conducting material It means ions will create an electrical pressure  Electric Potential The ability of a charged body to do the work is called electric potential Electric potential is expressed in Volts or J/C  Potential Difference The difference between the electric potentials of two charged bodies is called as potential difference Potential difference = A – B = 10 - 4 = 6 V. Due to this potential difference between two charged bodies, there is a continuous flow of current in the electric circuit The unit of potential difference is Volt indicated by the letter ‘v’  Electro-Motive Force (EMF) The electric pressure which sets the electrons in motion is called as EMF (OR) The Electrical Force which is required to over come the resistance of a conductor is called the EMF Cells, Batteries, Generators develop an EMF  EMF is the primary cause of the flow of electrons Potential difference is produced by the EMF EMF is the Cause whereas Potential difference is the Effect of EMF The unit of EMF is Volt  Voltage The EMF of an electric source (or ) the potential difference across a resistor is called the voltage (OR) Energy (work done) / unit charge is called as Voltage Voltage is the work done per unit charge V=W/Q Voltage is represented by letter V The unit of Voltage is Volt Voltage = Work done / Charge J/C One Volt : A ∞ x +1C “ Electric Potential at a point is 1 Volt if 1 Joule of work is done in bringing a unit +ve charge of 1 Coulomb from infinity to that point” 1 Volt = 1 J/C  Basic Concept for Current There are free flow of electrons available in all semi-conducting & Conducting materials These free electrons move randomly in all directions within the substance in the absence of external force (or) Voltage as in figure(1)  If a certain amount of voltage is applied across the material, all the free electrons move in one direction depending on the polarity of the applied voltage as Show in fig(2) Flow of electrons + - V Electric Current Flow of free electrons in a closed circuit is in one direction is called electric current It is denoted by letter I Current is measured in Amperes  Direction of Electric Current Green arrow head indicates the direction of flow of Electrons. (i.e., from +ve to -ve terminal of the battery) Red arrows indicates the direction of conventional current. (i.e., from -ve to +ve terminal of the battery) Direction of Electrons is in opposite to the direction of Conventional Current In circuits, we always represent the current direction In the - conventional form + V Fig(3)  The unit of current is Ampere (Amp.) I = charge / Time = Q / T C /sec Q = IT Current is the Rate of flow of charge. i = dq / dt  Ampere If one Coulomb of charge flows per one second, it is called 1 Ampere 1 Ampere = 1 Coulomb / 1 second. NB : 1 Coulomb = charge of 6.25 x 1018 electrons Energy The capacity to do work is called energy Energy = Rate of doing work x Time (in hours) = Power x Time = VI x t : watt-hr = (Joule / sec.) x 3600 sec = 3600 Joules The unit of energy is watt – hour or Joules Electrical energy is measured in KWH 1 KWH = 1000 Watt – hour = 1000 x 3600 Joules = 36 x 105 Joules Power The work done per unit time is called Power (or) The rate of doing work is called power Power is measured in watts or Joule/sec Power = Work done / Time P = VIt / t P = VI  Other formulae of power P = VI P = I2 R (since V=IR) P = V2 / R (since I=V/R) Watt : The unit of power is watt (i.e., The work is done at a rate of 1 joule/sec) 1 Watt = 1 Joule/Sec The higher unit of power are Kilowatts (KW) or Megawatts (MW) 1 KW = 103 Watts 1 MW = 106 Watts Ohm’s Law Temperature remaining Constant, “ The Current flowing in a circuit is directly proportional to the Potential difference driving it ” i.e, I = k x V where k is constant of proportionality and is equal to V/I Therefore V/ I = Constant = R (or) I=V/R  Conductor R I V=IR V Where, Fig(4) V = Voltage across the conductor (Volts) I = Current flowing through the conductor (Amp) R = Resistance of the conductor (Ohm) Resistance In Ohm’s law, the constant ‘R’ is called Resistance Resistance : Resistance is the property of a material which opposes the flow of current through it Resistance is represented by a letter ‘R’ It is measured in Ohms (Ω) It is denoted by a symbol How to Remember Ohm’s Law Formulae V V V I R I R I R V = IR I = V/R R = V/I  Limitations of Ohm’s Law Ohm’s Law cannot be applied to the Non-linear devices such as → Voltage regulators → Vacuum Devices → Gas filled Devices → Thermion Devices → Semi-Conductor Devices It cannot be applied to Arcing Devices  Ohm’s law cannot be applied to those electrolytes in which gases are liberated on the electrodes It cannot be applied to those conductors whose temperature changes due to the flow of current through them. Symbols:  Definition of Current(I)- Unit Ampere(A) Definition of Voltage(V)- Unit Volts(V) Definition of Power(P) – Watt(W) Definition of Energy(E) – Watt-hour(Wh or Kwh) Ohm’s Law V = IR I = V/R R = V/I V V V I R I R I R 74  Basic Circuit Elements Active Elements Passive Elements Sources 1. Resistors 2. Inductors 3. Capacitors Dependent Independent sources sources Voltage sources Current sources Ideal (DC) Practical (DC) Ideal (DC) Practical (DC) Voltage source Voltage source Current source Current source + - +- Vs Vs Rs Is Rs Is Vs B Tech. 1st Year Basic Electrical Technology  Active Elements - Sources VL Ideal Voltage Source (DC) VL = VS Vs Voltage Source: +- Vs ▪Ideal: IL oMaintains constant voltage irrespective of connected load oInternal resistance 𝑹𝒔 = 𝟎 Practical Voltage Source VL ▪Practical: Vs Rs oTerminal voltage changes based on the +- connected load VL = VS – IL RS oInternal resistance 𝑹𝒔 ≠ 𝟎 IL Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 76  Active Elements - Sources Ideal Current Source (DC) IL IL = IS Current Source: ▪Ideal: Is oMaintains constant current irrespective of the load connected VL oInternal resistance 𝑹𝒔 = ∞ ▪Practical: Practical Current Source IL oOutput current changes based on the connected load IL = IS – VL /RS Is Rs oInternal resistance 𝑹𝒔 < ∞ VL Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 77  Basic Circuit Elements Active Elements Passive Elements Sources 1. Resistors 2. Inductors 3. Capacitors Dependent Independent sources sources B Tech. 1st Year Basic Electrical Technology  Passive Elements Resistance (R): Property of opposition to flow of current The voltage across the resistor is proportional to the current flowing through it ; VR α I or VR = IR, ‘R’ = VR/I; Unit - Ohm Inductance (L): Property of opposition to the rate of change of current The voltage induced in the inductor is proportional to the rate of change of current flowing through it; eL = L (di/dt) = N (dФ/dt); L = N (dФ/di); Unit – Henry (H) Capacitance (C): Property which opposes the rate of change of voltage; The capacitive current is proportional to the rate of change of voltage across it ; ic = C (dv/dt); Unit – Farad (F). B Tech. 1st Year Basic Electrical Technology  Passive Element : The element which receives energy (or absorbs energy) and then either convert it into heat (R) or can be stored it in an electric (C) or magnetic (L) filed. Active Element : The elements that supply energy to the circuit. Example: Voltage & Current sources Transistor (can amplify power of a signal) Note: Transformer is an example of passive element. (Because it does not amplify the power level and power remains same both in primary and secondary sides) Bilateral Element : Conduction of current in both directions in an element with same magnitude. Example: R, C & L Unilateral Element : Conduction of current in one directions. Example: Diode, Transistor  Lumped Elements : Lumped Elements are those elements which are very small in size in which simultaneous action takes place for any given cause at the same instant of time. Typical lumped elements are capacitor, resistor, inductor and transformer Generally, the elements are considered as lumped when their size is very small compared to the wavelength of the applied signal. Distributed Element : Which are not electrically separable for analytical purposes Example: Transmission line Resistor ▪Passive electric device that dissipates energy ▪Resistance: Property which opposes flow of current oSymbol: R oUnit: Ohms (Ω) oPower Consumed = 𝑰𝟐 𝑹 ▪Conductance oReciprocal of resistance oSymbol: G oUnit – Siemens (S) Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 82  Resistors Series connection of Resistors Parallel connection of Resistors I1 R 1 R1 R2 R3 V1 V2 V3 I2 R 2 I I V I3 R 3 I I V Current (I) in the all the resistors remains same Voltage (V) is same 𝑉 = 𝑉1 + 𝑉2 + 𝑉3 𝐼 = 𝐼1 + 𝐼2 + 𝐼3 1 1 1 1 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 𝑅 =𝑅 +𝑅 +𝑅 𝑒𝑞 1 2 3 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 83 Inductor Energy Storing Element Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 85  Inductor ▪ Passive electric device that stores energy in its magnetic field when current flows through it ▪ A coil of wire wound on a core o Eg.: Air core Inductor, iron core inductor ▪ Inductance: property which opposes rate of change of current o Symbol: L o Unit: Henry (H) ▪ The voltage across inductor is proportional to the rate of change of current through it 𝒅𝒊 𝒗𝑳 = 𝑳 𝒅𝒕 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 86  Inductive Circuit For a coil uniformly wound on a non-magnetic core of uniform cross section, self-inductance is given by where, 𝜇0 𝐴𝑁 2 𝐿= 𝑙 = length of the magnetic circuit in meters 𝑙 𝐴 = cross sectional area in square meters 𝜇𝑜 = Permeability of free space = 4π × 10−7 H/m 𝑁 = 𝑁𝑜. 𝑜𝑓 𝑡𝑢𝑟𝑛𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑖𝑙 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 87 Equivalent Inductance Inductors in series L 1 L 2 L 3 𝑳𝒆𝒒 = 𝑳𝟏 + 𝑳𝟐 + … … + 𝑳𝒏 L1 L2 Inductors in Parallel L3 𝟏 𝟏 𝟏 𝟏 = + + …….+ 𝑳𝒆𝒒 𝑳𝟏 𝑳𝟐 𝑳𝒏 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 88 Energy Stored in an Inductor ▪ Instantaneous power, 𝒅𝒊 𝒑 = 𝒗𝑳. 𝒊 = 𝑳 𝒊 𝒅𝒕 ▪ Energy absorbed in ‘𝒅𝒕’ time is 𝒅𝒘 = 𝑳 𝒊 𝒅𝒊 ▪ Energy absorbed by the magnetic field when current increases from 𝟎 to 𝑰 amperes, is 𝑰 𝟏 𝑾 = ‫𝟐𝑰 𝑳 = 𝒊𝒅 𝒊 𝑳 𝟎׬‬ 𝟐 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 89 Capacitor Energy Storing Element Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 90  Capacitors ▪Passive electric device that stores energy in the electric field between a pair of closely spaced conductors ▪Capacitance: Property which opposes the rate of change of voltage oSymbol: C oUnit: Farad (F) ▪The capacitive current is proportional to the rate of change of voltage across it 𝒅𝒗𝒄 𝒊𝒄 = 𝑪 𝒅𝒕 ▪Charge stored in a capacitor whose plates are maintained at constant voltage: 𝑸 = 𝑪𝑽 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 91 Terminologies ▪ Electric field strength, 𝑽 𝑬 = 𝒗𝒐𝒍𝒕𝒔/𝒎 𝒅 ▪ Electric flux density, 𝑸 𝑫 = 𝑪/𝒎𝟐 𝑨 ▪ Permittivity of free space, 𝜺𝟎 = 𝟖. 𝟖𝟓𝟒 × 𝟏𝟎−𝟏𝟐 𝑭/𝒎 ▪ Relative permittivity, 𝜺𝒓 ▪ Capacitance of parallel plate capacitor 𝜺𝟎 𝜺𝒓 𝑨 𝑪= 𝒅 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 92 Equivalent Capacitance Capacitors in Series 𝟏 𝟏 𝟏 𝟏 = + + ……+ 𝑪𝒆𝒒 𝑪𝟏 𝑪𝟐 𝑪𝒏 Capacitors in Parallel 𝑪𝒆𝒒 = 𝑪𝟏 + 𝑪𝟐 + ….. + 𝑪𝒏 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 93 Energy stored in a Capacitor ▪ Instantaneous power 𝒅𝒗𝒄 𝒑 = 𝒗𝒄 × 𝒊 = 𝑪 𝒗𝒄 𝒅𝒕 ▪ Energy supplied during ‘𝒅𝒕’ time is: 𝒅𝒘 = 𝑪 𝒗𝒄 𝒅𝒗𝒄 ▪ Energy stored in the electric field when potential rises from 𝟎 to 𝑽 volts is, 𝑽 𝟏 𝑾 = ‫𝐬𝐞𝐥𝐮𝐨𝐉 𝟐𝑽𝑪 = 𝒄𝒗𝒅 𝒄𝒗 𝑪 𝟎׬‬ 𝟐 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 94  Quick Quiz 2 1. Reduce the network to its equivalent resistance between terminals A and B  Quick Quiz 2 2. Find the equivalent resistance RT for the figures given below. (R1 = R2 = R3 = R4 = R5 = R6 =10Ω)  Quick Quiz 2 3. Find resultant resistance in below circuit Quick Quiz 2 4. Find resultant resistance in below circuit 5. Find resultant resistance in below circuit  Answers to Quiz Questions Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 100  Answers to Quiz 2 1. 150 Ω 2. 1.6667 Ω 3. 75 KΩ 4. 9.329 Ω 5. 45.569 Ω Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 101  Quick Quiz 1 1. An inductor and a resistor opposes _____ & _____ respectively A) flow of current, rate of change of current B) rate of change of current, flow of current C) rate of change of current, rate of change of current D) flow of current, flow of current 2. What is the equivalent resistance across the terminals A & B in the network shown? Ans 1. B) Ans 2Ω Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 102  Quick Quiz 1 3. a) 15 resistors are connected as shown in the diagram. Each of the resistors has resistance 1 Ω. Find A B the equivalent resistance of the network between A & B. b) What will be the equivalent resistance of this network if the resistors arranged in the sequence extends to infinity? Ans a)1.875 Ω b)2 Ω Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 103 Topics for Discussion Source Transformation Current & Voltage Division Star- Delta Transformation Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 104 Source Transformation Practical Voltage source Practical Current source +a R +a + RL Is R RL Vs - -b 𝑽𝒔 -b 𝑰𝒔 = 𝑹 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 105 Source Transformation Practical Voltage source Practical Current source +a R +a + RL Is R RL Vs - -b -b 𝑽𝒔 = 𝑹 × 𝑰𝒔 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 106  The series combination seems to behave identically to the parallel combination. One can replace one configuration for the other where needed. And this switch is called source transformation. B Tech. 1st Year Basic Electrical Technology 1. Convert to voltage source to current source 2. Convert current source to voltage source Voltage Division (in Series Circuit) IR= IR1+IR2 I=V/Req I=V1/R1 I=V2/R2 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 116 Current Division (in Parallel Circuit) V= V1=V2 IReq= I1R1=I2R2 Req= (R1R2)/(R1+R2) Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 117 Current Division: ANS: I1=I2=I3=4A Voltage Division: Find voltage across resistor 20 and 40 Voltage Division: Voltage Division : Voltage Division :  The source voltage is ___ A) 10 V B) 20 V C) 30 V D) 40 V Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 128 Star & Delta Connections A A Ra R1 R2 Rb Rc R3 B C B C Delta Connection Star Connection Link for the formula derivation: https://nptel.ac.in/content/storage2/courses/108105053/pdf/L-06(GDR)(ET)%20((EE)NPTEL).pdf Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 129 Delta to Star Connections A A Ra R1 R2 R3 Rb Rc B C Delta Connection B C Star Connection Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 130  Star to Delta Connections A A Ra R1 R2 Rb Rc R3 B C B C Star Connection Delta Connection Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 131 Derivation Find the equivalent resistance Req? Find equivalent resistance Req?  Illustration 1 For the circuit shown, determine the total power supplied by the source using star-delta transformation 30  40  G 100  20  50  10  100 V Ans: Psupplied = 223.12 W Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 144 Illustration 2 A 4 6 Determine the equivalent resistance between the terminals A and B. 3 5 5 2 4 B Ans: 3.85 Ω Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 145  Illustration 3 Using Source Transformation method, deduce the circuit shown to a single voltage source and a resistor across terminals A and B Ans: 45 V, 6.25 Ω Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 146 Illustration 4 Find the current through 8 Ω resistor by source transformation method Ans: 1 A Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 147 Homework 1 Find the power dissipated in 1Ω resistor and the power delivered by 10V source using network reduction technique 1 3 2 10V 3 3 2 2 3 Ans: P1Ω = 11.65 W, P10V= 34.2 W Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 148 Homework 2 R R R R R R R A B R R 1. Reduce the network to its equivalent resistance between R R terminals A and B R Ans 1: 5R/6 Course Code: [ELE 1051] Dept. of Electrical & Electronics Engg., MIT - Manipal 149  Quiz 1. If 3Ω,3Ω & 3Ω resistances are connected in Delta. Then the equivalent resistances of star network are A) 1,1,1 B) 1,2,3 C) 3,3,3 D) None A) 1,1,1 9EE303.9 150 2) If three resistances each of R ohm are connected in Star, then the equivalent resistance in each arm of delta will be______ ohm A) R B) R/3 C) 3R D) None C) 3R 151  Quiz 3 ) If three resistances each of 1 ohm are connected in star, then the equivalent resistance in each arm of delta will be______ ohm A) 1 B) 2 C) 3 D) None C) 3R 152 4) Three resistances of 3Ω each, are connected in star. Find the value of each resistor in delta (a) 3Ω (b) 27Ω (c) 9Ω (d) 4Ω (c) 9Ω 9EE303.11 153