EE 151 Applied Electricity Notes - PDF

EE 151-APPLIED ELECTRICITY Facilitator:- F. B. Effah (PhD) Teaching Assistants: David Bawiina (0560456972) Tunteiya Alhassan (0249117004) Address: Department of Electrical & Electronic Engineering Faculty of Electrical & Computer Engineering Colleg...

EE 151-APPLIED ELECTRICITY Facilitator:- F. B. Effah (PhD) Teaching Assistants: David Bawiina (0560456972) Tunteiya Alhassan (0249117004) Address: Department of Electrical & Electronic Engineering Faculty of Electrical & Computer Engineering College of Engineering Email: [email protected] Office: Room 13, Bamfo Kwakye Building Office Hours: Monday - Friday, 9am - 5pm ▪ Target students: First-year students on courses offered by the Department of Electrical and Electronic Engineering. ▪ Aim: To introduce students to the physical principles underpinning Electrical and Electronic Engineering, including circuit theory and electric and magnetic fields.  LEARNING OUTCOMES On completion of this subject, students will be able to Knowledge and understanding LO1: Understand Kirchhoff’s laws, Norton and Thévenin equivalent circuits and be able to apply to simple circuits LO2: Understand the concept of phasors and apply them to simple AC circuits. Intended LO3: Understand the superposition principle and be able to apply it to Learning simple AC circuits. Outcomes Intellectual skills LO4: Reduce complex circuits to a simple form LO5: Perform analysis on linear and non-linear magnetic circuits Professional practical skills LO6: Apply appropriate methods to analyse various circuits  TEACHING ACTIVITIES ❖ Presentation of lecture notes ❖ Tutorials ❖ Examples in class  ASSESSMENT Distribution (%) LO1 LO2 LO3 LO4 LO5 LO6 Progress Test 10 ✓ ✓ ✓ ✓ ✓ ✓ Assessment Homework 5 ✓ ✓ ✓ ✓ ✓ ✓ Mid-Sem. Exam. 15 ✓ ✓ ✓ ✓ ✓ ✓ Final Exam. 70 ✓ ✓ ✓ ✓ ✓ ✓ Total 100  READING LIST 1. J. W Nilsson and S. A. Riedel, Electric circuits, Prentice Hall, 7th ed., 2006 2. R. Boylestad, Introductory circuit analysis, Prentice Hall, 11th ed., 2007 3. R C Dorf, Introduction to Electric Circuits 4. Johnson Johnson and Hilburn, Electric Circuit Analysis 5. R Powell, Introduction to Electric Circuits 6. Bell Whitehead and Bolton, Basic Electrical and Electronic Engineering 7. P. Y. Okyere, E. A. Frimpong, Fundamentals of Electric and Magnetic Circuits  ORGANISATION OF SEMESTER ❖ Unit 1: Circuits and Network Theorems Wk. 2 – Wk. 4 Kirchhoff's laws, Thevenin’s Theorem, Norton's Theorem, Superposition Theorem, Reciprocity Theorem and Delta-Star Transformation. ❖ Unit 2: Alternating current circuits Wk. 4 – Wk. 6 Determination of Average and RMS values, Harmonics, Phasors, impedance, current and power in AC circuits. ❖ Unit 3: Three-phase circuits Wk. 6 – Wk. 9 Connection of three-phase windings, three-phase loads, power in three-phase circuits, solving three- phase circuit problems. ❖ Unit 4: Magnetic circuits Wk. 9 – Wk. 11 Components and terminologies, solving magnetic circuit problems.  UNIT 1: CIRCUIT AND NETWORK THEOREMS  INTRODUCTION SI UNITS This is a standard set of units used by all scientists and engineers. Always use SI (System International) units in calculations.  BASIC QUANTITIES AND SI UNITS Physical quantity Quantity Symbol Basic SI Unit Unit Symbol Name length l, d, h, r meter m mass m kilogram kg time t second s Electric current I ampere A temperature T Kelvin K  IMPORTANT DERIVED QUANTITIES (1) Quantity Symbol Definition Unit (Symbol) Alternative SI expressions Force f Push or pull Newton (N) Kg.m/s2 Energy W, E Ability to do Joule (J) N.m work Electric Charge q Quantity of Coulomb (C) A.s electricity Power p Energy/unit of Watt (W) J/s = N.m/s time  IMPORTANT DERIVED QUANTITIES (2) Quantity Symbol Definition Unit (Symbol) Alternative SI expressions Voltage v Energy/unit of time Volt (V) W/A = N.m/s/A Electric field E Force/unit charge Volt/mater N/C=kg.m/s2/A.s strength (V/m) Electric charge Coulomb per cubic density 𝜌 meter A s m-3 Cm-3 Electric flux Coulomb per square density 𝜎 meter A s m-2 Cm-2  ELECTRICAL CHARGE The concept of electrical charge is the basis for describing all electrical phenomena. Let’s review some important characteristics of electric charge: The charge is bipolar, meaning that electrical effects are described in terms of positive and negative charges Nature’s basic carrier of negative charge is the electron Nature’s basic carrier of positive charge is the proton The SI unit of charge is the Coulomb (C) e=1.6x10-19 C  ELECTRICAL CHARGE (CONT’D) Charge is never created nor destroyed – it is conserved Like charges of the same sign repel each other. Unlike charges of the opposite sign attract each other Charge is quantized Charge always comes in an integral multiple of a basic unit, symbolized by e Electrons have a charge of –e Protons have a charge of +e  VOLTAGE AND CURRENT In circuit theory, the separation of charge creates an electric force (Voltage) The motion of charge creates an electrical fluid (Current) Whenever positive and negative charges are separated energy is expended.  Voltage is energy per unit charge created by separation: 𝑑𝑤 𝑣= 𝑑𝑞 v = the voltage in volts w =the energy in joules q=the charge in coulombs  The electrical effects caused by charges in motion depend on the rate of charge flow. The rate of charge flow is known as the electric current, which is expressed as: 𝑑𝑞 𝑖= 𝑑𝑡 i=current in amperes q=the charge in coulombs t=the time in second  CURRENT Is the movement of electrical charge in a conductor (e.g. in a wire) 𝜕𝑞 𝐼= 𝜕𝑡 When 1C of charge crosses an area in 1 second, a current of 1A flows Current density (J) is defined as current/unit area. Its unit is A/m2 𝐼 𝐽= 𝐴  SOME IMPORTANT NOTES Electrons move in the opposite direction of the current. Current is measured at one point and has a direction Voltage is measured across two points and has a polarity One volt is the potential difference between two points when one joule of energy is used to move one coulomb of charge from one point to the other  POWER AND ENERGY Recall from basic physics that power is the time rate of expending or absorbing energy 𝒅𝒘 𝒑= 𝒅𝒕 p=power in wat w=energy in joule t=the time in seconds Thus 1 W is equal to 1 J/s  𝒅𝒘 𝒅𝒘 𝒅𝒒 𝒑= = = 𝒗𝒊 𝒅𝒕 𝒅𝒒 𝒅𝒕 p=the power in watt v=voltage in volts i=the current in amperes Power is energy per unit time Sign indicates whether power is absorbed or delivered: If power is positive, it indicates absorption by circuit; if negative, delivery.  POWER SIGN CONVENTION The current should enter the positive voltage terminal. Consequence for P = IV Positive (+) power: element absorbs power Negative (-) power: element supplies power  VOLTAGE AND CURRENT SOURCES An electrical source converts non-electric energy to electric energy Ideal, independent and dependent sources The circuit symbol for an ideal The circuit symbol for an ideal independent voltage source independent current source  DEPENDENT VOLTAGE AND CURRENT SOURCES (a) Ideal dependent voltage controlled voltage source (b) Ideal dependent current controlled voltage source (c) Ideal dependent voltage controlled current source (d) Ideal dependent current controlled current source  Active element Is one that models a device capable of generating electric energy, e.g. Batteries, Generators, solar cells, transistor models Passive element Physical devices that can not generate electric energy, e.g. resistors, inductors and capacitors  ELECTRIC RESISTANCE: OHM’S LAW Resistors resist, or limit the flow of electrical current in a circuit. It is found that if a voltage V is applied between the ends of a wire, the current that flows, I, is proportional to V, the constant of proportionality is called resistance. ▪ Voltage across resistor is V = IR (Ohm’s Law) ▪ Power dissipated P = I2R = V2/R  Ohm’s law is widely obeyed, but there are some exceptions. The most obvious is that the temperature must be kept constant. Resistivity (𝝆) For a given material, it is found that the resistance of a length of wire is proportional to the length and inversely proportional to the area A. This gives rise to the definition of resistivity 𝜌, where 𝒍 𝑹= 𝝆 𝑨 Conductivity Is the inverse of resistivity 𝒍 Conductance = 𝝈𝑨  Power can also be expressed as: 2 𝑉 𝑝 = 𝑣𝑖 = 𝑖 𝑅 𝑖 = 𝑖 2 𝑅 = = 𝑣 2𝐺 𝑅 Reciprocal of resistance is conductance, G and measured in Siemens (S); G=1/R. Resistance dissipates energy; it gets hot. Note that electric stoves, toasters, water heaters and iron boxes make profitable use of the conductivity property. Example: (a) Calculate the values of v and I (b) Determine the power dissipated in each resistor  ELECTRIC FIELD If we have two plates at different voltages, a positively charged particle will experience a force, moving it to the lower potential. The gradient gives the force. Or V/d in the above case. The unit of field is E (Vm-1) The force experienced in a field of E by a charge q is qE (Fq =q.E) The electric field is a vector quantity  CAPACITANCE The simplest capacitor is made up of two parallel plates separated by a non-conducting medium, such as air. Q = CV (charge = capacitance*voltage) Unit of C is Farad (F) Typical values are pF to 𝜇F  CURRENT FLOW IN A CAPACITOR Current only flows into or out of a capacitor if the voltage across it is changed.  An ideal capacitor does not dissipate energy – it does not get hot. It stores energy by keeping +Q and –Q separated and creating an electric field between the plates. This energy is recoverable; e.g. if the battery is replaced by a resistor, current flows until the capacitor is discharged. When we connect the battery to the capacitor, current flows for time T until the capacitor is charged to the voltage V of the battery.  Then the energy transferred from the battery to the capacitor is given by: 𝑇 𝑇 𝑑𝑣𝑐 𝑊𝑐 = ‫׬‬0 𝑣𝑐 𝑖𝑐 𝑑𝑡 ➔ 𝑊𝑐 = ‫׬‬0 𝑣𝑐 𝐶 𝑑𝑡 𝑑𝑡 𝑉 1 2 𝑉 𝟏 𝟐 𝑊𝑐 = 𝐶 ‫׬‬0 𝑣𝑐 𝑑𝑣 ➔ 𝑊𝑐 = C 𝑣𝑐 0 ➔𝑾𝒄 = 𝐂𝑽 2 𝟐 The idea of an ideal capacitor is not very physical because a capacitor must have some series resistance  From the definition of power 𝑑𝑣 𝑝 = 𝑣𝑖 = 𝐶𝑣 𝑑𝑡  EXAMPLE The voltage pulse described by the following equations is impressed across the terminal of a 0.5 𝜇F capacitor. Derive the expression for the capacitor current, power, and energy. 0, 𝑡 ≤ 0 ▪ 𝑣(𝑡) = ቐ 4𝑡𝑉, 0 ≤ 𝑡 ≤ 1 4𝑒 −(𝑡−1) 𝑉, 1 ≤ 𝑡 ≤ ∞  EXAMPLE A triangular current pulse drives an uncharged 0.2 μF capacitor. The current pulse is described as: 0 𝐴, 𝑡 ≤ 0 5000𝑡 𝐴, 0 ≤ 𝑡 ≤ 20𝜇s 𝑖(𝑡) = 0.2 − 5000𝑡 𝐴, 20 ≤ 𝑡 ≤ 40𝜇s 0 𝐴, 𝑡 ≥ 40𝜇s Derive the expression for the capacitor voltage, power and energy for each of the four-time intervals needed to describe the current.  INDUCTOR Inductance is a consequence of a conductor linking a magnetic field Inductance is symbolized by the letter “L” Is measured in Henrys (H) Is represented graphically as a coiled wire When current passes in a wire, a magnetic field is created around the wire If the current is made to change, then the magnetic field changes, which produces a voltage across the wire that opposes the current source. This is often called a back emf.  Faraday’s law gives this self-induced emf as: 𝑑𝑖 𝑣=𝐿 𝑑𝑡 ▪ Where L is the inductance of the coil in Henrys (H) To maintain the current flow, the source must supply voltage equal to and opposite to this. The Henry is defined as the inductance of a circuit in which an emf of 1V is induced when the current varies uniformly at a rate of 1 A/s.  ENERGY STORED IN AN INDUCTOR An ideal inductor does not dissipate energy but stores it as the energy of the magnetic field. If the current stops by replacing the supply with a resistor, a current will flow through the resistor. If it flows for time T to build up a current I we have, 𝑇 𝑇 𝑑𝑖 𝑊𝐿 = න (𝑣. 𝑖)𝑑𝑡 = න 𝐿 × 𝑖𝑑𝑡 0 0 𝑑𝑡 𝑰 𝟏 𝟐 𝑾𝑳 = 𝑳 න 𝒊 𝒅𝒊 = 𝑳𝑰 𝟎 𝟐 The idea of an ideal inductor is not very physical because a coil of wire must have some resistance.  TERMINOLOGIES ❖Node – A point where two or more circuit elements join. Example: a, b, d, c, e, f, g ❖Essential node – A point where three or more circuit elements join. Example: b, c, e, g ❖Branch – A path that connects two nodes. Example: c-a; d-e; e-g, etc. ❖ Essential branch – A path that connects two essential nodes without passing through an essential node. Example: b-e; etc. ❖Loop – A path whose last node is the same as the starting node. Example: c-a-b-e-d-c; a-b-e-g-f-c-a; etc. ❖Mesh – A mesh is a special type of loop that does not contain any other loops within it. Example: a-b-e-d-c-a; c-d-e-g-f-c, etc. ❖c-a-b-e-g-f-c is not a mesh but it is a loop. ▪ Short-circuit– A branch of theoretically zero resistance. It diverts to itself all currents that would have flown in adjacent branches (branches hooked to the same node) except branches with sources. b R1 c R3 R1 a R2 R4 a d V R2 V R3 h g f e b Fig. 1 Fig. 2 ❖Short-circuit cont. ▪ SELF ASSESSMENT ▪ Which of the resistors in the circuit below have been short- circuited? ▪ ANS: R2 and R3 R1 V R2 R3 ▪ Open circuit – A branch of theoretically infinite resistance. It prevents current from flowing in its branch. R2 R3 R5 V R1 R4  CIRCUIT REDUCTION The equivalent circuit reduces a complicated circuit to a straightforward form. It contains fewer elements (Rs, Cs and Ls) but is similar to the former electrically. Two elements are in series if the current that flows through one must also flow through the other. Two elements are parallel if they are connected between (share) the same two (distinct) end nodes.  RESISTORS IN SERIES Series-connected circuit elements carry the same current 𝑅𝑒𝑞 𝑠𝑒𝑟𝑖𝑒𝑠 = σ𝑘𝑖=1 𝑅𝑖 = 𝑅1 + 𝑅2 + ⋯ 𝑅𝑘 Equivalent resistance is always larger than that of the largest resistor in series connection ▪ ➔ 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3  RESISTORS IN PARALLEL Parallel-connected circuit elements have the same voltage across their terminals 1 𝑘 1 1 1 1 = σ𝑖=1 = + + ⋯ 𝑅𝑒𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑅 𝑖 𝑅1 𝑅2 𝑅𝑘 Equivalent resistance is always smaller than the smallest resistor in the parallel connection. ▪ 1 ➔ 𝑅𝑒𝑞 = 1 1 1 + + 𝑅1 𝑅2 𝑅3  RESISTORS IN PARALLEL (CONT’D) Alternatively, if we replace the resistors with conductance (G=1/R) ▪ Note that when just two resistors are connected in parallel.  EXAMPLE Find the currents ix, i1, i2, and vx. Take notes!!  EFFECTIVE RESISTANCE OF A CIRCUIT ▪ Effective circuit resistance is found by identifying and putting together series and or parallel resistors. ▪ E.g. 1. Find the total resistance of the circuit below.  EFFECTIVE RESISTANCE OF A CIRCUIT ❖Solution 23 RT = (2 // 3) + 1 = 11 +1 =  2+3 5  EFFECTIVE RESISTANCE OF A CIRCUIT ▪ E.g. 2. Find the total resistance of the circuit below.  EFFECTIVE RESISTANCE OF A CIRCUIT ▪ Solution 2 2 RT = (2 // 2) + 1 = + 1 = 2 2+2  EFFECTIVE RESISTANCE OF A CIRCUIT Self Assessment 1 Find the total resistance of the circuit below. 2Ω 3Ω 6Ω 4Ω V ▪ Answer  4 6  22 66 RT = (4 // 6) + 2// 3 =  + 2 // 3 = // 3 =  4 + 6  5 37  EFFECTIVE RESISTANCE OF A CIRCUIT ▪ Self Assessment 2 ▪ Find the total resistance of the circuit below. ▪ Answer 1  RT = (1 // 1) + 1 // 1 // 1 =  + 1 // 1 // 1 = // =  3 1 3 2  2 2 8 INDUCTORS IN SERIES  INDUCTORS IN SERIES 3 𝐿𝑒𝑞 = ෍ 𝐿𝑖 = 𝐿1 + 𝐿2 + 𝐿3 𝑖=1 Inductors in series combine like resistors in series. INDUCTORS IN PARALLEL  INDUCTORS IN PARALLEL 1 1 1 1 ▪ = + + 𝐿𝑒𝑞 𝐿1 𝐿2 𝐿3 Inductors in parallel combine like resistors in parallel.  EXAMPLE ▪ Find the equivalent inductance (all values are in Henrys).  CAPACITORS IN PARALLEL 3 𝐶𝑒𝑞 = ෍ 𝐶𝑖 = 𝐶1 + 𝐶2 + 𝐶3 𝑖=1 Capacitors in parallel combine like resistors in series.  CAPACITORS IN SERIES 1 1 1 1 = + + 𝐶𝑒𝑞 𝐶1 𝐶2 𝐶3 Capacitors in series combine like resistors in parallel.  EXAMPLE ▪ Find the equivalent capacitance  VOLTAGE DIVIDER Consider two resistors in series with a voltage v(t) across them: find v1(t) and v2(t); Easy way to find the voltages v1(t) and v2(t) is given by:  EXAMPLE ▪ The resistors used in the voltage-divider circuit shown below have a tolerance of ±10%. Find the maximum and minimum voltages.  CURRENT DIVISION RULE ▪ The current division rule is applied to share current between parallel branches. Consider the circuits below: I R1 R2 RT = R1 + R2 I1 I2 R1 R2 V = IRT = I R1 R2 R1 + R2 V R1 R2 I V R1 + R2 R1 R2 1 R2 I1 = = =I  =I R1 R1 R1 + R2 R1 R1 + R2  CURRENT DIVISION RULE ▪ Similarly, I I1 I2 V R1 R2 R1 I2 = I R1 + R2  CURRENT DIVISION RULE Comparing currents: I I1 I2 V R1 R2 R2 R1 I1 = I I2 =I R1 + R2 R1 + R2  CURRENT DIVISION RULE ▪ Consider the figure below, I I1 I2 V R1 R2 R2 R1 I1 = I I2 = − I R1 + R2 R1 + R2  CURRENT DIVISION RULE ▪ Example 1 ▪ Find the values of I1 and I2 in the circuit below. 10A I1 I2 V 3Ω 2Ω  CURRENT DIVISION RULE ▪ Solution R2 2 I1 = I =  10 = 4 A R1 + R2 2+3 R1 3 I2 = − I =−  10 = −6 A R1 + R2 2+3  CURRENT DIVISION RULE ▪ Example 2 ▪ Find the value of I1 in the circuit below. 10A I1 V 3Ω 2Ω 2Ω  CURRENT DIVISION RULE ▪ Solution 10A I1 1 V 3Ω 1Ω I1 =  10 = 2.5 A 1+ 3  VOLTAGE DROP ❖Any time a voltage drives current through a resistor, some of the voltage drops across the resistor. ❖The magnitude of the drop is the product of the resistance and current. V1 V2 I R1 R2 V V2 = V − V1  VOLTAGE DROP ▪ Example ▪ Find the values of I and R in the circuit below. I 4V R 3 10V ▪ Solution ▪ Voltage across 3Ω resistor = 10 – 4 = 6V ▪ Current in 3Ω resistor = I = 6/3 = 2A ▪ Resistance R = 4V/I = 4/2 = 2Ω  REVISION EXERCISE ▪ Find the value of I in the circuit below. I 3 3.5 4 1 2 5 7 25V ▪ Solution I I 3 6  4 1 2 4 1 5 25V 25V  REVISION EXERCISE 12  13 1 25V 25 RT =  13 V 25 IT = = = 13 A RT 25 13  REVISION EXERCISE I 4 1 25V 6 I = − 5  13 = −3 A 6 +4 5  ASSIGNMENT 1 ▪ Find the value of the current in all resistors of the circuit below using total resistance and voltage drop principles. DO NOT use current division rule. 3.5 3 2 1 4 5 7 25V ▪ Submission date: God willing a week today ▪ Submission time: Before lecture starts ▪ Where to submit: Teaching Assistant’s office  KIRCHHOFF’S CURRENT LAW (KCL) ❖THE LAW ▪ The sum of currents entering a node equals the sum leaving the node. i5 i6 i1 i4 i3 i2 ▪ The sum of currents entering i1 + i3 + i5 ▪ The sum of currents Leaving i2 + i4 + i6 ▪ Applying KCL i1 + i3 + i5 = i2 + i4 + i6  KIRCHHOFF’S CURRENT LAW (KCL) ❖Example ▪ Find the value of i in the figure below. i 2 3 4 7 ▪ Solution i+2+3= 4+7 i + 5 = 11 i =6  KIRCHHOFF’S CURRENT LAW (KCL) ▪ SELF ASSESSMENT ▪ Find the value of i in the figure below. i 2 3 4 7 ▪ ANSWER i=2  KIRCHHOFF’S VOLTAGE LAW (KVL) ▪ THE LAW ▪ The algebraic sum of the voltages in a loop (closed path) equals zero. Alternatively, the algebraic sum of voltage sources equals the algebraic sum of voltage drops. R1 R2 R4 a I1 b I2 c I4 d I3 I5 R3 R5 V1 V2 h g f e ▪ Loop abgha V1=I1R1+I3R3 ▪ Loop adeha V1-V2=I1R1+I2R2-I4R4  KIRCHHOFF’S VOLTAGE LAW (KVL) R1 R2 R4 a I1 b I2 c I4 d I3 I5 R3 R5 V1 V2 h g f e ▪ Loop acfha V1=I1R1 + I2R2 – I5R5 ▪ Loop cbgfc 0 = -I2R2 + I3R3 +I5R5  KIRCHHOFF’S VOLTAGE LAW (KVL) ▪ Example 1 ▪ Find the current in all2Ωparts of the circuit 4Ω below. b c e I1 I2 32V 8Ω 20V I3 a d f 32 – 2I1 – 8I3 = 0 ❖Applying KVL to loop bcdab 32 = 2I1 + 8I3 (1) 20 – 4I2 – 8I3 = 0 ❖Applying KVL to loop ecdfe 20 = 4I2 + 8I3 (2)  KIRCHHOFF’S VOLTAGE LAW (KVL) 2Ω 4Ω b c e I1 I2 32V 8Ω 20V I3 a d f ❖Applying KCL to node c: I 3 = I1 + I 2 (3) ▪ Solving the equations simultaneously yields I1 = 4A, I2 = -1A and I3 = 3A  KIRCHHOFF’S VOLTAGE LAW (KVL) Example 2 Find the currents in all parts of the circuit below. 2Ω c 5Ω e 2Ω g b I1 I1-I2 I2 I2-I3 I3 7V 3Ω 2Ω 10 V a d f h 2Ω Solution Apply KVL to loop cefdc: 5I2 + 2 ( I2 – I3 ) + 2I2 - 3 ( I1 – I2 ) = 0  KIRCHHOFF’S VOLTAGE LAW (KVL) 0 = -3I1 + 12I2 - 2I3 (1) 2Ω c 5Ω e 2Ω g b I1 I1-I2 I2 I2-I3 I3 7V 3Ω 2Ω 10 V a d f h 2Ω 7 = 2I1 + 3(I1 – I2) ▪ Apply KVL to loop abcda: 7 = 5I1 – 3I2 (2) 10 = -2 ( I2 – I3 ) + 2I3 ▪ Apply KVL to loop ghfeg: 10 = -2I2 + 4I3 (3)  KIRCHHOFF’S VOLTAGE LAW (KVL) ▪ Solving the three equations: 0 = -3I1 + 12I2 - 2I3 (1) 7 = 5I1 – 3I2 (2) 10 = -2I2 + 4I3 (3) ▪ Simultaneously, I1 = 2.0A, I2 = 1.0A and I3 = 3.0A  THEVENIN’S THEOREM ▪ Theorem: ▪ Any linear circuit connected between two terminals can be replaced by a Thevenin’s voltage(VTH) in series with a Thevenin's resistance (RTH). ▪ VTH is the open-circuit voltage across the two terminals. ▪ RTH is the resistance seen from the two terminals when all sources have been deactivated R TH 12 Ω A B A 6Ω VTH 42 V 35 V B  THEVENIN’S THEOREM ▪ To find the current through a resistor in a circuit, the following steps are taken: 1. Remove the resistor from the circuit and mark the two terminals. 12 Ω 3 42 V 35 V 6Ω  THEVENIN’S THEOREM 2. Find the open-circuit voltage (VTH) across the two terminals by applying KVL. Treat VTH as a source. 12 Ω A B VTH 42 V 35 V 6Ω  THEVENIN’S THEOREM 3. Recall the circuit created before step 2 and deactivate all sources. Short-circuit voltage sources and Open-circuit current sources. 12 Ω A B 42 V 6Ω 35 V  THEVENIN’S THEOREM 4. Find the total resistance of the circuit resulting from step 3 as seen from the two terminals 12 Ω A B 6Ω  THEVENIN’S THEOREM 5. Reproduce the Thevenin’s equivalent circuit and connect the resistor whose current is to be found. RTH A VTH 3 B  THEVENIN’S THEOREM 6. Calculate the current in the circuit in step 5. This is the current being sought. RTH I A VTH i= VTH 3 RTH + R B  THEVENIN’S THEOREM Example 1 Using Thevenin’s theorem, determine the current in the 3-Ω resistor of the circuit below. 12 Ω 3Ω 42 V 35 V 6Ω Solution 12 Ω A B Steps 1 & 2 VTH 6Ω 42 V 35 V I  THEVENIN’S THEOREM 12 Ω A B VTH 6Ω 42 V 35 V I Applying KVL to loop dcbed: 35 + VTH = 6I (1) Applying KVL to loop fabef: 42 = (12 + 6) I 7 I = A 3  THEVENIN’S THEOREM 7 Substituting for I in equation 1: 35 + VTH = 6( ) 3 VTH = −21V Steps 3 & 4: 12 Ω RTH 12  6 A B RTH = 12 // 6 = = 4 12 + 6 6Ω  THEVENIN’S THEOREM Steps 5 & 6: RTH = 4Ω A I3 VTH= -21V 3Ω B VTH 21 I3 = =− = −3 A RTH + 3 4+3  THEVENIN’S THEOREM Example 2 Find the current in the 10-Ω resistor of the circuit below using Thevenin’s theorem. 5Ω 10 Ω 12 Ω 4V 15 Ω 8Ω 6V Solution 5Ω VTH 12 Ω Steps 1 & 2 A B 4V I1 15 Ω I2 8Ω 6V  THEVENIN’S THEOREM 5Ω VTH b c 12 Ω d a A B 4V I1 15 Ω I2 8Ω 6V h g e f ▪ Applying KVL to loop cbgfc: VTH = 15I1 - 8I2 1 ▪ Applying KVL to loop abgha:4 = (5+15)I1 I1 = A 5 ▪ Applying KVL to loop dcfed: 6 = (12 + 8) I 2 3 I2 = A 10  THEVENIN’S THEOREM Substituting for I1 and I2 in equation 1 yields: VTH  1  3  3 = 15  − 8  = V  5   10  5 Steps 3 & 4 5Ω RTH 12 Ω A B 15 Ω 8Ω 171 RTH = (5 // 15) + (12 // 8) =  20  THEVENIN’S THEOREM Steps 5 & 6 RTH A I VTH 10 Ω B VTH 3 3  20 I = = = = 0.032 A RTH + 10  171  5  371 5 + 10   20   ASSIGNMENT 2 ▪ Use Thevenin’s theorem to find the current in the 5Ω resistor of the circuit below. 2Ω c 5Ω e 2Ω g b 7V 3Ω 2Ω 10 V a d f h 2Ω ▪ Submission date: God willing, a week today ▪ Submission time: Before the lecture starts ▪ Where to submit: Teaching Assistant’s office  NORTON’S THEOREM ▪ THEOREM: ▪ Any linear circuit connected between two terminals can be replaced by a Norton’s current(IN) in parallel with a Norton's resistance (RN). ▪ IN is the short-circuit current between the two terminals ▪ RN is the resistance seen from the two terminals when all sources have been deactivated (RN = RTH) A 12 Ω A B IN RN 6Ω 42 V 35 V B  NORTON’S THEOREM To find the current through a resistor in a circuit, the following steps are taken: 1. Remove the resistor from the circuit and mark the two terminals. 12 Ω 3 42 V 35 V 6Ω  NORTON’S THEOREM 2. Find the short-circuit current (IN) through the two terminals by applying KVL. 12 Ω IN A B 42 V 35 V 6Ω  NORTON’S THEOREM 3. Recall the circuit created before step 2 and deactivate all sources. Short-circuit voltage sources and Open-circuit current sources. 12 Ω A B 42 V 35 V 6Ω  NORTON’S THEOREM 4. Find the total resistance of the circuit resulting from step 3 as seen from the two terminals 12 Ω A B 6Ω  NORTON’S THEOREM 5. Reproduce Norton’s equivalent circuit and connect the resistor whose current is to be found. A IN RN 3 B  NORTON’S THEOREM 6. Calculate the current in the circuit in step 5. This is the current being sought for. I A RN IN 3 i=  IN RN RN + 3 B  NORTON’S THEOREM Example 1 Using Norton’s theorem, determine the current in the 3-Ω resistor of the circuit below. 12 Ω 3Ω 42 V 35 V 6Ω I+IN 12 Ω IN Solution A B I Steps 1 & 2 42 V 6 Ω 35 V  NORTON’S THEOREM I+IN 12 Ω IN a b A B c I 6 Ω 42 V 35 V f e d ▪ Applying KVL to loop abefa: 42 = 12(I+IN) + 6I 42 = 18I + 12IN (1) ▪ Applying KVL to loop cbedc: 35 = 6I 35 I= A 6  NORTON’S THEOREM Substituting for I in equation 1:  35  42 = 18  + 12 I N  6 − 21 IN = A Steps 3 & 4 4 12 Ω RN A B 12  6 RN = 12 // 6 = = 4 6Ω 12 + 6  NORTON’S THEOREM Steps 5 & 6 I3 A − 21 4 A 3 B 4 − 21 I3 =  = −3 A 4+3 4  NORTON’S THEOREM ▪ Example 2 ▪ Determine the current in the load resistor RL A R1 R2 R3 R4 RL E2 ▪ Solution E1 E3 E4 B A I1 I2 I3 I4 R1 R2 R3 R4 IN E1 E2 E3 E4 B  NORTON’S THEOREM A I1 I2 I3 I4 R1 R2 R3 R4 IN E1 E2 E3 E4 B Solution Applying KCL: I N = I1 + I 2 + I 3 + I 4 E1 E2 E3 E4 = + + + R1 R2 R3 R4  NORTON’S THEOREM Finding RN A R1 R2 R3 R4 RN B 1 1 1 1 1 = + + + R N R1 R2 R3 R4  NORTON’S THEOREM Finding IL A IL IN RN RL B RN IL =  IN RN + RL  SOURCE TRANSFORMATIONS Voltage source in series with a resistor is equivalent to current source in parallel with the same resistor. Suppose RL is connected between node a, b. 𝑣 𝑅 𝑖𝐿 = 𝑠 AND 𝑖𝐿 = 𝑖𝑠 𝑅+𝑅𝐿 𝑅+𝑅𝐿 If the two circuits 𝑣𝑠 are equivalent, iL must be the same, therefore,𝑖𝑠 = 𝑅  SUPERPOSITION THEOREM ▪ THE THEOREM The current through(or the voltage across) any element in a multiple-source linear circuit can be found by taking the algebraic sum of the current through(or the voltage across) that element due to each individual source acting alone.  SUPERPOSITION THEOREM ▪ THE THEOREM I 12Ω 6Ω 3Ω 42V 35V IA IB 12Ω 12Ω 6Ω 3Ω 6Ω 3Ω 42V + 35V  SUPERPOSITION THEOREM ▪ Example 1 ▪ Use superposition theorem to find the current supplied by the 35V battery of the circuit below. I 12Ω 6Ω 3Ω 42V 35V ▪ Solution ▪ With the 42V battery acting alone, R = (3 // 6) + 12 = 14 A T I 12Ω 42 6Ω 3Ω IT = = 3A 14 6 42V IA = 3 = 2A 6+3  SUPERPOSITION THEOREM ▪ With the 35V battery acting alone, IB 12Ω 6Ω 3Ω 35V RT = (12 // 6) + 3 = 7 35 IT = = 5A 7 I B = IT = 5 A  SUPERPOSITION THEOREM ▪ With both batteries acting, I IA IB 12Ω ▪ 12Ω 6Ω 3Ω = 6Ω 3Ω + 12Ω 6Ω 3Ω 42V 35V 35V 42V I = IB − I A = 5 − 2 = 3A  RECIPROCITY THEOREM THE THEOREM An ideal ammeter and ideal voltage source when inserted in two different branches of a linear network can be interchanged without changing the reading of the ammeter R1 R R1 R2 2 + = + V(t) A A V(t) R3 R3 - -  RECIPROCITY THEOREM ▪ Similarly, An ideal voltmeter and ideal current source when connected across two different branches of a network can be interchanged without changing the reading of the voltmeter. R 3 R3 + + R1 R2 + R1 + i R4 - V = V R2 R4 i - - -  RECIPROCITY THEOREM ▪ Example 1 Jointly use superposition and reciprocity theorems to find the current supplied by the 35V battery of the circuit below. I 12Ω 6Ω 3Ω ▪ Solution 42V 35V ▪ With the 42V battery acting alone, IA RT = (3 // 6) + 12 = 14 12Ω 42 6Ω 3Ω IT = = 3A 14 6 42V IA = 3 = 2A 6+3  RECIPROCITY THEOREM The second circuit to be solved is: I B IA 12Ω 12Ω 6Ω 3Ω is comparable 6Ω 3Ω to 35V 42V Applying the reciprocity theorem IA IA 12Ω 12Ω = 6Ω 3Ω 6Ω 3Ω 42V 42V A B  RECIPROCITY THEOREM IB IA 12Ω 6Ω 3Ω 12Ω is comparable 6Ω 3Ω to circuit B 35V 42V ▪ Applying proportion, If 42V = IA =2A, 5 A Then 35V =(35*2)/42=5/3A 3 IB 5 12Ω 6Ω 3Ω 35 = 3I + 12  B 3 35V  I B = 5A RECIPROCITY THEOREM I = IB − I A = 5 − 2 = 3A  ASSIGNMENT 3 Use Norton’s theorem to find the current in the 2Ω resistor connected between e and f in the circuit below. 2Ω c 5Ω e 2Ω g b 7V 3Ω 2Ω 10 V a d f h 2Ω Submission date: God willing a week today Submission time: Before lecture starts Where to submit: Teaching Assistant’s office  DELTA-STAR TRANSFORMATION ❖The transformation is employed in situations where neither series nor parallel arrangements can be identified. ❖An arrangement of three(3) resistors where the resistors are connected to each other is a delta arrangement. A A A R2 R3 R2 R3 B R1 C B R1 C  DELTA-STAR TRANSFORMATION ▪ An arrangement of three(3) resistors where all resistors have a common point of connection through one terminal is a star(wye) arrangement. Ra Rc Rb Ra Rc Rb  DELTA-STAR TRANSFORMATION ▪ A delta arrangement can be changed to star and vice versa using the following relations: R2 R3 A Ra = R1 + R2 + R3 R3 R1 Ra Rb = R2 R3 R1 + R2 + R3 R1 R2 Rc Rb Rc = R1 + R2 + R3 B R1 C  DELTA-STAR TRANSFORMATION A Rb Rc R1 = Rb + Rc + Ra Ra Ra Rc R2 R3 R2 = Ra + Rc + Rb Rc Rb Ra Rb R3 = Ra + Rb + B R1 C Rc ▪ When all values are the same, delta values are 3 times star values  DELTA-STAR TRANSFORMATION Example 1 Determine the voltage V0 across the 6Ω resistor of the circuit below 6Ω 2Ω 2Ω a 10 V 2Ω 6Ω Vo Solution 6 b 2 Ra 2 Ra = Rb = Rc = 2  3 = 6 2 Rb 6 10 R c  DELTA-STAR TRANSFORMATION 6Ω 6Ω 10V 6Ω 6Ω 6Ω 3Ω a 3 6Ω 3Ω V0 =  10V = 5V 3+3 10V Vo b  ASSIGNMENT 4 Use Norton’s theorem to find the current in the 3Ω resistor connected between points A and B of the circuit below. A 3Ω 2.4Ω 3Ω 3Ω 5Ω B 3Ω 10v Submission date: God willing a week today Submission time: Before lecture starts Where to submit: Teaching Assistant's office THANK YOU