Lecture 1-Introduction to Circuit Theory Dr. Kevin Gallacher Contact Details ❑ Dr. Kevin Gallacher, MEng ❑ Lecturer in Electronic and Photonic Devices ❑ James Watt School of Engineering ❑ Rankine Building, Room 733 ❑ Email: Kevin.Gallacher@glasgow.ac.uk ❑ https://www.gla.ac.uk/schools/engineering/staff/kevingallacher/ 2 Recommended Reading List Free access to ebook version on the UofG library References: 1) Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. 2) Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 Learning Outcomes from lecture 1. Electrical circuit notation and how sources and elements are depicted 2. Electrical Resistance of materials 3. Describe flow of electrons (Current) through materials 4. What is Voltage 5. How Voltage, Current, and Resistance relate via Ohms law 6. Lastly, how Power is defined and calculated 4 Basic Circuit of Headlights in Automobile (a) Physical Configuration Image Source - Hambley, Allan R. et al. : Pearson Prentice Hall, 2008 (b) Circuit Diagram 5 Basic Components in an Electrical Circuit ❑ An electrical circuit consists of various types of circuit elements connected in closed paths by conductors ❑ Conductors correspond to connecting wires in physical circuits ❑ Conventional Flow Current ❑ Electron Flow ❑ Voltage sources create forces that cause charge (q) to flow through the conductors and other circuit elements. Image Source - Hambley, Allan R. et al. : Pearson Prentice Hall, 2008 6 Electron flow through a material Copper 29 Electrons 29 Protons 35 Neutrons 7 Electron flow through a material Random motion of free electrons in a material Electrons flow from negative to positive ❑ As a result, energy is transferred between the circuit elements, resulting in a current flow 8 Resistance and Resistor ❑ The property of a material that restricts the flow of electrons is called resistance (R) ❑ Resistance is expressed in ohms, symbolized by the Greek letter omega (Ω) ❑ A component that is specifically designed to have a certain amount of resistance is called a resistor ❑ The principal applications of resistors are:❑ to limit current in a circuit ❑ to divide voltage ❑ in certain cases, to generate heat, light 9 Resistance and Resistor Analogy of Electrical Resistance with Water Flowing in a pipe ❑ The reciprocal of resistance is conductance, symbolized by G ❑ Mathematically, 𝐺 = 1 𝑅 ❑ The unit of conductance is the siemens (S) Resistance Further reading - https://www.youtube.com/watch?v=cx9xLwa7Gco 10 Resistance and Resistor Source - https://www.falstad.com/circuit/ 11 Calculate the conductance of a resistor of 40 kΩ. 1 𝐺= 𝑅 1 𝐺= 40kΩ 𝐺 = 25 µS 12 Resistance and Resistor cross-sectional area 𝜌𝑙 𝑅= 𝐴 𝜌 – resistivity of material (constant value at a given temperature) l – length of the wire A - cross-sectional area 13 Find the resistance of a 100 m length of copper wire with a cross-sectional area of 2.5 mm2. The resistivity of copper is 𝟏. 𝟕𝟐 𝐱 𝟏𝟎−𝟖 Ω − 𝐦 𝜌𝑙 𝑅= 𝐴 Ω𝑚 × 𝑚 Ω= 𝑚2 1.72 𝑥 10−8 𝑥 100 𝑅= 2.5 𝑥 10−6 𝑅 = 0.688 Ω 14 Resistance and Resistor Resistor Linear Resistor Variable Resistor Fixed Resistor Carbon Film Thick Film Nonlinear Resistor Thermistor LDR Thin Film Potentiometer Image Source - http://www.schoolphysics.co.uk Rheostat 15 Resistor Colour Coding The color code is read as follows: 1. Start with the band closest to one end of the resistor. This band is the first digit of the resistance value. Do not begin with a gold or silver band. 2. The second band is the second digit of the resistance value. 3. The third band is the number of zeros following the second digit, or the multiplier. 4. The fourth band indicates the percent tolerance and is usually gold or silver. 16 Find the resistance value in ohms and the percent tolerance for each of the color-coded resistors? 1st band - Red 2, 2nd band - Violet 7, 3rd band - Orange 3 zeros, 4th band - Silver 10% tolerance R = 27,000 Ω ± 10% 1st band - Green 5, 2nd band - Blue 6, 3rd band - Green 5 zeros, 4th band - Gold 5% tolerance R = 5,600,000 Ω ± 5% Image Source - Hambley, Allan R. et al. : Pearson Prentice Hall, 2008 17 Electrical Current Form & Notations ia(t) = 2 A iab = −iba (a) Dc current ib(t) = 2 cos 2πt A (b.1) Ac current (b.2) Triangular current Time varying current (b.3) Square current 18 Electrical Current Form & Notations I = I 19 Electrical Current Form & Notations ❑ Electrical current (I) is the time rate of flow of electrical charge through a conductor or circuit element ❑ The unit of electrical current is amperes (A), equivalent to coulombs per second (C/s) ❑ Electrons are the basic unit of electric charge with a value of 1.602 × 10−19 C ❑ Mathematically, 𝐼= 𝑄 𝑡 𝑑𝑞 𝑡 𝐼= 𝑑𝑡 ❑ A constant current of one ampere means that one coulomb of charge passes through the cross section each second 20 20 coulombs of charge flow past a given point in a wire in 5 s. What is the current in amperes? 𝑄 𝐼= 𝑡 20 𝐼 = 5 𝐼=4A 21 Electrical Current Practical Current Sources Image Source - https://www.tek.com/ 22 Voltage ❑ Voltage (V) is a measure of the energy transferred per unit of charge when charge moves from one point in an electrical circuit to a second point. ❑ When charge moves through circuit elements, energy can be transferred. ❑ In the case of automobile headlights, stored chemical energy is supplied by the battery and absorbed by the headlights where it appears as heat and light. ❑ Mathematically, 𝑉 = 𝑊 also V 𝑄 = E.l W is energy in joules (J), and Q is charge in coulombs (C), E electric field and l is the length Note- Voltage is measured across the ends of a circuit element, whereas current is a measure of charge flow through the element. vab = −vba 23 23 Electrical Voltage Form & Notations + V + - + = V - = V - 24 Voltage Practical Voltage Sources 25 If 80 J of energy is required to move 20 C of charge, what is the voltage? 𝑊 𝑉= 𝑄 80 𝑉= 20 𝑉=4V 26 Ohm’s Law, Georg Ohm At constant temperature, the electrical current flowing through a fixed linear resistance is directly proportional to the voltage applied across it. 𝐼𝜌𝑙 𝑉 = 𝐼𝑅 = 𝐴 𝑉α𝐼 I 𝐼 𝑉= 𝐺 I R V Ideal Practical 0 V 0 Ohm’s law Further reading - https://www.youtube.com/watch?v=iLzfe_HxrWI R 27 Ohm’s Law V I V=I × R R V I= R V R= I 28 Ohm’s Law Image Source - https://www.build-electronic-circuits.com/ohms-law/ 29 Ohm’s Law 𝐼𝜌𝑙 𝑉 = 𝐼𝑅 = 𝐴 𝑉 𝐼𝜌 = 𝑙 𝐴 𝑉 𝑉 = 𝐸. 𝑙 𝑜𝑟 𝐸 = 𝑙 𝐸 = 𝜌𝐽 𝐽 = 𝐼/𝐴 1 σ E – Electric Field (V/m) J – Current Density (A/m2) 𝜎 – Conductivity (S/m) 𝜌 – resistivity (Ohm-m) =𝜌 𝐽 = 𝜎𝐸 30 History of Ohm’s Law, Georg Ohm Ohm’s law Further reading - https://www.youtube.com/watch?v=iLzfe_HxrWI 31 Power and Energy Source - https://japancarnews.info/relationship/power-energy-voltage-current-resistance-relationship.php 32 Summary ❑ Provided overview of circuit notation and element depiction ❑ Discussed Resistance, Current, and Voltage Terms introduced Quantity Symbol Unit Abbreviation Resistance R Ohm Ω Conductance G Siemens S Current A Amp A Electric charge Q Coulomb C Electric potential V Volt V Energy W Joule J Power P Watt W ❑ Test quiz on Moodle to practise before real quizzes start after next lecture ❑ Next lecture Kirchhoff's laws 33 Lecture 2 -Kirchhoff’s Laws & SeriesParallel Circuits Previous Lecture ❑ Electrical circuit notation ❑ What is Voltage, Current, and Resistance ❑ Ohm’s law 2 Learning Outcomes 1. Kirchhoff’s laws 1. KIRCHHOFF’S current law 2. KIRCHHOFF’S Voltage law 2. Resistors in series and parallel circuits 1. Voltage divider 2. Current divider 3. Power in series and parallel circuits References: 1) Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. 2) Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 3 Kirchhoff’s Laws Gustav Robert Kirchhoff KIRCHHOFF’S CURRENT LAW – ❑ The sum of the currents entering a node equals the sum of the currents leaving a node. ❑ A node in an electrical circuit is a point at which two or more circuit elements are joined together ❑ Mathematically Σ𝐼𝐼𝑁 = Σ𝐼𝑂𝑈𝑇 Node a: i1 + i2 − i3 = 0 Node b : i5 + i6 + i7 = 0 Image Source - Hambley, Allan R. et al. : Pearson Prentice Hall, 2008 IT = I1 + I2 + I3 Source - Thomas L. Floyd - Pearson Education Limited (2013). 4 Determine the current I2 through R2 ? IT = I1 + I2 + I3 I2 = IT - I1 - I3 I2 = 100 mA - 30 mA - 20 mA I2 = 50 mA Source - Thomas L. Floyd - Pearson Education Limited (2013). 5 Use Kirchhoff’s current law to find the current measured by ammeters A3 and A5? The total current into node X is 5 mA 5 mA - 1.5 mA - IA3 = 0 IA3 = 5 mA - 1.5 mA IA3 = 3.5 mA The total current into node Y is 3.5 mA 3.5 mA - 1 mA - IA5 = 0 IA5 = 3.5 mA - 1 mA IA5 = 2.5 mA Source - Thomas L. Floyd - Pearson Education Limited (2013). 6 Kirchhoff’s Laws Gustav Robert Kirchhoff KIRCHHOFF’S VOLTAGE LAW – ❑ In any closed loop electrical circuit, the algebraic sum of the voltages equals zero Loop 1: −va + vb + vc = 0 Loop 2: −vc − vd + ve = 0 Loop 3: va − vb + vd − ve = 0 7 Kirchhoff’s Laws Gustav Robert Kirchhoff Loop : -vs + 6.6V + 5.4V = 0 Source - Thomas L. Floyd - Pearson Education Limited (2013). Kirchhoff’s law Further reading-https://www.youtube.com/watch?v=V2yKWpzpavA 8 Determine the source voltage where the two voltage drops are given? -VS +VR1 + VR2= 0 VS = VR1 + VR2 VR1 = 5 V VR2 = 10 V Loop VS = 5 V + 10 V VS = 15 V 9 Find the value of R4 ? VR1 = IR1 = (10 mA)(100 Ω) = 1.0 V VR2 = IR2 = (10 mA)(470 Ω) = 4.7 V VR3 = IR3 = (10 mA)(1.0 k Ω) = 10 V Using Kirchhoff’s voltage (KVL) -VS + V1 + V2 + V3 + V4 = 0 V -50 V + 1.0 V + 4.7 V + 10 V + V4 = 0 V -34.3 V + V4 = 0 V V4 = 34.3 V 𝑅4 = Source - Thomas L. Floyd - Pearson Education Limited (2013). 𝑉4 34.3 𝑉 = = 3.43 kΩ 𝐼 10 𝑚𝐴 10 Energy Conservation Using Kirchhoff’s Law pa + pb + pc = 0 vai − vbi + vci = 0 va − vb + vc = 0 Element A: pa = vai P = IV Element B: pb = −vbi Element C: pc = vci 11 Series and Parallel Circuits Refs. Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 Parallel and Series Circuits Classic example of Ohms law and Kirchoff’s laws – Series and parallel resistors: ❑ Circuit elements are said to be in series if they are connected end to end. ❑ Elements are said to be in parallel if both ends of the elements are connected directly to corresponding ends of the other. Current in both the elements is same, but voltage across them is different Voltage across both elements is the same but the current splits between them Series Circuits Some examples of series circuits. Notice that the current is the same at all points because the current has only one path. 14 Resistor in Series In this case, current in both the elements is same, but voltage across them is different Using Ohms law Voltage across R1, 𝑉1 = 𝐼𝑅1 Voltage across R2, 𝑉2 = 𝐼𝑅2 Using KVL Total voltage 𝑉𝐵 = 𝑉1 + 𝑉2 = 𝐼𝑅1 + 𝐼𝑅2 = 𝐼 𝑅1 + 𝑅2 If you consider that combined equivalent resistance is RT Then, 𝑉𝐵 = 𝐼𝑅𝑇 = 𝐼 𝑅1 + 𝑅2 So, 𝑅𝑇 = 𝑅1 + 𝑅2 𝑅𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 𝑅1 + 𝑅2 15 Resistor in Series Also, for n resistors in series, just add all the resistance 𝑅𝑇 = 𝑅1 + 𝑅2 + …… +𝑅𝑛 Notice also that, (Voltage Divider) 𝑉𝐵 = 𝐼 𝑅1 + 𝑅2 𝑉𝐵 𝐼= 𝑅1 + 𝑅2 Current common in R1 and R2 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 𝑉𝑅2 𝑉𝐵 𝐼𝑅2 = = 𝑅2 𝑅1 + 𝑅2 16 Resistor in Series Also, for n resistors in series, just add all the resistance 𝑅𝑇 = 𝑅1 + 𝑅2 + …… +𝑅𝑛 Notice also that, (Voltage Divider) 𝑉2 𝑅2 = 𝑉𝐵 𝑅1 + 𝑅2 𝑉1 𝑅1 = 𝑉𝐵 𝑅1 + 𝑅2 𝑉1 𝑅1 = 𝑉2 𝑅2 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 17 Voltage Division – for 3 Resistor 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 Current is the total voltage divided by the equivalent resistance 𝐼= 𝑉𝑡𝑜𝑡𝑎𝑙 𝑅𝑒𝑞 𝑉𝑡𝑜𝑡𝑎𝑙 = 𝑅1 + 𝑅2 + 𝑅3 Voltage across R1 is 𝑅1 𝑉1 = 𝑅1𝐼 = 𝑉 𝑅1 + 𝑅2 + 𝑅3 𝑡𝑜𝑡𝑎𝑙 𝑅2 𝑉2 = 𝑅1𝐼 = 𝑉𝑡𝑜𝑡𝑎𝑙 𝑅1 + 𝑅2 + 𝑅3 𝑅3 𝑉3 = 𝑅1𝐼 = 𝑉𝑡𝑜𝑡𝑎𝑙 𝑅1 + 𝑅2 + 𝑅3 18 Find out voltages across different resistor? Series Circuit 𝑅1 𝑉1 = 𝑅1𝐼 = 𝑉𝑡𝑜𝑡𝑎𝑙 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 1000 𝑉1 = × 15 V = 1.5 V 1000 + 1000 + 2000 + 6000 𝑅4 𝑉4 = 𝑅4𝐼 = 𝑉𝑡𝑜𝑡𝑎𝑙 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 6000 𝑉4 = × 15𝑉 = 9 V 1000 + 1000 + 2000 + 6000 19 Voltage Sources in Series Voltage sources in series are similar to the resistors in series. But, here you have to be careful about the polarity of the sources under consideration. 20 Break in series circuit 21 Parallel Circuits Some examples of parallel circuits. Notice that the current splits at the points where the current has multiple path to flow. 22 Resistor in Parallel In this case, voltage across both elements is the same but the current splits, Current in R1, 𝐼1 = Current in R2, 𝐼2 = 𝑉𝐵 𝑅1 𝑉𝐵 𝑅2 Using Ohm’s law Total current is, 𝐼 = 𝐼1 + 𝐼2 = 𝑉𝐵 𝑅1 + 𝑉𝐵 𝑅2 Using KCL If you consider the total equivalent resistance to be RT 1 𝑅𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 1 1 = + 𝑅1 𝑅2 Then, I = 𝑉𝐵 𝑅𝑇 = 𝑉𝐵 𝑅1 𝑉 + 𝐵 𝑅2 = 1 𝑉𝐵 𝑅1 1 + 𝑅2 = 𝑉𝐵 𝑅2 𝑅1 + 𝑅1 𝑅2 𝑅2 𝑅1 Using lowest common denominator (LCD) Alternatively, 𝑅𝑇 = 𝑅1 𝑅2 𝑅1 +𝑅2 23 Resistor in Parallel Also, for n resistors in parallel, just add all the resistance 1 1 1 1 = + +⋯+ 𝑅𝑇 𝑅1 𝑅2 𝑅𝑛 Notice also that, (Current Divider) 𝐼𝑡𝑜𝑡𝑎𝑙 = 𝐼𝑅1 + 𝐼𝑅2 𝑉𝐵 𝑉𝐵 1 1 𝐼𝑡𝑜𝑡𝑎𝑙 = + = 𝑉𝐵 + 𝑅1 𝑅2 𝑅1 𝑅2 𝑅1 𝑅2 𝑉𝐵 = 𝐼𝑡𝑜𝑡𝑎𝑙 𝑅1 + 𝑅2 𝑉𝐵 𝐼𝑅2 = = 𝐼𝑡𝑜𝑡𝑎𝑙 𝑅2 𝑅1 𝑅1 + 𝑅2 Common voltage across R1 and R2 𝑅1 𝑅2 𝑅𝑇 = 𝑅1 + 𝑅2 24 Current Sources in Parallel Total current from a parallel connected current sources are the algebraic sum of all of them. One has to consider the direction of currents. 𝐼𝑇 = 1𝐴 + 2𝐴 + 2𝐴 = 5𝐴 𝐼𝑇 = 2𝐴 + 2𝐴 − 1𝐴 = 3𝐴 25 Break in parallel circuit Break in parallel circuit does not effect the whole circuit. It only effects the branch with the fault 26 Combination of Series and Parallel 27 Simplify the circuit ? A 100 Ω A 52 V + C 400 Ω 60 Ω 40 Ω B 400 Ω 100 Ω 52 V + B - 60 Ω 40 Ω C Chaotic, messy, hard to understand. (Let’s simplify) Restful, symmetric, easy to understand. Parallel and series connections are clear. 28 Power – in series and parallel Total power consumption can be calculated as, 𝑃𝑇 = 𝑉𝑆 × 𝐼 𝑃𝑇 = 𝐼 2 × 𝑅𝑇 𝑃𝑇 = 𝑉 2 Τ𝑅𝑇 Total power consumption in a series/parallel resistive circuit is, 𝑃𝑇 = 𝑃1 + 𝑃2 + 𝑃3 + 𝑃4 + 𝑃5 +……+ 𝑃𝑛 29 Facts - Circuit Geometry ❑ Although, we can often decompose a circuit into series and parallel elements, it is sometimes difficult to recognise ❑ If all the current from one component (or network) goes into one and only one other component (or network), then they are in series. ❑ All the current from one component goes into another component – series. ❑ If both ends of a particular component (or network) are connected to both ends of another components (or network) – with nothing in between – then they are in parallel. ❑ All the voltage across one component is also across the other component – parallel. ❑ Eventually you will get used to this and see what is in series and what is in parallel. In the meantime one technique which can help is to redraw the circuits in standard form. ❑ Start with the highest voltage at the top and the lowest at the bottom. At each node (junction point of components) draw a horizontal line representing a set of points at the same potential (voltage), connected to a branch. This method represent the cascade of voltage (electrochemical potential) as a cascade of height (gravitational potential), in a tree like structure. This picture appeals to the human animal, for some reason! 30 Summary ❑ Introduced Kirchhoff’s laws (KVL and KCL) ❑ Investigated resistors in series and parallel ❑ Complete first live Moodle quiz ❑ Next lecture Thevenin & Norton theorems 31 Lecture 3 -Thevenin & Norton Theorems Learning outcomes 1. Thévenin’s Theorem ❑ How complex network (circuit) can be simplified for load analysis ❑ Circuit consists of ideal voltage source and series resistance 2. Norton’s Theorem ❑ Circuit consists of ideal current source and parallel resistance 3. Thévenin / Norton equivalency ❑ How to convert between the two 2 Previous Lecture ❑ KCL, KVL, Resistors in Series and Parallel ❑ Voltage divider ❑ Current divider Series resistors 𝑅𝑇 = 𝑅1 + 𝑅2 + …… +𝑅𝑛 𝑉𝑅2 𝐼𝑅2 𝑅1 𝑅2 VB = VB 𝑉𝑅1 = 𝑅1 + 𝑅2 𝑅1 + 𝑅2 𝑉𝐵 = = 𝐼𝑡𝑜𝑡 𝑅2 𝑉1 𝑅1 = 𝑉2 𝑅2 𝑅1 𝑅1 + 𝑅2 Parallel resistors 1 1 1 1 = + + ⋯+ 𝑅𝑇 𝑅1 𝑅2 𝑅𝑛 𝑅1 𝑅2 𝑅𝑇 = 𝑅1 + 𝑅2 3 Example 1, Find the current i1 ? Circuit after combining R2 and R3 The current-division principle applies for two resistances in parallel. Therefore, first step is to combine R2 and R3: 𝑅𝑒𝑞 = The current-division principle: 𝑖1 = 𝑖𝑠 𝑅2 𝑅3 𝑅2 +𝑅3 = 𝑅𝑒𝑞 𝑅1 + 𝑅𝑒𝑞 30 × 60 30 + 60 = 15 = 20 Ω 20 10 + 30 = 10 A 4 Example 2 - Solving for equivalent resistance Circuit after combining R3 and R4 Circuit after combining R2 and Req1 Circuit after combining R1 and Req2 5 Example 3 - Find out currents in each branch? 𝑣2 𝑖2 = =2A 𝑅2 𝑖3 = 𝑣2 =1A 𝑅3 𝑅𝑒𝑞1 30 × 60 = = 20 Ω 30 + 60 𝑣1 = 𝑅1 . 𝑖1 = 30 V 𝑣2 = 𝑅𝑒𝑞 . 𝑖1 = 60 V 𝑣𝑠 𝑖1 = = 3A 𝑅𝑒𝑞 6 Example 3 - Find out currents in each branch? To determine currents, 1. Start from Figure (c) 2. Find out voltage V2 in Figure (b) 3. Figure out i2 and i3 applying ohm’s law where resistance and V2 is known. 4. Apply KCL at Node A, to determine, i1 = i2 + i3 (a) (b) (c) 7 Thévenin’s Theorem ❑ Previously been simplifying series and parallel circuits using KVL, KCL, Nodal analysis, and Ohm’s law ❑ Time-consuming process, especially if your circuit has a load value that changes (DC power network) ❑ Léon Charles Thévenin wanted to make complex circuit analysis easier and developed his now famous theorem ❑ Thévenin Theorem to find an equivalent circuit that contains only the load resistance of interest Load 8 Thévenin’s Theorem Thévenin Theorem states that, as far as any external load is concerned, any two port network may be replaced by a voltage source VT in series with a resistance RT A General Network B May be replaced by: + - A B VT (Thévenin voltage) is the open circuit voltage across A &B. RT (Thévenin Resistance) is the resistance that would be measured across the terminal A & B if the voltage source are replaced by short circuits and current source replaced by open circuits) Thévenin’s Theorem This is of practical use, as we have implicitly assumed that energy is supplied to a circuit by either (a) An ideal voltage source- creates a constant voltage or potential difference between its terminals irrespective of current flowing through it + DC voltage source V (a) An ideal current source-drives constant current irrespective of voltage across its terminal I Thévenin’s Theorem In reality, practical generators cannot sustain voltages and current regardless of the load because practical generators have internal resistance. A real generator can be modelled as an ideal generator and a resistor Thévenin’s Theorem In reality, practical generators cannot sustain voltages and current regardless of the load because practical generators have internal resistance. A real generator can be modelled as an ideal generator and a resistor Ro + RL Vo Vo= IRo + IRL (Kirchhoff’s Voltage law) VL= IRL = Vo - IRo The voltage which can be measured at the terminals is the ideal generator voltage minus the voltage dropped across the internal resistance due to the current taken from the generator. Thévenin Equivalent Circuit ❑ Thévenin equivalent consists of an independent voltage source in series with a resistance ❑ By definition, no current can flow through an open circuit, therefore no current flow through Rt ❑ Applying KVL, we conclude that Vt = voc ❑ Consider the short circuit case, the isc is the same as the original and Thévenin equivalent A General Network B 𝑣𝑜𝑐 𝑅𝑡 = 𝐼𝑠𝑐 Find the Thévenin equivalent for the circuit? Find the Thévenin equivalent for the circuit? + 𝑣𝑜𝑐 𝑖1 1. Analysis with an open circuit 𝑣𝑠 15 𝑖1 = = = 0.1 A 𝑅1 + 𝑅2 100 + 50 𝑣𝑜𝑐 = 𝑅2. 𝑖1 = 50 × 0.1 = 5 V Find the Thévenin equivalent for the circuit? 𝑖𝑠𝑐 𝑖1 1. Analysis with an open circuit 2. Analysis with short circuit 𝑣𝑠 15 𝑖1 = = = 0.1 A 𝑅1 + 𝑅2 100 + 50 𝑖𝑠𝑐 = 𝑣𝑠 𝑅1 = 15 100 = 0.15 A 𝑣𝑜𝑐 = 𝑅2. 𝑖1 = 50 × 0.1 = 5 V Find the Thévenin equivalent for the circuit? 𝑖𝑠𝑐 𝑖1 1. Analysis with an open circuit 2. Analysis with short circuit 𝑣𝑠 15 𝑖1 = = = 0.1 A 𝑅1 + 𝑅2 100 + 50 𝑖𝑠𝑐 = 3. Calculate the Thévenin equivalent 𝑣𝑠 𝑅1 = 15 100 𝑅𝑡 = = 0.15 A 𝑣𝑜𝑐 5 = = 33.3 Ω 𝑖𝑠𝑐 0.15 𝑣𝑜𝑐 = 𝑅2. 𝑖1 = 50 × 0.1 = 5 V Find the Thévenin equivalent for the circuit? 𝑅𝐿 = 10 Ω 𝑅𝐿 = 10 Ω 𝑅𝑒𝑞1 = 𝑅2 𝑅𝐿 𝑅2 +𝑅𝐿 = 50 ×10 50 +10 = 8.333 Ω 𝑅𝑒𝑞2 = 100 + 8.333 = 108.333 Ω 15 𝑖1 = = 0.13846A 108.333 𝑣2 = 𝑅𝑒𝑞1 . 𝑖1 = 1.154 V 1.154 𝑖2 = = 0.11 A 10 Thévenin equivalent 𝑉𝑅𝐿 5 × 10 = = 1.154 V 10 + 33.3 5 𝑖𝑅𝐿 = = 0.11 A 44.3 Thévenin Resistance ❑ Alternative approach to directly solve for the Thévenin Resistance ❑ If there are no dependent sources can zero all of the sources within the circuit ❑ Zero voltage source equivalent to a short circuit and zero current source equivalent to an open circuit ❑ Thévenin Resistance is equivalent to the resistance between the terminals 20 Find the Thévenin equivalent for the circuit? Circuit with sources zeroed 𝑅1 𝑅2 5 × 20 𝑅𝑡 = = =4Ω 𝑅1 + 𝑅2 5 + 20 21 Find the Thévenin equivalent for the circuit? Circuit with sources zeroed 𝑅1 𝑅2 5 × 20 𝑅𝑡 = = =4Ω 𝑅1 + 𝑅2 5 + 20 Circuit with short circuit 𝑖3 𝑖1 − 𝑖2 + 𝑖3 = 0 20 𝑣 = 𝑖1 𝑜𝑐= 𝐼𝑠𝑐=×4 𝑅 A𝑡 5 𝑖2 = 0 A 𝑖3 = 2 − 𝑖𝑠𝑐 𝑖𝑠𝑐 = 𝑖1 − 𝑖2 + 2 = 4 − 0 + 2 = 6 A 22 Find the Thévenin equivalent for the circuit? Circuit with sources zeroed 𝑅1 𝑅2 5 × 20 𝑅𝑡 = = =4Ω 𝑅1 + 𝑅2 5 + 20 Circuit with short circuit 𝑖3 𝑖𝑠𝑐 = 6 A 20 𝑣 = 𝑖1 𝑜𝑐= 𝐼𝑠𝑐=×4 𝑅 A𝑡 5 Thévenin equivalent 23 Thévenin’s circuit between different terminals Norton’s Theorem Any network with 2 accessible terminals, may, so far as external loads are concerned, replaced by an ideal current source in parallel with a resistor. Although this is usually called Norton’s theorem, it is not fundamentally different from Thévenin’s Theorem. So, as far as any external load is concerned any two port network may be replaced by a Norton current source,…. IN in parallel with a Norton resistance RN. Norton Equivalent May be replaced by Thevenin Equivalent Find the Thévenin / Norton equivalent for the circuit? 1. Perform two of these a) Determine the open-circuit voltage (𝑉𝑇 = 𝑣𝑜𝑐 ). b) Determine the short-circuit current (𝐼𝑁 = 𝑖𝑠𝑐). c) Zero the independent sources and find the Thévenin resistance RT looking back into the terminals. Do not zero dependent sources 2. Use the equation 𝑉𝑇 = 𝑅𝑇 𝐼𝑁 to compute the remaining value. 3. The Thévenin equivalent consists of a voltage source VT in series with RT. 4. The Norton equivalent consists of a current source 𝐼𝑁 in parallel with RT. Find the Thévenin / Norton equivalent for the circuit? 16 Ω 10 Ω A + 100 V 40 Ω 8Ω B Find the Thévenin / Norton equivalent for the circuit? 16 Ω 10 Ω A + 100 V 40 Ω 8Ω B 1. Analysis with an open circuit 𝑅𝑒𝑞1 40 × [16 + 8] = = 15 Ω 40 + [16 + 8] Find the Thévenin / Norton equivalent for the circuit? 10 Ω A + 100 V 15 Ω B 1. Analysis with an open circuit 𝑅𝑒𝑞1 40 × [16 + 8] 15 = = 15 Ω 𝑉𝑅15 = × 100 = 60 V 40 + [16 + 8] 10 + 15 Find the Thévenin / Norton equivalent for the circuit? 16 Ω 10 Ω A + 100 V 40 Ω 8Ω B 1. Analysis with an open circuit 𝑅𝑒𝑞1 40 × [16 + 8] 15 = = 15 Ω 𝑉𝑅15 = × 100 = 60 V 40 + [16 + 8] 10 + 15 8 𝑉𝑇 = × 60 = 20 V 16 + 8 Find the Thévenin / Norton equivalent for the circuit? 16 Ω 10 Ω A 40 Ω 8Ω B 1. Analysis with an open circuit 𝑅𝑒𝑞1 40 × [16 + 8] 15 = = 15 Ω 𝑉𝑅15 = × 100 = 60 V 40 + [16 + 8] 10 + 15 2. Zero the independent sources 𝑅𝑒𝑞2 10 × 40 = = 8Ω 10 + 40 8 × [16 + 8] 𝑅𝑇 = =6Ω 8 + [16 + 8] 8 𝑉𝑇 = × 60 = 20 V 16 + 8 Find the Thévenin / Norton equivalent for the circuit? 6Ω A Thévenin Equivalent circuit 20V B The Norton resistance is also 6 Ω. The Norton current is that current required to give 20 V over 6 Ω which is 3.33 A. 𝑉𝑇 𝐼𝑠𝑐 = 𝑅𝑇 Norton Equivalent circuit 3.33 A 6Ω As far as any external circuit is concerned, the General network may be replaced by either it’s Norton or Thévenin equivalent, but the Norton and Thévenin equivalents are NOT the same internally. Summary ❑ Discussed Thévenin / Norton theorems and their equivalency ❑ Motivation to reduce complex circuits to study the effect over a single load Terms introduced Thévenin voltage, Thévenin resistance, Norton current Open circuit voltage, Zeroed sources, Short circuit current Next ❑ Complete next ‘live’ Moodle quiz ❑ Next lecture → Superposition, Max Power, and Star-Delta 33 Lecture 4 Superposition, Max Power, Star-Delta Refs. Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 Lecture 4 - Max Power, Superposition Star-Delta Learning outcomes 1. Max Power ❑ How to calculate the optimum load resistance to obtain maximum power transfer 2. Superposition ❑ Circuit with multiple sources is equal to the sum of simplified circuits using just one of the sources 3. Star-Delta transformation ❑ Technique to simplify circuits that contain elements that cannot be classed as in series or parallel References: 1) Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. 2) Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 2 Previous Lecture 𝑹𝒕 = 𝒗𝒐𝒄 𝑰𝒔𝒄 Find the Thévenin / Norton equivalent for the circuit? 1. Perform two of these a) Determine the open-circuit voltage (𝑉𝑇 = 𝑣𝑜𝑐). b) Determine the short-circuit current (𝐼𝑁 = 𝑖𝑠𝑐). c) Zero the independent sources and find the Thévenin resistance RT looking back into the terminals. Do not zero dependent sources 2. Use the equation 𝑉𝑇 = 𝑅𝑇 𝐼𝑁 to compute the remaining value. 3. The Thévenin equivalent consists of a voltage source VT in series with RT. 4. The Norton equivalent consists of a current source IN in parallel with RT. 𝑰𝑵 = 𝑰𝒔𝒄 A B 3 Maximum Power Theorem Power in load is: 𝑉𝑡 𝑖𝐿 = 𝑅𝑡 + 𝑅𝐿 𝑃𝐿 = 𝑖𝐿2 × 𝑅𝐿 𝑉𝑡2 𝑅𝐿 𝑃𝐿 = 𝑅𝑂 + 𝑅𝐿 2 How do we calculate the maximum power ? 4 Maximum Power Theorem We need to know how the power, 𝑃𝐿 changes with RL. This looks like a job for Calculus! The rate of change of power with respect to RL is: 𝑑𝑃𝐿 𝑑 = 𝑑𝑅𝐿 𝑑𝑅𝐿 𝑉𝑡2 𝑅𝐿 𝑅𝑂 + 𝑅𝐿 2 =0 Now, for a very low resistance not much power will be dissipated in the load. For zero resistance there will be zero power. Also for a very high resistance there will be little power dissipated, for an infinite resistance this would be zero since the current would be zero. So increasing the load resistance from zero the power will go up, reach a maximum and then go down. Mathematically, if we can find the point where it just stops going up, and has not quite started going down, this will be the point of maximum power. That is the maximum power occurs when the rate of change of power is zero. 5 Maximum Power Theorem 𝑔 𝑥 = 𝑉𝑡2 𝑅𝐿 Quotient rule 𝑑𝑃𝐿 𝑑 = 𝑑𝑅𝐿 𝑑𝑅𝐿 𝑓′ 𝑥 = 𝑉𝑡2 𝑅𝐿 𝑅𝑡 + 𝑅𝐿 𝑉𝑡2 𝑅𝑡 + 𝑅𝐿 𝑓 𝑥 = 𝑉𝑡2 𝑥 ℎ 𝑥 − 𝑔 𝑥 ℎ′ 𝑥 ℎ(𝑥)2 − 2𝑉𝑡2 𝑅𝐿 𝑅𝑡 + 𝑅𝐿 𝑅𝑡 + 𝑅𝐿 4 1 𝑅𝑡 + 𝑅𝐿 −2 − 2𝑅𝐿 (𝑅𝑡 + 𝑅𝐿 )−3 ℎ 𝑥 = 𝑅𝑂 + 𝑅𝐿 2 𝑔′ 𝑥 = 𝑉𝑡2 ℎ′ 𝑥 = 2 𝑅𝑂 + 𝑅𝐿 2 𝑓 ′ 𝑥 = 𝑉𝑡2 𝑅𝑡 + 𝑅𝐿 ′ 2 𝑓′ 𝑥 = 𝑔′ This equals zero when 2𝑅𝐿 𝑅𝑡 + 𝑅𝐿 − 2𝑅𝐿 2 − = 𝑉𝑡 2 (𝑅𝑡 + 𝑅𝐿 )3 (𝑅𝑡 + 𝑅𝐿 )3 𝑅𝑡 + 𝑅𝐿 − 2𝑅𝐿 = 0 𝑹𝑳 = 𝑹𝒕 Under these conditions RL is said to be matched to the output resistance (better: output impedance) of the source. Often will wish to design circuits and systems to achieve this. 6 Find when the maximum power is dissipated in the load? R1 100 Ω + 400 Ω R2 Vs - RL Lets try an example using Thévenin equivalent circuit. Split the circuit into the components under the study and everything else. Find the Thévenin (or Norton, if convenient) equivalent of everything else. R1 + Vs - A 100 Ω 400 Ω R2 RL B 7 Find when the maximum power is dissipated in the load? To determine the Thévenin resistance, short-circuit the voltage source giving R1 and R2 in parallel. R1 + Vs - A 100 Ω 400 Ω R2 B Resistance between A and B, is the Thévenin resistance, 𝑅𝑡ℎ 𝑅1 × 𝑅2 = = 80Ω 𝑅1 + 𝑅2 80 Ω So, the original circuit is equivalent to, Vth + - RL 𝑹𝑳 = 𝑹𝒕 So, maximum power dissipation will occur at 80 Ω. Note that, we do not need to know the voltage of the source to work this out. 8 Find when the maximum power is dissipated in the load? The Thévenin voltage is the open circuit output voltage, which in this case will be, A + Vs 100 Ω 400 Ω 𝑉𝑡ℎ = - 400 4 𝑉𝑠 = 𝑉𝑠 100 + 400 5 B If 𝑉𝑠 =10V, for example, the V𝑡ℎ will be 8V. This is the last example from Lab 1 In this case the Maximum power will be, 𝑉𝐿2 80 P= = 𝑉 𝑅𝐿 80 + 80 𝑡ℎ 2 1 × = 0.2 𝑊 𝑅𝐿 9 Superposition principle In a “linear” circuit the response to independent sources is the sum of the response to each sources alone, with the other sources zeroed (infinite Ω for the current and 0 Ω for voltage sources). 𝑟𝑇 = 𝑟1 + 𝑟2 + …… +𝑟𝑛 10 Superposition principle R2 // R3 6A + 24V - 𝑅𝑒𝑞1 = 4A 1×2 2 = 1+2 3 + 4V - 2A + 4V - 𝑖1 𝑅𝑒𝑞1 = 0.667 Ω 𝑖1 = 6 A R1 R2 R3 V 24 4 4 Volts I 6 2 4 Amps R 4 2 1 Ohms 11 Superposition principle 1A - 4V + 𝑅𝑒𝑞1 R1 // R2 4×2 8 = = 4+2 6 3A - 3V + 2A + 4V - 𝑅𝑒𝑞1 = 1.333 Ω 𝑖1 𝑖1 = 3 A R1 R2 R3 V 4 4 3 Volts I 1 2 3 Amps R 4 2 1 Ohms 12 Superposition principle When superimposing the values of current and volage, we have to be very careful to consider the polarity (voltage drop) and direction (current flow), as these values are added algebraically 13 Superposition principle 5A 𝑉𝑅1 = 24 − 4 = 20 V 𝑉𝑅3 = 4 − 3 = 1 V 4A + 1V - + 8V - 𝑉𝑅2 = 4 + 4 = 8 V + 20V - 1A 𝐼𝑅1 = 6 − 1 = 5 A 𝐼𝑅2 = 2 + 2 = 4 A 𝐼𝑅3 = 4 − 3 = 1 A 14 Find the current through R2 using superposition? 𝐼𝑅2(𝑉𝑠) 𝑉𝑠 10 = = = 31.25 mA 𝑅1 + 𝑅2 320 𝐼𝑅2(𝐼𝑠) = 𝐼𝑆 𝑅1 𝑅1 + 𝑅2 220 = × 100 mA = 68.75 mA 320 𝐼𝑅2(𝑉𝑠) + 𝐼𝑅2(𝐼𝑠) = 100 mA 15 Delta and Star (Wye) Circuit Many circuits have elements which are neither in series or in parallel. For example, the circuit below can not be decomposed into series and parallel elements: T-network Pi-network Delta-network Star-network Delta – Star or Star - Delta conversion 16 Delta – Star (Wye) conversion 𝑅𝐴𝐵 = 𝑅1 ∥ 𝑅2 + 𝑅3 1. 𝑅1 × 𝑅2 + 𝑅3 = 𝑅1 + 𝑅2 + 𝑅3 𝑅𝐴 + 𝑅𝐵 = 𝑅1 × 𝑅2 + 𝑅3 𝑅1 + 𝑅2 + 𝑅3 𝑅𝐴𝐵 = 𝑅𝐴 + 𝑅𝐵 17 Delta – Star (Wye) conversion 1. 𝑅1 × 𝑅2 + 𝑅3 𝑅𝐴 + 𝑅𝐵 = 𝑅1 + 𝑅2 + 𝑅3 2. 𝑅2 × 𝑅1 + 𝑅3 𝑅𝐵 + 𝑅𝐶 = 𝑅1 + 𝑅2 + 𝑅3 3. 𝑅3 × 𝑅1 + 𝑅2 𝑅𝐶 + 𝑅𝐴 = 𝑅1 + 𝑅2 + 𝑅3 Equation 1+2+3 2 𝑅1 . 𝑅2 + 𝑅2 . 𝑅3 + 𝑅3 . 𝑅1 2 𝑅𝐴 + 𝑅𝐵 + 𝑅𝐶 = 𝑅1 + 𝑅2 + 𝑅3 4. 𝑅𝐴 + 𝑅𝐵 + 𝑅𝐶 𝑅1 . 𝑅2 + 𝑅2 . 𝑅3 + 𝑅3 . 𝑅1 = 𝑅1 + 𝑅2 + 𝑅3 𝑅𝐴𝐵 = 𝑅𝐴 + 𝑅𝐵 𝑅𝐵𝐶 = 𝑅𝐵 + 𝑅𝐶 𝑅𝐶𝐴 = 𝑅𝐶 + 𝑅𝐴 Subtract equations 1,2, and 3 from 4 𝑅3 × 𝑅1 𝑅𝐴 = 𝑅1 + 𝑅2 + 𝑅3 𝑅𝐵 = 𝑅𝐶 = 𝑅2 × 𝑅3 𝑅1 + 𝑅2 + 𝑅3 𝑅1 × 𝑅2 𝑅1 + 𝑅2 + 𝑅3 18 Star (Wye) - Delta conversion 𝑅3 𝑅1 𝑅𝐴 𝑅1 + 𝑅2 + 𝑅3 𝑅3 𝑅1 𝑅3 = = = 𝑅1 𝑅2 𝑅𝐵 𝑅1 𝑅2 𝑅2 𝑅1 + 𝑅2 + 𝑅3 𝑅3 = 𝑅𝐴 𝑅2 𝑅𝐵 𝑅𝐴 = 𝑅3 𝑅1 𝑅1 + 𝑅2 + 𝑅3 𝑅𝐵 = 𝑅1 𝑅2 𝑅1 + 𝑅2 + 𝑅3 𝑅𝐶 = 𝑅2 𝑅3 𝑅1 + 𝑅2 + 𝑅3 𝑅3 𝑅1 𝑅𝐴 𝑅1 + 𝑅2 + 𝑅3 𝑅3 𝑅1 𝑅1 = = = 𝑅2 𝑅3 𝑅𝐶 𝑅2 𝑅3 𝑅2 𝑅1 + 𝑅2 + 𝑅3 𝑅1 = 𝑅𝐴 𝑅2 𝑅𝐶 Substitute into RA to find expression for R2 𝑅𝐴 𝑅𝐵 + 𝑅𝐵 𝑅𝐶 + 𝑅𝐶 𝑅𝐴 𝑅1 = 𝑅𝐶 𝑅2 = 𝑅𝐴 𝑅𝐵 + 𝑅𝐵 𝑅𝐶 + 𝑅𝐶 𝑅𝐴 𝑅𝐴 𝑅3 = 𝑅𝐴 𝑅𝐵 + 𝑅𝐵 𝑅𝐶 + 𝑅𝐶 𝑅𝐴 𝑅𝐵 19 Delta – Star / Star - Delta conversion R3 and R1 are adjacent to RA: R1 is opposite to RC: 𝑅3 × 𝑅1 𝑅𝐴 = 𝑅1 + 𝑅2 + 𝑅3 𝑅𝐴 𝑅𝐵 + 𝑅𝐵 𝑅𝐶 + 𝑅𝐶 𝑅𝐴 𝑅1 = 𝑅𝐶 R2 and R1 are adjacent to RB: R2 is opposite to RA: 𝑅1 × 𝑅2 𝑅𝐵 = 𝑅1 + 𝑅2 + 𝑅3 𝑅2 = R3 and R2 are adjacent to RC: R3 is opposite to RB: 𝑅2 × 𝑅3 𝑅𝐶 = 𝑅1 + 𝑅2 + 𝑅3 𝑅𝐴 𝑅𝐵 + 𝑅𝐵 𝑅𝐶 + 𝑅𝐶 𝑅𝐴 𝑅3 = 𝑅𝐵 𝑅𝐴 𝑅𝐵 + 𝑅𝐵 𝑅𝐶 + 𝑅𝐶 𝑅𝐴 𝑅𝐴 20 Find the equivalent resistance between top and bottom terminal ? R2 B R3 R1 𝑅3 × 𝑅1 20 𝑅𝐴 = = = 1.667 Ω 𝑅1 + 𝑅2 + 𝑅3 12 𝑅1 × 𝑅2 15 𝑅𝐵 = = = 1.25 Ω 𝑅1 + 𝑅2 + 𝑅3 12 𝑅𝐶 = C A 𝑅2 × 𝑅3 12 = =1Ω 𝑅1 + 𝑅2 + 𝑅3 12 21 Find the equivalent resistance between top and bottom terminal ? Circuit now contains only series and parallel connections 𝑅𝑒𝑞 3.125 + 1.25 × 4 + 1 = + 1.66 = 4 Ω 3.125 + 1.25 + 4 + 1 22 Summary ❑ Derived the expression for maximum power transfer to a single load ❑ Utilised superposition theorem to analyse circuits that contain multiple sources ❑ Demonstrated Delta – Star (Wye) transformation for simplifying circuits Next ❑ Complete Moodle quiz ❑ Next lecture, Mesh-Current Analysis 23 Lecture 5 - Mesh-Current Analysis Maximum Power Previous Lecture 4 𝑹𝑳 = 𝑹𝒕 𝑑𝑃𝐿 𝑑 = 𝑑𝑅𝐿 𝑑𝑅𝐿 𝑉𝑡2 𝑅𝐿 𝑅𝑡 + 𝑅𝐿 2 𝑉𝑡2 𝑅𝑡 + 𝑅𝐿 − 2𝑅𝐿 (𝑅𝑡 + 𝑅𝐿 )3 Delta – Star conversion Superposition 2 Norton and Thevenin Questions for credit Long form Quizzes ❑ ❑ ❑ ❑ 5 quizzes (4 for credit and 1 practice) First live credit quiz is the “Norton and Thevenin” 12 questions in total Opened 21st October with deadline 31st October Be careful with units, answers for resistance in Ohms 3 Learning outcomes Mesh-current analysis ❑ Methodology ❑ Solving by determinants ❑ Examples and activities References: 1) Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. 2) Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 4 Mesh-current analysis If the source voltages and resistances are known and we wish to solve for the currents then Mesh-current analysis can be used. 6 Mesh-current analysis ❑ Loop currents are mathematical quantities, rather than actual physical currents, that are used to make circuit analysis easier. ❑ In general, the number of independent KVL (Kirchhoff’s Voltage Law) equations is equal to the number of open areas defined by the network layout. For instance, the above circuit has two open areas: ❑ One defined by vA , R1 and R3 ❑ Another defined by R3 , R2 and vB ❑ So for this network we can write 2 independent KVL equations 7 Mesh-current analysis Step 1: Assign loop currents The direction of an assigned loop current is arbitrary. Using a pattern in solving networks by the meshcurrent method helps to avoid errors. Part of the pattern that we usually choose the current variables to flow clockwise (CW) around the periphery of each of the open areas of the circuit diagram. ❑ This may not be the actual current direction, but it does not matter. ❑i1 and i2 are assigned in the CW direction as shown below. 8 Mesh-current analysis Step 2: Apply Kirchhoff’s voltage law around each loop Now, we write a KVL equation for each mesh, going around the meshes clockwise. In this case, the pattern in solving networks will be: ❑ Always to take the first end of each resistor encountered as the positive reference for its voltage. Thus, we are always adding the resistor voltages. ❑ We add a voltage if its positive reference is encountered first in traveling around the mesh, and we subtract the voltage if the negative reference is encountered first. 𝑅1 𝑖1 + 𝑅3 𝑖3 = 𝑉𝐴 −𝑅3 𝑖3 + 𝑅2 𝑖2 = −𝑉𝐵 9 Mesh-current analysis Step 2: Apply Kirchhoff’s voltage law around each loop Now, we write a KVL equation for each mesh, going around the meshes clockwise. In this case, the pattern in solving networks will be: ❑ Always to take the first end of each resistor encountered as the positive reference for its voltage. Thus, we are always adding the resistor voltages. ❑ We add a voltage if its positive reference is encountered first in traveling around the mesh, and we subtract the voltage if the negative reference is encountered first. 𝑖1 𝑖3 𝑖2 𝑅1 𝑖1 + 𝑅3 𝑖1 − 𝑖2 = 𝑉𝐴 −𝑅3 𝑖1 − 𝑖2 + 𝑅2 𝑖2 = −𝑉𝐵 𝑖3 = 𝑖1 − 𝑖2 10 Mesh-current analysis 𝑅1 𝑖1 + 𝑅3 𝑖1 − 𝑖2 = 𝑉𝐴 −𝑅3 𝑖1 − 𝑖2 + 𝑅2 𝑖2 = −𝑉𝐵 ❑ Keep in mind that, if two mesh currents flow through a circuit element, we consider the current in that element to be the algebraic sum of the mesh currents (e.g. the current flowing through R3) ❑ By using this pattern, we ‘add’ a term for each resistor in the KVL equation, consisting of the resistance times the current in the mesh under consideration minus the current in the adjacent mesh (if any). 11 Mesh-current analysis Step 3: Develop the loop equations Combine and rearrange the like terms in the equations into standard form for convenient solution so that they have the same position in each equation, that is, the i1 term is first and the i2 term is second. 𝑅1 𝑖1 + 𝑅3 𝑖1 − 𝑖2 = 𝑉𝐴 −𝑅3 𝑖1 − 𝑖2 + 𝑅2 𝑖2 = −𝑉𝐵 𝑅1 + 𝑅3 𝑖1 − 𝑅3 𝑖2 = 𝑉𝐴 −𝑅3 𝑖1 + 𝑅2 + 𝑅3 𝑖2 = −𝑉𝐵 ❑ Notice that for mesh 1, the total resistance in the mesh (R1+R3) is multiplied by its loop current (𝑖1 ). Also in the mesh 1 equation, the resistance common to both loops (R3) is multiplied by the other loop current (𝑖2 ), and subtracted from the first term. The same form is seen in the mesh 2 equation except that the terms have been rearranged. 12 Mesh-current analysis From these observations, a concise rule for applying steps 1 to 3 is as follows: (Sum of resistors in loop) × (loop current) − (each resistor common to both loops) × (adjacent loop current) = (source voltage in the loop) 𝑅1 + 𝑅3 𝑖1 − 𝑅3 𝑖2 = 𝑉𝐴 −𝑅3 𝑖1 + 𝑅2 + 𝑅3 𝑖2 = −𝑉𝐵 ❑ Once the loop currents are evaluated, all of the branch currents can be determined. ❑ Once you know the branch currents, you can find the voltages by using Ohm’s law. 13 Mesh-current analysis Step 4: Solve the loop equations Using substitution or determinants, solve the resulting equations for the loop currents. Example with numerical values R1 470 Ω R2 820 Ω R3 220 Ω VA 10 V VB 5V Loop 2 Loop 1 𝑅1 + 𝑅3 𝑖1 − 𝑅3 𝑖2 = 𝑉𝐴 −𝑖1 𝑅3 + 𝑅2 + 𝑅3 𝑖2 = −𝑉𝐵 470 + 220 𝑖1 − 220𝑖2 = 10 −220𝑖1 + 220 + 820 𝑖2 = −5 690𝑖1 − 220𝑖2 = 10 −220𝑖1 + 1040𝑖2 = −5 14 Solving simultaneous equations Different techniques available such as substitution, elimination, and matrix methods (Gaussian elimination and Determinants method) Elimination: (1) 690𝑖1 − 220𝑖2 = 10 (2) −220𝑖1 + 1040𝑖2 = −5 69 1 + 2 𝑥 22 3041.82𝑖2 = −5.682 −5.682 𝑖2 = = −1.87 mA 3041.82 Substitute 𝑖2 into (1) or (2) 𝑖1 = 13.9 mA 15 Solving by determinants ❑ The determinant method is a part of matrix algebra and provides a “cookbook” approach for solving simultaneous equations with two or three variables. ❑ A matrix is an array of numbers, and a determinant is effectively the matrix solution to a matrix, resulting in a specific value. ❑ Second-order determinants are used for two variables and third-order determinants are used for three variables. The equations must be in standard form for a solution. ❑ To illustrate the determinant method for second-order equations, let’s find the values of I1 and I2 in the following two equations expressed in standard form. The coefficients are resistance values in ohms, and the constants are voltage values in volts. 𝑅 × [𝐼] [𝑉] 10 5 I1 15 . = 2 4 I2 2 Cramer’s rule det(𝑅𝑖 ) 𝐼𝑖 = det(𝑅) 16 Solving by determinants First, form the characteristic determinant from the matrix of the coefficients of the unknown currents. ❑ The first column in the determinant consists of the coefficients of I1 and the second column consists of the coefficients of I2 ❑ The resulting determinant is: 2x2 An evaluation of this characteristic determinant requires three steps Step 1 Step 2 A = ad − bc Step 3 17 Solving by determinants ❑ Next, replace the coefficients of I1 in the first column of the characteristic determinant with the constants (fixed numbers) on the right side of the equations to form another determinant. Then, evaluate this I1 determinant as follows: ❑ Now solve for I1 by dividing the determinant by the characteristic determinant as follows: det(𝑅𝑖 ) 𝐼𝑖 = det(𝑅) 18 Solving by determinants ❑ To find I2 form another determinant by substituting the constants on the right side of the given equations for the coefficients of I2 in the second column of the characteristic determinant. 19 Mesh-current analysis Coming back to Step 4: i.e. Solve the loop equations Using substitution or determinants, solve the resulting equations for the loop currents. IA Loop 1 Loop 2 690𝑖1 − 220𝑖2 = 10 −220𝑖1 + 1040𝑖2 = −5 690 −220 I1 10 . = −220 1040 I2 −5 20 Mesh-current analysis ❑ Using determinants to find I1 : 690 −220 10 = −220 1040 −5 10 −220 10 1040 − −5 −220 104000 − 1100 102900 −5 1040 𝐼1 = = = = = 13.9 mA 690 −220 690 1040 − −220 −220 717600 − 48400 669200 −220 1040 ❑ Solving for I2 yields: 690 10 690 −5 − −220 10 −3450 − (−2200) −1250 −220 −5 𝐼2 = = = = = −1.87 mA 669200 669200 669200 669200 Negative sign on I2 means that its assigned direction is opposite to the actual current. Branch currents: ❑ I1 is the only current through R1 → IR1 = I1 =13.9 mA ❑ I2 is the only current through R2 → IR2 = I2 = – 1.87 mA (opposite direction) ❑ Both loop currents I1 and I2 are through R3 in opposite directions → → IR3= I1 – I2 = 13.9 mA – (– 1.87 mA ) = 15.8 mA 21 Mesh-current analysis Can reduce analysis complexity by ignoring potential voltage across resistor elements Loop 1: 𝑅1 𝑖1 + 𝑅3 𝑖1 − 𝑖2 = 𝑉𝐴 Loop 2: −𝑅3 𝑖1 − 𝑖2 + 𝑅2 𝑖2 = −𝑉𝐵 Loop 1: 𝑅1 + 𝑅3 𝑖1 − 𝑅3 𝑖2 = 𝑉𝐴 Loop 2: −𝑅3 𝑖1 + 𝑅2 + 𝑅3 𝑖2 = −𝑉𝐵 22 Mesh-current analysis Can reduce analysis complexity by ignoring potential voltage across resistor elements Loop 1: 𝑅1 𝑖1 + 𝑅3 𝒊𝟏 − 𝒊𝟐 = 𝑉𝐴 𝑅1 𝑖1 + 𝑅3 𝒊𝟏 − 𝒊𝟐 = 𝑉𝐴 Loop 2: −𝑅3 𝒊𝟏 − 𝒊𝟐 + 𝑅2 𝑖2 = −𝑉𝐵 𝑅3 𝒊𝟐 − 𝒊𝟏 + 𝑅2 𝑖2 = −𝑉𝐵 Loop 1: 𝑅1 + 𝑅3 𝑖1 − 𝑅3 𝑖2 = 𝑉𝐴 Loop 2: −𝑅3 𝑖1 + 𝑅2 + 𝑅3 𝑖2 = −𝑉𝐵 𝑅1 + 𝑅3 𝑖1 − 𝑅3 𝑖2 = 𝑉𝐴 −𝑅3 𝑖1 + 𝑅2 + 𝑅3 𝑖2 = −𝑉𝐵 23 Advantage of Mesh-current analysis ❑ The mesh-current analysis method can be systematically applied to circuits with any number of loops. ❑ Primary advantage over branch current method, solution requires fewer unknown values and simultaneous equations 24 Advantage of Mesh-current analysis ❑ The mesh-current analysis method can be systematically applied to circuits with any number of loops. ❑ Primary advantage over branch current method, solution requires fewer unknown values and simultaneous equations 5 unknowns and 5 simultaneous eqns 𝐼1 − 𝐼2 − 𝐼3 = 0 𝐼3 − 𝐼4 − 𝐼5 = 0 −𝐸𝐵1 + 𝐼2 𝑅2 + 𝐼1 𝑅1 = 0 −𝐼2 𝑅2 + 𝐼3 𝑅3 + 𝐼4 𝑅4 = 0 −𝐼4 𝑅4 + 𝐸𝐵2 − 𝐼5 𝑅5 = 0 25 Advantage of Mesh-current analysis ❑ The mesh-current analysis method can be systematically applied to circuits with any number of loops. ❑ Primary advantage over branch current method, solution requires fewer unknown values and simultaneous equations −𝑉𝐵1 + 𝑅2 𝐼1 − 𝐼2 + 𝐼1 𝑅1 = 0 𝑅2 𝐼2 − 𝐼1 + 𝑅4 𝐼2 − 𝐼3 + 𝑅3 𝐼2 = 0 𝑅4 𝐼3 − 𝐼2 + 𝑉𝐵2 + 𝐼3 𝑅5 = 0 3 unknowns and 3 simultaneous eqns 26 Solve for the current in each element of the following circuit? R4 R3 R1 VA R2 VB 𝑅1 𝑖1 − 𝑖3 + 𝑅2 𝑖1 − 𝑖2 = 𝑉𝐴 𝑅2 𝑖2 − 𝑖1 + 𝑅3 𝑖2 − 𝑖3 = −𝑉𝐵 𝑖3 𝑅4 + 𝑅3 𝑖3 − 𝑖2 + 𝑅1 𝑖3 − 𝑖1 = 0 𝑅1 + 𝑅2 𝑖1 − 𝑅2 𝑖2 − 𝑅1 𝑖3 = 𝑉𝐴 -𝑅2 𝑖1 + 𝑅2 + 𝑅3 𝑖2 − 𝑅3 𝑖3 = −𝑉𝐵 −𝑅1 𝑖1 − 𝑅3 𝑖2 + 𝑅1 + 𝑅3 + 𝑅4 𝑖3 = 0 Loop 1: Loop 2: Loop 3: 𝑅1 + 𝑅2 −𝑅2 −𝑅1 − 𝑅2 𝑅2 + 𝑅3 −𝑅3 −𝑅1 𝑖1 𝑉𝐴 −𝑅3 . 𝑖2 = −𝑉𝐵 𝑅1 + 𝑅3 + 𝑅4 𝑖3 0 27 Find when the maximum power is dissipated in the load? In matrix form: 𝑅1 + 𝑅2 −𝑅2 −𝑅1 − 𝑅2 𝑅2 + 𝑅3 −𝑅3 −𝑅1 𝑖1 𝑉𝐴 −𝑅3 . 𝑖2 = −𝑉𝐵 𝑅1 + 𝑅3 + 𝑅4 𝑖3 0 Often, we use R to represent the coefficient matrix, I to represent the column vector of mesh currents, and V to represent the column vector of the terms on the righthand sides of the equations in standard form. Then, the meshcurrent equations are represented as RI=V After we have obtained the equations in standard form, we can solve them by a variety of methods, including substitution, Gaussian elimination, and determinants. 28 Solve for the current in each element of the following circuit? R4 3x3 R3 R1 A = a ei − fh − b di − fg + c(dh − eg) 30 −10 −20 − 10 22 −12 −20 𝑖1 70 −12 . 𝑖2 = −42 𝑖3 46 0 VA R2 VB A = 26040 − 7000 + −11200 = 7840 29 Solve for the current in each element of the following circuit? R4 3x3 R3 R1 A = a ei − fh − b di − fg + c(dh − eg) 30 −10 −20 − 10 22 −12 −20 𝑖1 70 −12 . 𝑖2 = −42 𝑖3 46 0 VA R2 VB 30((22)(46) − (−12)(−12)) = 26040 A = 26040 − 7000 + −11200 = 7840 30 Solve for the current in each element of the following circuit? R4 3x3 R3 R1 A = a ei − fh − b di − fg + c(dh − eg) 30 −10 −20 − 10 22 −12 −20 𝑖1 70 −12 . 𝑖2 = −42 𝑖3 46 0 VA R2 VB 30((22)(46) − (−12)(−12)) = 26040 −10((−10)(46) − (−12)(−20)) = 7000 A = 26040 − 7000 + −11200 = 7840 31 Solve for the current in each element of the following circuit? R4 3x3 R3 R1 A = a ei − fh − b di − fg + c(dh − eg) 30 −10 −20 − 10 22 −12 −20 𝑖1 70 −12 . 𝑖2 = −42 𝑖3 46 0 VA R2 VB 30((22)(46) − (−12)(−12)) = 26040 −10((−10)(46) − (−12)(−20)) = 7000 A = 26040 − 7000 + −11200 = 7840 −20((−10)(−12) − (22)(−20)) = −11200 32 Solve for the current in each element of the following circuit? R4 3x3 R3 R1 A = a ei − fh − b di − fg + c(dh − eg) 30 −10 −20 − 10 22 −12 −20 𝑖1 70 −12 . 𝑖2 = −42 𝑖3 46 0 VA R2 VB 30((22)(46) − (−12)(−12)) = 26040 −10((−10)(46) − (−12)(−20)) = 7000 A = 26040 − 7000 + −11200 = 7840 −20((−10)(−12) − (22)(−20)) = −11200 33 Solve for the current in each element of the following circuit? A1 31360 𝐼1 = 70 −42 0 30 −10 −20 − 10 22 −12 70 −42 0 −20 −12 46 −20 −12 46 70 22 46 − −12 −12 A − 10 22 −12 70 −42 0 7840 = 4A = 60760 −10((−42)(46) − (−12)(0)) = 19320 A1 = 60760 − 19320 + −10080 = 31360 −20((−42)(−12) − (22)(0)) = −10080 30 −42 46 − −12 0 = −57960 70((−10)(46) − (−12)(−20)) = −49000 A2 7840 𝐼2 = = = 1A A 7840 A2 = −57960 − −49000 + 16800 = 7840 −20((−10)(0) − (−42)(−20)) = 16800 𝐼3 = 30 −10 −20 = 30((22)(0) − (−42)(−12)) = −15120 −10((−10)(0) − (−42)(−20)) = 8400 A3 15680 = = 2A A 7840 A3 = −15120 − 8400 + 39200 = 15680 70((−10)(−12) − (22)(−20)) = 39200 34 Solve for the current in each element of the following circuit? R4 IR1= 4 – 2 =2 A IR2= 4 – 1 =3 A R3 R1 VA IR3= 4 – 2 =2 A R2 VB IR4 =2 A 𝐼1 = 4 A 𝐼2 = 1 A 𝐼3 = 2 A 35 Mesh-current analysis Mesh Currents in Circuits Containing Current Sources ❑ Recall that a current source forces a specified current to flow through its terminals, but the voltage across its terminals is not predetermined. ❑ Instead, the voltage across a current source depends on the circuit to which the source is connected. ❑ A common mistake made by beginning students is to assume that the voltages across current sources are zero. ❑ Consider the following circuit: 36 Mesh-current analysis Mesh Currents in Circuits Containing Current Sources ❑ If we were to try to write a KVL equation for mesh 1, we would need to include an unknown for the voltage across the current source. However, we avoid writing KVL equations for loops that include current sources. ❑ In this circuit, we have defined the current in the current source as i1. However, we know that this current is 2 A. Thus, we can directly write: ❑ The second equation needed can be obtained by applying KVL to mesh 2: ❑ Notice that in this case the presence of a current source facilitates the solution. 37 Write the equations needed to solve for the mesh currents in the following circuit? Loop I1 𝑖1 = −5 A Loop I2 10 𝑖2 − 𝑖1 + 5𝑖2 − 100 = 0 15𝑖2 − 10𝑖1 = 100 15𝑖2 − (10 × −5) = 100 100 − 50 𝑖2 = = 3.33 A 15 38 Mesh-current method 1. Identify all of the individual meshes in the circuit. 2. Assign a mesh current to each mesh. Identify meshes in which the current is known because there is a current source in an outside branch of the mesh. 3. Assign voltages to all of the elements in the meshes with unknown currents. 4. Use KVL around each mesh to write loop equation. 5. Use Ohm’s law to write the resistor voltages in terms of the mesh currents. 6. Insert the voltage expressions into the KVL equations to arrive at a set of mesh current equations. If there are n unknown currents, there should be n equations relating them. 7. Solve the system of equations to find the mesh currents. 39 Summary ❑ Investigated Mesh current analysis to solve arbitrary circuits ❑ Can be applied to circuits that cannot be easily simplified into series / parallel ❑ Leads to fewer unknown variables and simultaneous equations Next ❑ Complete Moodle quiz ❑ Next lecture, Node-Voltage Analysis 40 Lecture 6- Node-Voltage Analysis & Wheatstone Bridge Previous Lecture Mesh Current Analysis 1. Identify all of the individual meshes in the circuit. 2. Assign a mesh current to each mesh. Identify meshes in which the current is known because there is a current source in an outside branch of the mesh. 3. Assign voltages to all of the elements in the meshes with unknown currents. 4. Use KVL around each mesh to write loop equation. 5. Use Ohm’s law to write the resistor voltages in terms of the mesh currents. 6. Insert the voltage expressions into the KVL equations to arrive at a set of mesh current equations. If there are n unknown currents, there should be n equations relating them. 7. Solve the system of equations to find the mesh currents. 2 Previous Lecture Mesh Current Analysis 3 Learning outcomes Node-voltage analysis ❑ Methodology ❑ Examples and activities Technique is a mirror image of the mesh current method References: 1) Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. 2) Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 4 Node-voltage analysis Consider the circuit. It looks easy, but we are quickly disappointed when we find that no shortcut methods work. Using KCL: 𝑖𝑅4 b a c 𝑖𝑅3 𝑖𝑉𝑆 𝑖𝑅2 𝑖𝑅1 a) 𝐼𝑉𝑆 = 𝐼𝑅1 + 𝐼𝑅4 b) 𝐼𝑅1 + 𝐼𝑅3 = 𝐼𝑅2 c) 𝐼𝑅4 + 𝐼𝑆 = 𝐼𝑅3 d) 𝐼𝑅2 = 𝐼𝑆 = 𝐼𝑉𝑆 5 unknowns and 4 simultaneous eqns d Imagine we have not yet learned mesh current method 5 Node-voltage analysis + 𝑣𝑅4 − Ohm’s law to write resistor currents in terms of the resistor voltages. (Pay attention to polarities. For the resistors, the voltage polarities must match the chosen current directions: a) 𝐼𝑉𝑆 = b) c) 𝑉𝑅4 𝑉𝑅3 + 𝐼𝑆 = 𝑅4 𝑅3 d) 𝑉𝑅2 = 𝐼𝑆 + 𝐼𝑉𝑆 𝑅2 c 𝑖𝑅1 𝑖𝑉𝑆 𝑉𝑅1 𝑉𝑅4 + 𝑅1 𝑅4 𝑉𝑅1 𝑉𝑅3 𝑉𝑅2 + = 𝑅1 𝑅3 𝑅2 a 𝑖𝑅4 + 𝑣𝑅1 − b − 𝑣𝑅3 + 𝑖𝑅3 + 𝑣𝑅2 − 𝑖𝑅2 d Still 5 unknowns and 4 simultaneous eqns 6 Node-voltage analysis + 𝑣𝑅4 − We can take another step and assign a voltage to each node. We can then write the resistor voltages as differences between the node voltages a) 𝐼𝑉𝑆 = b) 𝑉𝑎 − 𝑉𝑏 𝑉𝑎 − 𝑉𝑐 + 𝑅1 𝑅4 𝑉𝑎 − 𝑉𝑏 𝑉𝑐 − 𝑉𝑏 𝑉𝑏 − 𝑉𝑑 + = 𝑅1 𝑅3 𝑅2 c) 𝑉𝑎 − 𝑉𝑐 𝑉𝑐 − 𝑉𝑏 + 𝐼𝑆 = 𝑅4 𝑅3 d) 𝑉𝑏 − 𝑉𝑑 = 𝐼𝑆 + 𝐼𝑉𝑆 𝑅2 Still 5 unknowns and 4 simultaneous eqns 𝑖𝑅4 + 𝑣𝑅1 − vb − 𝑣𝑅3 + va vc 𝑖𝑅1 𝑖𝑉𝑆 𝑖𝑅3 + 𝑣𝑅2 − 𝑖𝑅2 vd Going round in circles? 𝑉𝑅1 = 𝑉𝑎 − 𝑉𝑏 𝑉𝑅2 = 𝑉𝑏 − 𝑉𝑑 𝑉𝑅3 = 𝑉𝑐 − 𝑉𝑏 𝑉𝑅4 = 𝑉𝑎 − 𝑉𝑐 7 Node-voltage analysis + 𝑣𝑅4 − Voltage, (like energy), is a relative quantity only differences matter. Can assign a voltage value to one node, then all other nodes are defined with respect to that chosen node voltage. We could assign any value but an obvious value is 0 V. When a particular node is chosen to have “0 volts”, we call it the ground node. a) 𝐼𝑉𝑆 = 𝑖𝑅4 + 𝑣𝑅1 − vb − 𝑣𝑅3 + va b) c) 𝑉𝑠 − 𝑉𝑐 𝑉𝑐 − 𝑉𝑏 + 𝐼𝑆 = 𝑅4 𝑅3 d) 𝑉𝑏 = 𝐼𝑆 + 𝐼𝑉𝑆 𝑅2 𝑖𝑅1 𝑖𝑉𝑆 𝑉𝑠 − 𝑉𝑏 𝑉𝑠 − 𝑉𝑐 + 𝑅1 𝑅4 𝑉𝑠 − 𝑉𝑏 𝑉𝑐 − 𝑉𝑏 𝑉𝑏 + = 𝑅1 𝑅3 𝑅2 vc Benefits 1. Remove unknowns 2. Va = Vs + 𝑣𝑅2 − 𝑖𝑅3 𝑖𝑅2 Vd = 0 Equation b) and c) now contain only two unknown voltages 8 Node-voltage analysis Node-voltage analysis is a general technique that can be applied to any circuit. Step 1: Selecting the Reference Node In principle, any node can be picked to be the reference node. However, the solution is usually facilitated by selecting one end of a voltage source as the reference node. For the example below, the circuit has four nodes. Let’s select the bottom node as the reference node. We mark the reference node by the ground symbol. 9 Node-voltage analysis Step 2: Assigning Node Voltages Next, we label the voltages at each of the other nodes. For example, the voltages at the three nodes are labelled va, vb, and vc ❑ The negative reference polarity for each of the node voltages is at the reference node 10 Node-voltage analysis Step 3: Writing KCL Equations in Terms of the Node Voltages After choosing the reference node and assigning the voltage variables, we write equations that can be solved for the node voltages. ❑ In this circuit, the voltage va is the same as the source voltage vS (in this case, one of the node voltages is known without any effort. This is the advantage in selecting the reference node at one end of an independent voltage source.) 𝑽𝒂 = 𝑽𝒔 11 Node-voltage analysis ❑ We need to determine the values of vb and vc, and we must write two independent equations. We usually start by writing current equations at each of the nodes corresponding to an unknown node voltage. For example, at node b, the current leaving through R2 is given by: 𝑉𝑏 𝑅2 ❑ Next, we see that the current flowing into node b through R3 is given by: ❑ And the current flowing into node b through R1 is given by: 𝑉𝑐 − 𝑉𝑏 𝑅3 𝑉𝑎 − 𝑉𝑏 𝑅1 12 Write equations that can be solved for the node voltages v1, v2, and v3? Node 1: 𝑖𝑅3 𝑣𝑅1 − 𝑖𝑅1 + 𝑖𝑅2 + 𝑣𝑅4 − + 𝑖𝑅4 𝑣𝑅3 − 𝑖𝑅5 + 𝑣𝑅2 − + Node 2: 𝑣𝑅5 − Node 3: 𝑣1 𝑣1 − 𝑣2 + + 𝑖𝑠 = 0 𝑅1 𝑅2 𝑣1 − 𝑣2 𝑣2 𝑣2 − 𝑣3 − + + =0 𝑅2 𝑅3 𝑅4 𝑣3 𝑣2 − 𝑣3 − − 𝑖𝑠 = 0 𝑅5 𝑅4 There is actually a slightly easier way to determine the nodal equations 13 Write equations that can be solved for the node voltages v1, v2, and v3? Easier way to determine nodal equations and automatically account for polarity (current direction) Assume current always flows out of node (exception is current sources!) Node 1: 𝑣1 𝑣1 − 𝑣2 + + 𝑖𝑠 = 0 𝑅1 𝑅2 Node 2: 𝑣1 − 𝑣2 𝑣2 𝑣2 − 𝑣3 − + + =0 𝑅2 𝑅3 𝑅4 Node 3: 𝑣3 𝑣2 − 𝑣3 − − 𝑖𝑠 = 0 𝑅5 𝑅4 𝑣1 𝑣1 − 𝑣2 + + 𝑖𝑠 = 0 𝑅1 𝑅2 𝑣2 − 𝑣1 𝑣2 𝑣2 − 𝑣3 + + =0 𝑅2 𝑅3 𝑅4 𝑣3 𝑣3 − 𝑣2 + − 𝑖𝑠 = 0 𝑅5 𝑅4 14 Node-voltage analysis To find the current flowing out of node n through a resistance toward node k, we subtract the voltage at node k from the voltage at node n and divide the difference by the resistance. Thus, if vn and vk are the node voltages and R is the resistance connected between the nodes, the current flowing from node n toward node k is given by: 𝑉𝑛 − 𝑉𝑘 𝑅 Of course, if the resistance is connected between node n and the reference node, the current away from node n toward the reference node is simply the node voltage vn divided by the resistance R. 𝑉𝑛 − 0 𝑅 15 Write KCL equations for each voltage node? Node 1: 𝑣1 − 𝑣3 𝑣1 − 𝑣2 + = 𝑖𝑎 𝑅1 𝑅2 Node 2: 𝑣2 − 𝑣1 𝑣2 𝑣2 − 𝑣3 + + =0 𝑅2 𝑅3 𝑅4 Node 3: 𝑣3 𝑣3 − 𝑣2 𝑣3 − 𝑣1 + + + 𝑖𝑏 = 0 𝑅5 𝑅4 𝑅1 Once we have written the equations needed to solve for the node voltages, we put the equations into standard form. We group the node-voltage variables on the left hand sides of the equations and place terms that do not involve the node voltages on the right-hand sides. (G = conductances) GV = I 16 Write KCL equations for each voltage node? 𝑅1 𝑣1 − 𝑣3 + 𝑣 − 𝑣2 = 𝑅1 𝑖𝑎 𝑅2 1 𝑅1 𝑅1 1+ 𝑣 − 𝑣3 − 𝑣2 = 𝑅1 𝑖𝑎 𝑅2 1 𝑅2 1 1 1 1 + 𝑣 − 𝑣 − 𝑣 = 𝑖𝑎 𝑅1 𝑅2 1 𝑅2 2 𝑅1 3 1 𝑅1 Node 1: 𝑣1 − 𝑣3 𝑣1 − 𝑣2 + = 𝑖𝑎 𝑅1 𝑅2 Node 2: 𝑣2 − 𝑣1 𝑣2 𝑣2 − 𝑣3 + + =0 𝑅2 𝑅3 𝑅4 Node 3: 𝑣3 𝑣3 − 𝑣2 𝑣3 − 𝑣1 + + + 𝑖𝑏 = 0 𝑅5 𝑅4 𝑅1 1 +𝑅 2 −1 𝑅2 −1 𝑅1 −1 𝑅2 1 1 1 + + 𝑅2 𝑅3 𝑅4 −1 𝑅4 −1 𝑅1 −1 𝑅4 1 1 1 + + 𝑅1 𝑅4 𝑅5 𝑣1 𝑖𝑎 𝑣2 = 0 𝑣3 𝑖𝑏 17 Write KCL equations for each voltage node? 1 𝑅1 + −1 𝑅2 −1 𝑅1 1 𝑅2 −1 𝑅2 1 1 1 + + 𝑅2 𝑅3 𝑅4 −1 𝑅4 −1 𝑅1 −1 𝑅4 1 1 1 + + 𝑅1 𝑅4 𝑅5 𝑣1 𝑖𝑎 𝑣2 = 0 𝑣3 𝑖𝑏 If circuit only contains resistances and independent current sources 1. Diagonal terms of G are the sums of the conductance connected to the node 2. Off-Diagonal terms are negatives of the conductance connected between nodes 3. Elements of I are the currents pushed into corresponding nodes 18 Write voltage node equations in matrix form for circuit? 1 1 + 4 5 −1 4 0 −1 4 1 1 1 + + 4 2.5 5 −1 5 0 −1 5 1 1 + 10 5 𝑣1 −3.5 𝑣2 = 3.5 𝑣3 2 ❑ One way to solve for the node voltages is to find the inverse of G and then compute the solution vector as: 19 Solve the node voltages to find ix? 𝑣1 𝑣1 − 𝑣2 𝑣1 − 𝑣3 + + =0 Node 1: 10 5 20 1 1 1 𝑣1 + 𝑣1 − 𝑣2 + 𝑣1 − 𝑣3 = 0 10 5 20 1 1 1 1 1 𝑣1 + 𝑣1 − 𝑣2 + 𝑣1 − 𝑣3 = 0 10 5 5 20 20 0.35𝑣1 − 0.2𝑣2 − 0.05𝑣3 = 0 1 1 1 1 1 + + 𝑣1 − 𝑣2 − 𝑣3 = 0 10 5 20 5 20 20 Solve the node voltages to find ix? Node 1: 𝑣1 𝑣1 − 𝑣2 𝑣1 − 𝑣3 + + =0 10 5 20 Node 2: 𝑣2 − 𝑣1 𝑣2 − 𝑣3 + = 10 5 10 Node 3: 𝑣3 𝑣3 − 𝑣2 𝑣3 − 𝑣1 + + =0 5 10 20 𝐺 × [𝑉] [𝐼] 0 0.35 −0.2 −0.05 𝑣1 −0.2 0.3 −0.1 𝑣2 = 10 0 −0.05 −0.1 0.35 𝑣3 21 Solve the node voltages to find ix? 3x3 𝐺 × [𝑉] [𝐼] 0 0.35 −0.2 −0.05 𝑣1 −0.2 0.3 −0.1 𝑣2 = 10 0 −0.05 −0.1 0.35 𝑣3 A = a ei − fh − b di − fg + c(dh − eg) 0.35((0.3)(0.35) − (−0.1)(−0.1)) = 0.03325 −0.2((−0.2)(0.35) − (−0.1)(−0.05)) = 0.015 −0.05 −0.2 −0.1 − 0.3 −0.05 A = 0.03325 − 0.015 − 0.00175 = 0.0165 = −0.00175 22 Solve the node voltages to find ix? 0 −0.2 −0.05 10 0.3 −0.1 0 −0.1 0.35 𝑣1 = A1 A = 0.0165 = 45.4545V 0.35 0 −0.2 10 −0.05 0 𝑣2 = A2 A = 0.0165 = 72.7273V 𝑣3 = A3 A = 0.0165 = 27.2727V −0.05 −0.1 0.35 0.35 −0.2 0 −0.2 0.3 10 −0.05 −0.1 0 0.75 1.2 0.45 𝑣1 − 𝑣3 45.4545 − 27.2727 𝐼𝑥 = = = 0.9091 A 20 20 23 Apply voltage node method to the following circuit? 7 complex nodes can be identified 24 Apply voltage node method to the following circuit? 𝑉𝑥 𝑉𝑦 𝑉𝑧 𝑁𝑜𝑑𝑒 𝑉𝑥 : 𝑣𝑥 − 𝑣𝑠 𝑣𝑥 𝑣𝑥 − 𝑣𝑦 + + =0 𝑅1 𝑅2 𝑅3 𝑁𝑜𝑑𝑒 𝑉𝑦 : 𝑣𝑦 − 𝑣𝑥 𝑣𝑦 − 𝑣𝑧 + 𝐼𝑠1 + =0 𝑅3 𝑅4 𝑣𝑧 − 𝑣𝑦 𝑣𝑧 𝑣𝑧 − 𝑣𝑠 𝑁𝑜𝑑𝑒 𝑉𝑧 : + + = 𝐼𝑠2 𝑅4 𝑅5 𝑅6 Assigning the ground to one of the bottom circuit nodes eliminates 4 unknown node voltages leaving only three unknowns 25 Apply voltage node method to the following circuit? 𝑁𝑜𝑑𝑒 𝑉𝑥 : 𝑣𝑥 − 𝑣𝑠 𝑣𝑥 𝑣𝑥 − 𝑣𝑦 + + =0 𝑅1 𝑅2 𝑅3 𝑁𝑜𝑑𝑒 𝑉𝑦 : 𝑣𝑦 − 𝑣𝑥 + 𝐼𝑠1 + 𝑣𝑦 − 𝑣𝑧 = 0 𝑅3 𝑁𝑜𝑑𝑒 𝑉𝑧 : Plugging in values of resistance 𝑅4 𝑣𝑧 − 𝑣𝑦 𝑣𝑧 𝑣𝑧 − 𝑣𝑠 + + = 𝐼𝑠2 𝑅4 𝑅5 𝑅6 1 1 1 1 1 + + 𝑣 − 𝑣 = − 𝑣𝑠 𝑅1 𝑅2 𝑅3 𝑥 𝑅3 𝑦 𝑅1 1 1 1 1 − 𝑣𝑥 + + 𝑣 − 𝑣 = −𝐼𝑠1 𝑅3 𝑅3 𝑅4 𝑦 𝑅4 𝑧 1 1 1 1 1 − 𝑣𝑦 + + + 𝑣 = 𝐼𝑠2 + − 𝑣𝑠 𝑅4 𝑅3 𝑅4 𝑅6 𝑧 𝑅6 0.0333 −0.00666 0 −0.00666 0.01 −0.0033 0 −0.0033 0.0166 𝑣1 0.8 𝑣2 = −0.2 𝑣3 0.4 26 Apply voltage node method to the following circuit? 𝑉𝑥 𝑉𝑦 0.0333 −0.00666 0 −0.00666 0.01 −0.0033 0 −0.0033 0.0166 𝑣𝑥 = A1 A = 0.000115555552444 0.000004444362219 = 26V 𝑣𝑦 = A1 A = 0.000044444444734 0.000004444362219 = 10V 𝑣𝑧 = A1 A = 0.000168888887671 0.000004444362219 = 38V 𝑉𝑧 𝑣1 0.8 𝑣2 = −0.2 𝑣3 0.6 27 The node-voltage method 1. Identify all of the nodes in the circuit. 2. Choose one node to be ground. In principle, the choice is arbitrary, but, if possible, choose a node that is connected to a voltage source. The chosen node is assigned a voltage of 0 (ground). 3. 4. Identify nodes for which the voltage is known due to sources. If possible, use short cuts to eliminate any non-essential nodes. 5. 6. Assign variables for the voltages at the remaining unknown nodes. Assign currents to all of the branches connected to the nodes. In principle, the direction is arbitrary. Label the 7. voltage polarity for any resistors. (Make sure that that voltage polarities match the current direction!) Write KCL equations relating the currents at each of the unknown nodes. 8. Use Ohm’s law to express resistor currents in terms of the (unknown) node voltages on either side of the resistor. 9. Substitute the resistor currents into the KCL equations to form the node voltage equations — a set of equations relating the unknown node voltages. 10. Do the math to solve the equations and determine the node voltages. 28 Node-voltage or mesh-current analysis? Deciding should be based on method that leads to easier maths Node Voltage number (N) 1. Count number of nodes in the circuit. 2. Subtract 1 for ground. 3. Subtract 1 for each voltage source which has a connection ( + or – ) to ground. 4. Add 1 for each voltage source which has no connection to ground. Mesh Current number (M) 1. Count number of meshes in the circuit. 2. Subtract 1 for each current source which is located in an outside branch of a mesh. 3. Add 1 for each current source which is located in an interior branch Is N > M ? 29 Node-voltage or mesh-current analysis? Node 1: 𝑣1 𝑣1 − 𝑣2 𝑣1 − 𝑣3 + + =0 10 5 20 Node 2: 𝑣2 − 𝑣1 𝑣2 − 𝑣3 + = 10 5 10 Node 3: 𝑣3 𝑣3 − 𝑣2 𝑣3 − 𝑣1 + + =0 5 10 20 𝐺 × [𝑉] [𝐼] 0 0.35 −0.2 −0.05 𝑣1 −0.2 0.3 −0.1 𝑣2 = 10 0 −0.05 −0.1 0.35 𝑣3 30 Node-voltage or mesh-current analysis? 𝑖3 Node-voltage method 𝑣1 = 45.4545V 𝐼𝑥 = 0.909 A 𝑣2 = 72.7273V 𝑣3 = 27.2727V 𝑖1 𝑖2 Mesh-current method Complicated by shared current source between meshes, adds extra unknown 31 Node-voltage or mesh-current analysis? 𝑖3 Node-voltage method 𝑣1 = 45.4545V 𝐼𝑥 = 0.909 A 𝑣2 = 72.7273V 𝑣3 = 27.2727V 𝑖1 𝑖2 Mesh-current method ❑ Create a Super Mesh that encompasses the shared current source and avoids the unknown voltage through it ❑ This works because KVL can be applied to any size of closed circuit loop! 32 Node-voltage or mesh-current analysis? 𝑖3 Node-voltage method 𝑣1 = 45.4545V 𝐼𝑥 = 0.909 A 𝑣2 = 72.7273V 𝑣3 = 27.2727V 𝑖1 𝑖2 Mesh-current method Apply KVL around Mesh Loop 1+2: Loop 1+2: 15𝑖1 + 15𝑖2 − 15𝑖3 = 0 Loop 3: −5𝑖1 − 10𝑖2 + 35𝑖3 = 0 10𝑖1 + 5 𝑖1 − 𝑖3 + 10 𝑖2 − 𝑖3 + 5𝑖2 = 0 Rearranged, three unknowns and two simultaneous eqns, need to eliminate i1 or i2 33 Node-voltage or mesh-current analysis? 𝑖3 Node-voltage method 𝑣1 = 45.4545V 𝐼𝑥 = 0.909 A 𝑣2 = 72.7273V 𝑣3 = 27.2727V 𝑖1 𝑖2 Mesh-current method Auxiliary equation using KCL at node (v2) between shared meshes Loop 1+2: Loop 3: 𝑖1 = 𝑖2 − 10 30𝑖2 − 15𝑖3 = 150 −15𝑖2 + 35𝑖3 = −50 34 Node-voltage or mesh-current analysis? 𝑖3 Node-voltage method 𝑣1 = 45.4545V 𝐼𝑥 = 0.909 A 𝑣2 = 72.7273V 𝑖1 𝑣3 = 27.2727V 𝑖2 Mesh-current method Solve these 2 simultaneous equations for 𝑖2 and 𝑖3 Loop 1+2: Loop 3: 30𝑖2 − 15𝑖3 = 150 −15𝑖2 + 35𝑖3 = −50 𝑖1 = 𝑖2 − 10 = −4.546 A 𝑖2 = 5.454 A 𝑖3 = 0.909 A 35 Summary ❑ Investigated Node-voltage analysis to solve arbitrary circuits ❑ Can be applied to circuits that cannot be easily simplified into series / parallel ❑ Leads to fewer unknown variables and simultaneous equations Next ❑ Complete Moodle quiz ❑ Next lecture, Wheatsone Bridge 36 Lecture 7- Mid Term Exam Prep Learning outcomes Examine types of DC questions that might appear in the mid-term exam ❑ Look at these together in an interactive setting (Live Tutorial) 1. Finding equivalent resistance 2. Thevenin and Norton Theorem including max power 3. Superposition 4. Mesh current and Node-voltage References: 1) Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. 2) Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 2 Previous Lecture ❑ KCL, KVL, Resistors in Series and Parallel Assuming current is constant in a series circuit ❑ Voltage divider ❑ Current divider Series resistors 𝑅𝑇 = 𝑅1 + 𝑅2 + …… +𝑅𝑛 Assuming potential difference (voltage) is the same at the top and bottom of a circuit: Parallel resistors 1 1 1 1 = + + ⋯+ 𝑅𝑇 𝑅1 𝑅2 𝑅𝑛 𝑅1 𝑅2 𝑅𝑇 = 𝑅1 + 𝑅2 3 Finding Equivalent Resistance for the circuit? 𝑅𝑒𝑞 = 20Ω + 9Ω ∥ 15Ω + 22Ω + 2Ω = 27.31 Ω Class1, 80% with correct answer Class2, 50% with correct answer 4 Finding Equivalent Resistance for the circuit? A 18Ω 𝑅𝑒𝑞 20Ω B 5 Finding Equivalent Resistance for the circuit? 20 × 5 𝑅𝑇 = =4 20 + 5 A 18Ω 𝑅𝑒𝑞 20Ω B 6 Finding Equivalent Resistance for the circuit? 4Ω A 18Ω 𝑅𝑒𝑞 20Ω B 7 Finding Equivalent Resistance for the circuit? 20 × 5 𝑅𝑇 = =4 20 + 5 4Ω A 18Ω 𝑅𝑒𝑞 20Ω 5Ω B 8 Finding Equivalent Resistance for the circuit? 4Ω A 18Ω 𝑅𝑒𝑞 20Ω 5Ω 4Ω B 9 Finding Equivalent Resistance for the circuit? 9×6 𝑅𝑇 = = 3.6 9+6 A 18Ω 𝑅𝑒𝑞 4Ω 6Ω B 10 Finding Equivalent Resistance for the circuit? 18 × 3.6 𝑅𝑇 = =3 8 + 3.6 A 18Ω 𝑅𝑒𝑞 3.6Ω 𝑅𝑒𝑞 = 8 + 3 = 11Ω B Class1, 50% with correct answer Class, 80% with correct answer 11 Previous Lecture 𝑹𝒕 = 𝒗𝒐𝒄 𝑰𝒔𝒄 Find the Thévenin / Norton equivalent for the circuit? 1. Perform two of these a) Determine the open-circuit voltage (𝑉𝑇 = 𝑣𝑜𝑐). b) Determine the short-circuit current (𝐼𝑁 = 𝑖𝑠𝑐). c) Zero the independent sources and find the Thévenin resistance RT looking back into the terminals. Do not zero dependent sources 2. Use the equation 𝑉𝑇 = 𝑅𝑇 𝐼𝑁 to compute the remaining value. 3. The Thévenin equivalent consists of a voltage source VT in series with RT. 4. The Norton equivalent consists of a current source IN in parallel with RT. 𝑰𝑵 = 𝑰𝒔𝒄 A B 12 Calculate the Thevenin Voltage for the circuit? 1000 𝑉𝑇 = × 5 = 0.5V 1000 + 9000 Class1, 70% with correct answer Class2, 70% with correct answer 13 Calculate the Thevenin Resistance for the circuit? Voltage source = short circuit Class1, 90% with correct answer 1000 × 9000 𝑅𝑇 = 1000 + = 1900Ω 1000 + 9000 Class2, 90% with correct answer 14 Calculate the maximum power that can be supplied to the load resistor? Maximum Power 𝑹𝑳 = 𝑹𝒕 𝑉𝑅𝐿 1900 = × 0.25𝑉 1900 + 1900 0.25 2 𝑃= = 32.89 μW 1900 Class2, 80% with correct answer 15 Maximum Power Previous Lecture 4 𝑹𝑳 = 𝑹𝒕 𝑑𝑃𝐿 𝑑 = 𝑑𝑅𝐿 𝑑𝑅𝐿 𝑉𝑡2 𝑅𝐿 𝑅𝑡 + 𝑅𝐿 2 𝑉𝑡2 𝑅𝑡 + 𝑅𝐿 − 2𝑅𝐿 (𝑅𝑡 + 𝑅𝐿 )3 Delta – Star conversion Superposition 16 What statement is correct about superposition? Class1, 50% with correct answer Class2, 40% with correct answer 17 Compute the current in the R = 42.3 Ω resistor by applying superposition? Moodle Quiz Question 𝐼𝑅1 𝑅𝑒𝑞1 1. Zero the current source (open circuit) 2. Zero the Voltage source (short circuit) 3. Add the linear responses (currents) 𝐼𝑅2 46.3 ∗ 27 = 47 + = 64.05 Ω 46.3 + 27 200 𝐼𝑇 = = 3.12 A 64.05 27 𝐼𝑅1 = = 0.368 A 73.3 𝑅𝑒𝑞2 𝐼𝑅2 47 ∗ 27 =4+ = 21.15 Ω 47 + 27 21.15 = ∗ 20 = 6.66 A 21.15 + 42.3 𝐼𝑅 = 𝐼𝑅1 + 𝐼𝑅2 = 7.028 A 18 Compute the current in the R = 42.3 Ω resistor by applying superposition? Class1, 40% with correct answer 1. Zero the current source (open circuit) 2. Zero the Voltage source (short circuit) 3. Add the linear responses (currents) Class2, 30% with correct answer 19 Previous Lecture Mesh Current Analysis 1. Identify all of the individual meshes in the circuit. 2. Assign a mesh current to each mesh. Identify meshes in which the current is known because there is a current source in an outside branch of the mesh. 3. Assign voltages to all of the elements in the meshes with unknown currents. 4. Use KVL around each mesh to write loop equation. 5. Use Ohm’s law to write the resistor voltages in terms of the mesh currents. 6. Insert the voltage expressions into the KVL equations to arrive at a set of mesh current equations. If there are n unknown currents, there should be n equations relating them. 7. Solve the system of equations to find the mesh currents. 20 True or False? Class1, 70% with correct answer Class2, 80% with correct answer 21 How Many Mesh Current loops to Solve the Circuit? Class1, 80% with correct answer Class2, 80% with correct answer 22 Calculate the loop currents for I2 and I3, assuming I1 = 1A? R4 16 R3 R1 VA 46 82 R2 VB 𝑅2 𝑖2 − 𝑖1 + 𝑅3 𝑖2 − 𝑖3 = −𝑉𝐵 𝑖3 𝑅4 + 𝑅3 𝑖3 − 𝑖2 + 𝑅1 𝑖3 − 𝑖1 = 0 -𝑅2 𝑖1 + 𝑅2 + 𝑅3 𝑖2 − 𝑅3 𝑖3 = −𝑉𝐵 −𝑅1 𝑖1 − 𝑅3 𝑖2 + 𝑅1 + 𝑅3 + 𝑅4 𝑖3 = 0 Loop 2: -𝑅2 𝑖1 + 𝑅2 + 𝑅3 𝑖2 − 𝑅3 𝑖3 = −𝑉𝐵 Loop 3: −𝑅1 𝑖1 − 𝑅3 𝑖2 + 𝑅1 + 𝑅3 + 𝑅4 𝑖3 = 0 23 Solve for the current in each element of the following circuit? Loop 2: -𝑅2 𝑖1 + 𝑅2 + 𝑅3 𝑖2 − 𝑅3 𝑖3 = −𝑉𝐵 Loop 3: −𝑅1 𝑖1 − 𝑅3 𝑖2 + 𝑅1 + 𝑅3 + 𝑅4 𝑖3 = 0 Loop 2: 22𝑖2 − 12𝑖3 = −36 Loop 3: −12𝑖2 + 48𝑖3 = 20 11 × 𝐿𝑜𝑜𝑝3 + 𝐿𝑜𝑜𝑝2 6 Loop 3: Class1, 30% with correct answer Class1, 40% with correct answer 76𝑖3 = 0.6667 𝑖3 = 8.77 mA 𝑖2 = −1.63 A 24 How many Node Voltages are there to solve after assigning a reference node? 25 Ω 15 Ω vb 10 Ω va=25V 25 V vc 3A 5Ω vd=0V 25 Calculate the node voltage for Vb and Vc? b) 𝑉𝑏 − 𝑉𝑠 𝑉𝑏 𝑉𝑏 − 𝑉𝑐 + + =0 𝑅1 𝑅2 𝑅3 c) 𝑉𝑐 − 𝑉𝑠 𝑉𝑐 − 𝑉𝑏 − 𝐼𝑆 + =0 𝑅4 𝑅3 − 1 1 1 1 1 + + 𝑣 − 𝑣 = 𝑣 𝑅1 𝑅2 𝑅3 𝑏 𝑅3 𝑐 𝑅1 𝑠 11 1 𝑣 − 𝑣 = 1.667 30 𝑏 10 𝑐 1 1 1 1 𝑣𝑏 + + 𝑣𝑐 = 𝑣𝑠 + 𝐼𝑆 𝑅3 𝑅3 𝑅4 𝑅4 1 7 − 𝑣𝑏 + 𝑣 =4 10 50 𝑐 25 Ω 15 Ω vb 10 Ω vc 25 V 5Ω 𝑉𝑏 = 15.32 V 𝑉𝑐 = 39.52 V 3A 26 Leader Board and Next Lecture Class1 Class2 27 Lecture 8- Mid Term Exam Prep Learning outcomes Examine types of DC questions that might appear in the mid-term exam ❑ Look at these together in an interactive setting (Live Tutorial) 1. Finding equivalent resistance 2. Thevenin and Norton Theorem including max power 3. Superposition 4. Mesh current and Node-voltage References: 1) Allan R. Hambley, Electrical Engineering - Principles and Applications, 6th Ed. Pearson, ISBN13 -978-0-13-311664-9. 2) Thomas L. Floyd, Principles of Electric Circuits - Conventional Current Version, 9th Ed. Pearson Int. Ed., ISBN 13: 978-1-292-02566-7 2 Previous Lecture ❑ KCL, KVL, Resistors in Series and Parallel Assuming current is constant in a series circuit ❑ Voltage divider ❑ Current divider Series resistors 𝑅𝑇 = 𝑅1 + 𝑅2 + …… +𝑅𝑛 Assuming potential difference (voltage) is the same at the top and bottom of a circuit: Parallel resistors 1 1 1 1 = + + ⋯+ 𝑅𝑇 𝑅1 𝑅2 𝑅𝑛 𝑅1 𝑅2 𝑅𝑇 = 𝑅1 + 𝑅2 3 Finding Equivalent Resistance for the circuit? 𝑅𝑒𝑞 = 20Ω + 9Ω ∥ 15Ω + 22Ω + 2Ω = 27.31 Ω Class1, 90% with correct answer Class2, 80% with correct answer 4 Finding Equivalent Resistance for the circuit? A 18Ω 𝑅𝑒𝑞 20Ω B 5 Finding Equivalent Resistance for the circuit? 20 × 5 𝑅𝑇 = =4 20 + 5 A 18Ω 𝑅𝑒𝑞 20Ω B 6 Finding Equivalent Resistance for the circuit? 4Ω A 18Ω 𝑅𝑒𝑞 20Ω B 7 Finding Equivalent Resistance for the circuit? 20 × 5 𝑅𝑇 = =4 20 + 5 4Ω A 18Ω 𝑅𝑒𝑞 20Ω 5Ω B 8 Finding Equivalent Resistance for the circuit? 4Ω A 18Ω 𝑅𝑒𝑞 20Ω 5Ω 4Ω B 9 Finding Equivalent Resistance for the circuit? 9×6 𝑅𝑇 = = 3.6 9+6 A 18Ω 𝑅𝑒𝑞 4Ω 6Ω B 10 Finding Equivalent Resistance for the circuit? 18 × 3.6 𝑅𝑇 = =3 8 + 3.6 A 18Ω 𝑅𝑒𝑞 3.6Ω 𝑅𝑒𝑞 = 8 + 3 = 11Ω B Class1, 50% with correct answer Class, 60% with correct answer 11 Previous Lecture 𝑹𝒕 = 𝒗𝒐𝒄 𝑰𝒔𝒄 Find the Thévenin / Norton equivalent for the circuit? 1. 𝑰𝑵 Perform two of these 𝑣𝑜𝑐). a) Determine the open-circuit voltage (𝑉𝑇 b) Determine the short-circuit current (𝐼𝑁 = 𝑖𝑠𝑐). c) Zero the independent sources and find the Thévenin resistance RT = = 𝑰𝒔𝒄 looking back into the terminals. Do not zero dependent sources 2. Use the equation 𝑉𝑇 = 𝑅𝑇 𝐼𝑁 to compute the remaining value. 3. The Thévenin equivalent consists of a voltage source VT in series with RT. 4. The Norton equivalent consists of a current source IN in parallel with RT. A B 12 Thévenin Equivalent Circuit ❑ Thévenin equivalent consists of an independent voltage source in series with a resistance ❑ By definition, no current can flow through an open circuit, therefore no current flow through Rt ❑ Applying KVL, we conclude that Vt = voc ❑ Consider the short circuit case, the isc is the same as the original and Thévenin equivalent A General Network B 𝑣𝑜𝑐 𝑅𝑡 = 𝐼𝑠𝑐 Thévenin’s circuit between different terminals Calculate the Thevenin Voltage for the circuit? 1000 𝑉𝑇 = × 5 = 0.5V 1000 + 9000 Class1, 70% with correct answer Class2, 70% with correct answer 15 Thévenin Resistance ❑ Alternative approach to directly solve for the Thévenin Resistance ❑ If there are no dependent sources can zero all of the sources within the circuit ❑ Zero voltage source equivalent to a short circuit and zero current source equivalent to an open circuit ❑ Thévenin Resistance is equivalent to the resistance between the terminals 16 Calculate the Thevenin Resistance for the circuit? Voltage source = short circuit Class1, 85% with correct answer 1000 × 9000 𝑅𝑇 = 1000 + = 1900Ω 1000 + 9000 Class2, 80% with correct answer 17 Maximum Power Theorem 𝑔 𝑥 = 𝑉𝑡2 𝑅𝐿 Quotient rule 𝑑𝑃𝐿 𝑑 = 𝑑𝑅𝐿 𝑑𝑅𝐿 𝑓′ 𝑥 = 𝑉𝑡2 𝑅𝐿 𝑅𝑡 + 𝑅𝐿 𝑉𝑡2 𝑅𝑡 + 𝑅𝐿 𝑥 = 𝑉𝑡2 𝑥 ℎ 𝑥 − 𝑔 𝑥 ℎ′ 𝑥 ℎ(𝑥)2 − 2𝑉𝑡2 𝑅𝐿 𝑅𝑡 + 𝑅𝐿 𝑅𝑡 + 𝑅𝐿 4 1 𝑅𝑡 + 𝑅𝐿 −2 − 2𝑅𝐿 (𝑅𝑡 + 𝑅𝐿 )−3 ℎ 𝑥 = 𝑅𝑂 + 𝑅𝐿 2 𝑔′ 𝑥 = 𝑉𝑡2 ℎ′ 𝑥 = 2 𝑅𝑂 + 𝑅𝐿 2 𝑓 ′ 𝑥 = 𝑉𝑡2 𝑅𝑡 + 𝑅𝐿 𝑓′ 2 𝑓′ 𝑥 = 𝑔′ This equals zero when 2𝑅𝐿 𝑅𝑡 + 𝑅𝐿 − 2𝑅𝐿 2 − = 𝑉𝑡 2 (𝑅𝑡 + 𝑅𝐿 )3 (𝑅𝑡 + 𝑅𝐿 )3 𝑅𝑡 + 𝑅𝐿 − 2𝑅𝐿 = 0 𝑹𝑳 = 𝑹𝒕 Under these conditions RL is said to be matched to the output resistance (better: output impedance) of the source. Often will wish to design circuits and systems to achieve this. 18 Calculate the maximum power that can be supplied to the load resistor? Maximum Power 𝑉𝑅𝐿 1900 = × 0.5𝑉 = 0.25𝑉 1900 + 1900 𝑹𝑳 = 𝑹𝒕 𝑉2 0.25 2 𝑃= = = 32.89 μW 𝑅 1900 19 What statement is correct about superposition? Class1, 55% with correct answer Class2, 50% with correct answer Ideal Voltage source + V I Ideal Current source 20 Compute the current in the R = 42.3 Ω resistor by applying superposition? 1. Zero the current source (open circuit) 2. Zero the Voltage source (short circuit) 3. Add the linear responses (currents) 21 Compute the current in the R = 42.3 Ω resistor by applying superposition? Moodle Quiz Question 𝐼𝑅1 𝑅𝑒𝑞1 = 4 + 42.3 = 46.3 Ω 𝑅𝑒𝑞1 = 47 + 1. Zero the current source (open circuit) 46.3 ∗ 27 = 64.05 Ω 46.3 + 27 200 = 3.12 A 64.05 Ohms law 𝐼𝑇 = Current division 27 = × 3.12𝐴 = 0.368 A 73.3 𝐼𝑅1 22 Compute the current in the R = 42.3 Ω resistor by applying superposition? Moodle Quiz Question 1. Zero the Voltage source (short circuit) 2. Add the linear responses (currents) 27 𝐼𝑅1 = = 0.368 A 73.3 𝐼𝑅2 𝑅𝑒𝑞2 𝐼𝑅2 47 ∗ 27 =4+ = 21.15 Ω 47 + 27 21.15 = ∗ 20 = 6.66 A 21.15 + 42.3 𝐼𝑅 = 𝐼𝑅1 + 𝐼𝑅2 = 7.028 A 23 Compute the current in the R = 42.3 Ω resistor by applying superposition? Class1, 65% with correct answer 1. Zero the current source (open circuit) 2. Zero the Voltage source (short circuit) 3. Add the linear responses (currents) Class2, 30% with correct answer 24 Previous Lecture Mesh Current Analysis 1. Identify all of the individual meshes in the circuit. 2. Assign a mesh current to each mesh. Identify meshes in which the current is known because there is a current source in an outside branch of the mesh. 3. Assign voltages to all of the elements in the meshes with unknown currents. 4. Use KVL around each mesh to write loop equation. 5. Use Ohm’s law to write the resistor voltages in terms of the mesh currents. 6. Insert the voltage expressions into the KVL equations to arrive at a set of mesh current equations. If there are n unknown currents, there should be n equations relating them. 7. Solve the system of equations to find the mesh currents. 25 Mesh-current analysis Can reduce analysis complexity by ignoring potential voltage across resistor elements (Sum of resistors in loop) × (loop current) − (each resistor common to both loops) × (adjacent loop current) = (source voltage in the loop) 26 True or False? Class1, 70% with correct answer Class2, 75% with correct answer 27 How Many Mesh Current loops to Solve the Circuit? Class1, 60% with correct answer Class2, 70% with correct answer 28 Calculate the loop currents for I2 and I3, assuming I1 = 1A? R4 16 R3 R1 VA 46 82 R2 VB (Sum of resistors in loop) × (loop current) − (each resistor common to both loops) × (adjacent loop current) = (source voltage in the loop) Loop 1: 1A Loop 2: -𝑅2 𝑖1 + 𝑅2 + 𝑅3 𝑖2 − 𝑅3 𝑖3 = −𝑉𝐵 Loop 3: −𝑅1 𝑖1 − 𝑅3 𝑖2 + 𝑅1 + 𝑅3 + 𝑅4 𝑖3 = 0 29 Calculate the loop currents for I2 and I3, assuming I1 = 1A? Loop 2: -𝑅2 𝑖1 + 𝑅2 + 𝑅3 𝑖2 − 𝑅3 𝑖3 = −𝑉𝐵 Loop 3: −𝑅1 𝑖1 − 𝑅3 𝑖2 + 𝑅1 + 𝑅3 + 𝑅4 𝑖3 = 0 With I1 = 1A Loop 2: -10x1 + 22 𝑖2 − 12𝑖3 = −46 Loop 3: -20x1 − 23𝑖2 + 48 𝑖3 = 0 Loop 2: 22𝑖2 − 12𝑖3 = −36 Loop 3: −12𝑖2 + 48𝑖3 = 20 11 × 𝐿𝑜𝑜𝑝3 + 𝐿𝑜𝑜𝑝2 6 Loop 3: Solve by elimination or substitution or determinants 76𝑖3 = 0.6667 𝑖3 = 8.77 mA 𝑖2 = −1.63 A 30 Node-voltage analysis To find the current flowing out of node n through a resistance toward node k, we subtract the voltage at node k from the voltage at node n and divide the difference by the resistance. Thus, if vn and vk are the node voltages and R is the resistance connected between the nodes, the current flowing from node n toward node k is given by: 𝑉𝑛 − 𝑉𝑘 𝑅 Of course, if the resistance is connected between node n and the reference node, the current away from node n toward the reference node is simply the node voltage vn divided by the resistance R. 𝑉𝑛 − 0 𝑅 31 How many Node Voltages are there to solve after assigning a reference node? 25 Ω By selecting Vd to be the reference node (GND = 0V) then Va = 25V 15 Ω va 25 V Class1, 5% with correct answer vb 10 Ω vc 3A 5Ω vd Class1, 8% with correct answer 32 Calculate the node voltage for Vb and Vc? 𝐼𝑠 = 3 𝐴 𝑉𝑠 = 𝑉𝑎 = 25𝑉 b) 𝑉𝑏 − 𝑉𝑠 𝑉𝑏 𝑉𝑏 − 𝑉𝑐 + + =0 𝑅1 𝑅2 𝑅3 1 1 1 1 1 + + 𝑣 − 𝑣 = 𝑣 𝑅1 𝑅2 𝑅3 𝑏 𝑅3 𝑐 𝑅1 𝑠 c) 𝑉𝑐 − 𝑉𝑠 𝑉𝑐 − 𝑉𝑏 − 𝐼𝑆 + =0 𝑅4 𝑅3 − 1 1 1 1 𝑣𝑏 + + 𝑣𝑐 = 𝑣𝑠 + 𝐼𝑆 𝑅3 𝑅3 𝑅4 𝑅4 1 1 1 1 1 + + 𝑣𝑏 − 𝑣𝑐 = 25𝑉 15 5 10 10 15 − 1 1 1 1 𝑣𝑏 + + 𝑣𝑐 = 25𝑉 + 3𝐴 10 10 25 25 25 Ω b) 15 Ω vb 10 Ω vc 25 V 5Ω c) 1 1 𝑣𝑏 − 𝑣𝑐 = 1.667 30 10 1 1 − 𝑣𝑏 + 𝑣 =4 10 35 𝑐 3A 𝑉𝑏 = −49.475 V 𝑉𝑐 = −33.162 V Solve by elimination or substitution or determinants 33 Summary ❑ Covered example questions for mid-term exam with previous lecture material ❑ DC part of the exam will include 2 questions ❑ Have to enter correct value (isn’t multiple choice) ❑ Please take image of your working and upload after the exam Next ❑ Mid-Term Exam (Moodle) ❑ Next lecture, Wheatsone Bridge 34
