Network Theory Past Paper PDF

NETWORK THEORY SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDYALAYA (University established under section 3 of UGC Act 1956)...

NETWORK THEORY SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDYALAYA (University established under section 3 of UGC Act 1956) (Accredited with ‘A’ Grade by NAAC) Enathur, Kanchipuram – 631 561 Course Material SUBJECT : NETWORK THEORY YEAR/SEM : II/II DEPARTMENT : ECE Prepared by, Dr.A.RAJASEKARAN, Assistant Professor Department of Electronics and Communication Engineering SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDYALAYA ENATHUR, KANCHIPURAM - 631561 Dept. of ECE SCSVMV Page 1  NETWORK THEORY NETWORK THEORY Prerequisite: Mathematics – II & Basic Electronics Engineering OBJECTIVES:  To learn techniques of solving circuits involving different active and passive elements.  To analyze the behavior of the circuit’s response in time domain.  To analyze the behavior of the circuit’s response in frequency domain.  To understand the significance of network function UNIT-1 (CIRCUIT ANALYSIS) KVL- KCL- circuit elements(R,L &C) in series and parallel- voltage and current divider rule-source transformation technique-duals and duality- mesh analysis-super mesh analysis-nodal analysis-super nodal analysis-network topology-definitions-incident matrix-fundamental cut set matrix-series and parallel resonance. UNIT-2(NETWORK THEOREMS) Superposition theorem, Thevenin’s theorem, Norton’s theorem, Maximum power Transfer theorem, reciprocity theorem, compensation theorem, and Tallegen's theorem as applied to DC and AC. Circuits UNIT-3(TWO PORT NETWORKS AND FILTERS DESIGN) Z parameter, Y parameter, h parameter, ABCD parameter, g parameter,Inter relationship of different parameters-inter connection of two port networks-classification of filters-constant k low pass and high pass filters-m-derived low pass and high pass filters-band pass filter-band elimination filter. UNIT-4(TRANSIENT AND S-DOMAIN ANALYSIS) Steady state and transient response-DC response of an R-L,R-C and R-L-C circuit-sinusoidal response of R-L,R-C and R-L-C circuit-concept of complex of frequency-poles and zeros of network function- significance of poles and zeros-properties of driving point and transfer function. UNIT-5(NETWORK SYNTHESIS) Hurwitz polynomial-positive real function, frequency response of reactive one port-synthesis of reactive one port by Foster’s Method &Cauer method- synthesis of R-L Network by Foster’s Method &Cauer method- synthesis of R-C Network by Foster’s Method &Cauer method. OUTCOMES: Understand the behavior of different circuits and their response using various circuit analysis tools and theorems Understand the analysis in time domain and frequency domain. Understand basic concepts regarding the system definition mathematically and associated network function. Understand the concept of Network synthesis. Text Books: 1. Sudhakar, A., Shyammohan, S. P.; “Circuits and Network”; Tata McGraw-Hill New Delhi, 1994 2. A William Hayt, “Engineering Circuit Analysis” 8th Edition, McGraw-Hill Education. Dept. of ECE SCSVMV Page 2  NETWORK THEORY Unit I -NETWORK ANALYSIS AIM: To create circuits involving different active and passive elements Pre-Requisites: Knowledge of Basic Mathematics – II & Basic Electronics Engineering Pre - MCQs: 1. Time constant of a capacitive circuit a. Increases with the decrease of capacitance and decrease of resistance b. Increases with the decrease of capacitance and increase of resistance c. Increases with the increase of capacitance and decrease of resistance d. Increase with increase of capacitance and increase of resistance 2. Which of the following is a correct statement of Ohm's law? a. I = R/V b. R = VI c. V = I/R d. I = V/R 3. A sinusoidal signal has a period of 40 ms. What is its frequency? a. 25 Hz b. 50 kHz c. 50 Hz d. 25 kHz 4. In a loss-free R-L-C circuit the transient current is a. Oscillating b. Square wave c. Sinusoidal d. Non-oscillating 5. In a circuit containing R, L and C, power loss can take place in a. C only b. L only c. R only d. All the above Dept. of ECE SCSVMV Page 3  NETWORK THEORY NETWORK ANALYSIS USING KVL & KCL INTRODUCTION: Today we live in a predominantly electrical world. Electrical technology is a driving force in the changes that are occurring in every engineering discipline. Circuit analysis is the foundation for electrical technology.Network is a system with interconnected electrical elements. Network and circuit are the same. The only difference being a circuit shall contain at least one closed path. Network analysis is the process of finding the voltages across, and the currents through every component in the network. Basic Circuit Elements Circuit: A circuit is a closed conducting path through which an electric current flows. Electric Network: A combination of various electric elements, connected in any manner is called an electric network. Electric Circuits consist of two basic types of elements. These are the active elements and the passive elements. An active element is capable of generating or supplying an electrical energy. Examples are voltage source (such as a battery or generator) and current source, oscillators etc.. A passive element is one which does not generate electricity but either consumes it or stores it. Resistors, Inductors and Capacitors are simple passive elements. Diodes, transistors etc. are also passive elements. These parameters may be lumped or distributed. Elements of a circuit, which are separated physically, are known as lumped elements. Ex:- L & C. Elements, which are not separable for analytical purposes, are known as distributed elements. Ex:- Transmission lines having R, L, C all along their length. Circuits may either be linear or non-linear A linear circuit is one whose parameter are constant i.e., they do not change with voltage or current. Linear elements obey a straight line law. For example, a linear resistor has a linear voltage v/s current relationship which passes through the origin (V = R.I). A linear inductor has a linear flux vs current relationship which passes through the origin (φ = LI) and a linear capacitor has a linear charge vs voltage relationship which passes through the origin (q = CV). [R, L and C are constants]. A Non linear circuit is one whose parameters change with voltage or current. Dept. of ECE SCSVMV Page 4  NETWORK THEORY Resistors, inductors and capacitors may be linear or non-linear, while diodes and transistors are always nonlinear. Circuits may either be Unilateral or Bilateral The circuit whose properties or characteristics change with the direction of its operation is said to be Unilateral. A diode rectifier is a unilateral, because it cannot perform rectification in both directions. A bilateral circuit is one whose properties or characteristics are the same in either direction. Examples are R, L & C. The usual transmission line is bilateral, because it can be made to perform its function equally well in either direction. Branch A branch represents a single element, such as a resistor or a battery. A branch is a part of the network which lies between two junctions Node A node is the point or junction in a circuit connecting two or more branches or circuit elements. The node is usually indicated by a dot (.) in a circuit Loop A loop is any closed path in a circuit, formed by starting at a node, passing through a number of branches and ending up once more at the original node. No element or node is encountered more than once. Mesh It is a loop that contains no other loop within it. Fig.1.1 For example, the circuit of Fig 1.1 has 4 nodes, 6 branches and 6 loops and 3 meshes. Resistance R [Unit: Ohm (Ω)] The relationship between voltage and current is given by v = R i, or i = G v, G = conductance = 1/R Dept. of ECE SCSVMV Page 5  NETWORK THEORY Power loss in a resistor = R i2. Energy dissipated in a resistor w = ∫ R.i2 dt There is no storage of energy in a resistor. Inductance L [Unit: Henry (H)] The relationship between voltage and current is given by v = N =L Energy stored in an inductor = ½ L i2 No energy is dissipated in a pure inductor. However as practical inductors have some wire resistance there would be some power loss. There would also be a small power loss in the magnetic core (if any). Capacitance C [Unit: Farad (F)] The relationship between voltage and current is given by i = =C Energy stored in an capacitor = ½ C v2 No energy is dissipated in a pure capacitor. However practical capacitors also have some power loss. Independent and Dependent sources Those voltage or current sources, which do not depend on any other quantity in the circuit, are called independent sources. An independent d.c. voltage source is shown in Fig.1.2 (a) whereas a time varying voltage source is shown in Fig.1.2 (b). The positive sign shows that terminal A is positive with respect to terminal B. In other words, potential of terminal A is v volts higher than the terminal B. Similarly, Fig.1.2 (c) shows an ideal constant current source whereas Fig.1.2 (d) depicts a time- varying current source. The arrow shows the direction of flow of the current at any moment under consideration. Fig.1.2: Independent voltage and current sources Dept. of ECE SCSVMV Page 6  NETWORK THEORY A dependent voltage or current source is one which depends on some other quantity which may be either a voltage or a current. Such a source is represented by a diamond shape as shown in Fig.1.3. There are four possible dependent sources. 1. Voltage-dependent voltage source [Fig.1.3 (a)] 2. Current- dependent voltage source [Fig.1.3 (b)] 3. Voltage- dependent current source [Fig.1.3 (c)] 4. Current - dependent current source [Fig.1.3 (d)] Such sources can also be either constant sources or time-varying sources. The constant of proportionality are written as α,r,g and β. The constants α and β have no units, r has the unit of ohms and g has the unit of seimens. Fig.1.3: Dependent voltage and current sources Fundamental Laws The fundamental laws that govern electric circuits are Ohm’s law and Kirchoff’s laws. Ohm’s Law Ohm’s law states that the voltage v across a resistor is directly proportional to the current i flowing through it. v ∝ i, v = R. i where R is the proportionality constant. A short circuit in a circuit element is when the resistance (and any other impedance) of the element approaches zero. [The term impedance is similar to resistance but is used in alternating current theory for other components] An open circuit in a circuit element is when the resistance (and any other impedance) of the element approaches infinity. In addition to Ohm’s law we need the Kirchoff’s voltage law and the Kirchoff’s current law to analyse circuits. Dept. of ECE SCSVMV Page 7  NETWORK THEORY Kirchoff’s Current Law Kirchoff’s Current Law states that the algebraic sum of the currents entering a node is zero. It simply means that the total current leaving a junction is equal to the current entering that junction. Σi=0 Consider the case of a few conductors meeting at a point A as in Fig.1.4. Some conductors have currents leading to point A, whereas some have currents leading away from point A. Fig.1.4 Assuming the incoming currents to be positive and the outgoing currents negative, we have i1 + i 2 − i3 + i 4 − i5 = 0 Kirchoff’s Voltage Law Kirchoff’s Voltage Law states that the algebraic sum of all voltages around a closed path (or loop) is zero. Σv=0 In other words,... round a mesh Consider a circuit as shown in Fig.1.5, we have − v1 + v2 + v3 + v4 = 0 depending on the convention, you may also write v1 − v2 − v3 − v4 = 0 Note: v1, v2 … may be voltages across either active elements or passive elements or both and may be obtained using Ohm’s law. Fig.1.5 Dept. of ECE SCSVMV Page 8  NETWORK THEORY Fig.1.6: Kirchoff’s analysis circuit Determination of Voltage sign In applying Kirchhoff’s laws to specific problems, for example, the circuit shown in Fig.1.6, particular attention should be paid to the algebraic signs of voltage drops and e.m.fs. Following sign conventions is suggested. (a) Sign of Battery E.M.F. A rise in voltage should be given a +ve sign and a fall in voltage a –ve sign. Keeping this in mind, it is clear that as we go from the –ve terminal of a battery to its +ve terminal as shown in Fig.1.7(a) there is a rise in potential, hence this voltage should be given a +ve sign. On the other hand, if we go from the +ve terminal of a battery to its -ve terminal) there is a fall in potential, hence this voltage should be preceeded by a -ve sign. It is important to note that the sign of the battery e.m.f is independent of the direction of the current through that branch. (a) (b) Fig.1.7: Voltage Sign Dept. of ECE SCSVMV Page 9  NETWORK THEORY (b) Sign of IR Drop Now, take the case of a resistor for Fig.1.7 (b). If we go through a resistor in the same direction as of the current, then there is a fall in potential because current flows from a higher to lower potential. Hence this voltage fall should be taken –ve. However, if we go in a direction opposite of the current, then there is a rise in voltage. Hence this voltage rise should be given a +ve. Consider the closed path ABCDA in Fig.1.8, as we travel around the mesh in clockwise direction, using KVL we get, Or Fig.1.8 Assumed Direction of Current The direction of current flow may be assumed either clockwise or anticlockwise. If the assumed direction of the current is not actual direction, then on solving the question, this current will be found to have a minus sign. If the answer is positive, then assumed direction is same as actual direction. Voltage Divider Fig.1.9: Voltage Divider Dept. of ECE SCSVMV Page 10  NETWORK THEORY The voltage divider circuit is shown in Fig.1.9. Ohm’s law gives …… (1) and we know that -……. (2) Substituting eqn (2) into eqn (1) gives eqn (3) (3) In general, if there are n resistors in series, the voltage across resistor Rx is given by Current Divider The two-resistor circuits shown in the circuit Fig.1.10 is a current divider circuit. The current through R1 is given by, Fig.1.10: Current Divider Series Circuits When elements are connected in series, from Kirchoff’s current law, i1 = i2 = i and from Kirchoff’s Voltage Law, v1 + v2 = v. Also from Ohm’s Law, v1 = R1 i1 , v2 = R2 i2 , v = R I ∴ R1 i + R2 i = R i, or R = R1 + R2 That is, in a series circuit, the total resistance is the sum of the individual resistances, and the voltage across the individual elements is directly proportional to the resistance of that element. Dept. of ECE SCSVMV Page 11  NETWORK THEORY Parallel Circuits When elements are connected in parallel, from Kirchoff’s current law, i1 + i2 = i and from Kirchoff’s Voltage Law, v1 = v2 = v. Also from Ohm’s Law, v1 = R1 i1 , v2 = R2 i2 , v = R i ∴ + = or = + or Also, = = = and = , =.................. current division rule In parallel circuits, the ratio of the current in one branch of a two-branch parallel circuit to the total current is equal to the ratio of the resistance of the other branch to the sum of the two resistances. Problems on KVL and KCL 1. What is the voltage VS across the open switch in the circuit shown in Fig. Q1? Fig. Q1 Dept. of ECE SCSVMV Page 12  NETWORK THEORY Solution: We will apply KVL to find VS. Starting from point A in the clockwise direction ∴ 2. Find the unknown voltage V1 in the circuit of Fig. Q2. Fig. Q2 Solution: Taking the outer closed loop ABCDEFA and applying KVL to it, we get ∴ 3. For the circuit shown in Fig. Q3, find VCE and VAG. Fig. Q3 Solution: Consider the two battery circuits of Fig Q3 separately. Current in the 20V battery circuit ABCD is. Similarly, current in the 40V battery circuit EFGH is Dept. of ECE SCSVMV Page 13  NETWORK THEORY For finding VCE , we will find the algebraic sum of the voltage drops from point E to C via H and B. ∴ The –ve sign shows that the point C is negative with respect to point E. For finding VAG , we will find the algebraic sum of the voltage drops from point E to C via H and B. 4. Using Kirchhoff’s Current Law and Ohm’s law, find the magnitude and polarity of voltage V in Fig. Q4 Fig. Q4 Solution: Applying KCL to node A, we have I1-I2+I3=22 ---(i) Applying Ohm’s law, we have I1 = V/2, I3= V/4, I2= -V/6 Substituting these values in eqn (i) ,we get V=24V I1=12A,I2=-4A,I3=6A The negative sign of I2 indicates that actual direction of its flow is opposite to that of shown in Fig. Q4. Dept. of ECE SCSVMV Page 14  NETWORK THEORY 5. Determine the branch currents in the network of Fig. Q.5. Fig. Q5 Solution: Apply KCL to the closed circuit ABDA, we get Similarly, circuit BCDB GIVES or From circuit ADCEA,we get or On solving we get and Current in branch AB = current in branch Current in branch ; current in branch AD = current in branch ; current in branch. Dept. of ECE SCSVMV Page 15  NETWORK THEORY Self Assessment 1. Use Kirchhoff’s laws to determine the values and directions of the currents flowing in each of the batteries and in the external resistors of the circuit shown in Fig. Q.6. Also determine the potential difference across the external resistors. Fig. Q. 6 2. Find ix in the circuit shown in Fig. Q.7 120mA R1 R2 10ix 12mA ix Fig. Q.7 Source Transformation In network analysis it may be required to transform a practical voltage source into its equivalent practical current source and vice versa which are depicted in Fig.1.11. These are as explained follows. R1 V1(t) V i1(t) i1(t) R1 V(t) Applying KVL, or Dept. of ECE SCSVMV Page 16  NETWORK THEORY Sources with equivalent terminal characteristics R1 V R1 V i1 i1 (v) Voltage source with parallel resistance (vi) Current source with series resistance Fig.1.11: Source Transformation Self Assessment 1.Using successive source transformation, simplify the network shown in Fig. Q8 betweenX &Y. X 2 1A 2 4V 3 1   6V 3V Y Fig.Q.8 Dept. of ECE SCSVMV Page 17  NETWORK THEORY Delta/Star Transformation In solving networks by the applications of Kirchhoff’s laws, one sometimes experiences great difficulty due to a large number of simultaneous equations that have to be solved. However such complicated network can be simplified by successively replacing delta meshes by equivalent star system and vice versa. A delta connected network of three resistances (or impedances) R12, R23, and R31 can be transformed into a star connected network of three resistances (or impedances) R1, R2, and R3 as shown in Fig.1.12 using following transformations Fig.1.12: Source Transformation Note: You can observe that in each of the above expressions, resistance of each arm of the star is given by the product of the resistances of the two delta sides that meet at its end divided by the sum of the three resistances. Star/Delta transformation This transformation can be easily done by the following equations The equivalent delta resistance between any two terminals is given by the sum of star resistances between those terminals plus the product of these two star resistances divide by the third star resistances. Dept. of ECE SCSVMV Page 18  NETWORK THEORY Problems 1. Calculate the equivalent resistance between the terminals A and B in the network shown in Fig. Q.9. Fig. Q.9 Solution: RCS= 16/9Ω, RES=24/9Ω and RDS=12/9Ω RAB= 4+(16/9)+(35/9)=87/9Ω Self Assessment 1. Calculate the current flowing through the 10Ω resistor of Fig. Q.10. Fig. Q.10 2. A network of resistances is formed as shown in Fig.Q.11. Compute the network resistance measured between (i) A and B (ii) B and C (iii) C and A. Fig. Q.11 Dept. of ECE SCSVMV Page 19  NETWORK THEORY Introduction to Nodal and Mesh Analysis When we want to analyse a given network, we try to pick the minimum number of variables and the corresponding number of equations to keep the calculations to a minimum. Thus we would normally work with either currents only or voltages only. This can be achieved using these two analyses. Mesh or Loop Analysis Mesh Analysis involves solving electronic circuits via finding mesh or loop currents of the circuit. This is done by forming KVL equations for respected loops and solving the equations to find individual mesh currents. This method eliminates a great deal of tedious work involved in the branch current method. We simply assume clockwise current flow in all the loops and find them to analyze the circuit. Also any independent current source in a loop becomes the loop current. No. of loops= No. of branches - (No. of nodes-1) Circuit with independent voltage sources Fig.1.13: Mesh analysis for independent voltage sources Using KVL for the circuit as shown in Fig.1.13, at loops 1 and 2, we form KVL equations using the current and components in the loops in terms of the loop currents. Important thing to look at it is the subtraction of the opposing loop current in the shared section of the loop. Equations: R1∙i1 +(i1 – i2) ∙ R3 = V1 R2 ∙ i2 + R3 ∙(i2 – i1) = -V2 i..e. (R1+ R3)∙i1 - i2 ∙R3 = V1 - R3 ∙ i1 + (R2 + R3)∙i2 = -V2 Note: i1 and i2 are mesh current. I1, I2 and I3 are branch current. Dept. of ECE SCSVMV Page 20  NETWORK THEORY I1 = i1; I2 = i2; I3 =( i1 - i2) Formalization: Network equations by inspection Use determinants and Cramer’s rule for solving network equations through manipulation of their co-efficients. Note: Solving equations with two unknowns Solving equations with three unknowns Dept. of ECE SCSVMV Page 21  NETWORK THEORY Problems 1. Determine the current supplied by each battery in the circuit shown in Fig. Q.12. Fig. Q.12 Solution: For loop 1 we get 20 - 5I1 - 3(I1-I2) – 5 = 0 or 8I1 - 3I2=15 For loop 2 we have -4I2 + 5 - 2(I2 – I3 ) + 5 + 5 - 3(I2 - I1 ) = 0 or 3I1 -9 I2 + 2 I3 = -15 Similarly, for loop 3, we get -8I3 – 30 – 5 - 2(I3 – I2 ) = 0 or 2I2 - 10I3 = 35 Dept. of ECE SCSVMV Page 22  NETWORK THEORY On solving, we get I1 = 765/299 A, I2 = 542/299 A and I3 = -1875/598 A So current supplied by each battery is B1=765/299 A B3 = I2+I3 = 2965/598 A B5 = 1875/598 A B2 = I1-I2=220/299 A B4= I2=545/299 A 2. Use mesh analysis to compute the voltage V10Ω in Fig. Q.13. Fig. Q.13 Solution: Fig. Q13.(a) On applying KVL to Fig. Q13.(a) , We have Mesh 1: 24i1-8i2-12i4-24-12=0 or 6i1-2i2-3i4=9 Mesh 2: -8i1+29i2-6i3-15i4=-24 Mesh 3: -6i2+16i3=0 or -3i2+8i3=0 Mesh 4: i4=10ix=10(i2-i3) or 10i2-10i3-i4=0 Dept. of ECE SCSVMV Page 23  NETWORK THEORY On solving, we get i1=1.94A i2=0.13A i3=0.05A i4=0.79A Now, we find V10Ω by ohm’s law, that is, V10Ω =10i3 = 10 * 0.05 = 0.5V 3. Using mesh analysis, find Io for the circuit shown in Fig. Q14. j4 4    I0 100 i1 i2 6-90 -j2    Fig. Q.14 Solution: On applying KVL, we have Mesh 1: 100 -4i1 + j2(i1-i2) = 0 or (2-j)i1 + ji2 = 5 Mesh 2: -j4i2 + j2(i2-i1) - 6/-90° = 0 or –j2i1 + (-j4+j2)i2 = 6-90  I0 = (i1 - i2) On solving, we get i1= 2+j0.5 i2=1-j0.5 I0=1+j=1.414/45° Nodal Analysis The node-equation method is based directly on KCL. In nodal analysis, basically we work with a set of node voltages. It provides a general procedure for analyzing circuits using node voltages as the circuit variables. Dept. of ECE SCSVMV Page 24  NETWORK THEORY For the application of this method, every junction in the network where three or more branches meet is regarded a node. One of these is regarded as the reference node or datum node or zero- potential node. Hence the number of simultaneous equations to be solved becomes (n-1) where n is the number of independent nodes. These node equations often become simplified if all voltage sources are converted into current sources. Then we write the KCL equations for the nodes and solve them to find the respected nodal voltages. Once we have these nodal voltages, we can use them to further analyze the circuit. Example Fig.1.14: Nodal analysis for independent current sources On applying KCL to the circuit shown in Fig.1.14, we get At node 1 1A = V1/2 + (V1-V2)/6 or 0.66V1-0.166V2 = 1A At node 2 (V1-V2)/6 = V2/7+4A or 0.166V1 - 0.309V2 = 4A On solving, we get V1 = -2.01V and V2 = -14.02V Dept. of ECE SCSVMV Page 25  NETWORK THEORY Problems 1. Using nodal analysis, find the node voltages V1 and V2 in Fig. Q.15 Fig. Q.15 Solution: Applying KCL to node 1, we get 8 – 1 - V1/3 - (V1-V2)/6 = 0 or 3V1 - V2 = 42 Similarly, applying KCL to node 2, we get 1 + (V1-V2)/6 – V2/15 – V2/10 = 0 or V1 - 2V2 = -6 Solving for V1 and V2, we get V1 = 18V and V2 = 12V 2. Use nodal analysis to determine the value of current i in the network of Fig. Q.16 Fig. Q.16 Dept. of ECE SCSVMV Page 26  NETWORK THEORY Solution: Applying KCL to node 1, we get 6= + + 3i As seen, i =. Hence, the above equation becomes 6= + +3 or 3V1-V2 = 24 Similarly, applying KCL to node 2, we get + 3i = or +3 = or 3V1 = 2V2 From the above two equations, we get V1 = 16V ∴ i = 16/8 = 2A 3. Find the value of the voltage v for the circuit of Fig. Q.17. Fig. Q.17 Solution: Application of KCL at Node A of the circuit below yields + = 2 or v - vx = 2 Also by KVL v = vx + 2vx and by substitution vx + 2vx – vx = 2 or vx = 1 and thus v = 3V Dept. of ECE SCSVMV Page 27  NETWORK THEORY Self Assessment: 1. Use mesh analysis to compute the current through the 6Ω resistor, and the power supplied (or absorbed) by the dependent source shown in Fig. Q18. Fig. Q18 2. Use mesh analysis to find V0 in the circuit of Fig. Q19. Fig. Q19 3. Use nodal analysis to compute the current through the 6Ω resistor and the power supplied (or absorbed) by the dependent source shown in Fig. Q20 Fig.Q20 Dept. of ECE SCSVMV Page 28  NETWORK THEORY Super Mesh Analysis: If there is only current source between two meshes in the given network then it is difficult to apply the mesh analysis. Because the current source has to be converted into a voltage source in terms of the current source, write down the mesh equations and relate the mesh currents to the current source. But this is a difficult approach.This difficulty can be avoided by creating super mesh which encloses the two meshes that have common current source Super Mesh: A super mesh is constituted by two adjacent meshes that have a common current source. Let us illustrate this method with the following simple generalized circuit. Solution: Step (1):Identify the position of current source. Here the current source is common to the two meshes 1 and 2. so, super mesh is nothing but the combination of meshes 1 and 2. Step (2):Apply KVL to super mesh and to other meshes Applying KVL to this super mesh (combination of meshes 1 and 2 ) we get R1.I1 + R3 ( I2 – I3) = V............. (1) Applying KVL to mesh 3, we get R3 ( I3 – I2) + R4.I3 = 0........... (2) Step (3):Make the relation between mesh currents with current source to get third equation. Third equation is nothing but he relation between I , I1 and I2 which is I1 - I2 = I........... (3) Step(4): Solve the above equations to get the mesh currents. Dept. of ECE SCSVMV Page 29  NETWORK THEORY Example(1): Determine the current in the 5 Ω resistor shown in the figure below. Solution: Step(1): Here the current source exists between mesh(2) and mesh(3).Hence, super mesh is the combination of mesh(2) and mesh(3).Applying KVL to the super mesh ( combination of mesh 2 and mesh 3 after removing the branch with the current source of 2 A and resistance of 3 Ω ) we get : 10( I2– I1) + 2.I2 + I3 + 5( I3 – I1) = 0 -15.I1 +12 I2 + 6.I3 = 0..................(1) Step (2): Applying KVL first to the normal mesh 1 we get : 10( I1 – I2) + 5( I1 – I3) = 50 15.I1 –10. I2 – 5.I3 = 50................... (2) Step (3): We can get the third equation from the relation between the current source of 2 A , and currents I2 & I3 as : I2 - I3 = 2 A.................. (3) Step (4): Solving the above three equations for I1, I2 and I3 we get I1 = 19.99 A I2 = 17.33 A and I3 = 15.33 A The current in the 5 Ω resistance = I1 - I3 = 19.99 - 15.33 = 4.66 A Example(2): Write down the mesh equations for the circuit shown in the figure below and find out the values of the currents I1, I2 and I3 Dept. of ECE SCSVMV Page 30  NETWORK THEORY Solution: In this circuit the current source is in the perimeter of the circuit and hence the first mesh is ignored. So, here no need to create the super mesh. Applying KVL to mesh 1 we get : 3( I2 – I1) + 2( I2 – I3) = -10 -3.I1 +5.I2 – 2.I3 = -10............. (1) Next applying KVL to mesh 2 we get : I3 + 2( I3– I2) = 10 -2.I2 +3.I3 = -10.............. (2) And from the first mesh we observe that....... I1 = 10 A............. (3) And solving these three equations we get : I1 = 10 A, I2 = 7.27 A, I3 = 8.18 A Nodal analysis: Nodal analysis provides another general procedure for analyzing circuits nodal voltages as the circuit variables. It is preferably useful for the circuits that have many no. of nodes. It is applicable for the both planar and non planar circuits. This analysis is done by using KCL and Ohm's law. Node: It is a junction at which two or more branches are interconnected. Simple Node: Node at which only two branches are interconnected. Principal Node: Node at which more than two branches are interconnected. Nodal analysis with example: Determination of node voltages: Dept. of ECE SCSVMV Page 31  NETWORK THEORY Procedure: Step (1): Identify the no. nodes, simple nodes and principal nodes in the given circuit. Among all the nodes one node is taken as reference node. Generally bottom is taken as reference node. The potential at the reference node is 0v. In the given circuit there are 3 principal nodes in which node (3) is the reference node. Step (2): Assign node voltages to the all the principal nodes except reference node and assign branch currents to all branches. Step (3): Apply KCL to those principal nodes for nodal equations and by using ohm's law express the node voltages in terms of branch current. Applying KCL to node (1) ---- 1=I2+I3 Using ohm's law, we get (V-V1)/R1 =(V2-0)/R2 +(V1-V2)/R3......... (1) Applying KCL to node (2) ---- 3=I4 +I5 Using ohm's law, we get (V1-V2)/R3 =(V4-0)/R4 +(V5-0)/R5.............. (2) Step(4): Solve the above nodal equations to get the node voltages. Example: Write the node voltage equations and find out the currents in each branch of the circuit shown in the figure below. Dept. of ECE SCSVMV Page 32  NETWORK THEORY Solution: The node voltages and the directions of the branch currents are assigned as shown in given figure. Applying KCL to node 1, we get: 5 = I10+ I3 5= (V1-0)/10 +(V1-V2)/3 V1(13/30) -V2(1/3) = 5...........(1) Applying KCL to node 2, we get: I3= I5 + I1 (V1-V2)/3 = (V2 -0)/5 + (V2-10) /1 V1(1/3)-V2(23/15) = -10............... (2) Solving the these two equations for V1 and V2 we get : V1 = 19.85 V and V2 = 10.9 V and the currents are : I10= V1/10 = 1.985A I3 = (V1-V2)/3 = (19.85-10.9)/3 = 2.98A I5 = V2/5 = 10.9/5 =2.18A I1 = (V2-10) = (10.9-10)/1 = 0.9A Super Node Analysis: If there is only voltage source between two nodes in the given network then it is difficult to apply the nodal analysis. Because the voltage source has to be converted into a current source in terms of the voltage source, write down the nodal equations and relate the node voltages to the voltage source. But this is a difficult approach.This difficulty can be avoided by creating super node which encloses the two nodes that have common voltage source. Super Node: A super node is constituted by two adjacent nodes that have a common voltage source. Example: Write the nodal equations by using super node analysis. Dept. of ECE SCSVMV Page 33  NETWORK THEORY Procedure: Step(1):Identify the position of voltage source.Here the voltage source is common to the two nodes 2 and 3.so, super node is nothing but the combination of nodes 2 and 3. Step (2):Apply KCL to super node and to other nodes. Applying KCL to this super node (combination of meshes 2 and 3 ), we get (V2-V1)/R2 + V2/R3 + (V3-Vy)/R4 + V3/R5 = 0.............. (1) Applying KVL to node 1 ,we get I = V1/R1 + (V1-V2)/R2.................(2) Step (3): Make the relation between node voltages with voltage source to get third equation. Third equation is nothing but the relation between VX , V2 and V3 which is V2 - V3 = Vx............. (3) Step (4): Solve the above nodal equations to get the node voltages. Example: Determine the current in the 5 Ω resistor shown in the circuit below Solution: Applying KCL to node 1: 10 = V1/3 + (V1-V2)/2 V1 [ 1/3 + 1/2 ] - V2 /2 = 10 0.83 V1 - 0.5 V2 = 10................ (1) Next applying KCL to the super node2&3 : (V2-V1)/2 + V2/1 + (V3-10)/5 + V3/2 = 0 -V1/2 + V2(1/2 + 1) V3(1/5 + 1/2) = 2 0.5 V1 + 1.5V2 + 0.7 V3 = 2....................... (2) and the third and final equation is: V2 - V3 = 20.................... (3) Dept. of ECE SCSVMV Page 34  NETWORK THEORY Solving the above three equations we get V3 = -8.42 V The current through the 5 Ω resistor I5 = [-8.42 -10 ] /5 = -3.68 A The negative sign indicates that the current flows towards the node 3. Assignment Questions: 3. Calculate the current flowing through the 10Ω resistor of Fig. Q.9. Fig. Q.9 2 Using Kirchhoff’s Current Law and Ohm’s law, find the magnitude and polarity of voltage V in Fig. Q10 Fig. Q10 3.A network of resistances is formed as shown in Fig.Q.10. Compute the network resistance measured between (i) A and B (ii) B and C (iii) C and A. Fig. Q.10 Dept. of ECE SCSVMV Page 35  NETWORK THEORY 6. Determine the current supplied by each battery in the circuit shown in Fig. Q.11. Fig. Q.11 5.Use nodal analysis to determine the value of current i in the network of Fig. Q.12 Fig. Q.12 Conclusion: In this topic learner will be able to apply knowledge of KVL.KCL, Star to delta and Delta to star to solve numerical based on network simplification and it will be used to analyze the same. Reference:.Sudhakar, A., Shyammohan, S. P.; “Circuits and Network”; Tata McGraw-Hill New Delhi,2000. A William Hayt, “Engineering Circuit Analysis” 8th Edition, McGraw-Hill Education 2004. Paranjothi SR, “Electric Circuits Analysis,” New Age International Ltd., New Delhi, 1996. Dept. of ECE SCSVMV Page 36  NETWORK THEORY Post Test MCQs: 1. In figure the voltage drop across the 10 Ω resistance is 10 V. The resistance R a. is 6 Ω b. is 8 Ω c. is 4 Ω d. cannot be found 2. The resistance of the circuit shown is figure is a. More than 6 Ω b. 5 Ω c. More than 4 Ω d. Between 6 and 7 Ω 3. For the network in figure, the correct loop equation for loop 3 is a. -I1 + 4I2 + 11I3 = 0 b. I1 + 4I2 + 11I3 = 0 c. -I1 - 4I2 + 11I3 = 0 d. I1 - 4I2 + 6I3 = 0 Dept. of ECE SCSVMV Page 37  NETWORK THEORY 4. A Thermister is used for a. over voltage protection b. automatic light control c. temperature alarm circuit d. none of the above 5. In figure, E = 1 V (rms value). The average power is 250 mW. Then phase angle between E and I is a. 90° b. 60° c. 45° d. 30° Dept. of ECE SCSVMV Page 38  NETWORK THEORY Unit-2 NETWORK THEOREM AIM: To learn techniques of solving circuits involving different active and passive elements. Pre-Requisites: Knowledge of Basic Mathematics – II & Basic Electronics Engineering Pre - MCQs: 1. What will be the value of Req in the following Circuit? a) 11.86 ohm b) 10 ohm c) 25 ohm d). 11.18 ohm 2. What will be the value of Va in the following circuit? a) -11 V b). 11 V c) 3 V d) -3 V 3. What will be the value of Va in the following circuit? a) 4.33 V b) 4.09 V c).8.67 V d) 8.18 V Dept. of ECE SCSVMV Page 39  NETWORK THEORY Network Theorems Introduction: Complex circuits could be analysed using Ohm’s Law and Kirchoff’s laws directly, but the calculations would be tedious. To handle the complexity, some theorems have been developed to simplify the analysis. Superposition Theorem The Superposition theorem states that in any linear bilateral network containing two or more independent sources (voltage or current sources or combination of voltage and current sources), the resultant current/voltage in any branch is the algebraic sum of currents/voltages caused by each independent sources acting alone, with all other independent sources being replaced meanwhile by their respective internal resistances as shown in Fig. 2.1 (a & b). Fig. 2.1: (a) A Linear bilateral network (b) The resultant circuit Procedure for using the superposition theorem Step-1: Retain one source at a time in the circuit and replace all other sources with their internal resistances. i.e., Independent voltage sources are replaced by 0 V (short circuit) and Independent current sources are replaced by 0 A (open circuit). Step-2: Determine the output (current or voltage) due to the single source acting alone using the mesh or nodal analysis. Step-3: Repeat steps 1 and 2 for each of the other independent sources. Step -4: Find the total contribution by adding algebraically all the contributions due to the independent sources. Note: Dependent sources are left intact because they are controlled by circuit variables. Example: Use the superposition theorem to find v in the circuit shown in Fig. 2.2. Dept. of ECE SCSVMV Page 40  NETWORK THEORY Fig. 2.2 Solution: Consider voltage source only as shown in Fig. 2.2(a) (current source 3A is discarded by open circuit) Fig. 2.2(a) Consider current source only as shown in Fig. 2.2(b) (voltage source 6V is discarded by short circuit) Fig. 2.2(b) Total voltage v = v1+v2 =10V Problems 1. Use the superposition theorem to find I in the circuit shown in Fig. 26 Fig. 26 Dept. of ECE SCSVMV Page 41  NETWORK THEORY Solution: Consider only 12V source Is1= = 4.2359 A ∴ Ir1= x 4.2359 = 0.3529 A Consider only 10V source Is2= =3.382A ∴ Ir2= x 3.38235 = 0.14706 A From superposition theorem, I= Ir1+Ir2 = 0.5A 2. Use the superposition theorem to find I in the circuit shown in Fig. 27 Fig. 27 Solution: Consider only 120A source Using the current divider rule, we get I1=120 x 50/200= 30 A Dept. of ECE SCSVMV Page 42  NETWORK THEORY Consider only 40A source I2=40 x 150/200= 30 A Consider only 10V source Using Ohm’s law I3 = 10/200 = 0.05 A Using superposition theorem, Since I1 and I2 cancel out, I=I3=0.05 A 3. Find the current i using superposition theorem for the circuit shown in Fig.28 Fig. 28 Dept. of ECE SCSVMV Page 43  NETWORK THEORY Solution: As a first step in the analysis, we will find the current resulting from the independent voltage source. The current source is deactivated and we have the circuit as shown in Fig. 28(a) Applying KVL clockwise around loop shown in Fig. 3.12, we find that 5i1+3i1-24=0 i1 = 3 A Fig. 28(a) As a second step, we set the voltage source to zero and determine the current i2 due to the current source as shown in Fig.28(b). Applying KCL at node 1, we get = and we get , On substituting for , we get Fig. 28(b) Thus, the total current i = i1 + i2 = 4. For the circuit shown in Fig.29, find the terminal voltage Vab using superposition principle. Fig.29 Dept. of ECE SCSVMV Page 44  NETWORK THEORY Solution: Consider 4V source Apply KVL, we get 4- 10 x 0 - 3Vab1 - Vab1 = 0 Vab1 = 1V Consider 2A source Apply KVL, we get -10 x 2 + 3Vab2 + Vab2 = 0 Vab2 = 5V According to superposition principle, Vab=Vab1+Vab2 = 6V Self Assessment 1. Find the current flowing in the branch XY of the circuit shown in Fig.30 by superposition theorem. Fig.30 (Ans: 1.33 A) Dept. of ECE SCSVMV Page 45  NETWORK THEORY 2. Apply Superposition theorem to the circuit of Fig.31 for finding the voltage drop V across the 5Ω resistor. Fig. 31 (Ans: 19 V) 3. Find the voltage V1 for the circuit shown in Fig. 32 using the superposition principle. Fig. Q32 (Ans: 82.5 V) 4. Find I for the circuit shown in Fig. 33 using the superposition theorem. 1 1   Vx   2Vx 2A 2V 1 i 2 Fig.33 (Ans: 2 A) Dept. of ECE SCSVMV Page 46  NETWORK THEORY Remarks: Superposition theorem is most often used when it is necessary to determine the individual contribution of each source to a particular response. Limitations: Superposition principle applies only to the current and voltage in a linear circuit but it cannot be used to determine power because power is a non-linear function. Thevenin’s theorem In section 2.1, we saw that the analysis of a circuit may be greatly reduced by the use of superposition principle. The main objective of Thevenin’s theorem is to reduce some portion of a circuit to an equivalent source and a single element. This reduced equivalent circuit connected to the remaining part of the circuit will allow us to find the desired current or voltage. Thevenin’s theorem is based on circuit equivalence. Fig.2.3: (a) A Linear two terminal network (b) The Thevenin’s equivalent circuit The Thevenin’s theorem may be stated as follows: A linear two–terminal circuit can be replaced by an equivalent circuit consisting of a voltage source Vt in series with a resistor Rt, Where Vt is the open–circuit voltage at the terminals and Rt is the input or equivalent resistance at the terminals when the independent sources are turned off or Rt is the ratio of open–circuit voltage to the short–circuit current at the terminal pair which is as shown in Fig. 2.3(a & b). 2.2.1 Action plan for using Thevenin’s theorem : 1. Divide the original circuit into circuit A and circuit B Dept. of ECE SCSVMV Page 47  NETWORK THEORY In general, circuit B is the load which may be linear or non-linear. Circuit A is the balance of the original network exclusive of load and must be linear. In general, circuit may contain independent sources, dependent sources and resistors or other linear elements. 2. Separate the circuit from circuit B 3. Replace circuit A with its Thevenin’s equivalent. 4. Reconnect circuit B and determine the variable of interest (e.g. current ‘i’ or voltage ‘v’) Procedure for finding Rt Three different types of circuits may be encountered in determining the resistance, R t (i) If the circuit contains only independent sources and resistors, deactivate the sources and find Rt by circuit reduction technique. Independent current sources, are deactivated by opening them while independent voltage sources are deactivated by shorting them. (ii) If the circuit contains resistors, dependent and independent sources, follow the instructions described below: (a) Determine the open circuit voltage voc with the sources activated. (b) Find the short circuit current isc when a short circuit is applied to the terminals a-b (c) (iii) If the circuit contains resistors and only dependent sources, then (a) voc = 0 (since there is no energy source) (b) Connect 1A current source to terminals a-b and determine vab (c) Dept. of ECE SCSVMV Page 48  NETWORK THEORY For all the cases discussed above, the Thevenin’s equivalent circuit is as shown in Fig. 2.4. Fig. 2.4: The Thevenin’s equivalent circuit Problems 1. Using the Thevenin’s theorem, find the current i through R = 2Ω for the circuit shown in Fig. 34. Fig. 34 Solution: Since we are interested in the current i through R, the resistor R is identified as circuit B and the remainder as circuit A. After removing the circuit B, circuit A is as shown in Fig.34(a). Dept. of ECE SCSVMV Page 49  NETWORK THEORY Fig.34(a) Referring to Fig. 34(a) Hence To find Rt, we have to deactivate the independent voltage source. Accordingly, we get the circuit in Fig.34(b). Fig.34(b) Thus, we get the Thevenin’s equivalent circuit which is as shown in Fig.34(c) Fig.34(c). Reconnecting the circuit B to the Thevenin’s equivalent circuit as shown in Fig.34(c), we get Dept. of ECE SCSVMV Page 50  NETWORK THEORY 2. Find V0 in the circuit of Fig.35 using Thevenin’s theorem. Fig.35 Solution: To find Voc : Since we are interested in the voltage across 2 kΩ resistor, it is removed from the circuit of Fig.35 and so the circuit becomes as shown in Fig.35(a) Fig.35(a) By inspection, Applying KVL to mesh 2, we get Solving, we get Applying KVL to the path 4 kΩ -> a-b -> 3 kΩ, we get On solving, Dept. of ECE SCSVMV Page 51  NETWORK THEORY To find Rt : Deactivating all the independent sources, we get the circuit diagram shown in Fig.35(b) Fig.35(b) Hence, the Thevenin equivalent circuit is as shown in Fig.35(c). Fig.35(c) Fig.35(d) If we connect the 2kΩ resistor to this equivalent network, we obtain the circuit of Fig.35(d). 3. Find the Theveni’s equivalent for the circuit shown in Fig.36 with respect to terminals a-b. Fig.36 Dept. of ECE SCSVMV Page 52  NETWORK THEORY Fig.36(a) Solution: To find = : Applying KVL around the mesh of Fig.36(a), we get On solving, Since there is no current flowing in 10Ω resistor, To find Rt : Since both dependent and independent sources are present, Thevenin’s resistance is found using the relation, Fig.36(b) Applying KVL clockwise for mesh 1 of Fig.36(b), we get Since Above equation becomes Dept. of ECE SCSVMV Page 53  NETWORK THEORY Applying KVL clockwise for mesh 2, we get Solving the above two mesh equations, we get Hence Self Assessment 1. For to the circuit shown in Fig.37, find the Thevenin’s equivalent circuit at the terminals a-b. Fig.37 (Ans: Voc = 10V, Rt = 3.33Ω) 2. For the circuit shown in Fig.38, find the Thevenin’s equivalent circuit between terminals a and b. Fig.38 (Ans: Voc = 32V, Rt = 4Ω) Dept. of ECE SCSVMV Page 54  NETWORK THEORY 3. For the circuit shown in Fig.39, find the Thevenin’s equivalent circuit between terminals a and b. Fig.39 (Ans: Voc = 12V, Rt = 0.5Ω) 4. Find the Thevenin’s equivalent circuit as seen from the terminals a-b. Refer the circuit diagram shown in Fig.40 Fig.40 Hint: Since the circuit has no independent sources, i = 0 when the terminals a-b are open. Therefore Voc = 0. Hence, we choose to connect a source of 1 A at the terminals a-b then, after finding Vab, the Thevenin resistance is, (Ans: Voc = 0 ; Rt = 3.8Ω) Norton’s theorem Norton’s theorem is the dual theorem of Thevenin’s theorem where the voltage source is replaced by a current source. Norton’s theorem states that a linear two-terminal network shown in Fig. 2.5(a) can be replaced by an equivalent circuit consisting of a current source iN in parallel with resistor RN, where iN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the independent sources are turned off. If one does not wish to turn off the independent sources, then RN is the ratio of open circuit voltage to short–circuit current at the terminal pair. Dept. of ECE SCSVMV Page 55  NETWORK THEORY Fig.2.5: (a) A Linear two terminal network (b) The Norton’s equivalent circuit Fig. 2.5(b) shows Norton’s equivalent circuit as seen from the terminals a-b of the original circuit shown in Fig. (a). Since this is the dual of the Thevenin’s circuit, it is clear that RN = Rt and. In fact, source transformation of Thevenin’s equivalent circuit leads to Norton’s equivalent circuit. Procedure for finding Norton’s equivalent circuit: (1) If the network contains resistors and independent sources, follow the instructions below: (a) Deactivate the sources and find RN by circuit reduction techniques. (b) Find iN with sources activated. (2) If the network contains resistors, independent and dependent sources, follow the steps given below: (a) Determine the short-circuit current iN with all sources activated. (b) Find the open-circuit voltage voc. (c) (3) If the network contains only resistors and dependent sources, follow the procedure described below: (a) Note that iN = 0. (b) Connect 1A current source to the terminals a-b and find vab. (c) Note: Also, since = and iN = isc Dept. of ECE SCSVMV Page 56  NETWORK THEORY Problems 1. Find the Norton equivalent for the circuit of Fig. 41 Fig. 41 Solution: As a first step, short the terminals a-b. This results in a circuit as shown in Fig. 41(a) Fig. 41(a) Applying KCL at node a, we get So To find RN, deactivate all the independent sources, resulting in a circuit diagram as shown in Fig. Q41(b). We find RN in the same way as Rt in the Thevenin’s equivalent circuit. Dept. of ECE SCSVMV Page 57  NETWORK THEORY Fig. 41 (b) Thus, we obtain Norton equivalent circuit as shown in Fig. Q41(c) Fig. 41(c) 2. Find i0 in the network of Fig.42 using Norton’s theorem. Fig. 42 Solution: We are interested in i0, hence the 2 kΩ resistor is removed from the circuit diagram of Fig. Q17. The resulting circuit diagram is shown in Fig. 42(a). Dept. of ECE SCSVMV Page 58  NETWORK THEORY Fig. 42(a) Fig. 42(b) To find iN or isc: Refer Fig. 42(b). By inspection, Applying KCL at node V2: Substituting V1, we get To find RN: Deactivate all the independent sources. Refer Fig. 42(c). Fig. 42(c) Dept. of ECE SCSVMV Page 59  NETWORK THEORY Hence, the Norton equivalent circuit along with 2 kΩ resistor is as shown in Fig. 42(d) Fig. 42(d) 3. Refer the circuit shown in Fig.43. find the value of ib using Norton’s equivalent circuit. Take R = 667 Ω. Fig.43 Solution: Since we want the current flowing through R, remove R from the circuit of Fig.43. The resulting circuit diagram is shown in Fig. 43(a). Fig.43(a) Since ia= 0A, Dept. of ECE SCSVMV Page 60  NETWORK THEORY To find RN: The procedure for finding RN is same that of Rt in the Thevenin equivalent circuit. To find voc, make use of the circuit diagram shown in Fig.43(b). Do not deactivate any source. Fig.43(b) Applying KVL clockwise, we get Therefore, The Norton equivalent circuit along with resistor R is as shown in Fig.43(c) Fig. 43(c) Dept. of ECE SCSVMV Page 61  NETWORK THEORY Self Assessment: 1. Find V0 in the circuit of Fig.45 Fig. 45 (Ans : V0 =258mV) 2. For the circuit shown in Fig.46, calculate the current in the 6Ω resistance using Norton’s theorem. Fig.46 (Ans : 0.5A from B to A) 3. Find the Norton equivalent to the left of the terminals a-b for the circuit of Fig.47 Fig.47 (Ans : isc =100mA, RN=50Ω) Dept. of ECE SCSVMV Page 62  NETWORK THEORY Maximum Power Transfer Theorem In circuit analysis, we are sometimes interested in determining the maximum power that a circuit can supply to the load. Consider the linear circuit A as shown in Fig. 2.6. Fig. 2.6: A Linear circuit Circuit A is replaced by its Thevenin’s equivalent circuit as seen from a and b as shown in Fig. 2.7. Fig. 2.7: Thevenin’s equivalent circuit is substituted for circuit A We wish to find the value of the load RL such that the maximum power is delivered to it. The power that is delivered to the load is given by (i) Assuming that Vt and RL are fixed for a given source, the maximum power is a function of RL. In order to determine the value of RL that maximizes p, we differentiate p with respect to RL and equate the derivative to zero. (ii) which yields To confirm that equation (ii) is a maximum, it should be shown that Hence, maximum power is transferred to the load when RL is equal to the Thevenin’s equivalent resistance Rt. The maximum power transferred to the load is obtained by substituting in equation (i). Accordingly, Dept. of ECE SCSVMV Page 63  NETWORK THEORY The maximum power transfer theorem states that the maximum power delivered by a source represented by its Thevenin equivalent circuit is attained when the load R L is equal to the Thevenin’s resistance Rt. Problems 1. Find RL for maximum power transfer and the maximum power that can be transferred in the network shown in Fig. 48. Fig.48 Solution: Disconnect the load resistor RL and deactivate all the independent sources to find Rt. The resultant circuit is as shown in the Fig.48(a) Fig.48(a) For maximum power transfer, Let us next find Voc or Vt. Refer Fig.48(b) By inspection, Fig.48(b) Dept. of ECE SCSVMV Page 64  NETWORK THEORY Applying KVL clockwise to the loop , we get On solving, The Thevenin’s equivalent circuit with load resistor RL is as shown in Fig.48(c) Fig.48(c) 2. Find the value of RL for maximum power transfer for the circuit shown in Fig.49. Hence find Pmax. Fig. 49 Solution: Removing RL from the original circuit gives us the circuit diagram shown in Fig.49(a) Fig.49(a) Dept. of ECE SCSVMV Page 65  NETWORK THEORY To find Voc : KCL at node A : Hence, To find Rt, refer Fig.49(b) we need to compute isc with all independent sources activated. KCL at node A: Hence, Hence, for maximum power transfer RL =Rt = 3Ω. Fig.49(b) The Thevenin’s equivalent circuit with R L = 3Ω inserted between the terminals a-b gives the network shown in Fig.49(c). Fig.49(c). Dept. of ECE SCSVMV Page 66  NETWORK THEORY Self Assessment 1. Find the value of RL for maximum power transfer in the circuit shown in Fig.50. Also find Pmax. Fig.50 (Ans: Pmax = 625mW) 2. Find the value of RL in the network shown in Fig.51 that will achieve maximum power transfer, and determine the value of the maximum power. Fig.51 (Ans: Pmax = 81mW) 3. Refer to the circuit shown in Fig.52 (a) Find the value of RL for maximum power transfer. (b) Find the maximum power that can be delivered to RL. Fig.52 (Ans: RL=2.5Ω, Pmax = 2250W) Dept. of ECE SCSVMV Page 67  NETWORK THEORY We have earlier shown that for a resistive network, maximum power is transferred from a source to the load, when the load resistance is set equal to the thevenin’s resistance with thevenin’s equivalent source. Now we extend this result to the ac circuits. Fig. 2.8: (a)Linear circuit (b) Thevenin’s equivalent circuit In Fig. 2.8(a), the linear circuit is made up of impedances, independent and dependent sources.This linear circuit is replaced by its thevenin’s equivalent circuit as shown in Fig. 2.8(b). In rectangular form, the thevenin impedance Z t and the load impedance ZL are and The current through the load is The phasors I and Vt are the maximum values. The corresponding RMS values are obtained by dividing the maximum values by. Also, the RMS value of phasor current flowing in the load must be taken for computing the average power delivered to the load. The average power delivered to the load is given by Our idea is to adjust the load parameters RL and XL so that P is maximum. To do this, we get and equal to zero. Dept. of ECE SCSVMV Page 68  NETWORK THEORY Setting gives (ii) and setting gives (iii) Combining equations (ii) and (iii), we can conclude that for maximum average power transfer, ZL must be selected such that and. That is the maximum average power of a circuit with an impedance Zt that is obtained when ZL is set equal to complex conjugate of Zt. Setting and in equation (i), we get the maximum average power as In a situation where the load is purely real, the condition for maximum power transfer is obtained by putting in equation (iii). That is, Hence for maximum average power transfer to a purely resistive load, the load resistance is equal to the magnitude of thevenin impedance. Maximum average power can be delivered to ZL only if. is the complex conjugate of ZL Problems 1. Find the load impedance that transfers the maximum power to the load for the circuit shown in Fig.53 Fig.53 Dept. of ECE SCSVMV Page 69  NETWORK THEORY Solution: We select, for maximum power transfer. Hence 2. For the circuit of Fig.54, what is the value of ZL that will absorb the maximum average power? Fig.54 Solution: Disconnecting ZL from the original circuit we get the circuit as shown in Fig.54(a). The first step is to find Vt. Fig.54(a). The next step is to find ZL. This requires deactivating the independent voltage source of Fig.54(b) Fig.54(b) Dept. of ECE SCSVMV Page 70  NETWORK THEORY The value of ZL for maximum average power absorbed is The Thevenin’s equivalent circuit along with ZL is as shown in Fig.54(c) Fig.54(c) Self Assessment 1. Find the load impedance that transfers the maximum average power to the load and determine the maximum average power transferred to the load ZL shown in Fig.55. Fig.55 (Ans: Pmax= 6W) Dept. of ECE SCSVMV Page 71  NETWORK THEORY 2.5 Reciprocity theorem The reciprocity theorem states that in a linear bilateral single source circuit, the ratio of excitation to response is constant when the positions of excitation and response are interchanged. Conditions to be met for the application of reciprocity theorem: (i) The circuit must have a single source. (ii) Dependent sources are excluded even if they are linear. (iii) When the positions of source and response are interchanged, their directions should be marked same as in the original circuit. Problems 1. In the circuit shown in Fig.57, find the current through 1.375 Ω resistor and hence verify reciprocity theorem. Fig.57 Solution: Apply KVL to Fig.57(a) KVL clockwise for mesh 1: KVL clockwise for mesh 2: KVL clockwise for mesh 3: Fig.57(a) Dept. of ECE SCSVMV Page 72  NETWORK THEORY Using Cramer’s rule, we get Dept. of ECE SCSVMV Page 73  NETWORK THEORY Verification using reciprocity theorem: The circuit is redrawn by interchanging the positions of excitation and response. The new circuit is shown in Fig.57(b) Fig.57(b) 1. TELLEGEN‟S THEOREM: Dc Excitation: Tellegen‘s theorem states algebraic sum of all delivered power must be equal to sum of all received powers.According to Tellegen‘s theorem, the summation of instantaneous powers for the n number of branches in an electrical network is zero. Are you confused? Let's explain. Suppose n number of branches in an electrical network have i1, i2, i3…. in respective instantaneous currents through them. These currents satisfy Kirchhoff's Current Law. Again, suppose these branches have instantaneous voltages across them are v1, v2, v3, vn respectively. Ifthese voltages across these elements satisfy Kirchhoff Voltage Law then, vk is the instantaneous voltage across the kth branch and ik is the instantaneous current flowing through this branch. Tellegen‟s theorem is applicable mainly in general class of lumped networks that consist of linear, non-linear, active, passive, time variant and time variant elements. This theorem can easily be explained by the following example. Dept. of ECE SCSVMV Page 74  NETWORK THEORY In the network shown, arbitrary reference directions have been selected for all of the branch currents, and the corresponding branch voltages have been indicated, with positive reference direction at the tail of the current arrow. For this network, we will assume a set of branch voltages satisfy the Kirchhoff voltage law and a set of branch current satisfy Kirchhoff current law at each node.We will then show that these arbitrary assumed voltages and currents satisfy the equation. And it is the condition of Tellegen’s theorem. In the network shown in the figure, let v1, v2 and v3 be 7, 2 and 3 volts respectively. Applying Kirchhoff Voltage Law around loop ABCDEA. We see that v4 = 2 volt is required. Around loop CDFC, v5 is required to be 3 volt and around loop DFED, v6 is required to be 2. We next apply Kirchhoff's Current Law successively to nodes B, C and D. At node B let ii = 5 A, then it is required that i2 = - 5 A. At node C let i3 = 3 A and then i5 is required to be - 8. At node D assume i4 to be 4 then i6 is required to be - 9. Carrying out the operation of equation, We get, Hence Tellegen’s theorem is verified. Conclusion: With help of Network Theorems one can find the choice of load resistance RL that results in the maximum power transfer to the load. On the other hand, the effort necessary to solve this problem-using node or mesh analysis methods can be quite complex and tedious from computational point of view. Reference: Dept. of ECE SCSVMV Page 75  NETWORK THEORY.Sudhakar, A., Shyammohan, S. P.; “Circuits and Network”; Tata McGraw-Hill New Delhi,2000. A William Hayt, “Engineering Circuit Analysis” 8th Edition, McGraw-Hill Education 2004. Paranjothi SR, “Electric Circuits Analysis,” New Age International Ltd., New Delhi, 1996. Post Test MCQs: 1.The equation where v1, v2... vb are the instantaneous branch voltages and i1, i2... ib are the instantaneous branch currents pertains to a. Compensation theorem b. Superposition theorem c. Tellegen's theorem d. Reciprocity theorem 2.The reading of the voltmeter in figure will be __________ volt. a. 100 b. 150 c. 0 d. 175 3.A circuit is replaced by its Thevenin's equivalent to find current through a certain branch. If VTH = 10 V and RTH = 20 Ω, the current through the branch a. will always be 0.5 A b. will always be less than 0.5 A c. will always be equal to or less than 0.5 A d. may be 0.5 A or more or less Dept. of ECE SCSVMV Page 76  NETWORK THEORY 4. Given Is = 20 A, Vs = 20 V, the current I in the 3 Ω resistance is given by a. 4 A b. 8 A c. 8 A d, 16 A 5. For maximum transfer of power, internal resistance of the source should be a. equal to load resistance b. less than the load resistance c. greater than the load resistance d. none of the above 6.Tellegen's theorem is applicable to a. circuits with passive elements only b. circuits with linear elements only c. circuits with time invariant elements only d..circuits with active or passive, linear or nonlinear and time invariant or time varying elements 7. Assertion (A): Millman's theorem helps in replacing a number of current sources in parallel by a single current source. Reason (R): Maximum power transfer theorem is applicable only for dc, circuits. a. Both A and R are true and R is correct explanation of A b. Both A and R are true and R is not the correct explanation of A c..A is true but R is false d. A is false but R is true 8. “Any number of current sources in parallel may be replaced by a single current source whose current is the algebraic sum of individual currents and source resistance is the parallel Dept. of ECE SCSVMV Page 77  NETWORK THEORY combination of individual source resistances”. The above statement is associated with. a. Thevenin’s theorem b..Millman’s theorem. c. Maximum power transfer theorem d. None of the above. 9. Superposition theorem can be applied only to circuits having a. Resistive elements b. Passive elements c. Nonlinear elements d..Linear bilateral elements 10. A nonlinear network does not satisfy a. Superposition condition b. Homogeneity condition c. Both homogeneity as well as superposition condition d. Homogeneity, superposition and associative condition. Dept. of ECE SCSVMV Page 78  NETWORK THEORY UNIT III TWO PORT NETWORKS AND FILTERS DESIGN AIM: To analyze the behavior of the circuit’s response in time domain Pre-Requisites: Knowledge of Basic Mathematics – II & Basic Electronics Engineering Pre - MCQs: 1.In figure, the power associated with 3 A source is a. 36 W b. 24 W c. 16 W d. 8 W Answer: Option B Explanation: Voltage across 4Ω resistance = source voltage = 8V. Power = 8 X 3 = 24 W. 2. In figure, the value of R should be a. 0.2 Ω b. 0.1 Ω c. 0.05 Ω d. 0.1 Ω Answer: Option B Explanation: Dept. of ECE SCSVMV Page 79  NETWORK THEORY 3. In figure, A is ideal ammeter having zero resistance. It will read __________ ampere. a. 2 b2.5 c. 3 d.4 Answer: Option C Explanation: INTRODUCTION: A pair of terminals through which a current may enter or leave a network is known as a port. Two-terminal devices or elements (such as resistors, capacitors, and inductors) result in one-port networks. Most of the circuits we have dealt with so far are two-terminal or one-port circuits, represented in Figure 2(a). We have considered the voltage across or current through a single pair of terminals—such as the two terminals of a resistor, a capacitor, or an inductor. We have also studied four-terminal or two-port circuits involving op amps, transistors, and transformers, as shown in Figure 2(b). In general, a network may