NETWORK THEORY Subject Code:3EC4-06 UNIT-1 Node and Prepared By Abhinandan Jain Assistant Professor, ECE

SKIT, Jaipur Abhinandan Jain, Asst. Professor, ECE, SKIT 1 Contents • Introduction • Classification of Network • Network Terminology • Kirchhoff’s Current Law(KCL) • Kirchhoff’s Law(KVL) • Nodal Analysis • Super Node Analysis • Mesh Analysis

Abhinandan Jain, Asst. Professor, ECE, SKIT 2

Contents……..

• Super Mesh analysis • Matrix approach of network containing voltage and current sources and reactance based circuits • • Duality

Abhinandan Jain, Asst. Professor, ECE, SKIT 3

Introduction of Subject

• Network Theory is a very basic subject which includes theorems and techniques to solve a complex network in a very simple form.

• An is composed of individual electronic components, such as , transistors, , and diodes, connected by conductive wires or traces.

Abhinandan Jain, Asst. Professor, ECE, SKIT 4

Continued………

• Network/ Circuit analysis is the process of finding the across, and the currents through, every component in the network.

• A useful procedure in network analysis is to simplify the network by reducing the number of components.

Abhinandan Jain, Asst. Professor, ECE, SKIT 5 Importance of Subject

• Network Theory is a core subject for undergraduate students in Electronics and Communication Engineering. • By learning Network Theory we will be able to analyze the different circuit and calculate the voltage, current and power delivered/ absorbed in the circuit elements.

• Knowledge gain by the Network Theory will help students in subjects like Electronics Devices, Analog Circuits, Signal and System and Control system etc. • Very important subject for competitive exams like IES, GATE and PSU’s entrance exams

Abhinandan Jain, Asst. Professor, ECE, SKIT 6 Applications

• Communications systems, Electronics circuits and power systems all consist of more or less complicated electric circuits which themselves are made up of a number of circuit elements.

• The devices and equipment may be represented by ‘equivalent circuits’ consisting of basic circuit elements, and an equivalent circuit must behave to all intents and purposes in the same way as the device or equipment which it represents.

Abhinandan Jain, Asst. Professor, ECE, SKIT 7 Syllabus Unit Contents 1 Node and Mesh Analysis, matrix approach of network containing voltage and current sources, and reactance's, source transformation and duality. 2 Network theorems: Superposition, reciprocity, Thevenin’s, Norton’s, Maximum power Transfer, compensation and Tallegen's theorem as applied to AC. circuits. 3 Trigonometric and exponential Fourier series: Discrete spectra and symmetry of waveform, steady state response of a network to non sinusoidal periodic inputs, power factor, effective values, Fourier transform and continuous spectra, three phase unbalanced circuit and power calculation.

4 Laplace transforms and properties: Partial fractions, singularity functions, waveform synthesis, analysis of RC, RL, and RLC networks with and without initial conditions with Laplace transforms evaluation of initial conditions. 5 Transient behavior, concept of complex frequency, Driving points and transfer functions poles and zeros of immittance function, their properties, sinusoidal response from pole-zero locations, convolution theorem and Two four port network and interconnections, Behaviors of series and parallel resonant circuits, Introduction to band pass, low pass, high pass and band reject filters.

Abhinandan Jain, Asst. Professor, ECE, SKIT 8

3EC4-06 Network Theory COURSE PLAN

UNIT UNIT NAME NO OF LECTURE REQUIRED 1 Node and Mesh Analysis 8

2 Network theorems 10

3 Trigonometric and exponential Fourier 10 series:

4 Laplace transforms 9

5 Transient behavior, Two port Network, 13 Network Function

Abhinandan Jain, Asst. Professor, ECE, SKIT 9 Marking Scheme 1. Internal There will be two internal exams (1st and 2nd Mid Terms) 1st Mid Term (covering nearly 50% syllabus) 2nd Mid Term (covering the remaining 50% syllabus) Two Mid Terms (32 Marks Each) Average of both Two Assignment (08 marks each) Average of both Final Marks- 40 marks internal 2. External University Exam: 160 marks (covering entire syllabus) 3. Total 40 + 160 =200 marks, Credits: 4

Abhinandan Jain, Asst. Professor, ECE, SKIT 10

Course Outcomes of Network Theory Students will be able to: • Apply the knowledge of basic circuital law and simplify the network using network • Apply the Fourier series techniques to solve different types of circuit problem. • Examine the transient and steady state behavior of the different circuits with the help of Laplace transform and differential method. • Evaluate different two port network and network functions

• Analyze the resonant behavior of different circuits comprising of Resister, and Abhinandan Jain, Asst. Professor, ECE, SKIT 11

TEXT BOOKS /REFERENCE BOOKS:

1. Network Analysis & Synthesis, Kuo, Wiley 2. Circuits and Network, Sudhakar & Shyam mohan, Tata McGraw-Hill 3. Sivanagaraju – Electrical circuit analysis, Cengage learning 4. Networks and Systems, Asfaq Hussain, Khanna Publishing House, Delhi 5. Networks and systems, D. Roy Choudhary, New Age International Publishers 6. Circuit Theory, A. Chakrabarti, Dhanpat rai and Co. Abhinandan Jain, Asst. Professor, ECE, SKIT 12

Classification of Network • Linear circuits: It is a circuit whose parameters remain constant with change in voltage or current. Examples are a resistance, inductance or capacitance.

• Non-linear Circuits: A linear circuit obeys ohm’s Law i.e current remains directly proportional with applied voltage, while in non linear circuit, ohm’s Law is not satisfied.

Abhinandan Jain, Asst. Professor, ECE, SKIT 13 Continued………

• Unilateral Circuit: when the direction of current is changed, the characteristics or properties of the circuit may change. Example: diode, transistors etc.

• Bilateral circuit: when with change in direction of current, the characteristics or properties of the circuit may not change, it is then called bilateral circuit. Abhinandan Jain, Asst. Professor, ECE, SKIT 14 Continued……… • Active Network: It is a network which contains one or more than one source of e.m.f. An active network consists of an active element like a battery or transistor.

• Passive Network: When a network does not contain any source of e.m.f., it is called passive network. A passive network consists of resistance, inductance or capacitance as passive element.

Abhinandan Jain, Asst. Professor, ECE, SKIT 15 Continued…………… • Lumped and Distributed Network: Physically separate network elements like R, L, or C are known as lumped elements.

• A transmission line or a cable is an example of distributed parameter network as throughout the line they are not physically separate. – If the network is fabricated with its element in lumped form, it is called lumped network and if in distributed form it is called as distributed network

Abhinandan Jain, Asst. Professor, ECE, SKIT 16 Network Terminology • Loop and mesh – A loop is any closed path of a circuit, while mesh is a loop which does not contain any other loop within it. Loop: (ABCDEFA & ABEF & BCDE) – The all the meshes are loops and a loop is not necessary a Mesh. Mesh: (ABEF and BCDE)

Abhinandan Jain, Asst. Professor, ECE, SKIT 17 Network Terminology • Node and Junction • A point where two or more branches meet is called a node, while a junction is a point at which three or more branches are joined together. • All the junction are nodes and a node is not necessarily a junction.

Abhinandan Jain, Asst. Professor, ECE, SKIT 18

Sign Assumption:

• When we are traverse from negative to positive(- to +) in supply the consider the Polarity of Voltage is Positive.

• When we are traverse from Positive to Negative (+ to -) in supply the consider the Polarity of Voltage is Negative.

• When we are traverse in the direction of current, consider negative sign of current(Due to current flow from high voltage to low voltage) • When we are traverse in the opposite direction of current the consider Positive sign of current(Due to current flow from Low voltage to High voltage)

Abhinandan Jain, Asst. Professor, ECE, SKIT 19

Kirchhoff’s Current Law(KCL) “The algebraic sum of all branch current leaving or entering a node is zero at all instant of time.”

“The algebraic sum of current entering a node must be equal to the algebraic sum of the current leaving a node.” It is based on conversation of charge.

Abhinandan Jain, Asst. Professor, ECE, SKIT 20

Ex.1: Find the currents i1,i2 and i3 in the given circuit.

Abhinandan Jain, Asst. Professor, ECE, SKIT 21

Solution:

Apply the KCL at node ‘A’: 퐸푛푡푒푟푖푛푔 푐푢푟푟푒푛푡 푎푡 푛표푑푒 퐴 = 퐿푒푎푣푖푛푔 푐푢푟푟푒푛푡 푎푡 푛표푑푒 퐴 푖1 = 1 + 6 = 7 퐴 풊ퟏ = ퟕ 푨 Apply the KCL at node ‘B’ 퐸푛푡푒푟푖푛푔 푐푢푟푟푒푛푡 푎푡 푛표푑푒 퐵 = 퐿푒푎푣푖푛푔 푐푢푟푟푒푛푡 푎푡 푛표푑푒 퐵 6 = 푖2 + 7 푖2 = 6 − 7 = −1 퐴 풊ퟐ = −ퟏ 푨

Abhinandan Jain, Asst. Professor, ECE, SKIT 22 Continued……… Apply the KCL at node ‘C’

퐸푛푡푒푟푖푛푔 푐푢푟푟푒푛푡 푎푡 푛표푑푒 퐶 = 퐿푒푎푣푖푛푔 푐푢푟푟푒푛푡 푎푡 푛표푑푒 퐶 7 = 2 + 푖3 푖3 = 7 − 2 = 5 퐴 풊ퟑ = ퟓ 푨

푆표 푡푕푎푡 푡푕푒 푐푢푟푟푒푛푡 푖1, 푖2 푎푛푑 푖3: 풊ퟏ = ퟕ푨, 풊ퟐ = −ퟏ 푨, 풊ퟑ = ퟓ푨

Abhinandan Jain, Asst. Professor, ECE, SKIT 23 Kirchhoff’s Voltage Law(KVL) • “The algebraic sum of voltage or voltage drop in any closed path of network that is traversed in a single direction is zero”

푉푠 − 퐼푅1 − 퐼푅2 = 0 푉푠 = 퐼(푅1+푅2) • It is based on conversation of energy. Abhinandan Jain, Asst. Professor, ECE, SKIT 24

Ex.2: Find the voltage Vs in the circuit if the current i=1.4 A.

Abhinandan Jain, Asst. Professor, ECE, SKIT 25 Solution:

• Apply the KVL in the loop “CDABC”: −10푖1 − 30푖1 + 10푖2 = 0 −40푖1 + 10푖2 = 0 … … … … … … . . (1) • Apply the KCL at node C: 푖 = 푖1 + 푖2 푖1 = 푖 − 푖2 … … … … … . (2) Abhinandan Jain, Asst. Professor, ECE, SKIT 26

Continued……… • Put the value of i1 from equation (2) to the equation (1) −40(푖 − 푖2) + 10푖2 = 0 50푖2 = 40푖 40 푖 = × 푖 2 50 • If i=1.4 A 40 푖 = × 1.4 = 1.12 퐴 2 50

Abhinandan Jain, Asst. Professor, ECE, SKIT 27 Continued………

Apply the KVL in the loop “CBEFC” −10푖2 − 6푖 + 푉푠 = 0 푉푠 = 10푖2 + 6푖 … … … (3)

Abhinandan Jain, Asst. Professor, ECE, SKIT 28 Continued……… Put the value of i2=1.12 A and i=1.4 A into the equation (3)

푉푠 = 10(1.12) + 6(1.4)

푉푠 = 11.2 + 8.4 = 19.6 푉

푽풔 = ퟏퟗ. ퟔ 푽

Abhinandan Jain, Asst. Professor, ECE, SKIT 29 Ex.3: Calculate the voltage V across 5KΩ in the given circuit

Abhinandan Jain, Asst. Professor, ECE, SKIT 30

Solution:

Apply the KVL in the loop “abcd” 36 − 20퐼1 = 0 퐼1=1.8 mA Apply the KCL at node “b” 퐼1 + 10퐼2 = 퐼2 Abhinandan Jain, Asst. Professor, ECE, SKIT 31 Continued………

1 1 퐼 = − 퐼 =− × 1.8 = −0.2 푚퐴 2 9 1 9 The voltage across the 5 KΩ 푉 = −5 × 10 × −0.2 = 10 푉

Abhinandan Jain, Asst. Professor, ECE, SKIT 32 Ex.4: Find the voltage across the capacitor .

Abhinandan Jain, Asst. Professor, ECE, SKIT 33 Solution: The given data: 0 푖푠 푡 = 18cos (2.5푡 + 60 ) 0 푖푠 푡 = 18∠60 R=20 Ohm, =1mH, C=0.1 F Calculate the reactance of inductance XL: ;3 푋퐿 = 푗푤퐿 = 푗 × 2.5 × 1 × 10 = 푗0.0025 Ω Calculate the reactance of capacitor XC: 1 1 푋 = = = −푗4 Ω 퐶 푗푤퐶 푗 × 2.5 × 0.1

Abhinandan Jain, Asst. Professor, ECE, SKIT 34

Continued………

Find the voltage Vx. Calculate the current I1 using current divider rule: 퐼푠 × (20 + 푗0.0025) 퐼 = 1 20 + 푗0.0025 − 푗4 18∠600 × (20 + 푗0.0025) 퐼 = 1 20 − 푗3.998 18∠600 × 20∠0.010 퐼 = = 17.647∠48.710 1 20.40∠11.300 Abhinandan Jain, Asst. Professor, ECE, SKIT 35 Continued………

So the voltage across the capacitor: 푣푥 = 퐼1 × (−푗4) 0 푣푥 = 17.647∠48.71 × (−푗4)

0 0 푣푥 = 17.647∠48.71 × 4∠ − 90 = 70.588∠ −41.290 0 푣푥 = 70.588∠ −41.29 The voltage Vx in time domain: 0 푣푥 푡 = 70.588 cos(2.5푡 − 41.29 )

Abhinandan Jain, Asst. Professor, ECE, SKIT 36 Practice Question:

Find the voltage across the R(50 ohm) in the given circuit.

Also calculate the power delivered by the source and power dissipation of all resister.

Abhinandan Jain, Asst. Professor, ECE, SKIT 37 Nodal Analysis • A point where two or more branches meet is called a node.

The Node Voltage Method breaks down circuit analysis into this sequence of steps.

Node Voltage Method Steps: • Assign a reference node (ground). • Assign node voltage names to the remaining nodes. • Solve the easy nodes first, the ones with a connected to the reference node. • Write Kirchhoff's Current Law for each node. • Solve the resulting system of equations for all node voltages.

Abhinandan Jain, Asst. Professor, ECE, SKIT 38

Ex.5: Using the Nodal method, find the current through r2 .

Abhinandan Jain, Asst. Professor, ECE, SKIT 39 Solution:

Abhinandan Jain, Asst. Professor, ECE, SKIT 40 Continued………

Abhinandan Jain, Asst. Professor, ECE, SKIT 41 Ex.6: Using nodal method, find the battery current I in the circuit.

Abhinandan Jain, Asst. Professor, ECE, SKIT 42 Solution: Let us consider V1 voltage at node ‘A’ and V2 at node ‘B’:

At node A 0 + 5 − 푉1 푉1 푉1 − 푉2 푉1 + 5 − 푉2 = + + = 0 2 + 2 1 2 4 푉1 − 5+4푉1+2푉1 − 2푉2+푉1 + 5 − 푉2=0 8푉1-3푉2=0……(1) Abhinandan Jain, Asst. Professor, ECE, SKIT 43

Continued………

At node ‘B’ 푉 ;푉 푉 :5;푉 푉 ;5;0 1 2 + 1 2= 2 2 4 4 −2푉1+3푉2=10………(2)

Abhinandan Jain, Asst. Professor, ECE, SKIT 44 Continued……… Put equation (1) into (2) 8 푉 = 푉 2 3 1 8 −2푉 +3× 푉 =10 1 3 1 6푉1 = 10 10 5 푉 = = 푉 1 6 3 40 푉 = 푉 2 9

Abhinandan Jain, Asst. Professor, ECE, SKIT 45 Continued………

The current I: 5 5 − 푉1 5 − 5 퐼 = = 3 = 퐴 4 4 6

Abhinandan Jain, Asst. Professor, ECE, SKIT 46 Practice Question

• By nodal analysis, find the current through 2 V source.

Abhinandan Jain, Asst. Professor, ECE, SKIT 47 SUPER NODE

• If any branch has only a voltage source then difficult to apply nodal analysis. So we applied the concept of super node.

• In this technique, the adjacent nodes those are connected by a voltage source are reduced to a single node and then the KCL can be applied as usual.

Abhinandan Jain, Asst. Professor, ECE, SKIT 48 Ex.7: Calculate the current in 1/6 ohm resistor for the circuit shown in figure.

Abhinandan Jain, Asst. Professor, ECE, SKIT 49 Solution:

Step 1: Apply KCL at node a: 푉2 − 푉1 푉1 4 + = + 퐼 … … … … . (1) 1/3 1/2 1 Apply KCL at node b: 푉2 − 푉1 푉2 퐼 + 9 = + 1 1/3 1/6

푉2 − 푉1 푉2 퐼 = + − 9 … … … … … … . (2) 1 1/3 1/6

Abhinandan Jain, Asst. Professor, ECE, SKIT 50

Continued………

Put the value of I1 from the equation (2) to the equation (1)

푉2 − 푉1 푉1 푉2 − 푉1 푉2 4 + = + + − 9 1/3 1/2 1/3 1/6

푉2 − 푉1 푉1 푉2 − 푉1 푉2 13 + = + + 1/3 1/2 1/3 1/6 푉1 푉2 13 = + 1/2 1/6 13 = 2푉1 + 6푉2 … … … … 3 Step 2: The voltage between node a and node b is given by 푉1 − 푉2 = 5 푉1 = 5 + 푉2 … … … (4) Abhinandan Jain, Asst. Professor, ECE, SKIT 51 Continued……… Step 3:

Put the value of V1 from the equation (4) to the equation (3) 13 = 2(5 + 푉2) + 6푉2 3 = 8푉2 푉2 = 0.375 푉 Put the value of 푉2 in equation (4) 푉1 = 5 + 0.375 = 5.375 So that the node voltage are 푽ퟏ = ퟓ. ퟑퟕퟓ 푽 풂풏풅 푽ퟐ = ퟎ. ퟑퟕퟓ 푽

Abhinandan Jain, Asst. Professor, ECE, SKIT 52

Continued……… Step 4: The current through 1/6 ohm register 푉2 퐼 = 2 1/6

0.375 퐼 = = 2.25 퐴 2 1/6 푰ퟐ = ퟐ. ퟐퟓ 푨

Abhinandan Jain, Asst. Professor, ECE, SKIT 53 Practice Question

Abhinandan Jain, Asst. Professor, ECE, SKIT 54 Mesh Analysis Method

• Mesh analysis is applicable only those network which are planner. • A circuit on a plane surface n such a way that no branch passes over or under any other branch, then the circuit is said to be planner circuit. • Mesh is a loop that does not contain any other loop within it.

Abhinandan Jain, Asst. Professor, ECE, SKIT 55

Mesh current method Steps The Mesh Current Method is based on loop currents flowing around meshes. The analysis is performed with this sequence of steps: • Identify the meshes,(the open windows of the circuit). • Assign a current variable to each mesh, using a consistent direction (clockwise or counterclockwise). • Write Kirchhoff's Voltage Law equations around each mesh. • Solve the resulting system of equations for all mesh currents. • Solve for other element currents and voltages you want using Ohm's Law.

Abhinandan Jain, Asst. Professor, ECE, SKIT 56 Ex.8: Find the current ia and ib if ic=-4A and also find Vo.

Abhinandan Jain, Asst. Professor, ECE, SKIT 57 Solution:

• Let us take a current 푖푎, 푖푐 푎푛푑 푖푏 in the mesh 1, mesh 2 and mesh 3 respectively The circuit diagram is:

Abhinandan Jain, Asst. Professor, ECE, SKIT 58 Continued………

The mesh equations are: Apply the KVL in mesh 1: 10 − 40 푖푎 − 푖푏 = 0 4푖푎 − 4푖푏 = 1 … … … … . . (1) • Apply the KVL in mesh 2: −10푖푏 − 15 푖푏 − 푖푐 − 40 푖푏 − 푖푎 = 0 −10푖푏 − 15푖푏 + 15푖푐 − 40푖푏 + 40푖푎 = 0 40푖푎 − 65푖푏 + 15푖푐 = 0 8푖푎 − 13푖푏 + 3푖푐 = 0 … … … … … . (2)

Abhinandan Jain, Asst. Professor, ECE, SKIT 59 Continued………

Apply the KVL in mesh 3: −150 − 30푖푐 − 15 푖푐 − 푖푏 = 0 15푖푏 − 45푖푐 = 150 3푖푏 − 9푖푐 = 30 … … … 3

Abhinandan Jain, Asst. Professor, ECE, SKIT 60

Continued………

• 푖푓 푖푐 = −4 퐴 푡푕푒푛 푓푖푛푑 푖푏, 푖푎 • Put the value of 푖푐 into equation (3) 3푖푏 − 9푖푐 = 30 3푖푏 − 9(−4) = 30 3푖푏 = −6 풊풃 = −ퟐ 푨 • Put the value of 푖푐 푎푛푑 푖푏 푖푛푡표 푒푞푢푎푡푖표푛 (2) 8푖푎 − 13푖푏 + 3푖푐 = 0 8푖푎 − 13(−2) + 3(−4) = 0

Abhinandan Jain, Asst. Professor, ECE, SKIT 61 Continued………

8푖푎 + 26 − 12 = 0 14 푖 = − = −1.75 퐴 푎 8 풊풂 = −ퟏ. ퟕퟓ 푨

Abhinandan Jain, Asst. Professor, ECE, SKIT 62

Ex.9: For the circuit shown in figure, determine the current I through the 10 ohm resistance by (1) Nodal analysis (2) Mesh analysis

Abhinandan Jain, Asst. Professor, ECE, SKIT 63 Solution:

Abhinandan Jain, Asst. Professor, ECE, SKIT 64

Continued………

Abhinandan Jain, Asst. Professor, ECE, SKIT 65 Continued………

Abhinandan Jain, Asst. Professor, ECE, SKIT 66 Practice Question

• Using Nodal and Mesh analysis, Find the current through the 10 V battery for the circuit shown in figure.

Abhinandan Jain, Asst. Professor, ECE, SKIT 67 SUPER MESH

• Some times, a branch in the circuit has a then it is slightly difficult to apply mesh analysis. • In this case, we may apply super mesh technique.

• In this technique, a super mesh is constituted by two adjacent loops that have a common current source.

Abhinandan Jain, Asst. Professor, ECE, SKIT 68 Ex.10: Find the mesh currents in the given circuit.

Abhinandan Jain, Asst. Professor, ECE, SKIT 69 Solution:

Applying KVL, In Super mesh (i.e. meshes 1 and 3): 3 − 1 푖1 − 푖2 − 4 푖3 − 푖2 − 1. 푖3 = 0 3 − 푖1 + 푖2 − 4푖3 + 4푖2 − 푖3 = 0 푖1 − 5푖2 + 5푖3 = 3 … … … … . . (1) In mesh 2: −2. 푖2 − 4 푖2 − 푖3 − 1 푖2 − 푖1 = 0 −2푖2 − 4푖2 + 4푖3 − 푖2 + 푖1 = 0 푖1 − 7푖2 + 4푖3 = 0 … … … . . (2) Abhinandan Jain, Asst. Professor, ECE, SKIT 70

Continued……… The current of common boundary of meshes 1 and 3 is given by: 푖1 − 푖3 = 2 푖1 = 2 + 푖3 … … … . (3) Put the equation (3) into (1) 2 + 푖3 − 5푖2 + 5푖3 = 3 5푖2 − 6푖3 = −1 … … … (4) Put the equation (3) into (2) 2 + 푖3 − 7푖2 + 4푖3 = 0 7푖2 − 5푖3 = 2 … … . . (5)

From the equation (4) and (5)

Abhinandan Jain, Asst. Professor, ECE, SKIT 71 Continued……… Multiply 5 to the equation (4) and multiply 6 to the equation (5) 푖2 = 1 푚퐴 Put the current i2 into equation (5) 7(1) − 5푖3 = 2 −5푖3 = −5 푖3 = 1푚퐴 Put the current i3 value into equation (3)

푖1 = 2 + 1 = 3 퐴

Abhinandan Jain, Asst. Professor, ECE, SKIT 72 Continued……… So that the meshes current is:

푖1 = 3 퐴, 푖2 = 1 푚퐴, 푖3 = 1푚퐴

Abhinandan Jain, Asst. Professor, ECE, SKIT 73 Practice Question

In the network configuration of fig. Find the current and voltage drops through 5 ohm resistor using mesh and Nodal analysis.

Abhinandan Jain, Asst. Professor, ECE, SKIT 74

Matrix approach of network containing voltage and current sources and reactance based circuits

Abhinandan Jain, Asst. Professor, ECE, SKIT 75

Ex.11: Find the current I1 and I2 in the given circuit using mesh technique.

Abhinandan Jain, Asst. Professor, ECE, SKIT 76 Solution:

Apply the mesh analysis in the Mesh 1 1 푉 − 퐼 1 + 푗4 − (1 + ) 퐼 − 퐼 = 0 1 1 푗4 1 2 푗 푗 푉 = 퐼 2 + 푗4 − + 퐼 (−1 + ) = 0 1 1 4 2 4

10 cos 4푡 = 퐼1 2 + 푗3.75 + 퐼2(−1 + 0.25푗) = 0 Abhinandan Jain, Asst. Professor, ECE, SKIT 77

Continued………

0 0 10 = 퐼1(4.25∠61.93 ) + 퐼2(1.03∠165.96 ) = 0 Apply the mesh analysis in the Mesh 2 푗 푉 + 퐼 1 + 푗4 + (1 − ) 퐼 − 퐼 = 0 2 2 4 2 1 푗 푗 푉 = 퐼 −2 − 푗4 + + 퐼 (1 − ) 2 2 4 1 4 0 20 cos 4푡 − 30 = 퐼1 1 − 0.25푗 + 퐼2 −2 − 3.75푗

Abhinandan Jain, Asst. Professor, ECE, SKIT 78 Continued………

0 0 0 20∠ − 30 = 퐼1(1.03∠345.96 ) + 퐼2(4.25∠241.93 ) = 0 Solve equation (1) and (2) 0 0 4.25∠61.93 1.03∠165.96 퐼1 10 0 0 = 0 1.03∠345.96 4.25∠241.93 퐼2 20∠ − 30

For I1 10 1.03∠165.960 0 0 퐼 = 20∠ − 30 4.25∠241.93 1 4.25∠61.930 1.03∠165.960 1.03∠345.960 4.25∠241.930

42.5∠241.930 − 20.6∠135.960 퐼 = 1 18.062∠303.860 − 1.0609∠511.920

Abhinandan Jain, Asst. Professor, ECE, SKIT 79 Continued………

20 + 푗37.5 − 14.81 + 푗14.32 5.19 + 푗51.82 퐼 = = 1 −10.06 + 15푗 − 0.94 + 0.5푗 −11 + 15.5 푗

52.12∠83.810 = 19.01∠125.360

0 퐼1 = 2.74∠ − 41.07

0 퐼1 = 2.74 cos (4푡 − 41.07 )

Abhinandan Jain, Asst. Professor, ECE, SKIT 80 Continued……… For I2

4.25∠61.930 10 0 0 퐼 = 1.03∠165.96 20∠ − 30 2 4.25∠61.930 1.03∠165.960 1.03∠165.960 4.25∠61.930

85∠31.930 − 10.3∠165.960 퐼 = 2 18.062∠123.860 − 1.0609∠331.920

72.14 + 푗44.96 + 9.99 − 푗2.5 82.13 + 42.46푗 퐼 = = 1 −10.06 + 15푗 − 0.94 + 0.5푗 −11 + 15.5 푗 92.46∠23.340 = 19.01∠125.360 0 퐼2 = 4.11∠ − 87.99 81 Abhinandan Jain, Asst. Professor, ECE, SKIT Continued………

We have taken I2 in opposite then 0 −퐼2= 4.11∠ − 87.99

0 퐼2 = 1∠180 × 4.11∠ − 87.99

0 퐼2 = 4.11∠92

0 퐼2 = 4.11 cos (4푡 + 92 )

Abhinandan Jain, Asst. Professor, ECE, SKIT 82

Ex.12:

Abhinandan Jain, Asst. Professor, ECE, SKIT 83 Solution:

• The current of common boundary of meshes 2 and 3 is given by: 0 퐼0 − 퐼1 = 4∠ − 90 0 0 퐼0 − 퐼1 = 4*cos −90 + 푗 sin(−90 )+ −퐼1 + 퐼0= −4푗 … … … . (1)

• In super mesh (mesh-2 and mesh-3) 12 − 8 퐼1 − 퐼2 − 1. 퐼0 − 퐼2 − −푗3 퐼0 − 퐼2 − 10퐼0 + 푗5퐼0 = 0 12 = 8 퐼1 − 퐼2 +. 퐼0 − 퐼2 + −푗3 퐼0 − 퐼2 + 10퐼0 − 푗5퐼0

Abhinandan Jain, Asst. Professor, ECE, SKIT 84 Continued……

8퐼1 − 8퐼2 + 퐼0 − 퐼2 − 푗3퐼0 + 푗3퐼2 + 10퐼0 − 푗5퐼0 = 12 8퐼1 + 퐼2 −9 + 푗3 + 퐼0 11 − 푗8 = 12 … … . (2) • Apply the KVL in mesh-1 −4퐼2 − 푗2퐼2 − −푗3 퐼2 − 퐼0 − 1. 퐼2 − 퐼0 − 8 퐼2 − 퐼1 = 0

Abhinandan Jain, Asst. Professor, ECE, SKIT 85 Continued……… 4퐼2 + 푗2퐼2 + −푗3 퐼2 − 퐼0 + 1. 퐼2 − 퐼0 + 8 퐼2 − 퐼1 = 0 4퐼2 + 푗2퐼2 − 푗3퐼2 + 푗3퐼0 + 퐼2 − 퐼0 + 8퐼2 − 8퐼1 = 0

−8퐼1 + 퐼2 13 − 푗 + 퐼0 −1 + 푗3 = 0 … … … … . . 3

Apply the Cramer rule: From equation (1), (2) and (3) −1 0 1 퐼1 −4푗 8 −9 + 푗3 11 − 푗8 퐼2 = 12 −8 13 − 푗 −1 + 푗3 퐼0 0

Abhinandan Jain, Asst. Professor, ECE, SKIT 86 Continued………

;1 0 ;4푗 8 ;9:푗3 12 ;8 13;푗 0 Find :퐼0 = ;1 0 1 8 ;9:푗3 11;푗8 ;8 13;푗 ;1:푗3

퐼0 −1 −9 + 푗3 0 − 12 13 − 푗 − 0 8 × 0 + 12 × 8 − 4푗*8 13 − 푗 + 8(−9 + 푗3)+ = −1 −9 + 푗3 −1 + 푗3 − 11 − 푗8 13 − 푗 − 0 8 × −1 + 푗3 + 11 − 푗8 × 8 + 1*8 13 − 푗 + 8(−9 + 푗3)+

퐼0 12(13 − 푗) − 4푗*104 − 8푗 − 72 + 푗24+ = −1* 9 − 푗27 − 푗3 − 9 − (143 − 11푗 − 104푗 − 8)+ + 1*104 − 8푗 − 72 + 푗24+

12(13 − 푗) − 4푗*32 + 푗16+ 퐼 = 0 −1* −푗30 − (135 − 115푗)+ + 1*32 + 푗16+

Abhinandan Jain, Asst. Professor, ECE, SKIT 87 Continued……… 156 − 12푗 − 128푗 + 64 퐼 = 0 푗30 + 135 − 115푗 + 32 + 푗16

220 − 140푗 퐼 = 0 167 − 69푗

140 2202 + (−140)2∠3600 − 푡푎푛;1 220 퐼 = 0 140 1672 + (−69)2∠3600 − 푡푎푛;1 220 260.77∠327.530 퐼 = 0 180.69∠337.550 ퟎ 푰ퟎ = ퟏ. ퟒퟒ∠ −ퟏퟎ. ퟎퟐ 푨 Abhinandan Jain, Asst. Professor, ECE, SKIT 88

Practice question:

Find the drop across 4 ohm and 10 ohm resistor using mesh current analysis.

Abhinandan Jain, Asst. Professor, ECE, SKIT 89 Source Transformation

Source transformation methods are used for circuit simplification to modify the complex circuits by transforming independent current sources into independent voltage sources and vice-versa.

Abhinandan Jain, Asst. Professor, ECE, SKIT 90

Conversion of Voltage Source to Current Source

Abhinandan Jain, Asst. Professor, ECE, SKIT 91

Conversion of Current Source to Voltage Source

Abhinandan Jain, Asst. Professor, ECE, SKIT 92

Ex.13: Find the output voltage using source transformation technique.

Abhinandan Jain, Asst. Professor, ECE, SKIT 93 Solution:

The 12 V Voltage source transfers into Current source: 12 퐼 = = 4푚퐴 3 Abhinandan Jain, Asst. Professor, ECE, SKIT 94

Continued………

The 3KΩ and 6 KΩ are parallel to each other so equivalent resistance: 3 × 6 = = 2퐾Ω 3 + 6

Abhinandan Jain, Asst. Professor, ECE, SKIT 95 Continued……… The equivalent circuit:

Abhinandan Jain, Asst. Professor, ECE, SKIT 96

Continued……… The reduced circuit diagram is:

Now 2K and 2K are in series:

Abhinandan Jain, Asst. Professor, ECE, SKIT 97 Continued……… The 8V voltage source convert into current source: 8 = = 2 퐴 4 The circuit diagram:

Abhinandan Jain, Asst. Professor, ECE, SKIT 98 Continued……… The both current source are in same direction so add the both current sources.

Abhinandan Jain, Asst. Professor, ECE, SKIT 99 Continued……… The 4mA current source transfer into a voltage source: = 4 × 4 = 16 푉

Abhinandan Jain, Asst. Professor, ECE, SKIT 100

Continued………

So calculate the output voltage across 8 ohm using voltage divider rule: 16 × 8 푉 = = 8 푉 0 8 + 8

Abhinandan Jain, Asst. Professor, ECE, SKIT 101 Practice Problem: For the circuit shown, use the source transformation and node analysis method to calculate vR4.

For the circuit, R1 = 20 Ω, R2 = 70 Ω, R3 = 35 Ω, R4 = 75 Ω, VS = 45 V, and IS = 0.2 A

Abhinandan Jain, Asst. Professor, ECE, SKIT 102 Duality

Duality: Two networks are said to be dual of each other when the mesh equations of one network are the same as the node equations of the other. Identical behavior pattern observed between voltages and currents between two independent circuits illustrate the principle of duality. Two circuits are said to be dual of each other, if the mesh equations characterize one of them has the same mathematical from as the nodal equations that characterize the other.

Abhinandan Jain, Asst. Professor, ECE, SKIT 103

The dual network are based on Kirchhoff Current Law and Kirchhoff Voltage Law

Abhinandan Jain, Asst. Professor, ECE, SKIT 104 Table of Dual Elements Element Element Voltage (V) Current (I)

Resistance (R) Conductance (G)

Inductor (L) Capacitor (C)

KVL KCL

Mesh Nodal

Open Circuit Short Circuit

Thevenin Norton

Switch in series(Getting Closed) Switch in parallel(Getting Opened)

Abhinandan Jain, Asst. Professor, ECE, SKIT 105

The following steps are involved in constructing the dual of a network: • Place a node inside each mesh of the given network. These internal nodes correspond to the independent nodes in the dual network.

• Place a node outside the given network. The external node corresponds to the datum node in the dual network.

• Connect all internal nodes in the adjacent mesh by dashed lines crossing the common branches.

Abhinandan Jain, Asst. Professor, ECE, SKIT 106 Continued………..

• Connect all internal nodes to the external node by dashed lines corresponding to all external branches..

• A clockwise current in a mesh corresponds to appositive polarity (with respect to the datum node) at the dual independent node.

• A voltage rise in the direction of a clockwise mesh current corresponds to a current flowing towards the dual independent node. Abhinandan Jain, Asst. Professor, ECE, SKIT 107

Ex.14: Draw the dual of the network shown in figure:

Abhinandan Jain, Asst. Professor, ECE, SKIT 108

Solution: For drawing the dual network, mark the node and dashed lines

Abhinandan Jain, Asst. Professor, ECE, SKIT 109 The dual network is shown in figure:

Abhinandan Jain, Asst. Professor, ECE, SKIT 110 Tutorial Problem: Problem:1

Abhinandan Jain, Asst. Professor, ECE, SKIT 111 Solution:

Abhinandan Jain, Asst. Professor, ECE, SKIT 112 Continued………..

Abhinandan Jain, Asst. Professor, ECE, SKIT 113 Continued………..

Abhinandan Jain, Asst. Professor, ECE, SKIT 114 Continued………..

Abhinandan Jain, Asst. Professor, ECE, SKIT 115 Problem:2

Using mesh analysis, find the magnitude of the current dependent voltage source and the current through the 2Ω resistor

Abhinandan Jain, Asst. Professor, ECE, SKIT 116 Solution:

From given fig., we see that

I1=-2 A

From mesh 2(dcefd), the equation is given by

-1(I3-I1)+5I-I3=0

-I3+I1+5I-I3=0

-2I3+I1+5I=0………..(1)

Abhinandan Jain, Asst. Professor, ECE, SKIT 117 Continued………..

From mesh 3(aeca), the equation is given by

-2I2-5I-(I2-I1)=0

-2I2-5I-I2+I1=0

-3I2+I1-5I=0………..(2) In the branch of cd, the current I is given by

I=I1-I3……………….(3)

Abhinandan Jain, Asst. Professor, ECE, SKIT 118 Continued……….. Put equation (3) into equation (1)

-2I3+I1+5(I1-I3)=0

6I1-7I3=0……………(4)

Put the value of I1 into equation 4

6(-2)-7I3=0 12 I =− =-1.71 A………..(5) 3 7 Put equation (3) into equation (2)

-3I2+I1-5(I1-I3)=0

-3I2-4I1+5I3=0……….(6)

Abhinandan Jain, Asst. Professor, ECE, SKIT 119 Continued………..

Put the value of I1 and I3 into equation (3)

-3I2-4(-2)+5(-1.71)=0

I2=-0.183 A------(7)

I=I1-I3=-2-(-1.71)=-0.29A----(8)

The current flow in 2Ω resistor is I2 =0.183 A, flowing anticlockwise direction. Use equation (8) find the magnitude of current dependent voltage source is=5I=5(-0.29)=-1.45 V

Abhinandan Jain, Asst. Professor, ECE, SKIT 120 Problem 3:

Find the node voltage in the network shown in figure.

Abhinandan Jain, Asst. Professor, ECE, SKIT 121

Abhinandan Jain, Asst. Professor, ECE, SKIT 122 Continued………..

Abhinandan Jain, Asst. Professor, ECE, SKIT 123 Problem:4

For the circuit shown in figure, Find the voltage across 4Ω resistor by using nodal analysis.

Abhinandan Jain, Asst. Professor, ECE, SKIT 124 Solution:

Abhinandan Jain, Asst. Professor, ECE, SKIT 125 Continued………..

Abhinandan Jain, Asst. Professor, ECE, SKIT 126 Problem :5

• In given figure, find the voltage drop across x-y terminals.

Abhinandan Jain, Asst. Professor, ECE, SKIT 127 :

Abhinandan Jain, Asst. Professor, ECE, SKIT 128 Continued………..

Abhinandan Jain, Asst. Professor, ECE, SKIT 129 Problem:6 Draw the dual of the network shown in figure:

Abhinandan Jain, Asst. Professor, ECE, SKIT 130 Solution: For drawing the dual network, mark the node and dashed lines

Abhinandan Jain, Asst. Professor, ECE, SKIT 131 The dual network is shown in figure:

Abhinandan Jain, Asst. Professor, ECE, SKIT 132

Abhinandan Jain, Asst. Professor, ECE, SKIT 133