USE OF ICT IN EDUCATION FOR ONLINE AND BLENDED LEARNING-IIT BOMBAY BIRLA INSTITUTE OF TECHNOLOGY MESRA, RANCHI ASSIGNMENT( MODULE NETWORK THEOREMS) Submitted by: RC_1331_Birla Institute of Technology, Mesra Dr. Deepak Kumar(Group Leader) Dr. Vikas Kumar Gupta

Statement of superposition theorem The response (voltage or current) in any branch of a bilateral linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, where all the other independent sources are replaced by their internal impedances. Superposition theorem can be explained through a simple resistive network as shown in fig. 1.9 .

Superposition Theorem 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. Step-2: Determine the output (current or voltage) due to the single source acting alone using the techniques discussed in previous lecture. 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.

Example: Using the superposition theorem, determine the voltage drop and current across the resistor 3.3K as shown in figure below. step:1 step:2 Step 1: Remove the 8V power supply from the original circuit, such that the new circuit becomes as the following and then measure voltage across resistor. Here 3.3K and 2K are in parallel, therefore resultant resistance will be 1.245K.Using voltage divider rule voltage across 1.245K will be V1= [1.245/(1.245+4.7)]*5 = 1.047V Step 2: Remove the 5V power supply from the original circuit such that the new circuit becomes as the following and then measure voltage across resistor. Here 3.3K and 4.7K are in parallel, therefore resultant resistance will be 1.938K. Using voltage divider rule voltage across 1.938K will be V2= [1.938/(1.938+2)]*8 = 3.9377V Therefore voltage drop across 3.3K resistor is V1+V2 = 1.047+3.9377=4.9847

Limitations of superposition Theorem Superposition theorem doesn’t work for power calculation. Power calculations involve either the product of voltage and current, the square of current or the square of the voltage. They are not linear operations. This statement can be explained with a simple example .

Thevenin’s Theorem Any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent voltage source Vth in series connection with an equivalent resistance Rth. The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. The equivalent resistance Rth is the resistance that the circuit between terminals A and B would have if all ideal voltage sources in the circuit were replaced by a short circuit and all ideal current sources were replaced by an open circuit. If terminals A and B are connected to one another, the current flowing from A to B will be Vth/Rth.

Thevenin’s Theorem Fig. 1.10(a) Circuit for Thevenin’s Theorem

Thevenin’s Theorem Fig. 1.10(b) Circuit for Thevenin’s Theorem

Thevenin’s Theorem Fig. 1.10(c) Circuit for Thevenin’s Theorem

Thevenin’s Theorem Fig. 1.10(d) Circuit for Thevenin’s Theorem

Thevenin’s Theorem Fig. 1.10(e) Circuit for Thevenin’s Theorem

Thevenin’s Theorem …..(1.9) …(1.10) ….(1.11)

The procedure for applying Thevenin’s theorem To find a current through the load resistance (as shown in fig. 1.10(a)) using Thevenin’s theorem, the following steps are followed: Step-1: Disconnect the load resistance from the circuit, as indicated in fig. 1.10(b) Step-2: Calculate the open-circuit voltage (shown in fig. 1.10(b)) at the load terminals (A & B). In general, one can apply any of the techniques (mesh-current, node-voltage and superposition method) learnt in earlier lessons to compute. Step-3: Redraw the circuit ( shown in fig. 1.10 (b)) with each practical source replaced by its internal resistance as shown in fig. 1.10(c) (note, voltage sources should be short-circuited and current sources should be open-circuited .

Thevenin’s Theorem Step-4: Look backward into the resulting circuit from the load terminals (A &B) , as suggested by the eye in fig.1.10(c). Calculate the resistance that would exist between the load terminals .The resistance is described in the statement of Thevenin’s theorem. Step-5: Place in Thevenin resistance series with Thevenin voltage to form the Thevenin’s equivalent circuit as shown in fig. 1.10(d) Step-6: Reconnect the original load to the Thevenin voltage circuit as shown in fig 1.10(e) ; the load’s voltage, current and power may be calculated by a simple arithmetic operation only.

Thevenin’s Theorem One great advantage of Thevenin’s theorem is this: once the Thevenin equivalent circuit has been formed, it can be reused in calculating load current , load voltage and load power for different loads using the equations . (ii) Also one can find the choice of load resistance that results in the maximum power transfer to the load..

Norton’s Theorem Norton’s theorem states that: Any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent current source INO in parallel connection with an equivalent resistance RNO. This equivalent current INO is the current obtained at terminals A-B of the network with terminals A-B short circuited. This equivalent resistance RNO is the resistance obtained at terminals A-B of the network with all its voltage sources short circuited and all its current sources open circuited. Norton’s theorem is a dual of Thevenin’s theorem. To find a current through the load resistance using Norton’s theorem, the following steps are followed:

Norton’s Theorem Fig. 1.11(a) Circuit for Norton’s Theorem

Norton’s Theorem Step-1: Short the output terminal after disconnecting the load resistance from the terminals A&B and then calculate the short circuit current . Step-2: Redraw the circuit with each practical sources replaced by its internal resistance ,voltage sources should be short-circuited and current sources should be open- circuited . Step-3: Calculate the resistance that would exist between the load terminals A&B Looking backward into the resulting circuit from the load terminals (A&B).

Norton’s Theorem Fig. 1.11(b) Circuit for Norton’s Theorem

Norton’s Theorem Fig. 1.11(c ) Circuit for Norton’s Theorem

Norton’s Theorem Step-4: form the Norton’s equivalent circuit (replacing the imaginary fencing portion or fixed part of the circuit with an equivalent practical current source) . Step-5: After Reconnecting the original load; the load’s voltage, current and power may be calculated by a simple arithmetic operation only.

Norton’s Theorem Fig. 1.11 (d) Circuit for Norton’s Theorem

Norton’s Theorem Norton’s theorem has also a similar advantage like the thevenin’s theorem over the normal circuit reduction technique. (ii) With help of either Norton’s theorem or Thevenin’s theorem one can find the choice of load resistance that results in the maximum power transfer to the load. (iii) Norton’s current source may be replaced by an equivalent Thevenin’s voltage source . In other words, a source transformation converts a Thevenin equivalent circuit into a Norton equivalent circuit or vice-versa.

Maximum Power Transfer Theorem The maximum power transfer theorem states that, to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals. The theorem states how to choose (so as to maximize power transfer) the load resistance, once the source resistance is given.

Maximum Power Transfer Theorem Let us consider an electric network as shown in fig.1.12(a), the problem is to find the choice of the resistance so that the network delivers maximum power to the load . This problem can be solved using nodal or mesh current analysis to obtain an expression for the power absorbed by Load resistance The derivative of this expression with respect to load resistance will establish the condition under what circumstances the  maximum power transfer occurs.

Maximum Power Transfer Theorem Fig. 1.12(a) Circuit for Maximum Power Transfer Theorem

Circuit for Maximum Power Transfer Fig. 1.12(b) Circuit for Maximum Power Transfer Theorem

Maximum Power Transfer Theorem ….(1.12) ….(1.13) ..(1.14)

….(1.15) ….(1.16) ….(1.17)

Thank You

EE 1x01 BASIC ELECTRICAL ENGINEERING L-T-P : 3-0-3 Credit : 5 1. Introduction : Electrical Elements and their Classification, KCL, KVL equation and node voltage method, D.C circuits- steady state analysis with independent and dependent sources, Series and parallel circuits, Star- delta conversion, Superposition Theorem, Thevenin’s Theorem, Norton’s theorem, Maximum Power Transfer Theorem. Lecture : 12 2 A.C Circuits : Common signals and their waveform, R.M.S and Average value, form factor and Peak factor of sinusoidal wave. Impedance of series and parallel circuits, Phasor diagram, Power, Power factor, Power triangle, Coupled circuits. Resonance ; series, parallel, Q-factor, Superposition, Thevenin’s and Norton’s, Maximum power transfer theorem for A.C circuits Lecture : 12 3 Three phase A.C Circuits : Star delta connection, line and phase relation, Power relation, Analysis of balanced and unbalanced 3-phase circuits. Lecture : 8 4 Magnetic circuits : Introduction; Flux, MMF, flux density, reluctance, Series & Parallel magnetic circuits. Analysis of Linear and non linear magnetic circuits. Energy storage, A.C. excitation. Eddy current and hysteresis losses. Lecture : 5 5 Basic indicating instruments for measurements Current Voltage, Power, Energy Insulation resistance. Lecture : 5 Text Book: 1.D.P. Kothari, I.J.Nagrath, Basic Electrical Engineering , 2nd Edition, TMH , New Delhi ,, 2010. Reference Books:Tata Fitzerald 1. Fitzerald, et al ,Basic Electrical Engineering by, Tata McGraw Hill, 2005 2. R. Prasad, Fundamentals of Electrical Engg., PHI Publication 3. Leonard S. Bobrow, Fundamental of Electrical Engg., Oxford