Faculty of Electrical & Electronics Basic Electrical Engineering Superposition theorem Superposition theorem states that: In a linear circuit with several sources the voltage and current responses in any branch is the algebraic sum of the voltage and current responses due to each source acting independently with all other sources replaced by their internal impedance. OR In any linear circuit containing multiple independent sources, the current or voltage at any point in the network may be calculated as algebraic sum of the individual contributions of each source acting alone The process of using Superposition Theorem on a circuit: To solve a circuit with the help of Superposition theorem follow the following steps: First of all make sure the circuit is a linear circuit; or a circuit where Ohm’s law implies, because Superposition theorem is applicable only to linear circuits and responses. Replace all the voltage and current sources on the circuit except for one of them. While replacing a Voltage source or Current Source replace it with their internal resistance or impedance. If the Source is an Ideal source or internal impedance is not given then replace a Voltage source with a short; so as to maintain a 0 V potential difference between two terminals of the voltage source. And replace a Current source with an Open; so as to maintain a 0 Amps Current between two terminals of the current source. Determine the branch responses or voltage drop and current on every branch simply by using KCL, KVL or Ohm’s Law. Repeat step 2 and 3 for every source the circuit has. Now algebraically add the responses due to each source on a branch to find the response on the branch due to the combined effect of all the sources. 1 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering The superposition theorem is not applicable for the power, as power is directly proportional to the square of the current which is not a linear function. Steps: 1) Select any one source and short all other voltage sources and open all current sources if internal impedance is not known. If known replace them by their impedance. 2) Find out the current or voltage across the required element, due to the source under consideration. 3) Repeat the above steps for all other sources. 4) Add all the individual effects produced by individual sources to obtain the total current in or across the voltage element. Problem 1 Find Ia in circuit shown below, where only the current source is kept in the circuit. The 5V is zeroed out yielding a 0V source, or a short. The 9V is zeroed out, making it a short also. Note that the 8K resistor is shorted out (that is, 8K in parallel with 0 yields 0. 2 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering Note that the 3mA flowing up through the 2K will split left and right at the top. Part of it will flow through the 1K and part of it will flow through the 4K. Let's use the label "I4" for the current flowing right through the 4K resistor. If we combine the parallel 6K and 7K (6K||7K = 3.2K) and then add the series 4K, the total resistance on the right is 7.2K. Now we can use a current divider to find that I4 = [1K / (1K + 7.2K)] * 3mA = 0.37mA. Note that the 2K does not enter into this computation because the entire 3mA flows through it. The 3mA does not split until it gets to the junction at the top of that branch. Now that we know I4, we can then split it again through the 6K and the 7K. Ix = [7K/(6K+7K)]*I4 = 0.20 mA. Problem 2 Using the superposition theorem, determine the voltage drop and current across the resistor 3.3K as shown in figure below. Solution: 3 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering 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 4 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering Therefore voltage drop across 3.3K resistor is V1+V2 = 1.047+3.9377=4.9847 Reciprocity Theorem The reciprocity theorem states that in a linear passive bilateral network by changing the voltage source from branch 1 to branch2, the current I in the branch 2 appears in branch 1. 1. Verify the reciprocity theorem for the given network. Find the current in branch AB RT = 2+[3 // (2+2 //2)] = 3.5Ω IT =V/RT =20/3.5 = 5.71A Apply current division technique for the circuit to find current through branch AB = 5.71 x 3/6 = 2.855A 5 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering Current in branch AB = 2.855 x 2/4 = 1.43A Interchanging the source to branch AB Total resistance RT = 3.23Ω IT = 20/3.23 = 6.19A Apply current division technique for the circuit to find current through branch CD = 6.19 x 2/5.2 = 2.38A Current in branch CD = 2.38 x 3/5 = 1.43A Hence reciprocity theorem verified. 2. Verify the reciprocity theorem for the given circuit diagram. Apply current division technique for the circuit to find current through branch 3Ω = 10 x 2/5 = 4A 6 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering Voltage across 3Ω = 4 x 3 = 12V Replace the current source and find the open circuit voltage Apply current division technique and find the current through 2Ω resistor. =10 x 3/5 = 6 Voltage across 2Ω resistor = Voltage across AB = 6 x 2 = 12V Thevenin’s theorems Objectives • To understand the basic philosophy behind the Thevenin’s theorem and its application to solve dc circuits. • Explain the advantage of Thevenin’s theorem over conventional circuit reduction techniques in situations where load changes. Introduction A simple circuit is considered to illustrate the concept of equivalent circuit and it is always possible to view even a very complicated circuit in terms of much simpler equivalent source and load circuits. Subsequently the reduction of computational complexity that involves in solving the current through a branch for different values of load resistance ( RL ) is also discussed. In many applications, a network may contain a variable component or element while other elements in the circuit are kept constant. If the solution for current ( I ) or voltage (V ) or power ( P ) in any 7 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering component of network is desired, in such cases the whole circuit need to be analyzed each time with the change in component value. In order to avoid such repeated computation, it is desirable to introduce a method that will not have to be repeated for each value of variable component. Such tedious computation burden can be avoided provided the fixed part of such networks could be converted into a very simple equivalent circuit that represents either in the form Let us consider the circuit shown in figure). Our problem is to find a current through RL using different techniques; the following observations are made. Find • Mesh current method needs 3 equations to be solved • Node voltage method requires 2 equations to be solved • Superposition method requires a complete solution through load resistance ( RL ) by considering each independent source at a time and replacing other sources by their internal source resistances. Suppose, if the value of RL is changed then the three (mesh current method) or two equations (node voltage method) need to be solved again to find the new current in RL . Similarly, in case of superposition theorem each time the load resistance RL is changed, the entire circuit has to be analyzed all over again. Much of the tedious mathematical work can be avoided if the fixed part of circuit or in other words, the circuit contained inside the imaginary fence or black box with two terminals A & B , is replaced 8 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering The procedure for applying Thevenin’s theorem To find a current IL through the load resistance RL using Thevenin’s theorem, the following steps are followed: 9 Department of Electrical and Electronics Engineering Faculty of Electrical & Electronics Basic Electrical Engineering Step-1: Disconnect the load resistance ( RL ) from the circuit, as indicated in Step-2: Calculate the open-circuit voltage VTH at the load terminals ( A& B ) after disconnecting the load resistance ( RL ). In general, one can apply any of the techniques (mesh-current, node-voltage and superposition method) to compute VTh (experimentally just measure the voltage across the load terminals using a voltmeter). Step-3: Redraw the circuit with each practical source replaced by its internal resistance .note, voltage sources should be short-circuited (just remove them and replace with plain wire) and current sources should be open-circuited (just removed).