Module 2 DC Circuit Version 2 EE IIT, Kharagpur Lesson 8 Thevenin’s and Norton’s theorems in the context of dc voltage and current sources acting in a resistive network Version 2 EE IIT, Kharagpur 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. • Maximum power transfer theorem and power transfer efficiency. • Use Norton’s theorem for analysis of dc circuits and study the advantage of this theorem over conventional circuit reduction techniques in situations where load changes. L.8.1 Introduction A simple circuit as shown in fig.8.1 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 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 of practical voltage source known as Thevenin’s voltage source ( , V magnitude of voltage source Th = int R ern Th = al ) or in the form of practical current source known as Norton’s current source ( , resis ce tan of the source NI magnitude of current source = int R erna N = l resis ce tan ). In true sense, this conversion will considerably simplify the analysis while the load resistance changes. Although the conversion technique accomplishes the same goal, it has certain advantages over the techniques that we have learnt in earlier lessons. of current source Let us consider the circuit shown in fig. 8.1(a). Our problem is to find a current through RL using different techniques; the following observations are made. Version 2 EE IIT, Kharagpur 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 (fig. 8.1(a)) or in other words, the circuit contained inside the imaginary fence or black box with two terminals A & B , is replaced by the simple equivalent voltage source (as shown in fig. 8.1(b)) or current source (as shown in fig.8.1(c)). Version 2 EE IIT, Kharagpur Thevenin’s Theorem: Thevenin’s theorem states that any two output terminals ( A & B , shown in fig. 8.2.(a)) of an active linear network containing independent sources (it includes voltage and current sources) can be replaced by a simple voltage source of magnitude in series with a single resistor (see fig. 8.2(d)) where is the equivalent resistance of the network when looking from the output terminals VTh RTh RTh A & B with all sources (voltage and current) removed and replaced by their internal resistances (see fig. 8.2(c)) and the magnitude of is equal to the open circuit voltage across the VTh A & B terminals. (The proof of the theorem will be given in section- L8. 5). Version 2 EE IIT, Kharagpur L.8.2 The procedure for applying Thevenin’s theorem To find a current L I through the load resistance RL (as shown in fig. 8.2(a)) using Thevenin’s theorem, the following steps are followed: Step-1: Disconnect the load resistance ( RL ) from the circuit, as indicated in fig. 8.2(b). Step-2: Calculate the open-circuit voltage (shown in fig.8.2(b)) at the load terminals ( VTH A& B ) after disconnecting the load resistance ( RL ). In general, one can apply any of the techniques (mesh-current, node-voltage and superposition method) learnt in earlier lessons to compute (experimentally just measure the voltage across the load terminals using a voltmeter). VTh Step-3: Redraw the circuit (fig. 8.2(b)) with each practical source replaced by its internal resistance as shown in fig.8.2(c). (note, voltage sources should be short-circuited (just remove them and replace with plain wire) and current sources should be open-circuited (just removed). Version 2 EE IIT, Kharagpur Step-4: Look backward into the resulting circuit from the load terminals ( A& B ) , as suggested by the eye in fig.L.8.2(c). Calculate the resistance that would exist between the load terminals ( or equivalently one can think as if a voltage source is applied across the load terminals and then trace the current distribution through the circuit (fig.8.2 (c)) in order to calculate the resistance across the load terminals.) The resistance is described in the statement of Thevenin’s theorem. Once again, calculating this resistance may be a difficult task but one can try to use the standard circuit reduction technique or transformation techniques. RTh Y or −Δ Δ−Y Step-5: Place in series with to form the Thevenin’s equivalent circuit (replacing the imaginary fencing portion or fixed part of the circuit with an equivalent practical voltage source) as shown in fig. 8.2(d). RTh VTh Step-6: Reconnect the original load to the Thevenin voltage circuit as shown in fig.8.2(e); the load’s voltage, current and power may be calculated by a simple arithmetic operation only. Version 2 EE IIT, Kharagpur Load current Th L Th L V I R R = + (8.1) Voltage across the load Th L L Th L V V RI R R = ×= + L × RL (8.2) Power absorbed by the load PL = 2 L L I × R (8.3) Remarks: (i) One great advantage of Thevenin’s theorem over the normal circuit reduction technique or any other technique is this: once the Thevenin equivalent circuit has been formed, it can be reused in calculating load current ( L I ), load voltage (VL ) and load power ( PL ) for different loads using the equations (8.1)-(8.3). (ii) Fortunately, with help of this theorem 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. L.8.3 Application of Thevenin’s theorem Example: L.8.1 For the circuit shown in fig.8.3(a), find the current through resistor ( branch) using Thevenin’s theorem & hence calculate the voltage across the current source ( ). R R L = =Ω 2 1 a b I − Vcg Version 2 EE IIT, Kharagpur Solution: Step-1: Disconnect the load resistance RL and redraw the circuit as shown in fig.8.3(b). Step-2: Apply any method (say node-voltage method) to calculate . VTh At node C: 2 0 1 2 (3 ) (0 ) 2 6 3 6 c c c I I V V V vo ++= − − + + ⇒= lt Now, the currents I1 2 & I can easily be computed using the following expressions. Version 2 EE IIT, Kharagpur 1 3 6 1 3 3 V V a c I A − − = = =− (note, current 1I is flowing from ‘c’ to ‘a’) 2 0 6 1 6 6 Vc I A − − = = =− (note, current 2 I is flowing from ‘c’ to ‘g’) Step-3: Redraw the circuit (fig.8.3(b) indicating the direction of currents in different branches. One can find the Thevenin’s voltage using KVL around the closed path ‘gabg’ (see fig.8.3.(c). VTh V V V volt Th ag bg = − =− = 321 Step-4: Replace all sources by their internal resistances. In this problem, voltage source has an internal resistance zero (0) (ideal voltage source) and it is short-circuited with a wire. On the other hand, the current source has an infinite internal resistance (ideal current source) and it is open-circuited (just remove the current source). Thevenin’s resistance of the fixed part of the circuit can be computed by looking at the load terminals ‘a’- ‘b’ (see fig.8.3(d)). RTh Version 2 EE IIT, Kharagpur R RR R Th = + =+ = Ω ( ) () 13 4 & & 3 4 2 1.555 Step-5: Place in series with to form the Thevenin’s equivalent circuit (a simple practical voltage source). Reconnect the original load resistance to the Thevenin’s equivalent circuit (note the polarity of ‘a’ and ‘b’ is to be considered carefully) as shown in fig.8.3(e). RTh VTh R R L = =Ω 2 1 1 0.39 ( ) 1.555 1 Th L Th L V I A a to b R R == = + + Step-6: The circuit shown in fig.8.3 (a) is redrawn to indicate different branch currents. Referring to fig.8.3 (f), one can calculate the voltage and voltage across the current source ( ) using the following equations. Vbg Vcg Version 2 EE IIT, Kharagpur 3 1 0.39 2.61 . 2.61 1.305 ; 1.305 0.39 0.915 2 4 0.915 2 1.305 6.27 . bg ag ab bg cb cg VVV volt I A I A V vol = − = −× = = = = −= =× +× = t Example-L.8.2 For the circuit shown in fig.8.4 (a), find the current L I through 6 Ω resistor using Thevenin’s theorem. Solution: Step-1: Disconnect 6 from the terminals ‘a’ and ‘b’ and the corresponding circuit diagram is shown in fig.L.8.4 (b). Consider point ‘g’ as ground potential and other voltages are measured with respect to this point. Ω Version 2 EE IIT, Kharagpur Step-2: Apply any suitable method to find the Thevenin’s voltage ( ) (or potential between the terminals ‘a’ and ‘b’). KVL is applied around the closed path ‘gcag’ to compute Thevenin’s voltage. VTh 42 8 4 30 0 1 , 30 4 34 ; 2 3 6 ag bg II IA Now V volt V volt − − − = ⇒= = += =×= . 34 6 28 t V V V V vol Th ab ag bg = = − = −= ( note ‘a’ is higher potential than ‘b’) Step-3: Thevenin’s resistance can be found by replacing all sources by their internal resistances ( all voltage sources are short-circuited and current sources are just removed or open circuited) as shown in fig.8.4 (c). RTh ( ) 8 4 14 8 4 2 2 4.666 12 3 RTh × = += += = & Ω Step-4: Thevenin’s equivalent circuit as shown in fig.8.4 (d) is now equivalently represents the original circuit (fig.L.8.4(a). Version 2 EE IIT, Kharagpur 28 2.625 4.666 6 Th L Th L V I A R R == = + + Example-L.8.3 The box shown in fig.8.5 (a) consists of independent dc sources and resistances. Measurements are taken by connecting an ammeter in series with the resistor R and the results are shown in table. Table R I 10Ω 2 A 20Ω 1.5 A ? 0.6 A Solution: The circuit shown in fig.8.5(a) can be replaced by an equivalent Thevenin’s voltage source as shown in fig.8.5(b). The current flowing through the resistor is expressed as R = + Th Th V I R R (8.4) Version 2 EE IIT, Kharagpur The following two equations are written from measurements recorded in table. 2 2 10 Th Th Th Th V V R R =⇒ − = + 20 (8.5) 1.5 1.5 30 20 Th Th Th Th V V R R =⇒ − = + (8.6) Solving equations (8.5) and (8.6) we get, 60 ; 20 V volt R Th = = Th Ω The choice of R that yields current flowing the resistor is 0.6 A can be obtained using the equation (8.4). 60 0.6 80 . 20 Th Th V I R RR R = = = ⇒= + + Ω L.8.4
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