# Appendix A Thevenin`s Theorem - Department of Physics | Oregon

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Appendix A Thevenin`s Theorem - Department of Physics | Oregon

Appendix A Theve nin's Theorem ,,- - -) !, , I - ----- - -- " ' , 13 :, ~_. _____ . ' Any network can be described by loop currents passing from volt age sources to the terminals AB, hence through a load resistor RL. An example is shown in figure A .I. In order to establish Thevenin's theorem, it is necessary to show that V AB = VEQ - REQ lAB, where VEQ and REQ do not depend on the load resistance RL· Around each current loop, there is a linear equation: = VI VA.B = V2 - VAB RllIl - RI2h R21Il - ... - Rlnln R2212 - . . , - R 2n l n Fig. A I. Loop currents through termInals AS. Because the voltage drops appearing on the right-hand side in the form ~j Ij stop short of the terminals AB , the load resistance RL does not appear explicitly anywhere in the equations. They can be solved for currents (by simple elimination and substitution), with the results = Y1I VI + Y12V2 + ... + Yin Vn 12 = Y21 VI + Y22 V 2 + .. . + Y2n Vn - It /1 VAB l /2 VAB where again none of the coefficients depend on RL (since ~j did not). Then hB = Li Ii can be written Bl VI + B2 V2 + ... + Bn Vn - CVAB . This is of the required form if C = l/ REQ and VEQ = (BI VI + B2 V2 + ... + En Vn)/C. In Chapter 6, it is shown that there is a linear relation between V and I for capacitors and inductors when the notation of com plex numbers is used. The linear relation between V and I is the foundation of Thevenin's theorem. Hence the proof carries over to include capacitors and inductors in AC circuits. Fourier's theorem expresses any waveform in terms of AC components, and Thevenin 's theorem is therefore generally valid for any network where the re lation between V and I is linear. 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T"-·K i I R,- i/ (~/ + f: If) A ~ ~CN TVlc.,,\r fY,(J "';. r-J .Q,~ \(j~ vcJ(·j v', \tc.. _, we..,. WQ j '. ~J R/ R' t R I; !he~io~~ I JeD • - - \ ~1 . -v -e..v-': --) "V ~ ~ "\J Q.\J- ~ A \tJ ~\' ® \ Lvru:;I lJ-eJ\ 7 ~GC(\.;u,Sv I-Jc;L-' ;J \ '-I0'~' h v., S1-" v ~ 1-..Y- C G c-~ ~~, -,1 tv ~ ~~ GJ.~s \ 'i 1). V_ \3 "'~ I> ( 0\.. V\). ~ "[ s s \"J,.. e.e\ C'IA (.L. v..-( (.j. ~I ~ t~ \>~ 1/ I q\\ h -\-l.-v s to ~R.A" ~1" w} Fl"x' "t -\ \...u '11 f v..~' ""t -\-"'-., \( , \0, \j' of\ ~ La...",:;· Anot her worked example Figure 1.19(a) shows a rather complicated circuit. As an illustra tlOn of all the methods which have been developed in this chapter we shall fi~d all the currents and the voltages VA and VB at node~ by three dIfferent methods. fig1 is fhe I kO O.6mA 15 I kO I kO B Rs R}R4 R)R Z R} + P4 p) + R2 cb t 2kO =3kO I} =1 kO + VB '2v O.SmA O.7S mA 2kO 12kO 6V IOV (a) 12kO (b) 14 =0 . I SmA 1}= -04SmA he ~r- Fi g. i . 19 Worked examp I e is , t - 10V (c) 12 =+0.2Sm A I) =- O.2SmA 14 Voltage, Current and Resistance Using superposition, the currents may be obtained from those due to the individual batteries. In (b), the 10 V battery is shorted out and R3 and R4 appear in parallel. The top three resistors of (b) provide 2 H2 in parallel with 4 kO between A and earth, i.e. ~ H2, so the contribution to II = 1.8 rnA. It is easy to see how this divides at A; then h splits between 0.15 mA through R4 and a contribution of -0.45 mA to h In (c) , the 6 V battery is shorted, so Rl and R2 appear in parallel. The arithmetic of the resulting currents is shown in the figure. The signs of the contributions to II and 12 are easy to follow from (a) and the sense in which the 10 V battery drives currents. Adding currents from (b) and (c), II = 1.55 mA, 12 = 1.45 mA, h = -0.7 mA , 14 = -0.6 mA and Is = 0.1 rnA. From these currents, it is simple to find VA = 6 - 3.1 = 2.9 V and VB = 10 + I4R4 = 2.8 V . Suppose instead the problem is to be solved using mesh cur rents . The ones to choose would be 11 , 14 and Is, as shown in (a). Applying Kirchhoff's current law, h = 12 + Is and 14 = h + Is · The values given in the previous paragraph satisfy these relations. Then applying Kirchhoff's voltage law to each loop in turn: 6 =2h +2(11 - h) = 4h - 21s 10 = - 1214 + 4(15 - 14) = 415 - 1614 o = lIs + 4(15 - 14) + 2(15 - h) = 715 - 414 - 211 with currents in mAo Solving these three simultaneous equations is tedious. It is however straightforward to substitute the values derived above and demonstrate that the equations are correctly satisfied . Using superposition is really a graphical way of eliminat ing variables from the simultaneous equations. The third alternative is to use node voltages V<\ and VB. Then current conservation at these nodes gives L Ii b c t, e. n u v' it ir a: Cl OJ v, E fc ti T (f. W. The solution of these two simultaneous equations is easy; a check is that the equations are satisfied by the values of VA and VB obtained above. ar T) 1.1 0· Non-linear Elements in a C ircuit Superposition is a valuable shortcut , but (as demonstrated below) it only works exactly for circuits containing linear components like resistors, where Vex: I. The next chapter develops other powerful shortcuts which again depend on linearity. However, many elec tronic devices such as diodes and transistors do not obey Ohm's 101 re ca sh Norton 's Tbeorem h 11 t e :- II f 27 (ii) Wi th the batteries replaced by short circuits, REq is the resist ance across AB , namely 5 kD in parallel with 10 kD. So REQ = 10/3 kD. (iii) As a check, consider the situation with AB open cir cuit. In this case, t here is a net voltage of 10 V in ( a), driving current I in t he direction of the arrow through the 5 and 10 kD resistors; I = 10/15 rnA and VAB 10 + 51 10 + 10/3 = 40/3 V. This agrees wit h hqREq from (i) and (ii). A circuit can often be simplified quickly and neatly by swopping backwards and forwards between Thevenin and Norton equivalent forms. T his is illustrated in figures 2.1l(b)-(d). It is a trick worth pr ac tising, since it often saves a great deal of algebra. The batteries and resistors of (a) are replaced by equivalent Norton circuits in (b); t hese are combined in parallel in (c) and then (d) converts back to the Thevenin equivalent form . An important warning is that you must not include the load resistor between terminals A and B in these manipulations : Thevenin's and Norton's theorems app ly to the circuits feeding terminals AB. = = Fig.2. 12. A worked exa mp l e Rs (al Another worked example F igure 2. 12 reproduces a fairly complicated example from Chap ter 1, fi gure 1. 19( a) . If all currents and voltages in the circuit are required , it is best to use one of the methods from Chapter 1. Sup pose , however, only current 12 is to be found. It can be obtained straightforwardly by application of Thevenin's and Norton's theo rems . The steps are shown in figure 2.1 3. In (b), V2 and Rt are replaced by their Norton equivalent. Then R3 and Rt are combined in par allel and (c) returns to the Thevenin equivalent form . W ith AB open circuit , R] R, R4 Vj V2 III Rs A B R] (bl RI W ith the batteries shorted out, R Eq is given by the parallel com bin ation of RI wi th R5 + R3R4/(R3 + R 4) (e ) i.e . 2 kn in parallel with 4 kn, so REq = t4/3) kD. As a check, the current through AB when shorted is 1Eq : I _ VI Eq - Rl V2 R 3 + R3 + ~ (R 5 R3~)-1 + R3 + Rt = 3 + 2.5/4 = 29/8 mA o Fig. 2 . 13. Worked exa mple 28 Tbevenin and Norton re This agrees with VEQ/REQ as it should . T he arithmetic and alge bra are sufficiently tortuous that this is a valuable crosscheck. Finally, the current h of figure 2.12 is V( di W. th go in agreement with the value obtained in t he previous chapter. Further examples are given in the exercises at the end of the chapter. If you can do question 6, you have m astered the vital points of Chapters 1 and 2 up to here . 2. St R: to 2.5 General Remarks on Thevenin's and Norton's Theore ms Fi g 2. 14 For const ant Vou t , R EO << R L' Fi g. 2 . 15. For constant Rr:o » R L lout, B (a) (b) Fig.. 2. 16. Wrong ch oi ces f or equiv alent circuit s. , (1) Suppose a constant voltage is required across a load R L , with as little variation as possible when RL is changed. From figure 2.14, REQ needs to be small compared wit h RL, so that mOlt of VEQ appears acrOls RL. Thus a constant voltage source should h ave a low output resistance or output impedance, as it is often called. (2) Conversely, suppose a constant output current is required, independent of load ; this is the case , for example, in supplying a m agnet or a motor . From figure 2.15, this demands REQ ~ RL, or high output resist ance. (3) When a circuit is meas ured with an Olcilloscope or voltmeter , it is desirable to disturb the circuit as little as possible , i.e. draw very little current. This requires the detector to have a high in put resistance or input impedance. Oscilloscopes and multimeters typically have input resistances of 10 6 - 10 7 n. On the other hand , if an ammeter is inserted into a circuit in order to measure current , we want to disturb the current as little as possible. Therefore an ammeter should have a low resistance. (4) Although T hevenin's and Norton's circuits are equivalent to any network in the sense of giving t he same output volt age and cur rent, they are not equivalent as regards power consumption within the equivalent circuit . You may easily verify that the power dissi pated in the Norton equivalent circuit of figure 2.10(b) is different from that dissipated in the Thevenin equivalent circuit (a) . T his is because power is non-linear in V or J. (5) Common student howlers are to draw equivalent circuits in the forms shown in figure 2.16 . It is wort h a. moment's thought to see as to why these must be wrong. In the former case, VAB = VEQ independent of load, which gives an absur d result if the terminals are shorted. In the second circuit, J AB = IEQ independent of load, and this is absurd if the terminals are open. (6) If you encounter a circuit like tha t in figure 2.1 6(a) where a resistor is applied directly across a bat tery, you can ignore the R: an Tl ml of no so sa pc ca th 2. In h m; to 1 -~t:.-===-=-_ --- w~~QJ dCJ'<\~ ~, \r" ~ Q l,JAt c ' e.,,\ G 0... "-' I 1::>v-~J~G - (\-"'- 01~\(V\J~ @ c- + . '" e- e,d IA.-C ..., ~VIOl" ~,~ ""I V\.e, ~ ~: c.S '~ -rv 1 t +~ ~"J,\/U \~ _~cJct-lAcel, +~ ;s IA.L clL"d f/o""y ~Lvoc {k Vv.<Ji:w ; leD.,.-r K -..5 k k, V l-t~f j v ~~ VA C = J \) ~ I 0- d 0 \ 1 c - K2.1o O.i~ Ie c f<~l~ 5 (,I v ; ~ /+ £ 1 1 D 'R 2. &. 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