Network Theory Comprehensive Notes, Lecture notes of Network Theory

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Senkeq and povallel. Com vinchiow "F eestor =r @ Hene , Cy Cask Lig O12 Cwnecdtod in Serie”. bal * Oe Q ta Guneted ; peraNet . 26 Circuits and Networks Hu Fig. 1.36 Solution From the circuit shown in Fig. 1.36, the current passing through the 8 ( resistor is 20V I, =——=2.5A * 80 The current passing through the 15 resistor is same as the ammeter reading. the current passing through the 15 2 resistor is _ 2.5x11 "11428 Reading of the ammeter = 0.71 A The voltage across the 28 © (13 ( in series with 15 () resistor is Vyg = 0.71 X 28 = 19.88 volts The voltage across the series arm (8 ( in series with 11 () resistor is Vig = 2.5 X 19 = 47.5 volts The emf of the battery = 19.88 + 47.5 = 67.38 volts =0.71A PROBLEM 1.3 An electric circuit has three terminals A,B,C. Between A and B is connected a 2) resistor, between B and C are connected a 7 resistor and a 5 Q, resistor in parallel, and between A and C is connected a 1 Q. resistor. A battery of 10 V is then connected between terminals A and C calculate (a) total current drawn from the battery, (b) voltage across the 2 Q, resistor, and (c) current passing through the 5 Q. resistor. Solution The circuit can be drawn as shown in Fig. 1.37 below. The current passing through the 1 ( resistor is 10 Tg =—=10A “A a) te | | The current passing through the series parallel branch between terminals A and C is [aa 10 the Lo= _ 20 24+)5) = Total current drawn from the battery is J; = 10 + 2 = 12A Fi ui = Voltage across the 2 2 resistor is V3 = 2 X 2 = 4 volts Be The current passing though the 5 () resistor is 2x7 Isq =) =1.17A 50 547 Circuit Elements and Kirchhoff’s Laws PROBLEM 1.4 27 Using Ohm’s law and Kirchhoff’s laws on the circuit given in Fig. 1.38, find V,,, V, and the power provided by the dependent source. Solution From the circuit shown in Fig. 1.38, applying Kirchhoff’s current law, we have 0 = bo + ho The current passing through the 4 Q branch is i; = 2+6=8A The voltage across the dependent source is Vij, =(4ig) = 4x8 =32V The voltage across the 2 2 resistor is V3g = 6 X 2 = 12V the voltage across each branch is V =V,,, -V, =32-12=20V The voltage across the 4 Q, branch is Vi=4Xi,=4X8 =32V According to Kirchhoff’s voltage law, V-V,+V5 =0 20—32+Vs =0 Vs =12V Similarly, Kirchhoff’s voltage law is applied to the 3 © branch. V-30+V;-V,,=0 20 - 30+ 6-V,,=0 From the above equation, the voltage Yn = ~4V The power provided by the dependent source is Pai = Vai, X igi, Pyi, = 4Xiy X6 = 4x8x6 = 192 watts PROBLEM 1.5. Find the power absorbed by each element in the circuit shown in Fig. 1.39. Fig. 1.39 30 Circuits and Networks Assume loop currents J, and J, as shown in Fig. 1.45. 4-2 =06A 10 1-2 =086A 14 V,, = Voltage drop across the 4 ( resistor = 0.6 X 4 = 2.4V Vz = Voltage drop across the 4 2 resistor = 0.86 X 4 = 3.44 V The voltage between points A and B is the sum of p voltages as shown in Fig. 1.46. Vip = -2.4 + 12 + 3.44 Fig. 1.46 = 13.04V day PROBLEM 1.9 Determine the current delivered by the source in the circuit shown in Fig. 1.47. Fig. 1.47 Solution The circuit can be modified as shown in Fig. 1.48, where Rj, is the series combination of R, and R3. r HN Ba xf ‘ vite s Fig. 1.48 Ry = Ri + Rs = 402 R;, is the series combination of R, and Rs. Ry =R, + Rs =30 Further simplification of the circuit leads to Fig. 1.49 where R,, is the parallel combination of Rj9 and Rg. Ry = (Rio || Ro) = 4 || 4) = 20 Circuit Elements and Kirchhoff’s Laws 31 Fig. 1.49 Fig. 1.50 Similarly, R,, is the parallel combination of R,, and Rg. Ry = Ry || Re) = G | 2) = 1.2.0 In Fig. 1.49 as shown, Rj, and R), are in series, which is in parallel with R, forming R,,. This is shown in Fig. 1.50. Rig = [Riz + Ris) || Ry] = [2 + 1.2)||2] = 1.230 Further, the resistances R,4 and Rg are in series, which is in parallel with R, and gives the total resistance Rr= [Rig + Ro)||Ri = [(1 + 1.23)||(2)] = 1.050 The current delivered by the source = 30/1.05 = 28.57A PROBLEM 1.10 Determine the current in the 10 Q. resistance and find V, in the circuit shown in Fig. 1.51. Fig. 1.51 Solution The current in the 10 ( resistance Jo = total current X (R7)/(Rp + Ryo) where R; is the total parallel resistance. Ty) =4x 7 =1.65A 17 Similarly, the current in the resistance R, is I, =4x19_=235A 10+7 or 4—1.65 =2.35A The same current flows through the 2 resistance. voltage across the 2 2 resistance, V, = I; X 2 =2.35X2=47V 32 Circuits and Networks PROBLEM 1.11 Determine the value of the resistance R and current in each branch when the total current taken by the circuit shown in Fig. 1.52 is 6 A. Solution The current in the branch ADB Ale fav —/H Typ = 50/25 + 5) = 1.66A Fig. 1.52 The current in the branch ACB, Iq , pg =50/(10 + R). According to Kirchhoff’s current law, Tr= Tyo + Tove 6A =166At]igup Tio 4p = 6 - 1.66 = 4.34 50 = 434 10+R 10+ R= ~e-=11.52 4.34 R=1520, PROBLEM 1.12 Find the power delivered by the source in the circuit shown in Fig. 1.53. Solution Between points C(£) and D, resistances R; and R, are in parallel, which gives Ry = (R3 || Ry) = 2.50 Circuit Elements and Kirchhoff’s Laws 35 When V, = 50 V, the current /, in the resistance R, becomes zero. h= 50—30 R where J, becomes the total current. _100-V, 100-50 — 5 5 — 20 20_ “1, 10 I, =10A R 20 PROBLEM 1.15 Determine the output voltage V,,,, in the circuit shown in Fig. 1.58. IAL 20 AA Rg it bs an Fig. 1.58 = As | jaa A Fig. 1.59 In Fig. 1.59, R, and R, are in parallel, R, and R, are in parallel. The complete circuit is a single-node pair circuit. Assuming voltage V, at the node A and applying Kirchhoff’s current law in the circuit, we have Vv, V; 10A—-4_-5A—_4_=0 4.43 2.67 1 1 =5A 4 astro 36 Circuits and Networks V {0.225 +.0.375]= 5 Vi=—-=8.33V 5 0.6 PROBLEM 1.16 Determine the voltage V4p in the circuit shown in Fig. 1.60. Rj 5 A | 2k Rs 4 Be | a0 eres 2 Ry 302 R | ALR Fig. 1.60 Solution The circuit in Fig. 1.60 can be redrawn as shown in Fig. 1.61 (a). 4 4 5k 4 R ! a oF 28 7 Riis, pu a Aph A tiy 1G ga [fy Rea ou Fig. 1.61 (a) At the node 3, the series combination of R; and Rg are in parallel with Re, which gives Ry = [(R; + Rg) ll RJ =30. At the node 2, the series combination of R; and R, are in parallel with R,, which gives R,y = [(R3 + Ry) IR] = 3 Q. It is further reduced and is shown in Fig. 1.61 (b). Simplifying further, we draw it as shown in Fig. 1.61 (c). . 1 Total current delivered by the source = T 100 ~ 13 //8) Current in the 8 Q resistor is 7, =20.2x 8 =12.5A 13+8 Circuit Elements and Kirchhoff’s Laws 37 Ayr Fig. 1.61 (b) Fig. 1.61 (c) Current in the 13 Q resistor is /,, =20.2x 8 =7.69 A 13+8 So Is = 12.5 A, and yy = 7.69 Current in the 4 (resistance, J, = 3.845 A Current in the 3 2 resistance, J, = 6.25 A Vag = Va-Ve where V,=1; X30 = 6.25 X 3 = 18.75 V Vz =1,X 40 = 3.845 X 4 = 15.38 V -. Vag = 18.75 — 15.38 = 3.37 V PROBLEM 1.17 Determine the value of R in the circuit shown in Fig. 1.62, when the current is zero in the branch CD. Fig. 1.62 Solution The current in the branch CD is zero, if the potential difference across CD is zero. That means, voltage at the point C = voltage at the point D. Since no current is flowing, the branch CD is open-circuited. So the same voltage is applied across ACB and ADB. 10 Vig =V 4x 10 A 15 R Va=V4x RAT OLR Vio = Vp and Vx =v, x R 15 20+R R=400 40 Circuits and Networks Solution The circuit shown in Fig. 1.67 is a parallel circuit and consists of a single node A. By assuming voltage V at the node A, we can find the current in each element. According to Kirchhoff’s current law, iz — 12 — 2i, — i, = 0 By using Ohm’s law, we have pa’ pel 37? 1 1 V\-+1+<|=12 ma V= 12 _ 656 1.83 i= OS 2.1874: n= 6963.28 Power absorbed by the 3 1) resistor = (+6.56) (2.187) = 14.35 W Power absorbed by the 12 A current source = (—6.56) 12 = — 78.72 W Power absorbed by the 2i, dependent current source = (-6.56) X 2 X (—3.28) = 43.03 W Power absorbed by the 2 10 resistor = (—6.52) (— 3.28) = 21.51 W PROBLEM 1.1 - * PROGRAM TO CALCULATE TOTAL CURRENT vS 1 0 DC30V RI 1 2 SOHM R2 2 0 40HM R3 2 0 20HM R4 2 0 40HM -OP -END OUTPUT #*** SMALL SIGNAL BIASSOLUTION TEMPERATURE = 27.000 DEG C  CHAPTER Useful Theorems in Circuit Analysis LEARNING oat MM ~~ ~~~ nnonooo orn After reading this chapter, the reader should be able to LOL Reduce the network by Star-Delta transformation LO? Explain the superposition theorem and apply it to solve the networks 103 Explain Thevenin's theorem and apply it to solve the networks LO4 Explain Norton's theorem and apply it te solve the networks LOS Explain the reciprecity theorem and apply It to solve the networks LOG Explain the compensation theorem and apply it to solve the networks LO? State the maximum power transfer theorem and apply it ta solve the networks LOS Explain the cancept of duals and the principle of duality LO9 Explain Tellegen’s theorem and apply it ta solve the networks LO 10 State Millman's theorem and apply it for solving the networks 3.1 | STAR-DELTA TRANSFORMATION In the preceding chapter, a simple technique called the source transformation £LO1 Beduce technique was discussed. The star-delta transformation is another technique useful the network in solving complex networks. Basically, any three circuit elements, i.e. resistive, « Sy Star-belta inductive or capacitive, may be i transformation connected in two different ways. One ° way of connecting these elements | is called the star connection, or the 2 . Y-connection. The other way of =p FA connecting these elements is called ] the delta (A) connection. The circuit is said to be in star connection, if three elements are connected as shown in Fig. 3.1 (a), when it appears like a star : (Y). Similarly, the circuit is said to be a in delta connection, if three elements "7 are connected as shown in Fig. 3.1 (b), when it appears like a delta (A). Fig. 3.1 Useful Theorems in Circuit Analysis 111 The above two circuits are equal if their respective resistances from the terminals AB, BC, and CA are equal. Consider the star-connected circuit in Fig. 3.1 (a); the resistance from the terminals AB, BC, and CA respectively are Ryg(V) = Ry + Rp Rac (Y) = Rg + Re Reg(Y) = Ret Ry Similarly, in the delta-connected network in Fig. 3.1 (b), the resistances seen from the terminals AB, BC, and CA respectively are R,(R, +R) Ryp(A) = | (R, +R) = ua(A)= Ri e+ R= Boe Ry (R, +R) Rgc(A) = R;|| (R, + 2) = 4 ac (A) = R||(R +R) R+R+R Ry (R, +R) A) = R, || (R, +R) = Re) = RVR +R) RR, Now, if we equate the resistances of star and delta circuits, we get R(R, +8. Ry +R, = RR +R) G.1) R +R, +R; R,(R, +R) Ry +Ro = SS 3.2 3 + Re R+R, +R; G2) Ry(R, +B: Ro+Ry= (RK, + Rs) (3) R,+R, +R; Subtracting Eq. (3.2) from Eq. (3.1), and adding Eq. (3.3) to the resultant, we have R, RB R,-=-—_12_ 3.4 A RAR, +R G4) RR. Similarly, R,;=——13— 3.5 YB RER TR (3.5) R, RB. and =——2 3 _ 3.6 Re R+R, +R, G6) Thus, a delta connection of R;, Rj, and R; may be replaced by a star connection of Ry, Rg, and Re as determined from Eqs (3.4), (3.5) and (3.6). Now if we multiply the Eqs (3.4) and (3.5), (3.5) and (3.6), (3.6) and (3.4), and add the three, we get the final equation as under: RR, Ry + RRR +R RR R, Ry +R, Re +R. R, = ns Ae FP AAG ‘ANB Bp Re + Re Ry (R +R, +R, In Eq. (3.7) dividing the LHS by R,, gives R,; dividing it by Rz gives R,, and doing the same with Ro, gives Rj. 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Grneckeck awTe A ond 2 3 then toto Cee daewn ‘oy Wn. cat, vetecmiae TT. = BV = BEs.mA (Ar) T ya KEL ' B and, Cwrvert torngh KEL Revrotey wAN Mee , ee et. = 0G MA Sed 10K SL nee | Curt oe Horumah SL with enn Ae p=  Gi) The NoHeg® awed sya it wh ee, —— = Ty K BRL = Qx” % BX 10? = 6V cite) Now YS Lima Va" av = Methodi CKVL? -$4+Gxa) +2 (G- yee oO = ST, - 27. =4 =” -Q) SrA) eh eae (a.-%) +32, -1=0 3 “29,759, =35 -- -=2G) frum ear) amd Gi) — > H=1P 2 D)= 1 arf 7-3 +(x 4) pae> aa \ 1S » Content sigh 42 awsisten (2) = Sree =448 4 cou cHo Te Fin voltage ot wede 9. 4 > \sv Nia, Applying KVL | =F -\K4+ SUT +507 +5T=0 ‘g = SP es] = T oe ‘O43 A = (15-505) = FR6 vor (Ay) 7 Goat a Alsonberd and Delivered rer =7 Grok * SO) Cuvted+ otens ivto petetwe 4fenmmal. oL voltage $8e1ce. —? fem 17 15 repenne.! em Alasorbed prredt s oun 34 Cuxrrek leaves {raw positive fenminat. ot vorag font —F Tham i+ 1 regfanre? OA Dalivened phen iF Re $store. ond locos s pee Thats ar for. Cesistor , Gatect Ate apres: he (+) polamntiotl to fener (-) potential. et i VA Detevmime prem elbtothbed om Deliveraol. Py Goch Kt alemants. =, + = =7 KVL epphed 5 + -140+2L+237r=0 = T= 2 awh: Noa, a . (Fan) =m = Onn = 8 ea = 19 AAt (Pevs) 7 FR = Ore And, porent. delivered “4 Hee voted SOerce , = 10*2 = 90 Wolt P eset 50, Qn +e eo cK, Tae (te404) - Ft (te4o!) | | | Aen, Giied pene. Gnservah & — fresno ore ‘Teliggens._ Beso :  a Additians Prtvlens Be Mesh Me Hy Fe ers > Rod Cvcrest sli 55) ngoistor - re Mrnere Kb +O Hesh 1 , -40 4 (1x2) + 3(3, -Ta) +6 @, -15) =0 =y 10T,-31,- 61, = 10-~ (i) Nov, . nae KVL to Mesh 2, — 45 4 3 (t.-1) +29, +57, =6 =p -$1,+10T, = -5 cosmengadk) and , Appvd Kv¥L tm Mesh $$, 7 20 +6 (33-ti) -5 +42, =0 = -6T, +0Ts =Q5 ------ Gi) Oy AEG ear.) Gi) and a q, =42tA 5 T,=OF8A | 32 cecil 59, Dea = t, = O-FRA ( 42) f 827 Find Curved singe O95 Manish Qu Mesh 4 5 dinech em o} Curtet. Sbwice ond measly Curette tT) ave Sane , aamieatll T= 6A =f ==) es Apelying Kye to Mesh 2 , -S6 412 (S.-4)) + dash © oy (QT, -12T, ~ S33 * 36 -.@ op Ve Gig S108 * T= 6A Applying KVL to Mesh S , 94+6 (33-1) + 34,491, =9 =p - G61, 47NT, = 79 @) Solving evr. (i) and Gy) , ss T= FA 5 I,=3A $0, Cure Cenk fee Ge WALSH a= t,.> Ty = 34 (n2). Circuits and Networks136 In Fig. 3.37 (a), i1 5 i2 5 2 A; i3 5 2 A and v1 5 –2 V, v2 5 –8 V, v3 5 10 V In Fig. 3.37 (b), i i i v v v 1 1 2 1 3 1 1 1 2 1 3 1 4 4 20 0 20 = = = =− = = A A and V V V ; ; ; Now ( ) (4) ( 8) (4) (10) (4) v i v i v i v iK K K 1 1 1 1 2 2 1 3 3 1 1 3 2 = + + = − + − + = ∑ = 0 and ( 20) (2) (0) (2) (20) (2) v i v i v i v iK K K 1 1 1 1 2 1 2 3 1 3 1 3 = + + = − + + = ∑ = 0 Similarly, v i v i v i v iK K K = + + = ∑ 1 1 2 2 3 3 1 3 = (–2) (2) + (–8) (2) + (10) (2) = 0 and v iK K K 1 1 1 3 20 4 0 4 20 4 0= − + + = = ∑ ( )( ) ( )( ) ( )( ) This verifies Tellegen’s theorem. Frequently Asked Questions linked to LO 9 rrr3-9.1 Verify Tellegen's theorem for the network shown in Fig. Q.1. [PTU 2009-10] rrr3-9.2 Verify Tellegen's theorem for the pair of networks. [PTU 2011-12] Fig. Q.1 Fig. Q.2 3.10 MILLMAN’S THEOREM Millman’s theorem states that in any network, if the voltage sources V1, V2, …, Vn in series with internal resistances R1, R2, …, Rn, respectively, are in parallel, then these sources may be replaced by a single voltage source V9in series with R9 as shown in Fig. 3.38. Fig. 3.37 (a) Fig. 3.37 (b) LO 10 - - .. - --- ' ..... - ~ - ' ' -- --- ..... ~ ~'4 ' ' '- ... - ~ ----........ , · .., .. " ' .... ... , ' -- ", ). ' ' " ' ..., .... , ' ... ~ \ ........ - "" ·~"""' ,, --- ~ ,, '- "" \ ,, ' ' ' \ ~ .. ' ' ,~, ' ' ' '- .... . ~ ~ • .. ~ "-' ~ . "' t: ' "' -~ ' ' "' " '" ' ~ ,, '\ ' .. "' ,\, ... ,, \. ' ---- '\ \ ~ ' , ~'\ \ ' • "- \ ~ I '\IAJ <\ \ 1~ p? .\ '1 N \. 2 \ :-=- \('I,\ N l I ~ -i\ro~'\ A 'l'- 1 I\ \ {:: /\ 1'N'-') :r')_ --;i: \ R, .\; J)\J\. I ~~¥\ ~ \ -~ --- - \ \\ ~ r(ltJ L1 1- )'\It'\ \ ~,p,-J t'\ ~1\ \'N\) ~v1. lc 'l ~ ~ ~ .\- J W L, - ')'/\I YI ,t 'D-,J 1.--,.. --1'}\-,J V j ~ t\) D .\- ,j1\l \..-1 c\-'JN ~ '}.f\"J Y\ --:C 7__ ~ '";.- \'----- 'J.. --- - --- Circuits and Networks128 The application of voltage V across AA9 produces current I at BB9. Now if the positions of the source and responses are interchanged, by connecting the voltage source across BB9, the resultant current I will be at terminals AA9. According to the reciprocity theorem, the ratio of response to excitation is the same in both cases. EXAMPLE 3.6 Verify the reciprocity theorem for the network shown in Fig. 3.23. Fig. 3.23 Solution  Total resistance in the circuit 5 2 1 [3 || (2 1 2 || 2)] 5 3.5 V The current drawn by the circuit (See Fig. 3.24 (a)). IT = = 20 3 5 5 71 . . V The current in the 2 V branch cd is I 5 1.43 A. Fig. 3.24 Applying the reciprocity theorem, by interchanging the source and response, we get Fig. 3.24 (b). Fig. 3.24 Total resistance in the circuit 5 3.23 V. Total current drawn by the circuit= = 20 3 23 6 19 . . A The current in the branch ab is I 5 1.43 A If we compare the results in both cases, the ratio of input to response is the same, i.e. (20/1.43) 5 13.99 Useful Theorems in Circuit Analysis 115 werk 3-1.5 Use the technique of D-Y conversion to find the equivalent resistance between terminals A-B of the circuit shown below. [AU April/May 2011] A 1Q. AQ 22 5Q B Fig. Q.5 wrtv*3-1.6 State and explain star-delta conversion in ac systems. [JNTU Nov. 2012] vex 3-1.7 Obtain the expressions for star-delta equivalence of an impedance network. [JNTU Nov. 2012] 3.2 | SUPERPOSITION THEOREM LO 2 Explain i the superposition ' theorem and apply it; The superposition theorem states that in any linear network containing two or more sources, the response in any element is equal to the algebraic sum of the responses caused by individual sources acting alone, while the other sources are non-operative; that is, while considering the effect of individual sources, other ideal voltage sources and ideal current sources in the network are replaced by short circuit and open circuit across their terminals. This theorem is valid only for linear systems. This theorem can be better understood with a numerical example. Consider the circuit which contains two sources as shown in Fig. 3.7. Now let us find the current passing through the 3 (0 resistor in the circuit. According to the superposition theorem, the current J, due to the 20 V voltage source with 5 A source open circuited = 20/(5 + 3) = 2.5A (see Fig. 3.8) to solve the networks 5Q 102 5Q 102 tH aL 20Vr 32 Ds A 20V5 Fig. 3.7 Fig. 3.8 The current /; due to the 5 A source with the 20 V source short circuited is 5 G+5) I; =5x =3.125A The total current passing through the 3 ( resistor is (2.5 + 3.125) = 5.625A Let us verify the above result by applying nodal analysis. 116 Circuits and Networks Sao , Fig. 3.9 Fig. 3.10 The current passing in the 3 ( resistor due to both sources should be 5.625 A. Applying nodal analysis to Fig. 3.10, we have v-20 V +—=5 5a "5 vita dasea 5.3 y=9x'e = 16.875 V The current passing through the 3 (0 resistor is equal to V/3, ie. [= se =5.625A So the superposition theorem is verified. Let us now examine the power responses. Power dissipated in the 3 Q resistor due to the voltage source acting alone Poy = (LR = (2.5)? 3 = 18.75 W Power dissipated in the 3 © resistor due to the current source acting alone Ps = (;)?R = (3.125)? 3 = 29.29 W Power dissipated in the 30, resistor when both the sources are acting simultaneously is given by P = (5.625)? X 3 = 94.92 W From the above results, the superposition of P) and P; gives Py + Ps = 48.04 W which is not equal to P = 94.92 W We can, therefore, state that the superposition theorem is not valid for power responses. It is applicable only for computing voltage and current responses. EXAMPLE 3.3 Find the voltage across the 2 V resistor in Fig. 3.11 by using the superposition theorem. 2.62 Circuit Theory and Networks—Analysis and Synthesis Solution Step I When the 10 V source is acting alone (Fig. 2.125) Applying KCL at the node, Ve r » VWy-> W-10 Vo 2 5 2 1 LU Ny > 5 2 2 Vé=1.67V =0 Step IT When the 1A current source is acting alone (Fig. 2.126) Applying KCL at the node, ve ” n Vi Yo" +14 Vo ¢— 2 LA yy 5 2 2 Vi'= 0.83 V Step IIT When the 4 V source is acting alone (Fig. 2.127) Applying KCL at the node, Ke" yg yen yn VO We" We" 2g 5 2 1 (+4+4 Ver=4 5°22 Wy" =3.33V Step IV By superposition theorem, 10V = Vo = V5 +Vo'+ Vo" = 1.67 —0.83 + 3.33 = 4.17 V PGR) THEVenin’s THEOREM 52 12 + Vy — vy S20 3 Fig. 2.125 5Q 19 + ve DIA wea Ne Fig. 2.126 5Q 4V 49 tk + Vy” wr S2 Wy Fig, 2.127 It states that ‘any two terminals of a network can be replaced by an equivalent voltage source and an equivalent series resistance. The voltage source is the voltage across the two terminals with load, if any, removed. The series resistance is the resistance of the network measured between two terminals with load removed and constant voltage source being replaced by its internal resistance (or if it is not given with zero resistance, i.e., short circuit) and constant current source replaced by infinite resistance, i.e., open circuit.’ 2.8 Thevenin’s Theorem 2.63 Fig. 2.128 Network illustrating Thevenin’s theorem Explanation Consider a simple network as shown in Fig. 2.129. R, Py A v= Ro R, B Fig. 2.129 Network R Rs For finding load current through R,, first remove the load oA resistor R, from the network and calculate open circuit voltage V,,, across points A and B as shown in Fig. 2.130. ve Po Vin R; Vp, =—— V ~ R,+R B For finding series resistance R,,,, replace the voltage source Fig. 2.130 Calculation of V,, by a short circuit and calculate resistance between points A and R, Ra Bas shown in Fig. 2.131. oA RiRo Ry = R3 + R, R. m= Rt ER 7 =< Ft Thevenin’s equivalent network is shown in Fig. 2.132. oB 1,-—® Fig. 2.131 Calculation of Ry, Rm +R, Pm If the network contains both independent and dependent A sources, Thevenin’s resistance R,, is calculated as, Vn Vin R Rr = | Ty L B where J, is the short-circuit current which would flow in a short circuit placed across the terminals A and B. Dependent sources Fig. 2.132 are active at all times. They have zero values only when the control voltage or current is zero. R,,, may be negative in some Thevenin’s equivalent network 2.64 Circuit Theory and Networks—Analysis and Synthesis cases which indicates negative resistance region of the device, i.e., as voltage increases, current decreases in the region and vice-versa. If the network contains only dependent sources then Vr, =0 Ty =0 For finding R,,, in such a network, a known voltage V is applied across the terminals A and B and current is calculated through the path AB. Vv Rm => . Fm or a known current source J is connected across the —W-—0 A terminals A and B and voltage is calculated across the terminals 4 and B. Vv Rm =— th = : {______» Thevenin’s equivalent network for such a network is shown in Fig. 2.133. Fig. 2.133 Thevenin’s equivalent network Steps to be Followed in Thevenin’s Theorem 1. Remove the load resistance R,. 2. Find the open circuit voltage V,,, across points A and B. 3. Find the resistance R,, as seen from points A and B. 4. Replace the network by a voltage source V,,, in series with resistance R,,. 5. Find the current through R, using Ohm’s law. 1, -—o_ Rm +R | eel (AeeM Determine the current through the 24 Q resistor in Fig. 2.134. 30Q 202 220V —— 502 52 Fig. 2.134 Solution StepI Calculation of V,,, (Fig. 2.135) 220 220V = 1 == = 2.75 A 30+50 22 Ih= 220 8.8A Ryn = (16.25 || 2.5) +2.5= 4.672 Step IT Calculation of J, (Fig. 2.145) 63 = = 2.55A 4.67+20 I, 2.8 Thevenin’s Theorem 2.67 252 l, Pr 16.25Q 63V = Fig. 2.145 | iS eTle(WMaey Find the current through the 10 Q resister in Fig. 2.146. 10Q Seennneed AA Gee 22 22 15V — 19 I 10V 1Q Fig. 2.146 Solution StepI Calculation of V,,, (Fig. 2.147) Applying KVL to Mesh 1, -15-2h -( -1,)-10-1f, =0 oi 4l, -In =-25 @ Applying KVL to Mesh 2, 10=-1(y — 1) -2y - Ly = 0 i) ei -h +4 =10 Solving Eqs (i) and (ii), h=-6A Ip=1A 12 2.68 Circuit Theory and Networks—Analysis and Synthesis Writing the V,,, equation, Voy +2Ly +21, =0 Voy = 21, + Qn = 2-6) +2(1) =-10V = 10 V(the terminal B is positive w.r.t. A) Step I Calculation of R,,, (Fig. 2.148) 22 ——o A, A B 22 Fig. 2.148 Converting the star network formed by resistors of 2Q,2Qand1Q into an equivalent delta network (Fig. 2.149), ——_ R,,o ——__ 2Q 22 R=242+72 _g9 Ry =2414 221240 2 19 2x1 R214 =4Q 2 19 Fig. 2.149 Simplifying the network (Fig. 2.150), Pry ° 3 r 8 AY 8B 82 BQ VW VW 4Q 4Q 0.82 0.82 (b) Fm g A B 12 12 1,33 Q (a) Fig. 2.150 2.8 Thevenin’s Theorem 2.69 Rm =1.33Q Step IT Calculation of J, (Fig. 2.151) 1.33Q A 1, =—o— = o88.a (ty tov we 100 1.33+10 B Fig. 2.151 I ZETA Find the current through the 1 Q resistor in Fig. 2.152. 1A Nw) 22 3Q 4v— 1a ao 3A Fig. 2.152 Solution StepI Calculation of V,,, (Fig. 2.153) 1A GC) + 2 ~32 aw AOt+ 4v— h Von O 3A Bo- ° I 22 30 Fig. 2.153 I A Writing the current equations for Meshes 1 and 2, R Th ° * Writing the V,,, equation, 4-21, - In) Vim =0 Fig. 2.154 Vin = 4-2 — In) = 4-2(-4) =12V Step IT Calculation of Ry, (Fig. 2.154) Ry =2Q 12v— Step IIT Calculation of J, (Fig. 2.155) acavyN 2+1 Fig. 2.155 Ih 2.72 Circuit Theory and Networks—Analysis and Synthesis | Set wAE I Obtain the Thevenin equivalent network of Fig. 2.165 for the terminals A and B. Vy, 19 mW —<>—0A 20 G 2A 2v tL | es Fig. 2.165 Solution StepI Calculation of V,, (Fig. 2.166) From Fig. 2.166, 2-2h -V, =0 V, =2-2h For Mesh 1, L=-2A V, =2-2(-2)=6V Writing the V,,, equation, 2-2h -0+4V, —Vy, = 0 2-2(-2)-0+4(6)-Vn, =0 Vr, =30V Step IT Calculation of I, (Fig. 2.167) From Fig. 2.167, V,=2-2h i) Meshes 1 and 2 will form a supermesh, Writing current equation for the supermesh h-l=2 di) Applying KVL to the outer path of the supermesh, 2-2-1) +4V, =0 2-21 -I +4(2-2h) =0 10, +1, =10 Solving Eqs (ii) and (iii), 1, =0.73A 1) =2.73A Iy =I, =2.73A Step I Calculation of R,, Vr _ 30 Rm = 2 = = 10.982 Th Ty 2.73 Step IV_ Thevenin’s Equivalent Network (Fig. 2.168) 30V— 10.98 Q Fig. 2.168 iii) A 2.8 Thevenin’s Theorem 2.73 | SETA Find the Thevenin equivalent network of Fig. 2.169 for the terminals A and B. 8, 12 <i 102 4 5V | Fig. 2.169 Solution ah 12 Step I Calculation of Vp, (Fig. 2.170) <P —aa Applying KVL to the mesh, 4 102 5-10%, -107, =0 109 Van 5 5V t=—=0.25A 20 Lj _______ss Writing the V,,, equation, Fig. 2.170 5-10K, +8, -0-V, = 0 Vin =5-2h, = 5-2(0.25)= 4.5 V Step I Calculation of I, (Fig. 2.171) 8, 19 Applying KVL to Mesh 1, A I 5-10/, -10(f, - I) =0 . 10a ” =i) 102 20K -101y =5 | | 1 2 Applying KVL to Mesh 2, Sv Ti B -10(, -h)+8h, -1hh =0 ii) 18%, -11/, =0 os Fig. 2.171 Solving Eqs (i) and (ii), TQ =1375A 1b =2.25A Ty = 1, =2.25A ——_AMA—__—0A Step IT Calculation of R,,, 45V = Vm _ 4.5 Rm =— => =22 Th Ty 2: Step IV_Thevenin’s Equivalent Network (Fig. 2.172) Fig. 2.172 2.74 Circuit Theory and Networks—Analysis and Synthesis I Sew a Ae Find V,,, and R,,, between terminals A and B of the network shown in Fig. 2.173. 1Q 22 I 12V— 4 ly 19 A Fig. 2.173 Solution Step I Calculation of V,,, (Fig. 2.174) I, =0 12V—— The dependent source 2/, depends on the controlling variable I, When J, = 0, the dependent source vanishes, i.e., 27, =0 as shown in Fig. 2.174. Writing the V,,, equation, Vin =12x a =6V wm 141 Step Il Calculation of /, (Fig. 2.175) 1Q 22 1Q Fig. 2.174 12 22 From Fig. 2.175, V, => 12V—— 2 Applying KCL at Node 1, Aa AA =21, 1 12 °* neneta-a(4) 2 2 W=8V Kw =F 8 Iyztaseaa v 2 Step HII Calculation of R,,, Vin _ 6 Rm = =—=15Q ™ Ty 4 Fig. 2.175 I SET LA Obtain the Thevenin equivalent network of Fig. 2.176 for the given network at terminals a and b . 2.8 Thevenin’s Theorem 2.77 Applying KVL to Mesh 2, —S(Ip —)-101, -101, =0 -5(Ip — 11) -10( — Ip) -101y = 0 51, +5, =0 iii) Solving Eqs (ii) and (iii), Q=1A In=-1A Ip =h-Ih=1-(-)=2A Writing the V,,, equation, 10J, -10—Vr, = 0 10(-1) -10-Vr,= 0 Vay, =-20V Step IH Calculation of J, (Fig. 2.186) 10 ly 10V From Fig. 2.186, + | L=h-h ...@i) I oll For Mesh 1, hel ay 14 a » 5Q » 109 » Iu Applying KVL to Mesh 2, -5([z 1) - 101, -10(Ip - 15) = 0 -5(Ip —1,)-10( — 1) -10(Iy - 5) = 0 Fig. 2.186 5h -5Ip +101; = 0 «iii Applying KVL to Mesh 3, -10(J5 -I)-10 = 0 101-107; =10 ..(iv) Solving Eqs (ii), (iii) and (iv), Q=1A 7100 1n=3A 13=2A -20V = Iy =1=2A Step IIT Calculation of R,,, U_______________5 B Rn = 28 = 20-100 Fig. 2.187 Ty 2 Step IV Thevenin’s Equivalent Network (Fig. 2.187) | Example 2.75 | Find Thevenin’s equivalent network at terminals A and B in the network of Fig. 2.188. 2.78 Circuit Theory and Networks—Analysis and Synthesis 42 2Q 4Q OA A + 4Vy % S52 B Fig. 2.188 Solution Since the network does not contain any independent source, Vey =0 20 Iy =0 But the R,, can be calculated by applying a known voltage source of 1 V at the terminals A and Bas shown 4x (,, in Fig. 2.189. Rm = v T sh From Fig. 2.189, Vz =5(h - hh) Applying KVL to Mesh 1, -4V,,- 2h, -5(; - In) = 0 -4[5() -h]-2h -5h +5 =0 -27f, +251, =0 Applying KVL to Mesh 2, -5(Iy -h)- 41, -1=0 5h, -91y =1 Solving Eqs (ii) and (iii), f=-0.21A I, =-0.23A Hence, current supplied by voltage source of 1 V is 0.23 A. 1 Rm =~ = 4.35Q Tm 0.23 Hence, Thevenin’s equivalent network is shown in Fig. 2.190. | Set ALA Find the current in the 9 Q resistor in Fig. 2.191. 6h, SS 42 62 Fig. 2.191 9Q di) iii) 4.359 Fig. 2.190 2.8 Thevenin’s Theorem 2.79 Solution 6l, Step I Calculation of V,,, (Fig. 2.192) iS A Applying KVL to the mesh, 40 * + 20-41, + 61, -61, =0 62 Vr, I,=5A - Writing the V,,, equation, . 20V _ 61, —Vrm = 0 °8 6(5)—Vim = 0 Fig. 2.192 Vr, =30V Step I Calculation of /, (Fig. 2.193). 6, From Fig. 2.193, 7 A I, I, =0 42 The dependent source 6/, depends on the controlling 6a In variable /,. When J, = 0, the dependent source vanishes, 20V ie., 62, = 0 as shown in Fig. 2.194. B 20 Iy == A Fig. 2.193 6h, A A 42 4a > ly 20V 20V B B (a) (b) Fig. 2.194 6a A Step IT Calculation of R,, Vm _ 30 Rm =o == =62 o_o mE TAS 30 V = » 92 Step IV Calculation of I, (Fig. 2.195) B 30 = e577 Fig. 2.195 | SET eAAAN © Determine the current in the 16 Q resistor in Fig. 2.196. 10Q 6Q 40V = } 08, 162 Fig. 2.196 2.82 Circuit Theory and Networks—Analysis and Synthesis Solution Vin 10V, Step I Calculation of V,,, (Fig. 2.206) a a3 ~ From Fig. 2.206, + V, =10xX5=50V 100 V —— 52S, Boa Writing the V,,, equation, - 100 —Vz, +10V, —V, = 0 100 -Vr, + VV, = 0 Fig. 2.206 100 -Vq, + 9(50) = 0 Vin = 550 V Step IT Calculation of I, (Fig. 2.207) A B 10V, From Fig. 2.207, yo V, = S(Iy +10) + Applying KVL to Mesh 1, 100V Ve 5a ® 10a 100+10V, -V, =0 | 100 Va = “9” Fig. 2.207 100 > =5Iy +50 550 Iy =-—— a) Step IIT Calculation of R,,, 550 fess 8 J 45 550 V = Step IV Calculation of /, (Fig. 2.208) L-= 550 _ «(110 1° 745+10 7 Fig. 2.208 [E53 norton’s THEOREM It states that ‘any two terminals of a network can be replaced by an equivalent current source and an equivalent parallel resistance.’ The constant current is equal to the current which would flow ina short circuit placed across the terminals. The parallel resistance is the resistance of the network when viewed from these open-circuited terminals after all voltage and current sources have been removed and replaced by internal resistances. Fig. 2.209 Network illustrating Norton’s theorem 2.106 Circuit Theory and Networks—Analysis and Synthesis Step I Calculation of J, (Fig. 2.301) 3h When a short circuit is placed across the 3 Q resistor, it <>} A gets shorted as shown in Fig. 2.302. I From Fig. 2.302, Leh-h eo JOA 19 6a ly For Mesh 1, B T,=10 di) Applying KVL to Mesh 2, Fig. 2.301 =M(y-h) +3; =0 3h “hh +h +3 -h)=0 i A 4-41 =0 «.(iii) Solving Eqs (ii) and (iii), wad) ) ie ) In h=10A ' ° 1h=10A 8 Iy =1,=10A Fig. 2.302 Step IIT Calculation of Ry, A Ry = = 74 2040 “ Iy 10 10A(4 242 162 Step IV Calculation of /, (Fig. 2.303) I, =10x 24 = 6A 8 2441.6 Fig. 2.303 ERT) maximum power TRANSFER THEOREM It states that ‘the maximum power is delivered from a source to a load when the load resistance is equal to the source resistance.” Proof From Fig. 2.304, Re Ro + Rr ) v= RA VR, ! Power delivered to the load R, = P=I 2 Rp= (Re+ Ri) To determine the value of R, for maximum power to be transferred Fig. 2.304 Network illustrating to the load, i maximum power transfer a 0 theorem aR, aed iV? a R, dR, dR, (R, + R,) _ PUR + RLY - QR (Re +R) (Re + Ri) 2.10 Maximum Power Transfer Theorem 2.107 (R, +R)? —2 Ri(R, + Ri) = 0 Ro +R, +2R,R, -2R,R, -2R? =0 R=R, Hence, the maximum power will be transferred to the load when load resistance is equal to the source resistance. Steps to be followed in Maximum Power Transfer Rm Theorem A 1. Remove the variable load resistor R,. 2. Find the open circuit voltage V,,, across points A and Vin PL= Fr B. I 3. Find the resistance R,,, as seen from points A and B B. 4. Find the resistance R, for maximum power Fig, 2.305 Thevenin’s equivalent network transfer. Ry = Ry 5. Find the maximum power (Fig. 2.305). _ Vm _ Vm Rm+R, 2m Vin Vin Prax = 1} Ry = 2x Ry = max = 47 Ay aR, Th aR, I, | CT ee For the value of resistance R, in Fig. 2.306 for maximum power transfer and calculate the maximum power. 152 ph 182 5Q 15Q 27Q 9Q 102 2092 272 Fig. 2.306 2.110 Circuit Theory and Networks—Analysis and Synthesis Solving Eqs (i) and (ii), 1 =3.43A Writing the V,,, equation, Vy - 20 (1, - In) 20 = 0 Vr, = 20(3.43-2) + 20 = 48.6 V Step HI Calculation of R,,, (Fig. 2.313) Ry, =15||20=8.572 Step IT Calculation of R, For maximum power transfer, R, = Rm =8.572 Step IV Calculation of P,,., (Fig. 2.314) = Vin _ 48.6)" Prax = = 68.9 W 4Rm 48.57 | ST] calculate the maximum power. 102 100 V —— 30Q Fig. 2.315 Solution StepI Calculation of V,,, (Fig. 2.316) 100 _ “10+30 1p = 8-166 20+40 1 Writing the V,,, equation, Vay +10 Ty — 20, =0 Ven = 201, -10/, = 20(1.66) -10(2.5) = 8.2 V 5Q 102 202 A Pon i B Fig. 2.313 8.572 A 48.6 V 8.5702 B Fig. 2.314 100V = mY For the value of resistance R ', in Fig. 2.315 for maximum power transfer and 2.10 Maximum Power Transfer Theorem 2.111 Step I Calculation of R,, (Fig. 2.317) Fig. 2.317 Redrawing the network (Fig. 2.318), 10Q 202 A B Rn = (10 || 30) + (20 || 40) = 20.83 Q 302 40Q Fig. 2.318 Step IIT Value of R, 20.83 Q For maximum power transfer, A Ry = Rm = 20.832 8.2V = 20.83 Q Step IV Calculation of P,,,, (Fig. 2.319) 2 2 B > a = Li 82) Lo giw 4Rm 420.83 Fig. 2.319 | ete CEM For the value of resistance R, in Fig. 2.320 for maximum power transfer and calculate the maximum power. 72vV— Fig. 2.320 2.112 Circuit Theory and Networks—Analysis and Synthesis Solution StepI Calculation of V,,, (Fig. 2.321) Applying KVL to Mesh 1, 72-61 -3(1; -n) =0 91, -3In =72 ...(i) 72V = Applying KVL to Mesh 2, -3(I, — hh) -2In - 41, = 0 30 +91, =0 ...(ii) Solving Eqs (i) and (ii), T=9A I1n=3A Writing the V,,, equation, Vr, -6F, — 21, =0 Vin = 61, + 2p = 6(9) + 2(3) = 60 V Step I Calculation of R,, (Fig. 2.322) 22 6Q A 4Q Prn B Fig. 2.322 Rm =[6 || 3)+2]|]4=22 Step III Calculation of R, ee A For maximum power transfer, Ry = Rm =2.0 60 Vv 20 Step IV Calculation of P,,.,, (Fig. 2.323) B = is _ (6 _ gsow me" ARm 4x2 Fig. 2.323 EXAMPLES WITH DEPENDENT SOURCES | Sele For the network shown in Fig. 2.324, find the value of R, for maximum power transfer. Also, calculate maximum power.