ProblemSolving Compnion To ccompny Bic Engineering Circuit Anlyi Eight Edition J. Dvid Irwin Auburn Univerity JOHN WILEY & SONS, INC.
Executive Editor Bill Zobrit Aitnt Editor Kelly Boyle Mrketing Mnger Frnk Lymn Senior Production Editor Jime Pere Copyright 5, John Wiley & Son, Inc. All right reerved No prt of thi publiction my be reproduced, tored in retrievl ytem or trnmitted in ny form or by ny men, electronic, mechnicl, photocopying, recording, cnning, or otherwie, except permitted under Section 7 or 8 of the 976 United Stte Copyright Act, without either the prior written permiion of the Publiher, or uthoriztion through pyment of the pproprite percopy fee to the Copyright Clernce Center, oewood Drive, Dnver, MA 9, (58) 7584, fx (58) 75 447. equet to the Publiher for permiion hould be ddreed to the Permiion Deprtment, John Wiley & Son, Inc., 65 Third Avenue, New York, NY 58, () 856, fx ()8568, emil: PEMEQ@WILEY.COM. ISBN 477468
TABLE OF CONTENTS Prefce. Acknowledgement... Chpter Problem... Solution... 4 Chpter Problem... 7 Solution... 9 Chpter Problem... 8 Solution... 9 Chpter 4 Problem... 5 Solution... 6 Chpter 5 Problem... Solution... Chpter 6 Problem... 4 Solution... 44 Chpter 7 Problem... 5 Solution... 5 Chpter 8 Problem... 66 Solution... 67 Chpter 9 Problem... 7 Solution... 74 Chpter Problem... 8 Solution... 8 Chpter Problem... 88 Solution... 89 Chpter Problem... 94 Solution... 96 Chpter Problem... Solution... 4 Chpter 4 Problem... Solution... Chpter 5 Problem... 7 Solution... 9 Chpter 6 Problem... 6 Solution... 7 Appendix Technique for Solving Liner Independent Simultneou Eqution...46
STUDENT POBLEM COMPANION To Accompny BASIC ENGINEEING CICUIT ANALYSIS, EIGHTH EDITION By J. Dvid Irwin nd. Mrk Nelm PEFACE Thi Student Problem Compnion i deigned to be ued in conjunction with Bic Engineering Circuit Anlyi, 8e, uthored by J. Dvid Irwin nd. Mrk Nelm nd publihed by John Wiley & Son, Inc.. The mteril trct directly the chpter in the book nd i orgnized in the following mnner. For ech chpter there i et of problem tht re repreenttive of the endofchpter problem in the book. Ech of the problem et could be thought of miniquiz on the prticulr chpter. The tudent i encourged to try to work the problem firt without ny id. If they re unble to work the problem for ny reon, the olution to ech of the problem et re lo included. An nlyi of the olution will hopefully clrify ny iue tht re not well undertood. Thu thi compnion document i prepred helpful djunct to the book.
CHAPTE POBLEMS. Determine whether the element in Fig.. i borbing or upplying power nd how much. A Fig... In Fig.., element borb 4W of power. I element borbing or upplying power nd how much. 6 Fig.... Given the network in Fig.. find the vlue of the unknown voltge X. 4 A 6A A 4A 8 Fig.. X
4 CHAPTE SOLUTIONS. One of the eiet wy to exmine thi problem i to compre it with the digrm tht illutrte the ign convention for power hown below in Fig. S.(b). A i(t) v(t) Fig. S.() Fig. S.(b) We know tht if we imply rrnge our vrible in the problem to mtch thoe in the digrm on the right, then p(t) i(t) v(t) nd the reultnt ign will indicte if the element i borbing ( ign) or upplying ( ign) power. If we revere the direction of the current, we mut chnge the ign nd if we revere the direction of the voltge we mut chnge the ign lo. Therefore, if we mke the digrm in Fig. S.() to look like tht in Fig. S.(b), the reulting digrm i hown in Fig. S.(c). Now the power i clculted A Fig. S.(c) P () () 4W () And the negtive ign indicte tht the element i upplying power.. ecll tht the digrm for the pive ign convention for power i hown in Fig. S.() nd if p vi i poitive the element i borbing power nd if p i negtive, power i being upplied by the element.
5 i v Fig. S.() If we now iolte the element nd exmine it, ince it i borbing power, the current mut enter the poitive terminl of thi element. Then P I 4 6(I) I 4A The current entering the poitive terminl of element i the me tht leving the poitive terminl of element. If we now iolte our dicuion on element, we find tht the voltge cro the element i 6 nd the current of 4A emnte from the poitive terminl. If we revere the current, nd chnge it ign, o tht the iolted element look like the one in Fig. S.(), then P (6) (4) 4W And element i upplying 4W of power.. By employing the ign convention for power, we cn determine whether ech element in the digrm i borbing or upplying power. Then we cn pply the principle of the conervtion of energy which men tht the power upplied mut be equl to the power borbed. If we now iolte ech element nd compre it to tht hown in Fig. S.() for the ign convention for power, we cn determine if the element re borbing or upplying power. i P i Fig. S.() For the ource nd the current through it to be rrnged hown in Fig. S.(), the current mut be revered nd it ign chnged. Therefore P () (6) 7W
6 Treting the remining element in imilr mnner yield P (4) (6) 4W P () () W P (8) (4) W P X ( X ) () X Applying the principle of the conervtion of energy, we obtin And 7 4 X X
7 CHAPTE POBLEMS. Determine the voltge nd in the network in Fig.. uing voltge diviion. v kω kω 4kΩ kω Fig... Find the current I nd I in the circuit in Fig.. uing current diviion. kω kω I 6kΩ kω 9mA I Fig... Find the reitnce of the network in Fig.. t the terminl AB. 8kΩ kω kω A kω kω 4kΩ kω 8kΩ 6kΩ B 6kΩ kω Fig...4 Find the reitnce of the network hown in Fig..4 t the terminl AB. A 4kΩ 6kΩ kω 8kΩ B kω Fig..4 kω kω
8.5 Find ll the current nd voltge in the network in Fig..5. kω A kω B I kω 6kΩ 48 4kΩ kω 4kΩ I I 5 I Fig..5.6 In the network in Fig..6, the current in the 4kΩ reitor i ma. Find the input voltge S. I 4 I 6 kω kω 9kΩ S kω 4kΩ ma 6kΩ kω Fig..6
9. We recll tht if the circuit i of the form CHAPTE SOLUTIONS Fig. S.() Then uing voltge diviion Tht i the voltge divide between the two reitor in direct proportion to their reitnce. With thi in mind, we cn drw the originl network in the form kω kω Fig. S.(b) 4kΩ kω The erie combintion of the 4kΩ nd kω reitor nd their prllel combintion with the kω reitor yield the network in Fig. S.(c). kω kω Fig. S.(c) Now voltge diviion cn be equentilly pplied. From Fig. S.(c). Then from the network in Fig. S.(b) k k k 6
k k 4k. If we combine the 6k nd k ohm reitor, the network i reduced to tht hown in Fig. S.(). kω I 9mA Fig. S.() kω 4kΩ The current emnting from the ource will plit between the two prllel pth, one of which i the kω reitor nd the other i the erie combintion of the k nd 4kΩ reitor. Applying current diviion I 9 k k ma k ( k 4k) Uing KCL or current diviion we cn lo how tht the current in the kω reitor i 6mA. The originl circuit in Fig. S. (b) indicte tht I will now be plit between the two prllel pth defined by the 6k nd kω reitor. I ma kω 6mA kω 9mA 6kΩ kω I Fig. S.(b) Applying current diviion gin 6k I 6k k 6k k 8k ma I I Likewie the current in the 6kΩ reitor cn be found by KCL or current diviion to be ma. Note tht KCL i tified t every node.
. To provide ome reference point, the circuit i lbeled hown in Fig. S.(). 8k A' k A" k A B 4k 6k 8k k B' k B" Fig. S.() Strting t the oppoite end of the network from the terminl AB, we begin looking for reitor tht cn be combined, e.g. reitor tht re in erie or prllel. Note tht none of the reitor in the middle of the network cn be combined in nywy. However, t the righthnd edge of the network, we ee tht the 6k nd k ohm reitor re in prllel nd their combintion i in erie with the kω reitor. Thi combintion of 6k k k i in prllel with the kω reitor reducing the network to tht hown in Fig. S.(b). 8k A' k A" A 4k k k k (6k k k) 8k 6k B B' k B" Fig. S.(b) epeting thi proce, we ee tht the kω reitor i in erie with the kω reitor nd tht combintion i in prllel with thekω reitor. Thi equivlent 6kΩ reitor (k k) k i in erie with the kω reitor nd tht combintion i in prllel with the 8kΩ reitor tht (6k k) 8k 6kΩ nd thu the network i reduced to tht hown in Fig. S.(c). 8k A' A k 6k k B 4k 6k B' Fig. S.(c) 6k
At thi point we ee tht the two 6kΩ reitor re in erie nd their combintion in prllel with the 4kΩ reitor. Thi combintion (6k 6k) 4k kω which i in erie with 8kΩ reitor yielding A totl reitnce AB k 8k kω..4 An exmintion of the network indicte tht there re no erie or prllel combintion of reitor in thi network. However, if we redrw the network in the form hown in Fig. S.4(), we find tht the network hve two delt bck to bck. A 4k 6k k 8k B k k k Fig. S.4() If we pply the Y trnformtion to either delt, the network cn be reduced to circuit in which the vriou reitor re either in erie or prllel. Employing the Y trnformtion to the upper delt, we find the new element uing the following eqution illutrted in Fig. S.4(b) 8k 6k k Fig. S.4(b) ( 6k) ( 8k) Ω 6k k 8k k ( 6k) ( k) Ω 6k k 8k k ( k) ( 8k) 6 Ω 6k k 8k k The network i now reduced to tht hown in Fig. S.4(c).
A 4k k k 6k B k k Fig. S.4(c) k Now the totl reitnce, AB i equl to the prllel combintion of (k k) nd (6k k) in erie with the remining reitor i.e. AB 4k k (4k 8k) k 6.875kΩ If we hd pplied the Y trnformtion to the lower delt, we would obtin the network in Fig. S.4(d). 4k A 6k 8k 4k 4k B k Fig. S.4(d) 4k In thi ce, the totl reitnce AB i AB 4k (6k 4k) (8k 4k) 4k k 6.875kΩ which i, of coure, the me our erlier reult..5 Our pproch to thi problem will be to firt find the totl reitnce een by the ource, ue it to find I nd then pply Ohm lw, KCL, KL, current diviion nd voltge diviion to determine the remining unknown quntitie. Strting t the oppoite end of the network from the ource, the k nd 4k ohm reitor re in erie nd tht combintion i in prllel with the kω reitor yielding the network in Fig. S.5().
4 A k B 48 I k I 4k 6k I I 4 k Fig. S.5() Proceeding, the k nd k ohm reitor re in erie nd their combintion i in prllel with both the 4k nd 6k ohm reitor. The combintion (k k) 6k 4k kω. Therefore, thi further reduction of the network i hown in Fig. S.5(b). k 48 I Fig. S.5(b) k Now I nd cn be eily obtined. And by Ohm lw 48 I ma k k or uing voltge diviion ki 4 k 48 k k 4 once i known, I nd I cn be obtined uing Ohm lw I I 4 4k 4k 4 6k 6k 6mA 4mA I 4 cn be obtined uing KCL t node A. A hown on the circuit digrm. I I I I 4
5 k 6 k k 4 k I 4 I 4 ma The voltge i then ki 4 4 k k 4 ( ) or uing voltge diviion k k k 4 6 4 Knowing, I 5 cn be derived uing Ohm lw nd lo I 5 k 4 ma I 6 k 4k ma current diviion cn lo be ued to find I 5 nd I 6. nd I 5 I 4 4 k 4k k 4k k ma
6 I 6 I 4 k k k 4k ma Finlly cn be obtined uing KL or voltge diviion nd ki 6 4 k 8 k 4k 4k k 8.6 The network i lbeled with ll current nd voltge in Fig. S.6. 4 I A 5 I B k I k 9k I 4 S k 4k k 6k k I Fig. S.6 Given the ma current in the 4kΩ reitor, the voltge k ( 4k) Now knowing, I nd I cn be obtined uing Ohm lw Applying KCL t node B I ma 6k 6k I ma 9k k k
7 Then uing Ohm lw I I I k 6mA KL cn then be ued to obtin i.e. I (k) 6 Then 6 8 I 4 k 9mA And I 5 4 I I 6 9 k k 5mA uing Ohm lw 4 (k) I 5 nd finlly S 4 48
8 CHAPTE POBLEMS. Ue nodl nlyi to find in the circuit in Fig... ma kω kω kω kω Fig... Ue loop nlyi to olve problem.. Find in the network in Fig.. uing nodl nlyi. kω ki kω X kω IX Fig...4 Ue loop nlyi to find in the network in Fig..4. I X 4mA kω kω kω I X Fig..4 kω
9 CHAPTE SOLUTIONS. Note tht the network h 4 node. If we elect the node on the bottom to be the reference node nd lbel the remining nonreference node, we obtin the network in Fig. S.(). k k k k k Fig. S.() emember the voltge, nd re meured with repect to the reference node. Since the ource i connected between node nd the reference, regrdle of the voltge or current in the reminder of the circuit. Therefore, one of the linerly independent eqution required to olve the network (N, where N i the number of node) i The remining linerly independent eqution re obtined by pplying KCL t the node lbeled nd. Summing ll the current leving node nd etting them equl to zero yield k k k Similrly, for the node lbeled, we obtin k k k The linerly independent eqution cn be quickly reduced to k k k k k k
or 4 6 Solving thee eqution uing ny convenient method yield nd. 7 7 We cn quickly check the ccurcy of our clcultion. Fig. S.(b) illutrte the circuit nd the quntitie tht re currently known. 4 7k A 4 84 7 7 I 6 k I k 7 k k I Fig. S.(b) I 4 All unknown brnch current cn be eily clculted follow. 84 4 I 7 7 k 4 4 I 7 A k 7k 4 6 I 7 7 k 6 8 I 7 4 A k 7k KCL i tified t every node nd thu we re confident tht our clcultion re correct.. The network contin window pne nd therefore linerly independent loop eqution will be required to determine the unknown current nd voltge. To begin we rbitrrily ign the loop current hown in Fig. S.. 44 7k 4 7k A A
k A I kω kω kω I I kω Fig. S. The eqution for the loop current re obtined by employing KL to the identified loop. For the loop lbeled I nd I, the KL eqution re nd k(i I ) k (I I ) k(i I ) k(i I ) ki In the ce of the rd loop, the current I goe directly through the current ource nd therefore Combining thee eqution yield I k ki ki 4 ki 4kI Solving thee eqution uing ny convenient method yield I Then i imply 58 A nd I 7k 8 A. 7k ki 6 7 Once gin, quick check indicte tht KCL i tified t every node. Furthermore, KL i tified round every cloed pth. For exmple, conider the pth round the two window pne in the bottom hlf of the circuit. KL for thi pth i k(i I ) k(i I ) ki
58 4 8 4 8 k k k 7k 7k 7k 7k 7k. The preence of the two voltge ource indicte tht nodl nlyi i indeed vible pproch for olving thi problem. If we elect the bottom node the reference node, the remining node re lbeled hown in Fig. S.(). ki X k k k I X Fig. S.() The node t the upper right of the circuit i clerly, the output voltge, nd becue the ource i tied directly between thi node nd the one in the center of the network, KL dictte tht the center node mut be e.g. if 4, then the voltge t the center node would be. Finlly, the node t the upper left i defined by the dependent ource ki X. If we now tret the ource nd it two connecting node upernode, the current ki X leving the upernode to the left i, the current down through the center k leg of the network i nd the current leving the upernode on the right edge i k. Therefore, KCL pplied to the upernode yield k ki X k k k Furthermore, the control vrible I X i defined combining thee two eqution yield I X k 6 7
The voltge t the remining nonreference node re 6 7 6 7 84 7 48 7 And ki X 48 k k 7 k k 4 7 The network, lbeled with the node voltge, i hown in Fig. S.(b) 4 48 7 7 6 7 I k I I k k Then Fig. S.(b) 4 48 7 7 I k 7k 48 4 I 7 A k 7k 6 I 7 k 6 7k Note crefully tht KCL i tified t every node..4 Becue of the preence of the two current ource, loop nlyi i vible olution method. We will elect our loop current (we need ince there re window pne in the network) o tht of them go directly through the current ource hown in Fig. S.4(). A
4 I X I 4 k k k k k I I I X Fig. S.4() Therefore, two of the three linerly independent eqution needed re Applying KL to the third loop yield combining eqution yield And then And I I X (I I ) 4 I k k(i I ) k(i I ) ki I ma 4 I 4mA Uing thee vlue, the brnch current re hown in Fig. S.4(b) 4 k k 4 k k Fig. S.4(b) Although one brnch of the network re no current, KCL i tified t every node. k
5 CHAPTE 4 POBLEMS 4. Derive the gin eqution for the nonidel noninverting opmp configurtion nd how tht it reduce to the idel gin eqution if i nd A re very lrge, i.e. greter thn 6. 4. Determine the voltge gin of the opmp circuit hown in Fig. 4.. v 5kΩ 5kΩ 5kΩ kω 6kΩ 5kΩ v o Fig. 4. 4. Uing the idel opmp model how tht for the circuit hown in Fig. 4., the output voltge i directly relted to ny mll chnge. v v o Fig. 4. 4.4 Given n opmp nd even tndrd kω reitor, deign n opmp circuit tht will produce n output of v v v
6 CHAPTE 4 SOLUTIONS 4. The noninverting opmp circuit i hown in Fig. S4.(). v N v o F I Fig. S4.() The nonidel model i v N v v e i v I F o Av e Fig. S4.(b) or v N v e v i F I o Av e v o Fig. S4.(c) The node eqution for thi circuit re v v i N v I v v F o
7 Av v v v o e o F o N e v v v or i N o F F I i v v v o N o o F o F Av v v A Following the development on pge 4 of the text yield o F F o F F I i o N F I i i N o F o A Av v A v uming i, the eqution reduce to o F F o F F I o F I N o A A v v Now dividing both numertor nd denomintor by A nd uing A yield I F F o F I o N o v v which i the idel gin eqution.
8 4. The network in Fig. 4. cn be reduced to tht hown in Fig. S4.() by combining reitor. 5kΩ 5kΩ v 75kΩ kω v o Fig. S4.() v i determined by the voltge divider t the input, i.e. 75k v v 5k 75k 4 v The opmp i in tndrd noninverting configurtion nd the gin i 5k 6. k Therefore v ( 6)( ) 4 o v nd v v o 9.5 4. The node eqution for the circuit in Fig. 4. re v v v o v v v v v v Then
9 v v v v v v o v v v o v ( ) v o v v ( ) ( )( ) v o v v v o 4.4 A weightedummer circuit hown in Fig. S4.4() cn be ued to produce n output of the form v o v v. v v v o Fig. S4.4() Note tht nd Therefore if
4kΩ (two kω reitor in erie) kω 48kΩ (four kω reitor in erie) then the deign condition re tified.
CHAPTE 5 POBLEMS 5. Find in the circuit in Fig. 5. uing the Principle of Superpoition. 6kΩ 8kΩ 6mA 4kΩ Fig. 5. 5. Solve problem 5. uing ource trnformtion. 5.. Find in the network in Fig. 5. uing Thevenin Theorem. 4mA 6kΩ 4kΩ kω kω Fig. 5. 5.4 Find I in the circuit in Fig. 5.4 uing Norton Theorem. kω kω 6kΩ ma I Fig. 5.4 5.5 For the network in Fig. 5.5, find L for mximum power trnfer nd the mximum power tht cn be trnferred to thi lod.
X kω X 6kΩ L Fig. 5.5
CHAPTE 5 SOLUTIONS 5. To pply uperpoition, we conider the contribution tht ech ource independently mke to the output voltge. In o doing, we conider ech ource operting lone nd we zero the other ource(). ecll, tht in order to zero voltge ource, we replce it with hort circuit ince the voltge cro hort circuit i zero. In ddition, in order to zero current ource, we replce the current ource with n open circuit ince there i no current in n open circuit. Conider now the voltge ource cting lone. The network ued to obtin thi contribution to the output i hown in Fig. S5.(). 6kΩ 6mA 8kΩ 4kΩ Fig. S5.() Then (only prt of ) i the contribution due to the ource. Uing voltge diviion 4k 4k 6k 8k 8 The current ource contribution to i obtined from the network in Fig. S5.(b). I 6k 6 k 8k 4k Fig. S5.(b) Uing current diviion, we find tht Then I 6 6k k 6k 8k 4k ma
4 Then uperpoition tte tht 4kI 8 8 8 5. ecll tht when employing ource trnformtion, t pir of terminl we cn exchnge voltge ource S in erie with reitor S for current ource I p in prllel with reitor p nd vice ver, provided tht the following reltionhip mong the prmeter exit. 6 S I p S p S Now the originl circuit i hown in Fig. S5.(). 6kΩ 6mA 8kΩ 4kΩ Fig. S5.() Note tht we hve ource in erie with 6kΩ reitor tht cn be exchnged for current ource in prllel with the reitor. Thi pper to be vible exchnge ince we will then hve two current ource in prllel which we cn dd lgebriclly. Performing the exchnge yield the network in Fig. S5.(b). k 6k 6 k 8k 4k Fig. S5.(b) Note tht the voltge ource w poitive t the bottom terminl nd therefore the current ource point in tht direction. Adding the two prllel current ource reduce the network to tht hown in Fig. S5.(c).
5 4 k 6k 8k 4k Fig. S5.(c) At thi point we cn pply current diviion to obtin olution. For exmple, the current in the 4kΩ reitor i I 4k 4 6k k 6k 8k 4k 4 ma Then ( I ) ( 4k) 4k 6 However, we could lo trnform the current ource nd the prllel 6kΩ reitor into voltge ource in erie with the 6kΩ reitor before completing the olution. If we mke thi exchnge, then the network become tht hown in Fig. S5.(d). 4 6k 8k 4k Fig. S5.(d) Then uing voltge diviion 4k 4 4k 6k 8k 6 5. Since the network contin no dependent ource, we will imply determine the open circuit voltge,, nd with the ource in the network mde zero, we will look into the c
6 open circuit terminl nd compute the reitnce t thee terminl, TH. The open circuit voltge i determined from the network in Fig. S5.(). 4 k I 4k 6k I k C Fig. S5.() Note the current nd voltge lbeled in the network. Firt of ll, note tht C Therefore, we need only to determine thee voltge. Clerly, the voltge i I (4k) 6 However, to find we need I. KL round the loop I yield 6k (I I ) ki or 6k (I 4 ) k I ki 4 k 4mA Now C 4kI 8 ki The Thevenin equivlent reitnce i found by zeroing ll ource nd looking into the open circuit terminl to determine the reitnce. The network ued for thi purpoe i hown in Fig. S5.(b).
7 6k 4k k TH Fig. S5.(b) From the network we ee tht the 6k nd k Ohm reitor re in prllel nd tht combintion i in erie with the 4kΩ reitor. Thu TH 4k k 6k 6kΩ Therefore, the Thevenin equivlent circuit conit of the 8 ource in erie with the 6kΩ reitor. If we connect the kω reitor to thi equivlent network we obtin the circuit in Fig. S5.(c). 8 6kΩ kω Fig. S5.(c) Then uing voltge diviion k 8 k 6k 7 5.4 In thi network, the kω reitor repreent the lod. In pplying Norton Theorem we will replce the network without the lod by current ource, the vlue of which i equl to the hortcircuit current computed from the network in Fig. S5.4(), in prllel with the Thevenin equivlent reitnce determined from Fig. S5.4(b). k 6k k I SC Fig. S5.4()
8 k TH 6k Fig. S5.4(b) with reference to Fig. S5.4(), ll current emnting from the ource will go through the hortcircuit. Likewie, ll the current in the ma current ource will lo go through the hortcircuit o tht I SC ma k k If thi ttement i not obviou to the reder, then conider the circuit hown in Fig. S5.4(c). I ISC Fig. S5.4(c) Knowing tht the reitnce of the hortcircuit i zero, we cn pply current diviion to find I SC I SC I I indicting tht ll the current in thi itution will go through the hortcircuit nd none of it will go through the reitor. From Fig. S5.4(b) we find tht the k nd 6k Ohm reitor re in prllel nd thu TH k 6k kω Now the Norton equivlent circuit conit of the hortcircuit current in prllel with the Thevenin equivlent reitnce hown in Fig. S5.4(d). ma kω Fig. S5.4(d)
9 emember, t the terminl of the kω lod, thi network i equivlent to the originl network with the lod removed. Therefore, if we now connect the lod to the Norton equivlent circuit hown in Fig. S5.4(e), the lod current I cn be clculted vi current diviion I k k k k ma k k k Fig. S5.4(e) I 5.5 The olution of thi problem involve finding the Thevenin equivlent circuit t the terminl of the lod reitor L nd etting L equl to the Thevenin equivlent reitnce TH. To determine the Thevenin equivlent circuit, we firt find the open circuit voltge hown in Fig. S5.5(). X k X 6k C Fig. S5.5() We employ the prime nottion on the control vrible x ince the circuit in Fig. S5.5() i different thn the originl network. Applying KL to the left ide of the network yield Then the open circuit voltge i x x x 4 C X 8 ince there i no current in the 6kΩ reitor nd therefore no voltge drop cro it.
4 Becue of the preence of the dependent ource we cnnot imply look bck into the open circuit terminl, with ll independent ource mde zero, nd determine the Thevenin equivlent reitnce. We mut determine the hortcircuit current, I SC nd determine TH from the expreion TH I C SC I SC i found from the circuit in Fig. S5.5(b). X k X 6k I SC Fig. S5.5(b) Once gin, uing KL x x x 4 Then, ince the dependent ource x 8 i connected directly cro the 6kΩ reitor nd I X 6k SC ma TH C 8 kω I SC k Hence, for mximum power trnfer L TH kω And the reminder of the problem involve finding the power borbed by the kω lod, P L. From the network in Fig. S5.5(c) we find tht
4 P L I L L 8 4k. mw kω ( k) I L 8 kω Fig. S5.5(c)
4 CHAPTE 6 POBLEMS 6. If the voltge cro µf cpcitor i hown in Fig. 6., derive the wveform for the cpcitor current. 4 4 6 8 Fig. 6. t (m) 6. If the voltge cro mh inductor i hown in Fig. 6., find the wveform for the inductor current. v(t) (m) 4.. t() Fig. 6. 6. Find the equivlent cpcitnce of the network in Fig. 6. t the terminl AB. All cpcitor re 6µF. A C eq B Fig. 6. 6.4 Find the equivlent inductnce of the network in Fig. 6.4 t the terminl AB. All inductor re mh.
4 A B Fig. 6.4
44 CHAPTE 6 SOLUTIONS 6. The eqution for the wveform in the 4 two milliecond time intervl re lited below. v () t mt b t 4 6 t t t m t 4m 4 t 6m 6 t 8m t <, t > 8m Note tht within ech intervl we hve imply written the eqution of tright line uing the expreion y mx b or equivlently v(t) mt b where m i the lope of the line nd b i the point t which the line interect the v(t) xi. The eqution for the current in cpcitor i i (t) dv( t) C dt Uing thi expreion we cn compute the current in ech intervl. For exmple, in the intervl from t m 6 () t ( ) ma d t dt d dt d dt i 6 () t ( ) ( ) i 6 () t ( ) i ma 6 () t ( ) ma d 4 6 dt i t t t m t 4m 4 t 6m 6 t 8m The wveform for the cpcitor current i hown in Fig. S6..
45 4 4 i(t) (ma) 4 6 8 Fig. S6. t(m) 6. The generl expreion for the current in n inductor i t ( t) i( t ) v( x) i t In order to evlute thi function we need the eqution of the voltge wveform in the two time intervl t. nd. t.. In the firt ce, the voltge function i tright line nd the function pe through the origin of the grph. The eqution of tright line on thi grph i v(t) mt b where m i the lope of the line nd b i the point t which the line interect the v(t) xi. 4 Since the lope i, the eqution of the line i. v 4. () t t where v(t) i meured in volt nd time i meured in econd i.e., the lope h unit of volt/ec. Therefore, i L 4. dx t () t i ( ) dx ince there i no initil current in the inductor i(t) nd L or i t ( t) 4 dx
46 i () t t x.4 dx.4.t A t ma Since the initil current for the econd time intervl i determined by the vlue of the current t the end of the firt time intervl we clculte ( t) i t ma t. t. ma Therefore, in the time intervl. t. i t () t i (.) v ( x) t dx. L Note tht in thi intervl v(x) i contnt m or. Hence, i t () t ( ). ( 4 t)ma If we now plot the two function for the current within their repective time intervl we obtin the plot in Fig. S6.. t. dx. Fig. S6.. t() 6. To begin our nlyi we firt lbel ll the cpcitor nd node in the network hown in Fig. S6.().
47 A C C C C D C 4 C 5 C 6 B Fig. S6.() Firt of ll, the reder hould note tht ll the node hve been lbeled, i.e., there re no other node. A we exmine the topology of the network we find tht ince C nd C 5 re both connected to node D the network cn be redrwn hown in Fig. S6.(b). C D A C C C 4 B C 6 Fig. S6.(b) C 5 C Clerly, C 5 nd C 6 re in prllel nd their combintion we will cll C 56 C 5 C 6. Combining thee two cpcitor reduce the network to tht hown in Fig. S6.(c). C C C A C 4 D C 56 C B Fig. S6.(c) At thi point we find tht C nd C 4 re in prllel nd their combintion, which we cll C 4 C C 4, reduce the network to tht hown in Fig. S6.(d). C C C4 D A B C 56 Fig. S6.(d) C
48 If we now ue the given cpcitor vlue, the network become tht hown in Fig. S6.(e). A 6µF µf B µf Fig. S6.(e) 6µF Strting t the oppoite end of the network from the terminl AB nd combining element we find tht 6µF in erie with µf i 4µF nd thi equivlent cpcitnce i in prllel with µf yielding 6µF, which in turn i in erie with 6µF producing totl cpcitnce of C eq 6µ F 6µ F 4.6µ F 6.4 To id our nlyi, we will firt lbel ll inductor nd node hown in Fig. S6.4(). L L C A L L 5 L 4 L 6 B Fig. S6.4() Note crefully tht ll the node hve been lbeled. Once reder recognize tht there re no other node, they re well on their wy to reducing the network ince thi node recognition provide dt indicting which element re in erie or prllel. For exmple, ince one end of L 4 i connected to node B, the network cn be redrwn hown in Fig. S6.4(b). L L C A L 5 L B L 4 Fig. S6.4(b) L 6
49 Thi digrm clerly indicte tht L nd L 5 re in prllel. In ddition, L 4 nd L 6 re in prllel. Therefore, if we combine element o tht L 5 L L 5 nd L 46 L 4 L 6, then the circuit cn be reduced to tht in Fig. S6.4(c). L L 5 C A L L 46 B Fig. S6.4(c) However, we note now if we did not ee it erlier tht L 5 i in prllel with L 46 o tht the network cn be reduced to tht hown in Fig. S6.4(d). A L C L L 456 B Fig. S6.4(d) Where L 456 L 5 L 46. Since ll inductor re mh, L 456 mh which i in erie with mh nd tht combintion i in prllel with mh yielding L AB mh 5mH 6.66mH
5 CHAPTE 7 POBLEMS 7. Ue the differentil eqution pproch to find i ( t) for t > in the circuit in Fig. 7. nd plot the repone including the time intervl jut prior to opening the witch. kω t kω Fig. 7. kω 5µF 7. Ue the differentil eqution pproch to find i(t) for t > in the circuit in Fig. 7. nd plot the repone including the time intervl jut prior to opening the witch. 5kΩ t kω 5mA kω mh Fig. 7. 7. Ue the tepbytep technique to find v ( t) for t > in the circuit in Fig. 7.. 6kΩ 5µF 6kΩ i ( t) kω i 6kΩ ( t) kω v ( t) 6kΩ t 6kΩ Fig. 7. 7.4 Ue the tepbytep method to find v ( t) for t > in the network in Fig. 7.4.
5 t Ω H 4Ω Ω Fig. 7.4 v ( t) 7.5 Given the network in Fig. 7.5, find () the differentil eqution tht decribe the current i(t) (b) the chrcteritic eqution for the network (c) the network nturl frequencie (d) the type of dmping exhibited by the circuit (e) the generl expreion for i(t) v S ( t) i ( t) H 4Ω.5F Fig. 7.5 7.6 Find i () t for t > in the circuit in Fig. 7.6 nd plot the repone including the time intervl jut prior to cloing the witch. F t 4Ω i ( t) 4Ω.4H 4Ω Fig. 7.6
5 CHAPTE 7 SOLUTIONS 7. We begin our olution by redrwing the network nd lbeling ll the component hown in Fig. S7.() t C i X ( t) 4 i Fig. S7.() In order to determine the initil condition of the network prior to witch ction, we mut determine the initil voltge cro the cpcitor. A circuit, which cn be ued for thi purpoe, i hown in Fig. S7.(b). ( t) 5 i X kω kω v C 5µF 6 4kΩ Fig. S7.(b) Where we hve combined the reitor t the right end of the network o tht 6 4 5 k k 6k 4kΩ In the tedytte condition before the witch i thrown, the cpcitor look like n opencircuit nd therefore v C () i the voltge cro the prllel combintion of nd 6. Uing voltge diviion, the ource will produce the voltge v C ( ) 6 6 k 6 k k Now tht the initil voltge cro the cpcitor i known, we cn find the initil vlue of the current () t. From Fig. S7.(b) we ee tht i i ( ) v C ( ) 6 4k x 6.5mA
5 Then, uing current diviion hown in Fig. S7.(), i ( ) ( ) ( ) i x 4 k k 6k 5 5 ( 6k) ma The prmeter for t < re now known. For the time intervl t >, the network i reduced to tht hown in Fig. S7.(c). i X kω Applying KCL to thi network yield v C ( t ) 5µF Fig. S7.(c) 6 4kΩ or uing the prmeter vlue dv C C dt ( t) v ( t) v ( t) dv C dt ( t) C 9 v C C 6 () t The olution of thi differentil eqution of the form v C t τ () t k k e Since the differentil eqution h no contnt forcing function, we know tht k. τ Therefore, ubtituting v () t k e into the eqution yield C t t τ t t τ τ k e k e 9 nd τ 9 ec.
54 In ddition, ince v C () 6 k e k 6 Thu ecll tht v C t 9 () t 6e i () t i () t x 4 5 5 nd x () t i v C ( t) 6 Then i () t v C t ( t) 6 5 4 5 9 e ma t > ma t < 7. The network cn be redrwn hown in Fig. S7.(). t 5 I S k A 5kΩ k i ( t) ma k Fig. S7.() In the tedytte time intervl prior to witch ction, the inductor look like hortcircuit. Therefore, in thi time period t <, the initil inductor current i i L () I S 5mA At t the witch chnge poition nd hence for t > the network reduce to tht hown in Fig. S7.(b).
55 k i ( t) i mh Fig. S7.(b) k If we let then the differentil eqution for the inductor current i ( t) di L i dt The olution of thi eqution i of the form () t t e τ () t k k i The differentil eqution h no contnt forcing function nd hence k. Subtituting t τ () t k e i into the eqution for the current yield t k τ t t k τ τ k e k e 4 where we hve ued the circuit prmeter vlue in the eqution, i.e., Ω 4k. Thi eqution produce τ vlue of L H nd k 4 τ µ ec. Furthermore, ince i ( ) ma nd i ( ) k e we find tht k 5mA Therefore,
56 ( t) i 5mA, 5e 5 7.5 t ma, t < t > 7. The circuit i redrwn for convenience in Fig. S7.(). 6k 6k C 5µF 4 6k t 6k v ( t) Fig. S7.() Before we begin our nlyi, we note tht reitor nd 4 re in prllel nd o we firt reduce the network to tht hown in Fig. S7.(b). 6k v v C C 5µF k t 6k v ( t) Fig. S7.(b) Now tht the network h been implified, we begin our nlyi 6k v v C 6k k Fig. S7.(c)
57 6k 6 6k v ( ) k Fig. S7.(d) 6k TH 6k k Fig. S7.(e) t e τ Step v () t k k Step In tedytte prior to witch ction, the cpcitor look like n opencircuit nd the ource i directly cro the reitor kω. A hown in Fig. S7.(c) the voltge v cro i equl nd oppoite to v C. Since the voltge of the ource i divided between nd we cn ue voltge diviion to find v hence v C v 6 ( ) v 6 v ( ) Step The new circuit, vlid only for t i hown in Fig. S7.(d). Once gin, uing voltge diviion, v ( ) v ( ) C 4 C Step4 For the period t > 5τ, the cpcitor look like n opencircuit nd the ource i diconnected. With no ource of energy preent in the network v ( )
58 Step5 The Thevenin equivlent reitnce obtined by looking into the network from the terminl of the cpcitor with ll ource mde zero i derived from the circuit in Fig. S7.(e) Then the time contnt of the network i TH (6k) (6k k).6kω τ TH C.8 ec. Step6 Evluting the contnt in the olution, we find tht k v v ( ) ( ) v ( ) 4 k Therefore, t 8 v. () t 4e 7.4 We begin our nlyi of the network with Step The output voltge will be of the form v t τ () t k k e Step In the tedytte prior to the time the witch i thrown, the inductor ct like hortcircuit nd hort out the 4Ω reitor. The network for thi itution i hown in Fig. S7.4(). i L ( ) 4Ω Ω Ω v Fig. S7.4() Under thee condition, ( ) i L i the current from the ource t the left ide of the network, through the hortcircuit, with return pth through the Ω reitor t the
59 output. Wht i the contribution of the ource in the center of the network? No contribution! Why? If we pplied uperpoition nd treted ech ource independently, we would quickly find tht when the leftmot ource w replced with hortcircuit, ll the current from the other ource would be diverted through thi hortcircuit. Therefore, i L ( ) 6A i ( ) Step The new network, vlid only for t 6A L, i hown in Fig. S7.4(b). Ω 4Ω Ω Fig. S7.4(b) v ( ) If we employ uperpoition, we find tht v ( ) 6 ( ) 4 4 4 where in thi eqution we hve ued firt voltge diviion in conjunction with current diviion to obtin the voltge. The two network employed re hown in Fig. S7.4(c) nd (d). 6A 4Ω Ω Ω v ( ) Ω 4Ω Ω v ( ) Fig. S7.4(c) Fig. S7.4(d) Step4 For t > 5τ, the inductor gin look like hortcircuit nd the network i of the form hown in Fig. S7.4(e).
6 Ω 4Ω Ω Fig. S7.4(e) v ( ) A imple voltge divider indicte tht the output voltge i v ( ) 6 Step5 The Thevenin equivlent reitnce obtined by looking into the circuit from the terminl of the inductor with ll ource mde zero i derived from the network in Fig. S7.4(f). TH 4Ω Ω Ω Clerly Then the time contnt i Fig. S7.4(f) TH 4 ( ) Ω τ L ec. 6 Step6 The olution contnt re then k v v ( ) 6 ( ) v ( ) ( 6) 9 k Hence, v 6t ( t) 6 9e
6 7.5 () Applying KL to the cloed pth yield v t C t () t i () t i ( x) dx S differentiting both ide of the eqution we obtin dv S dt di( t) L dt () t di ( t) i ( t) d i ( t) dt C L By rerrnging the term, the eqution cn be expreed in the form d i L dt dt ( t) di ( t) i ( t) dv ( t) dt C S dt or d i dt () t di ( t) L dt i C () t L dv S dt ( t) Uing the circuit component vlue yield d i dt () t di ( t) 7 dt i () t (b) The chrcteritic eqution for the network i 7 dv S dt (c) The network nturl frequencie re the root of the chrcteritic eqution. The qudrtic formul could be ued to obtin thoe root or we cn imply recognize tht the eqution cn be expreed in the form ( ) ( 5) 7 Therefore, the network nturl frequencie re 5 (d) Since the root of the chrcteritic eqution re rel nd unequl, the network repone i overdmped. (e) Bed upon the bove nlyi, the generl expreion for the current i ( t)
6 i t 5t () t k k e k e A where k i the tedytte vlue nd the contnt k nd k re determined from initil condition. 7.6 The network i relbeled hown in Fig. S7.6(). v C ( t) t where ll 4Ω, L.4H nd C Fig. S7.6() C network t three criticl point in time. i ( t) i ( t) L ( t) v F. Conider now the condition of the At t, i.e., the tedytte condition prior to witch ction, the cpcitor ct like n opencircuit, the inductor ct like hortcircuit nd hence v C ( ), i L ( ), i nd v ( ). ( ) At t, i.e., the intnt the witch i thrown, the condition on the torge element (L nd C) cnnot chnge intntneouly nd therefore v C ( ), i L ( ), i ( ) A nd v ( ). At t, i.e., the tedytte condition fter the witch i thrown, the cpcitor ct like n opencircuit, the inductor ct like hortcircuit nd hence v C ( ), i ( ) L A, i ( ) nd v ( ). Now pplying KCL to the network in the time intervl t >, we obtin v () t d ( v ( t) ) C dt L v t ( x) dx v ( t)
v () t i expreing v ( t) in term of ( t) where () t reduce the eqution to i () t 5 di dt ( t) t i i nd uing the component vlue ( x) dx i ( t) Combining term nd differentiting thi expreion yield d i dt ( t) di ( t) dt 5 i () t Therefore, the chrcteritic eqution for the network i 5 Fctoring thi eqution uing the qudrtic formul or ny other convenient men yield, 5 ± j5 σ ± jω Since the root of the chrcteritic eqution re complex conjugte, the network i i t i underdmped nd the generl form of the current ( ) i σt () t k e ( A co ωt B in ωt) 5t k e ( A co 5t B in 5t) where k i the tedytte term reulting from the preence of the voltge ource in the time intervl t. We cn now evlute the contnt k, A nd B uing the known condition for the network. For exmple, nd Therefore, k nd A. i ( ) k A ( ) k i 6
64 We need nother eqution in order to evlute the contnt B. If we return to our originl eqution nd evlute it t time t, we hve dv ( t) t dt where v ( ), the integrtion intervl i zero nd the derivtive function i our unknown. Therefore, dv dt ( t) t 6 or di dt The generl form of the olution i ( t) t.5 i () 5t t e co 5t B in 5t Then di dt () t 5e co 5t e 5 in 5t 5e B in 5t e 5t 5t 5t 5t 5B co 5t nd di dt ( t) t 5 5B Therefore, 5.5 5B or The generl olution i then B
65 i ( t) e 5t co 5t t < t > A plot of thi function i hown in Fig. S7.6(b)..6 I (t) (A).4....4.6.8 t (ec) Fig. S7.6(b)
66 CHAPTE 8 POBLEMS 8. Find the frequency domin impednce Z, hown in Fig. 8.. Ω Z Ω jω jω Ω Fig. 8. 8. If the impednce of the network in Fig. 8. i rel t f 6Hz, wht i the vlue of the inductor L? Ω L Z Ω mf Fig. 8. 8. Ue nodl nlyi to find in the network in Fig. 8.. jω Ω Ω Ω j4ω Fig. 8. 8.4 Find in the network in Fig. 8.4 uing () loop nlyi (b) uperpoition nd (c) Thevenin Theorem. Ω jω jω 4Ω A Fig. 8.4
67 CHAPTE 8 SOLUTIONS 8.. To begin our nlyi, we note tht the circuit cn be lbeled hown in Fig. S8.. Z Z Z Z Fig. S8. In thi ce, Z conit of Ω reitor, Z i the erie combintion of Ω reitor nd jω inductor nd Z conit of jω cpcitor in erie with Ω reitor. Therefore, Z Ω Z jω Z jω Strting t the oppoite end of the network from the terminl t which Z i clculted we note tht Z nd Z re in prllel nd their combintion i in erie with Z. Hence Z Z Z Z Ω ( j)( j) j j 8. The generl expreion for the impednce of thi network i Z jωl jωc In order for Z to be purely reitive, the term However, ince Z LC cn be written Z LC LC j j ωl mut be rel, i.e. jωc Z LC C j ωl ω
68 it i clerly n imginry term nd LC. Therefore, in order for Z to be reitive or ω L ωc L ω C ( 77) ( ) 7.6µ H 8. The preence of the voltge ource indicte tht nodl nlyi i vible pproch to thi problem. The voltge ource nd it two connecting node form upernode hown in Fig. S8.. jω Ω Ω Ω j4ω Fig. S8. Note tht there re three nonreference node, i.e.,, nd. Becue the voltge ource i tied directly between node nd,. Thi contrint condition i one of our three eqution required to olve the network. The two remining eqution re obtined by pplying KCL t the upernode nd the node lbeled. For the upernode, KCL yield j j4 At the node lbeled, KCL yield Therefore, the three eqution tht will provide the node voltge re
69 j j 4 Subtituting the firt eqution in for the two remining eqution nd combining term yield j 4 5 j6 Solving for in thi lt eqution nd ubtituting it into the one bove it, we obtin nd hence (.4 j.5).4 j6.57 6. 8.4 () Since the network h two loop, or in thi ce two mehe, we will need two eqution to determine ll the current. Conider the network lbeled in Fig. S8.4(). Ω I jω 4Ω jω I A Fig. S8.4() Note tht ince I goe directly through the current ource, I mut be A. Hence, one of our two eqution i I If we now pply KL to the loop on the left of the network, we obtin ( j ) ( I I ) ( 4 j) I
7 Thee two eqution will yield the current. Subtituting the firt eqution into the econd yield nd then Finlly, ( j 4 j) ( 4 j) I j4 I.5.85 A 6 j ( I I ) 4 j4 4 6 j 5.4 4.57 (b) In pplying uperpoition to thi problem, we conider ech ource cting lone. If we zero the current ource, i.e., replce it with n open circuit, the circuit we obtin i hown in Fig. S8.4(b). Uing voltge diviion Ω jω Fig. S8.4(b) 4Ω jω 4 j 48 6 j 4 j Now, if we zero the voltge ource, i.e., replce it with hort circuit, we obtin the circuit in Fig. S8.4(c).
7 jω Ω I X 4Ω jω Fig. S8.4(c) A Employing current diviion, the current I X i I X 4 j A 6 j j j 4 j Then, 4I X 6 j8 6 j And finlly, 48 6 j8 6 j 6 j j8 6 j 5.4 4.57 (c) In pplying Thevenin Theorem, we firt brek the network t the lod nd determine the opencircuit voltge hown in Fig. S8.4(d). Ω jω jω A C Fig. S8.4(d)
7 Note tht there exit only one cloed pth nd the current in it mut be A. Note lo tht there i no current in the inductor nd therefore no voltge cro it. Hence i lo the voltge cro the current ource. Hence, C 8 j ( j) The Thevenin equivlent impednce found by zeroing the independent ource nd looking into the network t the terminl of the lod cn be determined from the circuit in Fig. S8.4(e). Ω jω jω OC Z TH Thi network indicte tht Fig. S8.4(e) Z TH j j jω If we now form the Thevenin equivlent circuit nd reconnect the lod, we obtin the network in Fig. S8.4(f). C 8j Ω Z TH jω 4Ω Fig. S8.4(f) Applying voltge diviion yield 4 ( 8 j) 4 j8 6 j 5.4 4.57 j
7 CHAPTE 9 POBLEMS 9. Determine the verge power upplied by ech ource in the circuit in Fig. 9.. jω jω Ω A Fig. 9. 9. Given the circuit in Fig. 9., determine the impednce Z L for mximum verge power trnfer nd the vlue of the mximum verge power trnferred to thi lod. jω Ω Ω 6 Fig. 9. ZL 9. Clculte the rm vlue of the wveform hown in Fig. 9.. v(t) () 6 4 5 6 7 t() Fig. 9. 9.4 Determine the ource voltge in the network hown in Fig. 9.4..Ω j.5ω 6 kw S 4.85 pf lgging Fig. 9.4 rm 4 kw.78 pf lgging 9.5 A plnt conume 75 kw t power fctor of.7 lgging from 4 rm 6 Hz line. Determine the vlue of the cpcitor tht when plced in prllel with the lod will chnge the lod power fctor to.9 lgging.
74 CHAPTE 9 SOLUTIONS 9. Becue the erie impednce of the inductor nd cpcitor re equl in mgnitude nd oppoite in ign, from the tndpoint of clculting verge power the network cn be reduced to tht hown in Fig. S9.. I S Ω Fig. S9. I CS A The generl expreion for verge power i P I co ( θ θ ) In the ce of the current ource, I CS A, θ nd θ I. Therefore, the verge power delivered by the current ource i I P CS 8.66 W ( )( ) co ( ) In order to clculte the verge power delivered by the voltge ource, we need the current I S. Uing KCL I S or I S 8. 6.9 A Now P S 4.4 W ( )( 8.) co ( ( 6.9 )) Therefore, the totl power generted in the network i P T P CS P S 5 W
75 Let u now clculte the verge power borbed by the reitor. We know tht the verge power borbed by the reitor mut be P m 5 W In ddition, the verge power borbed by the reitor cn lo be determined by P I m However, we do not know the current in the reitor. Uing KCL. Now I I S I CS 8.66 6.9 A P 5 W ( ) ( ) Thu, we find tht the totl verge power generted i equl to the verge power borbed. 9. We will firt determine the Thevenin equivlent circuit for the network without the lod ttched. The opencircuit voltge, C, cn be determined from the network in Fig. S9.(). jω Ω 6 Fig. S9.() Ω C I Thi opencircuit voltge cn be clculted in number of wy. For exmple, we cn compute the current I
76 I ( ( 6 )) j 8 A j Then uing KL, C I 6 6j j or, we could ue voltge diviion to determine the voltge cro the Ohm reitor on the right, i.e., Then, once gin [ ( 6 )] 8 j j C 6 6j j 9.49 7.56 The Thevenin equivlent impednce i obtined by looking into the opencircuit terminl with ll ource mde zero. In thi ce, we replce the voltge ource with hort circuit. Thi network i hown in Fig. S9.(b). jω Ω Ω Z TH Fig. S9.(b) Note tht the Ohm reitor on the left i horted nd thu the Z TH i Z TH ( ) ( j) j j Ω j Ω j
77 Hence, for mximum verge power trnfer or * Z Z L TH Z L j Ω Therefore, the network i reduced to tht hown in Fig. S9.(c). j Ι 9.49 7.56 j Then Fig. S9.(c) 9.49 7.56 I j j 9.49 7.56 A nd the mximum verge power trnferred to the lod i P L ( 9.49) 9 W 9. In order to clculte the rm vlue of the wveform, we need the eqution for the wveform within ech of the ditinctive intervl. In the intervl t, the wveform i tright line tht pe through the origin of the grph. The eqution for tright line in thi grph i v(t) mt b Where m i the lope of the line nd b i the v(t) intercept. Since the line goe through the origin, b. The lope m i
78 6 m Therefore, in the intervl t, v(t) t The wveform h contnt vlue in the intervl t nd t 4, i.e., v(t) 6 v(t) t t 4 Since the wveform repet fter 4, the period of the wveform i T 4 Now tht the dt for the wveform i known, rm T 4 v () t dt Therefore, in thi ce rm 4 4 4 ( 5) 4 ( t) dt ( 6) dt ( ) [ t 6t ] ( 4 6).87 rm dt 9.4 We begin our nlyi by lbeling the vriou current nd voltge in the circuit hown in Fig. S9.4. I S S.Ω j.5ω I I 6 kw 4 kw.85 pf.78 pf L 4 rm lg lg Fig. S9.4
79 Our pproch to determining S i tright forwrd: We will compute the current I nd I ; dd them uing KCL to find I S ; determine the voltge cro the line impednce nd finlly ue KL to dd the line voltge nd lod voltge to determine the ource voltge. The mgnitude of the current I i I L P ( pf ) 6, 4 ( )(.85) 94. A rm. And the phe ngle i θ I co.79 (.85) The negtive ign i reult of the fct tht the power fctor i lgging. Thu I 94..79 A rm. The mgnitude of the current I i I L P ( pf ) 4, 4 ( )(.78).68 A rm. And the phe ngle i θ I co 8.74 (.78) Thu I.68 8.74 A rm. Uing KCL
8 I I I S 94..79.68 8.74 54. 4.5 A rm. Then I ( ) S. j.5 4 ( 54. 4.5 )(.5 78.7 ) 57.4 44.44 4 46.7. rm. 4 9.5 Since the originl power fctor i.7 lgging the power fctor ngle i Then θ OLD co (.7) 45.57 Hence Q OLD P OLD tn θ OLD 75, tn 45.57 76.5 kvr S OLD 75, j76,55 7.4 45.57 ka The new power fctor ngle we wih to chieve i Then θ NEW co (new power fctor) co (.9) 5.84 Q NEW P OLD tn θ NEW 75, tn 5.84 6,4 kvr Now the difference between Q NEW nd Q OLD i chieved by the cpcitor, i.e., Q CAP Q NEW Q OLD 6,4 76,55 4,9 kvr
8 And ince Q CAP ω C Then 4,9 C ( 77)( 4) 85.8µ F
8 CHAPTE POBLEMS. Find in the network in Fig... Ω jω jω A Ω jω jω Ω Fig... Determine the impednce een by the ource in the circuit in Fig... Ω jω jω Ω j4ω jω Ω jω Ω jω Fig... Determine I, I, nd in the circuit in Fig... Ω : Idel Ω I I Fig...4 Given the circuit in Fig.., determine the two network obtined by replcing () the primry nd the idel trnformer with n equivlent circuit nd (b) the idel trnformer nd the econdry with n equivlent circuit.
8 CHAPTE SOLUTIONS. Our firt tep in the olution of thi problem i to pply ource trnformtion to the leftend of the network nd trnform the A ource in prllel with the Ω reitor into ource in erie with the Ω reitor hown in Fig. S.(). jω Ω Ω jω jω Fig. S.() jω Ω Let u redrw the network hown in Fig. S.(b). jω jω Ω I Ω I Fig. S.(b) The eqution for thi network re I I ( j) We now write the eqution for the mutully coupled coil. In order to force the vrible in thi circuit into our tndrd form for mutully coupled inductor, we mut revere the ign on, I nd I. Therefore, the eqution tht relte nd to I nd I, in thi ce, re Combining the eqution yield j(i ) j(i ) j(i ) j(i ) ( j) I j I j I ( j) I Solving for I in the econd eqution nd ubtituting it into the firt eqution yield or [( j) ( j) j] I
84 I j.894. A And finlly I.894.. Let u firt determine the totl impednce on the right ide of the circuit hown in Fig. S.(). Ω A the figure indicte Z L Z L jω Ω Fig. S.() 6 jω ( j) ( j) ( j)( j) j j jω The originl network cn now be redrwn in the following form hown in Fig. S.(b). jω Ω jω 6Ω j4ω jω I I jω Ω Fig. S.(b) The two KL eqution for the network in Fig. S.(b) re (4 j) I (6 j) I
85 In order to force the vrible in thi circuit into our tndrd form for mutully coupled inductor, we mut revere the ign on. Therefore, the eqution tht relte nd to I nd I, in thi prticulr ce, re j4 I j I j I j I Combining ll of thee eqution reult in the following two eqution. (4 j) I j I j I 6 I Solving the econd eqution for I nd ubtituting thi vlue into the firt eqution yield 4 j I 6 Then, the impednce een by the ource i Z S 4.67 jω I. The KL eqution for thi network re ( ) I I If we now force the vrible in thi circuit into our tndrd form for the idel trnformer, we mut revere the ign on nd I. Therefore, the eqution tht relte to nd I to I, in thi prticulr ce, re I ( I ) Solving the lter eqution for nd I nd ubtituting thee vlue into the firt eqution yield Solving thee eqution produce I I
86 Then, the trnformer reltionhip yield I.4 8 A.4 8 I I Therefore, I.57 8 A.84.4 A hown in the previou problem, the idel trnformer eqution re I ( I ) Thee two eqution nd the eqution for reflecting impednce from the primry of the trnformer to the econdry i.e., Z p N N ZS 4 Z S re the necery eqution for developing the equivlent circuit. () If we reflect the primry to the econdry, we note tht And Z S 4Z p Therefore, the voltge ource in the primry become ( ) 8
87 And Z S 4() 4Ω Therefore, the equivlent circuit in thi ce i hown in Fig. S.4(). I 4Ω Ω Fig. S.4() (b) Once gin, uing the idel trnformer eqution to reflect the econdry to the primry we obtin the network in Fig. S.4(b). Ω Ω 4 Fig. S.4(b)
88 CHAPTE POBLEMS. In threephe blnced wyewye ytem, the ource i n bcequence et of voltge with n 4 rm. The per phe impednce of the lod i j8ω. If the line impednce per phe i.6 j.4ω, find the line current nd lod voltge.. An bcequence et of voltge feed blnced threephe wyewye ytem. If n 44 4 rm, AN 4 9 rm nd the line impednce i.5 j.ω, find the lod impednce.. In blnced threephe wyedelt ytem, the ource h n bcphe equence nd n rm. The line nd lod impednce re.6 j.4ω nd 4 jω, repectively. Find the delt current in the lod..4 A blnced threephe ource erve two lod: Lod : ka t.85 pf lgging. Lod : ka t.6 pf lgging. The line voltge t the lod i 8 rm t 6Hz. Determine the line current nd the combined power fctor t the lod..5 In threephe blnced ytem n bcequence wyeconnected ource with n rm upplie power to wyeconnected lod tht conume 6 kw of power in ech phe t pf of.75 lgging. Three cpcitor, ech with n impednce of j.ω, re connected in prllel with the originl lod in wye configurtion. Determine the power fctor of the combined lod een by the ource.