Magnetically Coupled Networks: Problems with Solution, Exercises of Electronic Circuits Design

Download Magnetically Coupled Networks: Problems with Solution and more Exercises Electronic Circuits Design in PDF only on Docsity! Chapter Ten: Magnetically Coupled Networks  Chapter Ten: Magnetically Coupled Networks 831 gee 10.8 Given the network in Fig. P10.3, 0 © (a) find the equations for v,(1) and v,(4). () find the equations for v.(7) and w,(1). . M ye I(t) /\ in(t) bie. =P . + + _ : @ : vt) Lys § Ly p(t) vg) * - _ + Figure P10.3 a SOLUTION: : ay Ueuds - bide, Madde Uy lee -thddy - Lg de ak ae chic ae \ by vets Ly day pM dae Valth= MA ply dae ala dh dk dk 832 Irwin, Basic Engineering Circuit Analysis, 8/E 70.4 Given the network in Fig. P16.4, (a) write the equations for v,(1) and v,(t \ je (b)} write the equations for v.(¢) and v,{r). nO AK ato ~ - + A. oft) Ly: EL vet) vf) a as e + + ~ Figure P10.4 SOLUTION: AY Dealt s - ld LM de Vy = - Ma . be dle ote aE ook chk by ow ikys uy Ak fae Hay, ug ys Mate the dae Chapter Ten: Magnetically Coupled Networks 833 10.5 Find the voltage gain Vv Fig. PI0.5, <8 /V 5 of the network shown in OF 20 WM ARO so jee Lig Ae it | -O + te ok sjso J Vs © fo™ 2g so™ 205 Vo Lh ef ‘le L NZ . ‘ _ oO Figure P10.5 SOLUTION: | We Ble Q gah bor Re RCea5) Ver 2% OTs Ver jete oe ye stieF) 2G, CeO) A4Ke “ees . 3 . Fy? Ys. Ve = f2Ns_ Wes Ae. Gry Sry Vs SYS 836 70.8 Find V, in the circuit in Fig. PLO.8. irwin, Basic Engineering Circuit Analysis, 8/E NG 20 vo 190 AA Ae 6 y s rye . oo y ro: v(4) ~ jea co 198 Vv, \ . xr, [ -O Figure P10.8 SOLUTION, VS Rlzyesco TYE, =» Ips ve¢ieime A Chapter Ten: Magnetically Coupled Networks 837 10.9 Find the voltage gain V,/Ws of the network shown in Fig. P10.9. No QE, FP Nget Cig pt, # -[2t+T(styeieo Let Vee I Let¥. Then Ve Vp sar. 1. bese 42 e, \ le ee ie we ls ‘] 2 T= oi LSude A Lose sayz PEs 183 838 irwin, Basic Engineering Circuit Analysis, 8/E 70.970 Find V, in the network in Fig. P1O.10. rafoev(S ad i) a i) 2 « Figure P10.70 SOLUTION: Chapter Ten: Magnetically Coupled Networks 844 10.12 Find V, in the network in Fig. P10.13. oo. po 10 to oo i Wh o + oy ft : joforv (+ 192 203 § pa Vv 4 2 je g fia o Oy °4 + 20 \ rz O Figure P10.13 SOLUTION: E,Cpej + jtexo 4 ~so]0° =X, + tl say) 4 Vou 20, 41019 les 5} x, 6 +5 J 1 : =) ts 2.a3/ 72° A gio By UFe} bL-e ~e om 842 Irwin, Basic Engineering Circuit Analysis, 8/E 16.14 Find V, im the network in Fig. PLO.14, : : so ; AM hw PN io. r Ha | ° : . Ft . : Ce | | toev(t) ase! pp fod po A 208 y, t Ree, | » Fe e Ly ; i i bonne ne new wnewenwuwed Figure P10.14 SOLUTION: ese . +f aa ? yee, +E (+f Ls =O € pte +35 ( 3ej1s o r | |» Ver OE, rz tle % om lo 4 Wes ete bouet v | ce ee Chapter Ten: Magnetically Coupled Networks 843 10.15 Find ¥,, in the network in Fig. PIQ.15. ofa Figure P10.13 SOLUTION: ZelePs Tlew) -2% f -et en (ey) -ptye f mr (zysjeo r te oe 24“\ 2 © 2 Vo HC) -2 eye aft op j}9 2.47 | o 4h 24s fo | eee ]\e> ZA7 is 846 Irwin, Basic Engineering Circuit Analysis, 8/E wenger VO.4@ Find I, in the circuit in Fig, P10.18. 4 jaa yio 20 3¢ 200 ri K j ¥ « s fooak f j2o 32, ev) Oo fq iy x, =, 20 Figure P10.18 SOLUTION: B2beu TAS thse) - 45. -2T = 32f0= 4 (452) 2X, (-2- fh) Oe-T Gh 256 R(ztegues) @ OF TCP) Ele) Chapter Ten: Magnetically Coupled Networks 847 10.19 Write the mesh equations for the network in Fig. P10.19. Ry lors oo oo “1 0 — PPB t \ i ib viG) NY joly§ NW folo [fk iam" Figure P10.19 SOLUTION: ° al Nome ~ Ny T, CB, 45 4Ly) jews, Ge 7 yum Dye Ty (jaeyritts) 848 trwin, Basic Engineering Circuit Analysis, 6/E 10.20 Write the mesh equations for the network in Fig. P10.20. _! a jot fol Ie a Vv oO ~ Joli lc oo Tw "lena dad Figure P70.20 SOLUTION: = rq, (je- 4% 2) - jouk, ~ yuk Ey Oe 7 femtys x, ( ju (Lith, ) “sce ) ~ eb yy Chapter Ten: Magnetically Coupled Networks 851 78.22 Write the mesh equations for the network in Fig, P10.23. Figure P10.23 SOLUTION: ~ . , -y . ve wy Rt \ Liabytishen “ire, ) [= just at - peks Ts feg. i) + juts (Fy-t,) + jury @.-z,) vyum, (Py 52) vjems (2, -1y) 52 tl (ut ew, Yee Vewl t -:el.t oe Ty Li (wbyrwl,- Koc, Pprgeli © - pel, w Cege) ee ce ee i . foe ae yee, CFT) t jbom (Ty Fs) xjomy (rr) 4 JemgiZ,-¥, )  . ot ve 0 [e+ Cralytisky Zest 3 “2D +H [ye ( Piss m-m a) | r 4. + Eg] mm eMs obs | ie oe t Vm (mye fy -,- uy eel (Cit, tam) ~ Yooh +t] 4 (+ My- Mans -i2) | ae on h Lye yey $y us) + Tp] ja(- PonPgndty Ly) | +Ey | be & je (larly + zma) | 852 Irwin, Basic Engineering Circuit Analysis, 8/E 10.24 Find V, in the network in Fig. P10.24. JZ feo uv eo, ic ° i. ® fh xn 228 A fen : £/22 7 fia Vo x x 4/0°A - O : Figure P10.24 SOLUTION: - Chapter Ten: Magnetically Coupled Networks 855 {0.27 Find V, in the network in Fig. P10.27. 8S¥ fig 10 joo fo p20 Pe 7 v ; o yn ~j2.2 © 24/O°V fio Xe 19% Vo 2 i | 5 Figure P10.27 SOLUTION, tti—<‘i—SSS—S Oe - zel@eOCpayy +t & we Oe 856 lrwin, Basic Engineering Circuit Analysis, B/E 10.28 Find V, in the network in Fig. P10.28. ©5 joo m 6 + @) 100/0° V 200/0° V e J e rn fy jsQ 3d 2 jan p Vo | + cy “ ' 100 ‘ —yid 19) 3 Figure P10.28 SOLUTION: tao [@ - 2e0l/@ = T lito js) 4+jleT, = -loo 1% Zoo 12 = jot +t, (-js)- 422, 0 + ~§e Tarts (2452) | Chapter Ten: Magnetically Coupled Networks 857 10.29 Find V, in the network in Fig. P10,29. ; fio [j/--= iw ° @ 1a/orv oe 24 /30°V wf, y a joo? © joo A 10 Vo / d - s e e Seat: jan q 3 Sp Q J te. L Oo Figure P10.29 SOLUTION: ~ sa LF 2b [Bes (2 ape) pyr, 2-878 yr ZH [52° 44+ EB (itpe) -j Ty Ge Ve Or -jT,+Fy (e059 zeje “yh oo Va] [278 “jie <j Leja - rE’ ae rie 6 Hi ee “lle 2 ~ : ™ — ~ 7 Tazuesleve qf we ies Lowey] \ Neen 860 Irwin, Basic Engineering Circuit Analysis, 8/E 10.32 Find L, im the circuit in Fig. PIO.32, jo x NAG, e . e « Mg A. fj2a § j2a 4 by “ (1) 4/PA S49 | ph q £09 ’ S10 4 By | Figure P710.32 SOLUTION: Tye4le $F ~ Xe Rlaeje)-jTe Tso ard Oe Fptel- a) * x, { z +52) \ 8 & [, | 4 | wt 24yh -behh Fert} po | O -t-jh Bai? x, | | Chapter Ten: Magnetically Coupled Networks 861 70.34 Find I, in the circuit in Fig. P10.33. fio ha 4A | et. ; HK j2o0% : ; lf. K , in %. , e ; Ss 4/o° A D 3 S12 1 "a. gia Figure P10.33 SOLUTION: Teglea F os ToGo) 4 Ts (j!+j2) Tye, 862 Ipwin, Basic Engineering Circuit Analysis. 8/E 70.34 Find V, in the network in Fig. P10.34. 2 fi ; Non t \ : 3 . NO? sya C) 4frA $10 S10 @ V, | 20, SOLUTION: {se Scura trans femating [ere nim BaBattins’ + A ¥ Figure P10.34 Voz qt Cee Loe Alor Ty Cass )-TaCz4ji0 & -¢ les -T, lass) +5, (3) Bejlo -e-fl yy 4 = ») The, -196° rs 3 le “b pe tssJoine7 A Chapter Ten: Magnetically Coupled Networks 865 1G.57 Determine the impedance seen by the source in the net- work shown in Fig. P10.37, : f2o ~j10 12 +o - . ® 2G a fi _ s 90 42/0°V *) ry Oy a j20n jen a y | x, * tL, Feo Figure P10.37 SOLUTION: 32 lo = El Lage} - (2X, f o- ar Eps] be Ly And yeh, +8 lr gsileo oe ©, an alizds A LS) Le] 866 Irwin, Basic Engineering Circuit Analysis, 8/E 10.38 Determine the impedance seen by the source in the net- work shown in Fig. P10.38. Ao —AO ‘ i ds hed h, Ses = my we aforv (+) A it Figure P10.38 SOLUTION: Z4¢lo2 DCieg- je, & ox Fer (G) +t anh ATge Ts (lejed so @p= z4le /x, eh pho TR) ea) yj) jh Uh FP fefo fe Tae TREES HEA _6 jh tegen le] Lo Pees 1 sos Aes ene rie ann gh Chapter Ten: Magnetically Coupled Networks 867 4 5 SS 3% Determine the input impedance Z;, in the network in Fig. P10.39. Aro : weo™N, A oe o @. s. 10 fA® faa Lin ; / . ~y2 0 18 3.0 Oo Figure P10.39 SOLUTION, é 4 ¢ q £ Suk Veni le? V , 55 bab eo rie Bau | . Ue | aus We, Ve am Cimjed jet, . mp4 : ete, ay ty en eee Way | OF VENI A Sg tT 52 jt 2 ag belo oO = jt, +) Ey, + Ty (a+53) 5 | 3 je Bays =, | io | . 4} | @7 96 Ee 6.43 ? i 322° 870 irwin, Basic Engineering Circuit Analysis, 8/E 70.42. Analyze the network in Fig. P10.42 and determine whether a value of X¢ can be found such that the out- put voltage is equal to twice the input voltage. 19 pea oe 8 ore, ep fle fon © po Finca Vo L 3 Figure P10.42 SOLUTION: Ve DCisse)-j20, yt Vie Sy (tage) ~§%, OF wpekie F (5 l4-x)) 4) 7+ {CD-R Y2 of fT yt, Cenk) wy Te Tz C2-Ke) me, Ve afer) -p] 9 Vem [anes tsa] Bur, Ves -j%e Ty Sey Ve. [ke 4 = 2+i§o V 2-% 44 (3-24) jt. = Zt5e 0 Hjl3-*) Te faves. 2-450 K=2 and Ve. Chapter Ten: Magnetically Coupled Networks 871 T@.42 Two coils in a network are positioned such that there is 100% coupling between them. If the inductance of one coil is 10 mH and the mutual inductance is 6 mH, compute the inductance of the other coil, =" SOLUTION:  872 trwin, Basic Engineering Circuit Analysis, 8/E if i ® 1.44 The currents in the nctwork in Fig, PIG.44 a a wy Lys w be alt) 10 ‘ S % i Figure P10.44 oF oshys PSE SL Ke Fhe = 7s mMakVEE, = eH wM= Gosst Ty = be {-Bo mA i Dz 2 20 [45 mA Chapter Ten: Magnetically Coupled Networks 875 76.47 Determine the energy stored in the coupled inductors in Problem 10.46 ait = ims. ©° SOLUTION: wy itis £ LL, Ale) + dba ty LE) #4 1) 4) U8) “by ible F dee (377 b~ te%) mA UE) den (897E- 509) HA Lyx Bia Laci kote Ss Meo ped Atk testing, leis Pooma fz Ce)= 3,52 mA 876 Irwin, Basic Engineering Circuit Analysis, 8/E 70.48 Determine 1,. 1,, Vy, and V> in the network in Fig. P10.48. 1O if a4 lb Af — , = + + s 12/30° V © Vi oe Vv S10 * Ideal Figure P10.4& SOLUTION: ~ 2) Be 2 COE, ey, $ i, sOT, 4 -Vie avy $ 2,2 -%, tg 48s DT, -B¥, ee Lp -2r,2= 2 FUE, r)« i2Je, [ey GE ATE) [vz 5 48 jo) S0° v] ere Chapter Ten: Magnetically Coupled Networks 877 10.4% Find all currents and voltages in the network in Fig. Pi0.49. Aaa ja ideal Figure P10.49 SOLUTION: 422 ate i2ie FOE, FV, ¥ Yo= Z20se70 $ VYee2¥, # 2,2 z%, Irwin, Basic Engineering Circuit Analysis, 8/E 880 V> in the network in @.52 Determine ¥,, 14, V). and 4 & Fig. P1052. 20 ty > b 40 AAP 132 NA + + a e 5 i . . a, 10/30° V ©) “fia Vv 3 I: Ve 3 ja Ideal Figure P10,52 SOLUTION: yo 2 24 tole se Tale +) Ty oa [Tq Ty( 5 Tas joe jo L3o= Ip (pa jeaide i (i453) Y= Tp Cte «~ Ipa/ Ts 3. ly Lethe? 4 | - TAs 1.58 138% 4 | Chapter Ten: Magnetically Coupled Networks 887 70.53 Determine I), I,, Vy. and V2 in the network in Fig. P10.53. Y, 10 x ona? Aime 10 + + Lb © oy it 20 : av) V1 iI V> ‘ CQ iaia @ Ideal Figure P10.83 SOLUTION: eau ~v2 2 1 Mee Vv = x 4 a sea eh dus ya “210 = Te(44+l42) 420112) tDe-nt, Tgzo.si J 10° A Tz + Te +/ Le Y= YA Emotes Lora 882 irwin, Basic Engineering Circuit Analysis, 8/E 10.84 Find Fin the network in Fig. P1054. 2:4 aa | 12 1 S40 2 el. 245" A q [ il: oe jr ® + oorv 5 AQ e ro 1 , Ideal Figura P1064 SOLUTION, n= '% fa ( (a PO — a = A die +htsa CG) i as -jia G) ussty ' a jie t Ll fe 4 tefhe hth 4p La - Ahem mony Ive 4 nO i ah ak as q Cp LisePVy tr me My _ e “e Le Chapter Ten: Magnetically Coupled Networks 885 40.57 Determine I,. In, V,, and V1 im the network in Fig. P10.57. Figure P10.87 SOLUTION: ~ : 4A. Ae Ty, . AN manne | wad aaiey (E) Yo = yon, G)eqie® Te, = 48 lo- 24 Le ~ £Ft8 6 4, Loo A tot oa Ve» 46/0 -ABI, = Nz= 259 f° V¥ [ Tot ep Let yeh LeV | ig = > O.% / OPA Vee 25.9 fo 886 Irwin, Basic Engineering Circuit Analysis, 8/E 10.28 Find V, in the network in Fig. P10.S8. ideal ! \ tf It : iv i o ta jo * ) I* feo + . HE Wise Ve 3 * a Ty, B20 | 5 Figure P10.58 SOLUTION: - 4,2; TT, yeve ‘ e410 = T, (ape) 2m ev, 1 abies TlinpOey, my Vee ret eala-j J W,2 Tle jn) z¢joeu, lich) +t leyd= Tlsys) & - Pg: £2 3ce Leva - } Vy = 2, \ No 2i32 £45" «| : Chapter Ten: Magnetically Coupled Networks 887 18.59 Find the current [in the network in Fig. P1059, (@< ti2 ig ANS | QO QO 220 [ a , ¢ e . a aeasa 4) 2 40 I) 120° v ee : me NP £f2Q é [ I ldeal Figure P10.59 SOLUTION: neg 2 ok - tuVe fov() Che, hi o & t * aj 8S0 irwin, Basic Engineering Circuit Analysis, 8/E 10.62 Find V, in the network in Fig. P10.62. fy 4 J {é $———O 40 ~ja a + @ J é sev (2) aq sas, * if * L O N ideal 2 — Ideal —je o Lohr & 62% jt29 rr, Figure P10.62 SOLUTION; ™,= ‘2. Mee ta ~S8e We | | ‘ re le O én ¢ Ne ]|}- “jen S4a jt8e Vy 27 itele Lg) | ve 1.44 Lizge v 1 ebt2d 44( 49-8 - 2) ~ Chapter Ten: Magnetically Coupled Networks 891 10.80 Find V, inthe circuit in Fig. P10.63. © 8 W122 iL K nN ° 20 4 “ia ine A 19s Vv, 240° V © 205 £20 Figure P70.62 SOLUTION: First, a. Singhal, an. 2a. . in aa ae [aie ahh 2 ~ ax i LG. wbod Zz, zésk i. . a pe B22 (2Ho a eb fot Pate 452 _S ga -~j8 ~-}' Co [afte Lea 2tdovV 894 irwin, Basic Engineering Circuit Analysis, 8/E 76.66 Determine the input impedance seen by the source in the network shown in Fig. P10.66. Luh 434 i 1. 2:1 200 4h 140 402 6 @ ® 8 “ (+ ate ane sO Oh iI i203 I 3 Ideal ideal Figure P70.66 SOLUTION: Ry) « YF ng ete ZO os | t Zot 42 and J ee . . OL. y Py 5 die s boa Zo = S26a J LA ss Ve s Chapter Ten: Magnetically Coupled Networks 895 10.87 Determine the input impedance seen by the source in q ip y the network shown in Fig. P10.67. “fia jt a Vt 2:1 ) K e € e ‘. ad 48.0 = v2aev(t) il: ile q WS —J32 O aE 12 AY ideal ideal Figure P10.67 SOLUTION: alu 896 irwin, Basic Engineering Circuit Analysis, 8/E L L i i 10.58 The output stage of an amplifier in an old radio is to be matched to the impedance of a speaker, as shown in Fig. P10.68. 1f the impedance of the speaker is 8 O and the amplifier requires a load impedance of 3.2 kQ, determine the turns ratio of the ideal transformer. SOLUTION: ny be . = Be # Ws $200 SPERKER. SPE AE 5 nit fase too | [yk ae | i  Chapter Ten: Magnetically Coupled Networks 893 70.77 Given that V, = 48 /30° V in the circuit shown in Fig. PIO.71, determine Vs. 2% i, 69 Tee ey If : » ~O + + + i @ yr 2 : ¥ s§ Vy 2 |: V> 240 > V, : | O ideal Figure P'10.71 SOLUTION: - t V2=Ve = 48/80" Vie -Vp = - 24) 30% = 2h [-ise®V e Tae Ve fey 2LE A Eye-e Ty a-4 [Be A 2 4 (aye Vs = t (6-56) +¥, [ vss 587 (-177 V 906 lrwin, Basic Engineering Circuit Analysis, 8/E 70.72 In the circuit in Fig. P10.72, WT, = 4 {30° A find V,. Vo — : jen Ah ne 20 4G ° e Vs © =He aad Ideal Ideal Figure P10.72 SOLUTION: Be 2 + Ve [t AN pe Minter a Bel ge ~y 4.52 25 £ ~ f” oe f Vg = ty (ieyh= 4¥e Lois’ v peepee  Chapter Ten: Magnetically Coupled Networks 907 10.73 In the nerwork in Fig. P10.73, if ¥, O° A, find V5 A 2:1 it 4:2 ig . of “flO * e —j40 : . ZS n Ge) ‘if S19 |: gos 7 ‘ ® i, ideal Ideal Figure P10.73 SOLUTION: - - ~ pst he “je Vy Oe Ty, “4 i Ana erl ee B@ 3 Fan, 2 Tye] Ve lOT 2 eLety Te ts SAG = pee 4 Togs D+ td, = 7-42/2.73° a 904 Irwin, Basic Engineering Circuit Analysis, 8/E TOPE-' In the network in Fig. 10PFE-1, find the impedance seen by the source, 08 M ety . nN 40 100 mF & e 24008 (2f + O°) V & 1H 4H Sided Figure 70PFE-1 SOLUTION: ig = z ris ie ke YUE, = tH a ow SPM si pe Fe 24lev © a jen fen QQ Bes Lo a 2¢ fos DT (4ejpey jet, > (“2 le Pet] cs 47 O 4 -setyeT, (s4j3) “je segy SUS] [>| Eye 4.92 fia am Lin ? Chapter Ten: Magnetically Coupled Networks $05 YOFE-2 Jn the circuit in Fig. 1OPPE-2, select the value of the transformer’s turns ration = N>/N, to achieve impedance matching for maximum power transfer. Using this value of n, calculate the power absorbed by the 3-2 resistor. Ny : N> yo ft [32.0 : 4gQ fs2e ~j2 0 e & woaev(®) i ns Ideal Figure 10PFE-2 —o me - / SOLUTION: agnta j3en? -y2 | Bye | We rn | — Vaz l2on {O° y i Vv Bau Zz ee EL a a ee tramsthe, tn = SQ pe Ent 4 i (Sane - 2\=3 J +4 ne la > Ory 2 $40 45 (24 -2y= Ba rl is aed H aa 906 Irwin, Basic Engineering Circuit Analysis, 8/E VEPE-S Tn the circuit in Fig. |OFE-3. select the turus ratio of the ideal transformer that will match the output of the transistor amplifier to the speaker represented by the 16-0 Toad. é 1605 Ane {speaker <. ideal Figure 16PFE-3 SOLUTION: ~ thd Book “ Ding bof row : po Spee ——s L | + Vee s/By\ 4 fie) @ eG) Biske Vow Ge) CRS 0% \ ee