HW_3_solutions

HW_3_solutions - Irwin, Basic Engineering Circuit Analysis,...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 3.3 Calculate the current in the induumr alum-'1] in Fig. Flt-1.3 if the unllagc input is; (a) Elli?) = CDHIIET-Ir + I”, {'3} 1'2“) : 5 5m {jar}: _ 90..) V Gm: tho uuh'wera in bath thus llrm‘ uml ['j'cqucncy' Llnrnains. L=1mH Figure FEE SOLUTION: (0») 1‘ ’9 V(t) L: ImH Mt)?- vmou: Mi) : :o J'Cosumt +qs°)ax I'm 4U): ~19” Am (3771+L450) Imam) J”) : 26'53 Slfl(377t+w5°) km) : 2653 (05 (3171— use) A Chapter 8: AC Steady-State Analysis Problem 8.8 2 lrwin, Basic Engineering Circuit Analysis, 9/E gm : 13-26 Cos (3711+4O") A 1‘: i326 <QO°A Problem 8.8 Chapter 8: AC Steady-State Analysis Inivin, Basic Engineering Circuit Analysis, 9/E 3+9 Fiml 1hr: frequency—ileumaijn impedance. 2-. in the rwlwm'i; in Fig. PM}. o—uW—fl I —- i 11 n —_i2 fl Figure F83; SOLUTION: 10 Chapter 8: AC Steady-State Analysis Problem 8.9 Irwin, Basic Engineering Circuit Analysis. 9/E 1 8.11 Find 1hr: ['i'r‘qucnuy'flrimuiu inipmlmiuc. I. as; sham-n in Fig. PtH ]. Figure P3 .11 SOLUTION: Z _: 2.4331(2on 2mo° ~r 2L0“ 2‘: itinsUz _______—______——_____—_—_._—————— Chapter 8: AC Steady-State Anaiysis Problem 8.11 Irwin, Basic Engineering Circuit Analysis, 9/E 3.12 Find the impedance. 1-. mam-n in Fig. P3]? in n frequency of Evil HI. 10InH :11 Figure P842 SOLUTION: 10mH 20. 10MF 2L: J(217)(iom) :J'3-77_0_ Z-C : __i________~_ : 726512512 j(377)(iou) j 347.0. 2,0. Z r [I llz‘j265.25:) +j3-77 Chapter 8: AC Steady-State Analysis Problem 8.12 Irwin, Basic Engineering Circuit Analysis, 9/E — z : /(2— 1265-25) +J' 3n77 I+2*j265d§ Problem 8.12 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 3+1] Fiml ‘i' in the rrclwm'i; in Fig. Phil} O—J‘N‘ 1 5 Jill 5 1‘2 3 I? 5 2 5 —__I'1 3 Figure P813 SOLUTION: -+J_+J_+4_ ’Ji 2 J2. —._'-—= 210° 7 37: __L§ '2. Chapter 8: AC Steady-State Analysis Problem 8.13 Irwin, Basic Engineering Circuit Analysis, 9/E 343 Fiml the i]]1]‘ICLILi[1L:C.I..Hhi‘M-‘flitlFig. Pam at a t'i'tqucncy ut' -—L|II|]I Hz. 10mH 2E1 Figurfi P818 SO UTION: 10mH 2n 10|.lF LL): Zn @406) : 25/3/3 fidd/J EL: J(25i3~3)(iom) 2 325-1312. 2} : J = ‘J'3q‘7Q—Q‘ X25); axiom )ZS-BJ‘L 21L 2 1.0. "7" "J 3’q'7‘i—n- Chapter 8: AC Steady-State Analysis Problem 818 2 Inivin, Basic Engineering Circuit Analysis. 9/E Problem 8.18 Chapter 8: AC Steady-State Analysis Inn/in, Basic Engineering Circuit Analysis, 9/E 3.21 The impedance of thc circu it in Fig. PHE] i5 real at f = 50 HI.1'r"-"h:.i[ islhc win: 01' L'?‘ L I—-* l 2:1 1U mF Figure P821 SOLUTION: L Z——> 2.9 10 mF E 29%: 211’:ij “ht—J J‘Luc J‘JJLLL Mu»: be Sid/siAtive LDC WC Rea/IO and (my; UJC x4071, m.§twe L: / : i W25 @7D‘CI0W7) Chapter 8: AC Steady-State Analysis Problem 8.21 Irwin, Basic Engineering Circuit Analysis. 91E 1 3.22 Fiml me value ut' the uupzmiluncc. II". Shawn in 111:: circuit in Fig. P333 an that firm will he in phase with this Maura: mlmgc. 31' r} ink]- : I303 [251% + 309] ‘-.-’ 4O mH Figure FEM: SOLUTION: HI \7: 60‘130°\/ M‘ [5.0. U'XC JIO-n— Chapter 8: AC Steady-State Analysis Problem 8.22 2 Irwin, Basic Engineering Circuit Analysis. 9/E Se): W Imcugl’nw [9w equal 225i + to; {to —__I__) = 0 WC LDC we go: 250 _______—_—_____——.__.——-———-—-——-— Problem 8.22 Chapter 8: AC Steady-State Analysis ...
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