IRWIN 9e 6_40

# IRWIN 9e 6_40 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis. 9/E 1 6.40 The mdumnr in Fig. Pawn-:1 is 4.? pH with z: IR‘IJL‘TUEIUC ni' HIE-é: Given the current waveform in Fig. b.4011. graph the wilt-age vﬂrj fair the minimum aml maximum induc— Inr values. .-—-. C H.» .-—-. h— H.» --... 010 20 30 4E] 50 I30 3’0 80 Time {ms} [b] Figure P640 SOLUTION: VG) = L Hgdx n ’ +09» 0 § t <20Yh5 Vmow (k): ("DC H'7U)( V2,.) : 2'?2U\/ Vmgy,(t) :(0‘8)CL+7HJ( 11):I'38L1v 40h Zovmsg ts Barns VWI 1: ("1) (—I'S):-g7-L/6UV vm-ntt) =(o~8’)( V'UUGI'S): ~5-ueuvé Chapter 6: Capacitance and Inductance Problem 6.40 Problem 6.40 Irwin, Basic Engineering Circuit Analysis. 9/E 50h ZOmS St S warm, qu1 (t): ("1)(‘4'UJHo-5) 1 ’2'82MV mec1) 2(0‘?) C wwxvos): —i~88uv Vim Harms 1S 5mm Vmaﬂt) 3 0'2) (V714) (I)= S‘gq‘uv Vmiy, (7H 40'3XV'7MCI): Brieuv Jrcm 50m; S t; < 60mg v02) : 0v 4pm soms s ts worm vmmut) *3 (m (wiuX!) : 56% UV me (t) - (0‘3)CL4’7J_1)(U = 3v‘leuv Chapter 6: Capacitance and Inductance —i-88 ~2‘Y2 66% 4744! Irwin, Basic Engineering Circuit Analysis. 9/E Chapter 6: Capacitance and Inductance i Problem 6.40 ...
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