Lec 38 - QW “QM- Jrl'MQ—v M 3144 450 MN EKPSPIQZQJ, p...

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Unformatted text preview: QW “QM- Jrl'MQ—v M 3144 450 MN EKPSPIQZQJ, p 23-25 'Tf—‘l'l M. 7.115 7.46:. E ...-= 2:; ~75 a. higééwz’lmsv m PH-B = 243.952; (4%.?qu C23 A-l‘o =23897M7° (3) me?) : zqq. [mama Q0. A/ :— -565 £300 (‘23. 3*3 — MA = c1427. Lies (—138.75? (a) A—B = $3.427 7.882“ (33 ME = 11200. bl’FI-E'b“ (43 A/13 = -337 (CR-130° 03) Am HERB = 84:35:79 (-szz“ m = 21% (1925(9mt- 1 m, 555") M H Milmkwflflh as 1% $3 V2. (21:3 +3 mXfl \3 = \DU} (223% so) Ix :- 1.01% 1:4 :10 fig \f”i’3"w 1 ’ Nadia, «pH-0.92. Mafia) . Neale e.qu . Dawmbkc Veebe. eguoiiérm . vatm scores— 255m 7.16: Use the nodewoltage method to find the steady-state expression for v (t) in the circuit shown. The sinusoidal sources are is = 10 cos wt A and vs = 100 sin wt V, where a) = 50 krad/s. ' "I. No N059- \Jol:met, METHQI} . V1= ELM—5 EILEEN {wWHM.. “"55 = D763 3 “(023 (10%(‘5qmoi—m569‘5) V ‘7. ‘7 Use the mesh—current method to find the phasor cur- ‘rent I in the circuit shown. [20 Chap 6 Responses of RLC circuits Parallel RLC neper and resonant frequencies overdamped critically damped underdamped Series RLC neper and resonant frequencies overdamped critically damped underdamped Step reSponses of parallel or series RLC final value Chap 7 Sinusoidal Steady—State Analysis Time function: Ud‘: \/m C53 (wit +Cb} V Phasor: V a VM [EL V Impedances: Z = R + j X 1% = wL xc = —1/wc Admittances: Y = G + j E BL = -1/WL BC = we nms for sinusoid: V‘f‘mfi : V in. \1—5: A C Analysis KC L KVL Ohm's Law 'Series - Parallel impedances Source transformations Thevenin circuit Norton circuit Node voltages Mesh currents u- erciu4- Noam: 0.358? _...OZU¢+VI =0 Goo-K} I50 got; 50 . ___. 14c: v Soldib-u v = VTF 15 (am-1° v SW lei+ a :0 .02.“; = O "‘51—‘39" = .. I25£°A ISL = 6004-3 [50 —\)l50 1.43 ...
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This note was uploaded on 04/02/2008 for the course EE 201 taught by Professor Kruempel during the Fall '06 term at Iowa State.

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Lec 38 - QW “QM- Jrl'MQ—v M 3144 450 MN EKPSPIQZQJ, p...

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