Homework8Key - Irwin Basic Engineering Circuit Ana|ysis 9/E...

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Unformatted text preview: Irwin, Basic Engineering Circuit Ana|ysis, 9/E 1 I53 Draw the wavafm'm for the ClllTEfll in a 121;]: capacilm' when the capacitor voltage is as described in Fig. 136.9. 'HUHV) 12 —a Figure P63 SOLUTION: 1(1) ~ C/dv (if Mr) : IZM [2110‘] 1(1): ZVA aeon 6M8 S 1:510:13 4(1) : 1214 [—5 mos] £0) = '60A Chapter 6: Capacitance and Inductance Problem 6.9 mam Irwin, Basic Engineering Circuit Analysis. 9/E 2 Mat): 29A ostééus véoA éus ét$lOlJ5 “DA [0113 S {5/6/13 0A 1- > [Sus L(;t) vs t 30 ~— —' 10 a M __#__,W_,_ u- w , i0 new A ‘7 O _____ _ m -.,__ -Io ' ____._.7.5. w _______ 9-- “W J5 A 20 ~20 __ V w , —3o _qo WM H _ -50 - -60 ____._ M. 40 k Mus) —______——__—__—_—————-————— Problem 6.9 Chapter 6: Capacitance and Inductance Irwin, Basic Engineering Circuit Analysis, 9/E 1 $.10 Thc voltage amass a Efi—uF capacitor is shown in Fig. PEJIJ. Detclminc Ihc curl'cnl waveform. iii?) (V) FigurePEa-a SOLUTION: t, : O'Z'mS i2:O'L“I’YY7$ 13: 0ng J74: 'TfiS is: #27718 ' : C CL! A d1 3:40 ) :o ,and 4,:0 Chapter 6: Capacitance and Inductance Problem 6.10 2 Irwin, Basic Engineering Circuit Analysis, 9/E J: 15MEIO5] = ~2v5A i3$i <11. , v=O 3L=O v: —I2oX/05r ) 1: 25M [/05] : MA 0 x< 0 1(1): 25 OS 1' <0-2ms O 0~2ms$ :t< O'Hms ’25 0ng S gt< oVYms O 0ng \< i- < Inns 2'5 has K j: < i2ms O J 7/ iamsi ___—_—____—.__._————_——————-————-—-— Problem 6.10 Chapter 6: Capacitance and Inductance Irwin, Basic Engineering Circuit Analysis. 9/E 6.23 The voltage acmss a lfl—mH inductm' is shown in Fig. P4523. Determine the wavefm'm for the inductor current. ”(1‘) ("W3 ‘Ifll— r {ms} Figure PIE-.23 SOLUTION: {5091, O \< i’ S iYYWS. VLif) : iOtWV L “(H : IOOf/Ojdx mt) , 5‘00 #1 7“” #0” im5$ t g ms VLU') : —i05t+20 wt) : (oofl-Iotno) wt ‘ _,_ . L dL(i’) : 500* ”0°C“: ”A Chapter 6: Capacitance and Inductance Problem 6.28 2 _ irwin, Basic Engineering Circuit Analysis. 9/E JLUMmA) 05 l 1 Hms) Problem 6.28 Chapter 6: Capacitance and Inductance Irwin, Basic Engineering Circuit Analysis, 9/E 1 6+2? The current in a lfl-mH inductor is Sl’lflWl’l in Fig. 136.29. Determine the waveform th' the veltnge across the inductor. itt) (me) Figure P649 SOLUTION: 40%. 1. > (mm D V(x,) :0 i <0 -gomv CDC :45 gums V‘t: < 6m5 6omv Hms< i\ 0 i 7 6ms Chapter 6: Capacitance and Inductance Problem 6.29 Inn/in. Basic Engineering Circuit Analysis, 9/E 1 6.5-: Find the: [cm] capacitance Cr cf the: nctwm'ls; in Fig P650. 4 #FJ: CT _.. 112 pF 1 pF 2 HF 3 [LP Figure PIS+5u SOLUTION: Qedraw :HKQ Grew ; HMF (—CT 1 iZUF i/LLF QuF 3w: q, Q 1 CU” 2m aux HM) iu+ 244+“; +Hu c': 6A.:(H/U) :1un 614+ MM C1 = Ci'i' mu : ”Li/“Vin“ CT : [HM 1UP Chapter 6: Capacitance and Inductance Problem 6.50 Irwin, Basic Engineering Circuit Analysis, 9/E ‘ 1 l5+62 Determine the inductance at termirmls A—B in the nrctwurk in Fig. P4162. 1mH Figurr.I Fifi: SOLUTION: L, 2 im H L2: Hum L”: Q'MH Le: HmH Lug: H mm] H [(LiiILQJquium [em] H [mow flmrrmmm Lag meHlm Lag: [m] II [37“ “3* 3““ ’ Lag: 3m+ 3m Lee: 6mH Chapter 6: Capacitance and Inductance Problem 6.62 ...
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