IRWIN 9e 6_38

# IRWIN 9e 6_38 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis. 9/E 1 ———————.__—___—_____ 6.33 The uapuuiiur in Fig. PEER-:1 is 51 HF with 21 lulerancc ui' [Iii-é: Given the wit-age wawt'ui'm in Fig. 6.3%. graph the current fir} fur the minimum and maximum capaci— tnr values. sir} ELF} viri {V} Figure P638 SOLUTION: at): CM cu \$0h O<t Mr): 0A 6091, Irm < 1' S 2m: (CI) =:. M(5m) (WW): 2.24 hnA WON My): O'Q(5in) ( LHue"): 1.2mm Chapter 6: Capacitance and Inductance Problem 6.38 2 Irwin, Basic Engineering Circuit Analysis, 9/E 50% 2mg 31\$ 3mg Imag(t): I'M'S’ln) (rm/0‘0 :W'UB’mA immm ’-'- O'Ci (5m) (’57X to”): 3-mmA 4071. 3mg<t < Hms Mt) =QA (30h Harm < ts 5m: [moat/d = N (5M) (qxno‘1)=2.wmA im'mm): 0““ 5m) (WNW): I'XHmA 40h 5MB <I\$ 67713 1(1): OA 1(1) (mix) l8“! t(m3) l.|C_ 3'6]: WM --—-- O'C'C Problem 6.38 Chapter 6: Capacitance and Inductance ...
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## This note was uploaded on 11/08/2011 for the course EE 2120 taught by Professor Aravena during the Fall '08 term at LSU.

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