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Unformatted text preview: Engineering Circuit Analysis. Tﬂ' Edition 22. Chapter Five Solutions 10 March 2006 (a) [120 cos 400i] 60 = 2 cos 4001* A. 60 120 = 40 Q.
[2 cos 40m] (40) = 80 cos 400! V. 40 — 10 = 50 Q.
[80 cos 40031:" 50 = 1.6 cos 4003‘ A. 50 50 = 259. 1.6 cos 4003 A (b) 2k 3k+6k=12kg 7.2k 12k = 4.5m QDmﬂ (20x45) = 90 v. PROPRIETARY MATERLLXL. '1C 200? The McGrawHill Companies. Inc. Limited distribution permitted only to
teachers and educators for course preparation. If you are a student using this Manual. you are using it Without permission. Engineering Circuit Analysis‘ Tﬂ' Edition Chapter Five Solutions 10 March 2006 25. (100 kQ)(6 111A) = 0.6 V
1 ME! 200 m 47‘” Ml 470k 300k = 133.11g
(3 —I1}.6).."300x103 = 12 uA
(133.1 kQ)(12 uA) = 9.197 v
1M!) 133.1142. 9 V o o 2.193 V (—1 Solving. 9 — 1183.1x1031— 2.197 = 0.501: 6.750 uA. Thus. PMQ =12  106 33.06 uW. PROPRIETARY MATERLLXL. '1C 200? The McGrawHill Companies. Inc. Limited distribution permitted only to
teachers and educators for course preparation. If you are a student using this Manual. you are using it Without permission. Engineering Circuit Analysis‘ Tﬂ' Edition Chapter Five Solutions 10 March 2006 30. To begin, note that (1 11L'51X9 Q) = 9 mV. and 3 4 = 2.222 Q. The above circuit may not be further simpliﬁed using only source transformation
techniques. PROPRIETARY MATERLLXL. '1C 200? The McGraWHill Companies. Inc. Limited distribution permitted only to
teachers and educators for course preparation. If you are a student using this Manual. you are using it Without permission. ...
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 Spring '06
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