HW 13 Solutions

HW 13 Solutions - Engineering Circuit Analysis. Tfl'...

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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 McGraw-Hill 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 McGraw-Hill 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 McGraW-Hill 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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HW 13 Solutions - Engineering Circuit Analysis. Tfl'...

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