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Thévenin and Norton Equivalents for Dependent Source Circuits: Corrected Problem 5.90, Slides of Electrical Circuit Analysis

The solution to problem 5.90 in electrical engineering, which involves finding the thévenin and norton equivalents for a dependent source circuit. The document corrects a typo in the problem statement, specifying that the independent source should have units of volts. Several diagrams for reference.

Typology: Slides

2012/2013

Uploaded on 04/30/2013

devguru
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Download Thévenin and Norton Equivalents for Dependent Source Circuits: Corrected Problem 5.90 and more Slides Electrical Circuit Analysis in PDF only on Docsity! Prob 5.90 Solution Thévenin & Norton Equivalent for Dependent Src Ckt Docsity.com Correct Text Typo  The INDEPENDENT SOURCE in Problem 5.90 Should have Units of VOLTS • The Corrected Ckt Diagram °∠30V10 Docsity.com Titus Te CKT Looks 41k Ee TAS © WHEN @9b ARE SHORTED . BY OM: 348 &) Tx rev 2Be CG (e+ 58)a x= fev 23 Jo~L £53.13? 1OVL 30° Dez (A £23135 TWEN Ls 2 WS Le =-15 (14 £-23./3°) Lserz 4 SA L/sG.B>° Fersye Bm = Yee / Ese Baz JOVESO = 607306787" AS& LISEB2° — 6 E79 LS53.13° ok [20 — 6472 LSBRIZ | Bn = Bocarjsael MODINE Var= You ¢€ tn ~fse p@aw CKTS Docsity.com J6N LarSO° ® THEN EENLAY C7 “ee SA LISG e27° INIORTON CKT Bruce Mayer, PE Engingering Instructor Chabot Cotlege 25555 Hesperian Bivd Hayward, CA 94545 eMail: bmaver@chabotcollege.edu ZS MAI (A Docsity.com PS.90 ALTERNATIVE © nee Voe/Zse POLARITY 5 PuT "b" "AT THE TOP” ee 2 i fb + 10VL30° Zx Vee —— a ee Fan BY KCL AT NaDeE-a Z Dor =o = Ixt0.SLe = SL OL “SIx zo = tie = & FHLS HAVE V-DROPS ACBeSS The of Ft OR 64 emPepAnces THs | Mec = rev 120°] Now Ww mHs case Ts. RONS ba g82 Docsity.com
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