Electric Circuits 8th Edition 140

Electric Circuits 8th Edition 140 - headed arrow emphasizes...

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116 Techniques of CncujtAnatysis objective 3-Deciding betw€en the node-vottage and mesh-current methods 4.13 Find the power delivered by the 2A current source in the circuit showr. 4.14 Find the power delivered by the 4A current source in the circuit shown. 15() NOTE: AIso try Chapter Problens 4.54 and 1.56. Figure 4.36 r. Source transtormatioN. Answer: 40 W: G) rj 4.9 S*ure* Tyaslsf*r$fi a"iioris Even though t1re node-voltage and mesh-current methods are powcrful tcchniques lor solving cilcuits,we are still irterestedin methods that can be used to simplify circuits. Sedes-parallel reductions and I{o-Y transfbrma tions are already on our list of simplifying techniques. We begin cxpandnrg this lisl with source transformations. A solrlc€ transformation. shown in Fig.4.36,allows avoltage source in serieswith a resistor to be replacedby a
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Unformatted text preview: headed arrow emphasizes tiai a source transformation is bilateral that is, lve can startwith either configuration and dedve the other. We need to find the relationship bctwcorl o, and ir thal guamntees the two configurations in Fig. 4.36 are equivalent with respect 1() iodes a,b. Equivalence is achieved if an_v resistor R, experiences the same current flo\l:, and thus the same voltage drop, whether connected between nodes a,b in Fig.4.36(a) or Fig.4.36(b). Suppose R, is connecled betlveen nodes a,b in Fig. 4.36(a). Using Ohm\ law. thc current in n, is 14.52) Now suppose thc same resistor R, is connected between nodes a,b in Fig.4.36(b). Using current division, lhe cunent in R, is If the lrvo circuits in Fig. 4.36 are equivalent. these resistor currents must be the same. Equating ihe righlhand sides ofEqs.4.52 and:1.53 and sirnplitying, , ' = R 2 A 2 5 V 14.54)...
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This note was uploaded on 07/05/2010 for the course EE 100 taught by Professor Boser during the Spring '07 term at University of California, Berkeley.

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