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Unformatted text preview: the presence
of the current sources reduces the number of equations. Consider the
following two possible cases.
4Ω 10 V +
− 6Ω i1 Figure 3.22 C A S E 1 When a current source exists only in one mesh: Consider the
circuit in Fig. 3.22, for example. We set i2 = −5 A and write a mesh
equation for the other mesh in the usual way, that is, 3Ω i2 5A A circuit with a current source. −10 + 4i1 + 6(i1 − i2 ) = 0 ⇒ i 1 = −2 A (3.17) C A S E 2 When a current source exists between two meshes: Consider
the circuit in Fig. 3.23(a), for example. We create a supermesh by excluding the current source and any elements connected in series with it,
as shown in Fig. 3.23(b). Thus, | v v A supermesh results when two meshes have a (dependent or independent)
current source in common. | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3
6Ω Methods of Analysis 93 10 Ω
6Ω 10 Ω 2Ω
20 V +
− i1 i2 4Ω 6A i1 0 i2 (a) Figure 3.23 20 V +
− i1 i2 4Ω (b) Exclude these
elements (a) Two meshes havi...
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This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.
- Spring '12