18 we apply kcl to a node in the branch where the two

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Unformatted text preview: ng a current source in common, (b) a supermesh, created by excluding the current source. As shown in Fig. 3.23(b), we create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh.) Why treat the supermesh differently? Because mesh analysis applies KVL—which requires that we know the voltage across each branch—and we do not know the voltage across a current source in advance. However, a supermesh must satisfy KVL like any other mesh. Therefore, applying KVL to the supermesh in Fig. 3.23(b) gives −20 + 6i1 + 10i2 + 4i2 = 0 or 6i1 + 14i2 = 20 (3.18) We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 in Fig. 3.23(a) gives i2 = i1 + 6 (3.19) Solving Eqs. (3.18) and (3.19), we get i1 = −3.2 A, i2 = 2.8 A (3.20) Note the following properties of a supermesh: 1. The current source in the supermesh is not completely ignored; it provide...
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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.

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