8 v 10 v3 3 v2 2 24 24 v 10 24 24 v 10 as

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Unformatted text preview: − 12 0 0 12 0 3 −4 2 3 −4 3 −2 −4 7 2 −3 3 −2 −4 7 −1 −1 1 −1 −1 + + + 12 0 0 12 0 = 0 + 0 − 24 − 0 − 0 + 48 = 24 = 0 + 144 + 0 − 168 − 0 − 0 = −24 + + + Thus, we find v1 = 1 = 48 = 4.8 V, 10 v3 = 3 = v2 = 2 = 24 = 2.4 V 10 −24 = −2.4 V 10 as we obtained with Method 1. PRACTICE PROBLEM 3.2 2Ω 3Ω 1 Find the voltages at the three nonreference nodes in the circuit of Fig. 3.6. Answer: v1 = 80 V, v2 = −64 V, v3 = 156 V. 4 ix 2 3 ix 4Ω 10 A Figure 3.6 6Ω For Practice Prob. 3.2. Electronic Testing Tutorials 3.3 NODAL ANALYSIS WITH VOLTAGE SOURCES We now consider how voltage sources affect nodal analysis. We use the circuit in Fig. 3.7 for illustration. Consider the following two possibilities. C A S E 1 If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. In Fig. 3.7, for example, v1 = 10 V (3.10) | v v Thus our analysis is somewhat simplified by thi...
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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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