# 35a solution the circuit in this example has three

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Unformatted text preview: MPLE 3.2 Determine the voltages at the nodes in Fig. 3.5(a). Solution: The circuit in this example has three nonreference nodes, unlike the previous example which has two nonreference nodes. We assign voltages to the three nodes as shown in Fig. 3.5(b) and label the currents. 4Ω 4Ω ix 1 2Ω i1 8Ω 2 v1 3 3A 4Ω 3A 2Ω 2ix v2 8Ω i2 i1 v3 i3 ix ix i2 4Ω 3A 2ix 0 (a) Figure 3.5 (b) For Example 3.2: (a) original circuit, (b) circuit for analysis. At node 1, v 1 − v2 v1 − v 3 + 4 2 Multiplying by 4 and rearranging terms, we get 3 = i1 + ix ⇒ 3= 3v1 − 2v2 − v3 = 12 (3.2.1) At node 2, v 1 − v2 v2 − v3 v2 − 0 = + 2 8 4 Multiplying by 8 and rearranging terms, we get ix = i2 + i3 ⇒ −4v1 + 7v2 − v3 = 0 (3.2.2) At node 3, v 2 − v3 2(v1 − v2 ) v 1 − v3 + = 4 8 2 Multiplying by 8, rearranging terms, and dividing by 3, we get i1 + i2 = 2ix ⇒ 2v1 − 3v2 + v3 = 0 (3.2.3) We have three simultaneous equations to solve to get the node voltages v1 , v2 , and v3 . We shall solv...
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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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