313b for loop 1 v1 20 v2 0 v1 v2 20 343 for

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Unformatted text preview: 2Ω i1 i3 v3 i3 v4 i5 i4 4Ω 10 A Loop 3 1Ω + v1 +− + + 6Ω v2 v3 − Loop 1 − +− − (a) Figure 3.13 3vx i3 20 V Loop 2 + v4 − (b) Applying: (a) KCL to the two supernodes, (b) KVL to the loops. We now apply KVL to the branches involving the voltage sources as shown in Fig. 3.13(b). For loop 1, −v1 + 20 + v2 = 0 ⇒ v1 − v2 = 20 (3.4.3) For loop 2, −v3 + 3vx + v4 = 0 But vx = v1 − v4 so that 3v1 − v3 − 2v4 = 0 (3.4.4) For loop 3, vx − 3vx + 6i3 − 20 = 0 But 6i3 = v3 − v2 and vx = v1 − v4 . Hence −2v1 − v2 + v3 + 2v4 = 20 (3.4.5) We need four node voltages, v1 , v2 , v3 , and v4 , and it requires only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although the fifth equation is redundant, it can be used to check results. We can eliminate one node voltage so that we solve three simultaneous equations instead of four. From Eq. (3.4.3), v2 = v1 − 20. Substituting this into Eqs. (3.4.1) and (3.4.2), respectively, gives 6v1 − v3 2v4 = 80 (3.4.6) 6v1 − 5v3 − 16v4 = 40 (3.4.7) and | v v Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as 3 −1 −2 v1 0 6 −1 −2 v3 = 80 6...
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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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