Chapt04PP - 2009 CHAPTER 4 P.P.4.1 i2 i1 iS 4 12 + vo 8 4 1...

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© 2009 CHAPTER 4 P.P.4.1 12 Ω 8 Ω i 2 + v o i 1 i S 4 Ω By current division, s s 2 i 6 1 i 8 12 4 4 i = + + = s 2 0 i 3 4 i 8 v = = When i s = 15A, = = ) 15 ( 3 4 v 0 20V When i s = 30A, = = ) 30 ( 3 2 v 0 40V P.P.4.2 Let v 0 = 1. Then i = 8 1 and 5 . 2 ) 8 12 ( 8 1 v 1 = + = giving v s = 2.5V. If v s = 30V, then v 0 = (30/2.5)(1) = 12 V 12 Ω 8 Ω v 1 + v o 5 Ω V S = 30 V +
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P.P.4.3 Let v 0 = v 1 + v 2 , where v 1 and v 2 are contributions to the 10–V and 4–A sources respectively. To get v 1 , consider the curcuit in Fig. (a). (2 + 3 + 5)i = 10 i = 10/(10) = 1A v 1 = 2i = 2V To get v 2 , consider the circuit in Fig. (b). i 1 = i 2 = 2A, v 2 = 2i 2 = 4V Thus, v = v 1 + v 2 = 2+4 = 6 V 5 Ω 2 Ω i 10 V + 3 Ω + v 1 (a) 5 Ω i 2 + v 2 2 Ω i 1 4 A (b) 3 Ω
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P.P.4.4 Let v x = v 1 + v 2 , where v 1 and v 2 are due to the 10-V and 2-A sources respectively. To obtain v 1 , consider Fig. (a). 0 4 0 v 20 20 v v 1 . 0 1 1 1 = + + v 1 = 5 V For v 2 , consider Fig. (b). –4 – 0.1v 2 + 0 4 0 v 20 0 v 2 2 = + v 2 = 20 v x = v 1 + v 2 = 25 V 20 Ω v 1 4 Ω 20 V + (a) 0.1v 1 4 Ω 4 A (b) 20 Ω 0.1v 2 v 2
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P.P.4.5 Let i = i 1 + i 2 + i 3 where i 1 , i 2 , and i 3 are contributions due to the 16-V, 4-A, and 12-V sources respectively. For i 1 , consider Fig. (a), A 1 8 2 6 16 i 1 = + + = For i 2 , consider Fig. (b). By current division, 5 . 0 ) 4 ( 14 2 2 i 2 = + = For i 3 , consider Fig. (c), A 75 . 0 16
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This note was uploaded on 10/25/2009 for the course ECE 280 taught by Professor Staff during the Spring '08 term at George Mason.

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Chapt04PP - 2009 CHAPTER 4 P.P.4.1 i2 i1 iS 4 12 + vo 8 4 1...

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