Chapt05PP

# Chapt05PP - © 2009 CHAPTER 5 P.P.5.1 The equivalent...

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Unformatted text preview: © 2009 CHAPTER 5 P.P.5.1 The equivalent circuit is shown below: At node 1, 3 1 3 1 6 1 s 10 x 40 v v 10 x 5 v 10 x 2 v v − + = − v 1 = 451 v 50 v S + (1) At node 2, 3 3 1 d 13 x 20 v 10 x 40 v v 50 v Av = − + − But v d = v 1- v S . [2 x 10 5 (v 1- v S ) - v ] 4000/(5) + v 1- v = 2v 1600 x 10 5 (v S- v 1 ) + 803v ≅ ( 2 ) Substituting v 1 in (1) into (2) gives 1.5914523 x 10 8 v S- 17737556v = 0 = = 17737556 10 x 5964523 . 1 v v 8 S 9.00041 If v S = 1 V, v = 9.00041 V, v 1 = 1.0000455 v d = v S- v 1 = - 4.545 x 10-5 Av d = - 9.0909, i = = − 50 v Av d 657 μ A- v d + + v- + v 1-- + v s + Av d- 2 M Ω 40 k Ω 20 k Ω 5 k Ω 50 Ω i 0 2 1 P.P.5.2 At node 1, 20 v v 10 v v 1 1 S − = − But v 1 = v 2 = 0, 20 v 10 v S − = = S v v –2 i = 3 3 10 x 20 v 10 x 20 v − = − When v s = 2V, v = -4, i = 20 10 x 4 3 − = 200 μ A P.P.5.3 v = ( ) = − = − mV 30 3 120 v R R i 1 2 –1.2 V i = = − k 120 v 10 μ A P.P.5.4 (a) i S = R v − R i v S − =- +- + V S i V 1 V 2 + V- 20 k Ω 10 k Ω (b)...
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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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Chapt05PP - © 2009 CHAPTER 5 P.P.5.1 The equivalent...

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