ECE2040D_HW3_Soln

# ECE2040D_HW3_Soln - ECE 2040D Solutions to HW#3 - Spring...

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Unformatted text preview: ECE 2040D Solutions to HW#3 - Spring 2009 P4.3-3 va 10 va va 8 .03 0 va 100 100 100 7V P4.3-4 va 8 + ( va 8 12 va 12 va 500 125 250 500 va 4 V 0 P4.3-5 The power supplied by the voltage source is v v v v 12 9.882 12 5.294 va i1 i 2 va a b a c 12 4 4 6 4 12(0.5295 1.118) 12(1.648) 19.76 W Page 1 / 7 ECE 2040D Solutions to HW#3 - Spring 2009 P4.4-2 va - 6 v v 4va + a a 1000 2000 3000 va 12 V ib va 4va 3000 12 mA 0 P4.4-4 Apply KCL to the supernode of the CCVS to get 12 10 14 10 1 i b 0 i b 2 A 4 2 2 Next 10 12 1 ia 2 V 4 4 2 r 1 A r i a 12 14 2 P4.4-5 First, express the controlling current of the v2 CCVS in terms of the node voltages: i x 2 Next, express the controlled voltage in terms of the node voltages: v2 24 v2 12 v 2 3 i x 3 V 2 5 so ix = 12/5 A = 2.4 A. Page 2 / 7 ECE 2040D Solutions to HW#3 - Spring 2009 P 4.5-1 2 i1 + 9 (i1 i 3 ) 3(i1 i 2 ) 0 15 - 3 (i1 i 2 ) 6 (i 2 i 3 ) 0 - 6 (i 2 i 3 ) 9 (i1 i 3 ) 21 0 or 14 i1 + 3 i 2 9 i 3 - 3 i1 9 i 2 6 i 3 - 9 i1 6 i 2 15 i 3 0 15 21 so i1 = 3 A, i2 = 2 A and i3 = 4 A. P4.5-4 KCL loop 1: 25 ia 2 250 ia 75 ia 4 100 (ia ib ) 450 ia 100 ib KCL loop 2: 100(ia ib ) 4 100 ib 100 ib 8 200 ib 100 ia 500 ib ia 4 9.3 mA 6.5 mA , ib 0 0 2 P4.6-1 mesh 1: i1 i2 ib 1 A 2 mesh 2: 75 i2 10 25 i2 0.1 A i1 i2 = 0.6 A 0 Page 3 / 7 ECE 2040D Solutions to HW#3 - Spring 2009 P4.6-3 Express the current source current as a function of the mesh currents: i1 i2 0.5 i1 i2 0.5 Apply KVL to the supermesh: 30 i1 20 i2 10 0 30 (i2 0.5) 20i2 50 i2 15 i1 .4 A and v2 10 10 i2 20 i2 2V 5 50 .1 A P4.6-4 Express the current source current in terms of the mesh currents: ib ia 0.02 Apply KVL to the supermesh: 250 ia 100 (ia 0.02) 9 ? ia vc .02 A 20 mA 4 V 100(ia 0.02) 0 Page 4 / 7 ECE 2040D Solutions to HW#3 - Spring 2009 P4.6-6 Mesh equation for right mesh: 4 i 2 2 i 6 i 3 0 12 i 8 18 0 i -10 12 A Page 5 / 7 ECE 2040D Solutions to HW#3 - Spring 2009 P4.7-6 (a) vo gRL v and v v v o i vi 1 R2 2 g RL R2 R1 R2 (b) ? vo vi 5 u103 103 g 1.1u103 170 g 0.0374 S Page 6 / 7 ECE 2040D Solutions to HW#3 - Spring 2009 Page 7 / 7 ...
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## This note was uploaded on 09/17/2009 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Institute of Technology.

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