hw 3 - 2 V 1 = 2 . 75 , V 1 = 2 . 75 2 = 1 . 375 V, V Th =...

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EECE 519 Homework 3 (Due Thursday, February 7) 1. Find the power delivered by the 14V source. - + 14 V 2Ω + 1Ω = 3Ω , || 3Ω = 1 . , 4Ω + 1 . 5Ω + 5Ω = 10 . , 10 . || 6Ω = 3 . 818Ω , R eq = 3 . 818Ω+7Ω = 10 . 818Ω , I = 14 V 10 . 818Ω = 1 . 29 A, P = (14 V )(1 . 29 A ) = 18 . 06 W 2. Find I 3 using current divider principle. - + 90 V I s I 1 10Ω I 3 Equivalent resistance as seen by the 90V source = 3 + 2 = 5Ω, The current supplied by the source = I s = 90 5 = 18 A , From current divider principle, the current through resistor = I 1 = 3 3+6 × I s = 3 9 × 18 = 6 A , Current through 10Ω resistor = I 3 = - 10 10+10 I 1 = - 1 2 (6) = - 3 A 3. Find and draw the Thevenin and Norton equivalent circuits with respect to terminals a,b. a b - + 3 V 0 . 0 . 0 . 25Ω 0 . 0 . 5 A R Th = 0 . 5 + 0 . 25 + 0 . 5(0 . 5) 0 . 5+0 . 5 = 1Ω R Th = 1Ω ( V 1 - 3 0 . 5 + V 1 0 . 5 + 0 . 5 = 0) × 0 . 5 , V 1 - 3 + V 1 + 0 . 25 = 0 ,
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Unformatted text preview: 2 V 1 = 2 . 75 , V 1 = 2 . 75 2 = 1 . 375 V, V Th = V 1-. 25(0 . 5) = 1 . 375-. 125 = 1 . 25 V , V Th = 1 . 25 V I N = V Th R Th = 1 . 25 1 = 1 . 25 A I N = 1 . 25 A Note: Short circuit current I N can also be computed by shorting a,b. ( V 1-3 . 5 + V 1 . 5 + V 1-V 2 . 25 = 0) . 5 , V 1-3 + V 1 + 2 V 1-2 V 2 = 0 , 4 V 1-2 V 2 = 3 , ( V 2-V 1 . 25 + 0 . 5 + V 2 . 5 = 0) . 5 , 2 V 2-2 V 1 + 0 . 25 + V 2 = 0 , 3 V 2-2 V 1 =-. 25 ,-4 V 1 + 6 V 2 =-. 5 Adding the two equations, 4 V 2 = 2 . 5 , V 2 = 0 . 625 V, I N = V 2 . 5 = . 625 . 5 = 1 . 25 A...
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This homework help was uploaded on 04/07/2008 for the course EECE 519 taught by Professor Chandra during the Spring '08 term at Kansas State University.

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