HW#1_solution

HW#1_solution - HW1 1. At node A, By KCL 1A V1 10 V + 10...

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HW1 1. 20V 20 10 V 10 10 5 20 i 1 V 2 V Node A Node B Node C 1A At node A, By KCL 11 10 10 0 20 5 VV i −+ ++ = (1) At node B, By KCL 1 12 10 10 10 V VV i + + = (2) At node C, By KCL 21 2 10 20 VVV + = (3) By Eq (1), (2) and (3), We obtain 4.783 , 3.478 , 0.304 V Vi A = −=− 2. . 5V 4 7 4 10V Loop A A I B I 2A At loop A, By KVL 4 7 10 5 0 AB II + − −= (1) 2 += (2) By Eq. (1), (2) 0.091 , 2.091 IA = 3. 1 2 1 4 2V + - x V 1.0 x V Node A V Node B At node A, by KCL 2 0 x x V V + −= (1) At node B, by KCL 2 0 1 14 x xx V −− + (2) By Eq. (1) and (2) 2.909 , 4.545 x V V = The power associated with the dependent source
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(4.545) ( 2.909) 13.22 x PV V W = ×− = × − = − 4. a. 10V 10 20 10V 10 5 20 + - o V 2A 10V 10||20 10V 15 20 + - o V 2A 10V 10||20 20 + - o V 15 2A 10/15 A 10V 10||20 20||15 + - o V 2A 10/15 A 10V 20/3 60/7 + - o V 2/3 A 40/3 V 20/3 60/7 + - o V 2/3 A 70/3 V 20/3 60/7 + - o V 2/3 A 7/2 A 20/3 60/7 + - o V 17/6 A 15/4 + - o V 17/6 A Figure 1. We can obtain the simple circuit using source transformation. Figure 1 represents the process to get the simple circuit. 17 15 85 10.625 64 8 o VV =×= =
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b. 10 10 20 20 5 20V + - ' o V Circuit (a) 10+5 20V 10||20||20 + - ' o V Circuit (b) We remove 2 voltage source and 1 current
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This note was uploaded on 10/16/2010 for the course EE 201 taught by Professor Kruempel during the Spring '06 term at Iowa State.

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HW#1_solution - HW1 1. At node A, By KCL 1A V1 10 V + 10...

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