240_homework_2_soln

240_homework_2_soln - ECE 240 Electrical Engineering...

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ECE 240 – Electrical Engineering Fundamentals Solutions to Homework #2 Try drawing the circuit diagrams that correspond to each of the steps described in these solutions for better understanding. 1. + - 24 1 3 4 2 + - 24 4||(3+1)=2 2 i 1 i 1 i 2 i 1 = 24 2 + 2 = 6 i 2 = 4 4 + 4 i 1 = 3 P 1 Ω = (1)(3) 2 = 9 W 2. 2 4 2 i 1 8 + V o - 6 2 i 1 8 i 1 = 2 2 + 6 (8) = 2 V 0 = 2 i 1 = 4 V 3. + - V 3 1 5 2 i 1 i 2 I o = 4 V 5 Ω = (5)(4) = 20 i 2 = 20 4 = 5 i 1 = I o + i 2 = 4 + 5 = 9 V = 2 i 1 + 5 I o = 18 + 20 = 38 V
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4. Redrawing the circuit: 3 3 3 3 3 3 3 Thus, R eq = 3 + 0 + 3 = 6 Ω [ Note the right 3 Ω resistors are all in parallel with a short so the equivalent resistance is 0 Ω ] 5. 1 ! 12 2 2 2.5 2 3 6 12 2||2||1=0.5 2.5 6||3||2=1 12 4 Thus, R eq = 12||4 = 3 Ω 6. 4 6 8 4 8 6 4 4 4 4 6 + 6 = 12 4 4 (4+8)||4=3 (4+8)||4=3 12||(3+3) = 4 4 4 Thus, R eq = 4 + 4 + 4 =12 Ω
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7. R eq 2 2 1 2 2 2 22 2 2 2 1 2 2 R eq 6 1 2 2 2 2 6 2 1 2 R eq 6||(1+2) = 2 2 2 2 2 6||(1+2) = 2 R eq 2+2+2=6 2+2+2=6 Now, the two 6 Ω equivalent resistors are in parallel, so R eq = 6||6 =3 Ω 8. I s 11 2 4 6 3 i 1 + v 1 - i 2 i 3 i 4 + v 2 - 32 = 2 i 1 2 i 1 = 4 v 1 = 2(4) = 8 i 2 = 8 4 = 2 i 3 = 2 + 4 = 6 v 2 = 6 i 3 + v 1 = 6(6) + 8 = 44 i 4 = 44 11 = 4
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This note was uploaded on 02/27/2010 for the course ECE 240 taught by Professor Drsharlenekatz during the Spring '10 term at CSU Northridge.

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240_homework_2_soln - ECE 240 Electrical Engineering...

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