Solution 5 - ECE 205 Section A, B, and C. Homework #5...

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ECE 205 Section A, B, and C. Homework #5 Solutions 2.34 40//(10 + 20 + 10)= 20 , 40//(8+12 + 20) = 20 20 20 40 eq R 2 12 12/ 40, 3.6 W 40 eq V I P VI R 2.38 20//80 = 80x20/100 = 16, 6//12 = 6x12/18 = 4 The circuit is reduced to that shown below. (4 + 16)//60 = 20x60/80 = 15 15//15 5 12.5 eq R   40 3.2 A o eq i R  2.45(b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5.
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This note was uploaded on 02/03/2010 for the course ECE 201 taught by Professor C during the Spring '10 term at Miami University.

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Solution 5 - ECE 205 Section A, B, and C. Homework #5...

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