W09 314 HW01 solutions

W09 314 HW01 solutions - EECS 314 Winter 2009 Homework set...

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EECS 314 Winter 2009 Homework set 1 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2009 Alexander Ganago Page 1 of 7 Solution by Idong Ebong Problem 3 Voltage and current division The Assignment Consider 3 circuits built of the same 3 resistors with resistances equal to 10 Ω , 20 Ω , and 40 Ω . Depending on how the resistors are connected, the distribution of total power among them varies. In each case, it can be found using the formulas for voltage and current division. Part 1 (15 points) For the circuit shown on this diagram, calculate the percentage of total power absorbed by each resistor. Record your answers in the table below. Show your work. Use additional pages as needed. Resistor 10 Ω 20 Ω 40 Ω Percentage of the total power absorbed 14.29% 28.57% 57.14% SOLUTION There are multiple ways to do this problem. Since this problem is under the voltage division section, a voltage divider method will be used to solve it. Vs is supplied to 10 , 20 , and 40 resistors connected in series. Vs is dropped to an equivalent resistance of 70 (the sum of all the series connected resistors). The voltage dropped on the 10 resistor is 10/70*Vs = Vs/7 The voltage dropped on the 20 resistor is 20/70*Vs = 2Vs/7 The voltage dropped on the 40 resistor is 40/70*Vs = 4Vs/7 Power dissipated by a resistor is the square of its voltage drop divided by its resistance. Part 1 2 3 Total Score
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EECS 314 Winter 2009 Homework set 1 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2009 Alexander Ganago Page 2 of 7 Solution by Idong Ebong The power consumed by the 10 is ( ) Ω = Ω 2 2 002041 . 0 10 7 / s s V V The power consumed by the 20 is ( ) Ω = Ω × 2 2 004082 . 0 20 7 / 2 s s V V The power consumed by the 40 is ( ) Ω = Ω × 2 2 008163 . 0 40 7 / 4 s s V V Total power consumed is (0.002041+0.004082+0.008163) Ω 2 s V = 0.014286 Ω 2 s V The power percentages can be calculated by dividing the power consumed by the total power and multiplying by 100. For the 10 case, % 29 . 14 100 014286 . 0 002041 . 0 = × For the 20 case, % 57 . 28 100 014286 . 0 004082 . 0 = × For the 40 case, % 14 . 57 100 014286 . 0 008163 . 0 = × Another way to do the problem would be to use P=I 2 R instead of P= R V 2 . Because of the series configuration, we know that the same current is flowing through all the resistors. From the P=I 2 R equation, I (current) is the same for all the resistors so the power difference between the resistors is dependent on R (resistance). Power of 10 = I 2 *10 = P 10 Power of 20 =I 2 *20 = 2*P 10 Power of 40 =I 2 *40 = 4*P 10 Total power = P 10 + 2*P 10 + 4*P 10 = 7*P 5 Percent of power absorbed by 10 = % 29 . 14 100 7 10 10 = × × Ω Ω P P Percent of power absorbed by 20 = % 57 . 28 100 7 2 10 10 = × × × Ω Ω P P
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EECS 314 Winter 2009 Homework set 1 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible
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W09 314 HW01 solutions - EECS 314 Winter 2009 Homework set...

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