hw13-solutions

# hw13-solutions - Reilly(mar3978 hw13 turner(56725 Power(W...

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Reilly (mar3978) – hw13 – turner – (56725) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A variable resistor is connected across a con- stant voltage source. Which oF the Following graphs represents the power P dissipated by the resistor as a Function oF its resistance R ? 1. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) Power (W) 2. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) 3. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) 4. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) 5. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) correct 6. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) 7. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) Explanation: The power dissipated in the resistor has several expressions P = E I = E 2 R = I 2 R , where the last two are simply derived From the frst equation together with the application oF the Ohm’s law. Since the resistor is connected to a constant voltage source E = constant P = E 2 R = constant R , tells us that the power is inversely propor- tional to the resistance p P 1 R P . 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) 002 (part 1 of 3) 10.0 points A 1120 Ω resistor is rated at 4 . 82 W. What is the maximum current through this resistor? Correct answer: 0 . 0656016 A. Explanation: Let : R = 1120 Ω and P = 4 . 82 W .

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Reilly (mar3978) – hw13 – turner – (56725) 2 The power rating of the resistor is the maxi- mum allowed power dissipation of the resistor and corresponds to a maximum current. P = I V = I 2 R I = r P R = r 4 . 82 W 1120 Ω = 0 . 0656016 A . 003 (part 2 of 3) 10.0 points If the maximum current has been passing through the resistor for 25 . 1 minutes, how many Coulombs of charge passes through the resistor in this period? Correct answer: 98 . 796 C. Explanation: Let : t = 25 . 1 minutes . Current is I = Q t Q = I t = (0 . 0656016 A) (25 . 1 minutes) p 60 s min P = 98 . 796 C . 004 (part 3 of 3) 10.0 points Denote the amount of charge in part 2 by Q . Consider the passage of the same maxi- mum current as in Part 2 through two 1120 Ω resistors connected in series. How much charge passes through any cross section in this resistor series in 25 . 1 minutes? 1. Q = 2 Q 2. Q = Q 2 3. Q = Q 4 4. Q = 4 Q 5. Q = 2 Q 6. None of these 7. Q = Q 2 8. Q = Q correct Explanation: Since Q = I T , then when I is Fxed and T is Fxed, Q is correspondingly Fxed. So Q = Q . 005 10.0 points We estimate that there are 268 million plug-in electric clocks in the United States, approxi- mately one clock for each person. The clocks convert energy at the average rate of 9 . 2 W. To supply this energy, how many metric tons of coal are burned per hour in coal-Fred electric-generating plants that are, on aver- age, 25% e±cient? The heat of combustion for coal is 33 MJ / kg. Correct answer: 1075 . 9 metric ton. Explanation: Let : R = 9 . 2 W / clock , L = 33 MJ / kg = 3 . 3 × 10 7 J / kg , N = 2 . 68 × 10 8 clocks , and Δ t = 1 h .
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hw13-solutions - Reilly(mar3978 hw13 turner(56725 Power(W...

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