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Unformatted text preview: Reilly (mar3978) – hw13 – turner – (56725) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find 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 1 2 3 4 5 Resistance (Ω) Power(W) 2. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 3. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 4. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 5. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) correct 6. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 7. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 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 first 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 parenleftbigg P ∝ 1 R parenrightbigg . 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 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 . 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 = radicalbigg P R = radicalbigg 4 . 82 W 1120 Ω = . 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) parenleftbigg 60 s min parenrightbigg = 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 fixed and T is fixed, Q is correspondingly fixed. So Q = Q ◦ . 005 10.0 points We estimate that there are 268 million plugin 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....
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 Spring '10
 TURNER,J
 Resistance, Resistor, Correct Answer, Electrical resistance, Series and parallel circuits, Reilly

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