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Unformatted text preview: marout (ssm979) 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 () Power(W) correct 2. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () Power(W) 3. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () Power(W) 4. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () Power(W) 5. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () Power(W) 6. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () Power(W) 7. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () 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 Ohms 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 () Power(W) 002 (part 1 of 3) 10.0 points A 1180 resistor is rated at 5 . 15 W. marout (ssm979) hw13 turner (56725) 2 What is the maximum current through this resistor? Correct answer: 0 . 0660637 A. Explanation: Let : R = 1180 and P = 5 . 15 W . 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 5 . 15 W 1180 = . 0660637 A . 003 (part 2 of 3) 10.0 points If the maximum current has been passing through the resistor for 39 minutes, how many Coulombs of charge passes through the resis tor in this period? Correct answer: 154 . 589 C. Explanation: Let : t = 39 minutes . Current is I = Q t Q = I t = (0 . 0660637 A) (39 minutes) parenleftbigg 60 s min parenrightbigg = 154 . 589 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 1180 resistors connected in series. How much charge passes through any cross section in this resistor series in 39 minutes? 1. Q = 2 Q 2. Q = 4 Q 3. None of these 4. Q = Q 2 5. Q = Q 4 6. Q = 2 Q 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 254 million plugin electric clocks in the United States, approxi mately one clock for each person. The clocks convert energy at the average rate of 2 . 9 W....
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This note was uploaded on 10/24/2010 for the course PHY 56735 taught by Professor Turner,j during the Fall '10 term at University of Texas at Austin.
 Fall '10
 TURNER,J

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