hw13-solutions

# Hw13-solutions - marout(ssm979 hw13 turner(56725 1 This print-out should have 15 questions Multiple-choice questions may continue on the next

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Unformatted text preview: marout (ssm979) hw13 turner (56725) 1 This print-out should have 15 questions. Multiple-choice 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 plug-in 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.

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Hw13-solutions - marout(ssm979 hw13 turner(56725 1 This print-out should have 15 questions Multiple-choice questions may continue on the next

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