hw13-solutions

hw13-solutions - howard(cah3459 hw13 turner(56705 Power(W...

This preview shows pages 1–3. Sign up to view the full content.

howard (cah3459) – hw13 – turner – (56705) 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) correct 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) 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 1 R ² . 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 880 Ω resistor is rated at 6 . 68 W.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
howard (cah3459) – hw13 – turner – (56705) 2 What is the maximum current through this resistor? Correct answer: 0 . 0871258 A. Explanation: Let : R = 880 Ω and P =6 . 68 W . The power rating of the resistor is the maxi- mum allowed power dissipation of the resistor and corresponds to a maximum current. P = IV = I 2 R I = ± P R = ± 6 . 68 W 880 Ω = 0 . 0871258 A . 003 (part 2 of 3) 10.0 points If the maximum current has been passing through the resistor for 19 . 4 minutes, how many Coulombs of charge passes through the resistor in this period? Correct answer: 101 . 414 C. Explanation: Let : t = 19 . 4 minutes . Current is I = Q t Q = It = (0 . 0871258 A) (19 . 4 minutes) ² 60 s min ³ = 101 . 414 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 880 Ω resistors connected in series. How much charge passes through any cross section in this resistor series in 19 . 4 minutes? 1. None of these 2. Q = Q 2 3. Q =2 Q 4. Q = 2 Q 5. Q = Q 2 6. Q = Q correct 7. Q =4 Q 8. Q = Q 4 Explanation: Since Q = IT , 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 265 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 8 . 1 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, 27 . 5% e±cient? The heat of combustion for coal is 33 MJ / kg. Correct answer: 851 . 504 metric ton. Explanation: Let : R =8 . 1W / clock , L = 33 MJ / kg = 3 . 3 × 10 7 J / kg , N . 65 × 10 8 clocks , and Δ t = 1 h .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 10

hw13-solutions - howard(cah3459 hw13 turner(56705 Power(W...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online