solution_pdf 14

# solution_pdf 14 - malsam(wgm329 – hw13 – turner...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: malsam (wgm329) – hw13 – turner – (56705) 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 (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) 6. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) correct 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 990 Ω resistor is rated at 5 . 36 W. What is the maximum current through this resistor? Correct answer: 0 . 0735808 A. Explanation: malsam (wgm329) – hw13 – turner – (56705) 2 Let : R = 990 Ω and P = 5 . 36 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 . 36 W 990 Ω = . 0735808 A . 003 (part 2 of 3) 10.0 points If the maximum current has been passing through the resistor for 38 . 6 minutes, how many Coulombs of charge passes through the resistor in this period? Correct answer: 170 . 413 C. Explanation: Let : t = 38 . 6 minutes . Current is I = Q t Q = I t = (0 . 0735808 A) (38 . 6 minutes) parenleftbigg 60 s min parenrightbigg = 170 . 413 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 990 Ω resistors connected in series. How much charge passes through any cross section in this resistor series in 38 . 6 minutes? 1. Q = √ 2 Q ◦ 2. Q = Q ◦ √ 2 3. Q = Q ◦ correct 4. Q = 2 Q ◦ 5. Q = Q ◦ 4 6. None of these 7. Q = 4 Q ◦ 8. Q = Q ◦ 2 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 259 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 7 . 2 W....
View Full Document

{[ snackBarMessage ]}

### Page1 / 10

solution_pdf 14 - malsam(wgm329 – hw13 – turner...

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

View Full Document
Ask a homework question - tutors are online