exam #2-solutions - Version 035 exam #2 Erskine (56905) 1...

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Unformatted text preview: Version 035 exam #2 Erskine (56905) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A multimeter in an RL circuit records an rms current of 0 . 326 A and a 60 Hz rms generator voltage of 157 V. A wattmeter shows that the average power delivered to the resistor is 7 . 8 W. Find the impedance in the circuit. 1. 182.815 2. 467.606 3. 500.0 4. 109.677 5. 254.386 6. 189.341 7. 210.28 8. 481.595 9. 284.848 10. 173.502 Correct answer: 481 . 595 . Explanation: Let : I = 0 . 326 A , and V = 157 V . Since V = I Z , the impedance is Z = V I = 157 V . 326 A = 481 . 595 . 002 (part 2 of 3) 10.0 points Find the resistance. 1. 15.8678 2. 18.3655 3. 19.0806 4. 55.8881 5. 12.4392 6. 123.334 7. 57.971 8. 28.5065 9. 73.3938 10. 16.7573 Correct answer: 73 . 3938 . Explanation: Let : P = 7 . 8 W . The power dissipated is P = I 2 R, so the resistance is R = P I 2 = 7 . 8 W (0 . 326 A) 2 = 73 . 3938 . 003 (part 3 of 3) 10.0 points Find the inductance. 1. 0.594566 2. 0.503981 3. 1.26255 4. 0.58635 5. 0.46782 6. 0.714368 7. 0.662649 8. 0.976468 9. 0.5714 10. 0.559212 Correct answer: 1 . 26255 H. Explanation: Let : f = 60 Hz . In an RL circuit, Z = radicalBig R 2 + X 2 L , or X L = radicalbig Z 2- R 2 = radicalBig (481 . 595 ) 2- (73 . 3938 ) 2 = 475 . 97 , so the inductance is L = X L 2 f = 475 . 97 2 (60 Hz) = 1 . 26255 H . 004 (part 1 of 2) 10.0 points Consider the setup of a Lloyd mirror shown in the figure. The interference pattern on Version 035 exam #2 Erskine (56905) 2 the vertical screen is due to the superposition of a direct ray and a reflected ray where at reflection there is a change in the phase angle of . Assume the wavelength of the light rays is . viewing y L h S viewing screen Mirror P O Find the path difference between the two rays which corresponds to the third maximum from the center of the screen (see the above figure). 1. = 3 2 2. = 5 3. = 3 4. = 2 5. = 6 6. = 3 4 7. = 4 8. = 5 2 correct 9. = 7 2 10. = 1 2 Explanation: The condition for interference maxima is that the total phase difference between the two rays be an even multiple of . In this case, the total phase difference between the two rays is k + . In other words, we have , 2 , 4 , 6 , = k + . Since must be a positive quantity, the third maximum corresponds to a phase difference of 6 ; i.e. , k = 5 . Therefore, = 5 k = 5 2 = 5 2 . 005 (part 2 of 2) 10.0 points Let L be the distance from the source to the screen and h be the distance from the source to the mirror....
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exam #2-solutions - Version 035 exam #2 Erskine (56905) 1...

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