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exam #2-solutions - Version 035 exam#2 Erskine(56905 This...

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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(part1of3)10.0points 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 Ω. Explanation: Let : I = 0 . 326 A , and V = 157 V . Since V = I Z , the impedance is Z = V I = 157 V 0 . 326 A = 481 . 595 Ω . 002(part2of3)10.0points 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 Ω. 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 Ω . 003(part3of3)10.0points 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 Ω) 2 - (73 . 3938 Ω) 2 = 475 . 97 Ω , so the inductance is L = X L 2 π f = 475 . 97 Ω 2 π (60 Hz) = 1 . 26255 H . 004(part1of2)10.0points Consider the setup of a Lloyd mirror shown in the figure. The interference pattern on
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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 π . Assume the wavelength of the light rays is λ . viewing y L h S screen Mirror P O Find the path difference δ between the two rays which corresponds to the third maximum from the center of the screen (see the above figure). 1. δ = 3 2 λ 2. δ = 5 λ 3. δ = 3 λ 4. δ = 2 λ 5. δ = 6 λ 6. δ = 3 4 λ 7. δ = 4 λ 8. δ = 5 2 λ correct 9. δ = 7 2 λ 10. δ = 1 2 λ Explanation: The condition for interference maxima is that the total phase difference between the two rays be an even multiple of π . In this case, the total phase difference between the two rays is k δ + π . In other words, we have 0 , 2 π, 4 π, 6 π, · · · = k δ + π . Since δ must be a positive quantity, the third maximum corresponds to a phase difference of 6 π ; i.e. , k δ = 5 π . Therefore, δ = 5 π k = 5 π 2 π λ = 5 2 λ . 005(part2of2)10.0points Let L be the distance from the source to the screen and h be the distance from the source to the mirror. Find the y coordinate of the first minimum above the horizontal plane (use the small an- gle approximation). 1. y = 3 L λ 4 h 2. y = L h λ 3. y = h λ L 4. y = 3 h λ L 5. y = L λ h 6. y = L λ 4 h 7. y = 2 h λ L 8. y = λ 2 9. y = L λ 2 h correct 10. y = L λ 8 h Explanation: r 2 r 1 y L d S 1 S 2 θ = tan 1 parenleftBig y L parenrightBig viewing screen δ d sin θ r 2 - r 1 P O S 1 is the source and S 2 is its image. negationslash S 2 Q S 1 90 d = 2 h Q R Minima are obtained whenever the phase difference is an odd multiple of π ; i.e. , π, 3 π, 5 π, · · · = k δ + π . The minimum that corresponds to a phase difference of π is
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Version 035 – exam #2 – Erskine – (56905) 3 located at the intersection of the screen and the mirror. The first minimum above the plane corresponds to a phase difference of 3 π ,
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