Midterm 3-solutions

Midterm 3-solutions - Version 083 Midterm 3 Shih(58250 This...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Version 083 – Midterm 3 – Shih – (58250) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Suppose that the polarizers are “crossed” so that no light can be transmitted through the second polarizer. Now a third polarizer is in- serted between the crossed polarizers with its transmission axis at α = 30 to the transmis- sion axis oF the frst polarizer ( Please ignore the frst sketch and see the second sketch below. ). IF initially the light is unpolarized with intensity I 0 , aFter passing through all three polarizers, what is the fnal intensity? Note: The order in which light passes through the polarizers is: #1, then #3, and fnally #2. α #1 #2 #3 #1 #2 θ 2 θ 1 1. I 0 16 2. 3 8 I 0 3. 1 3 I 0 4. I 0 8 5. I 0 2 6. I 0 4 7. 3 32 I 0 correct 8. 2 15 I 0 9. 5 8 I 0 10. 1 9 I 0 Explanation: See the sketch below. AFter the frst polar- izer (Polarizer # 1): I 1 = I 0 2 AFter the second polarizer (Polarizer # 3, which was inserted): I 3 = I 1 cos 2 30 = 3 4 I 1 AFter the third polarizer (Polarizer # 2, which was Formerly the 2nd polarizer beFore Polar- izer # 3 was inserted): I 2 = I 3 cos 2 60 = 1 4 I 3 = 1 4 × 3 4 × 1 2 I 0 = 3 32 I 0 #1 #2 #3 Ι Ι Ι 0 1 Ι 2 3 002 (part 1 oF 2) 10.0 points In a certain series RLC circuit, the rms cur- rent is 7 . 2A , the rms voltage is 240 V and the current leads the voltage by 37 . What is the total resistance oF the circuit? 1. 15.2021 2. 8.63041 3. 12.6184 4. 6.94996
Background image of page 1

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

View Full DocumentRight Arrow Icon
Version 083 – Midterm 3 – Shih – (58250) 2 5. 13.6909 6. 7.7962 7. 12.9366 8. 26.6212 9. 17.7108 10. 10.2466 Correct answer: 26 . 6212 Ω. Explanation: Let : I rms =7 . 2A , V rms = 240 V , and φ = - 37 . The average power delivered by the generator is dissipated as heat in the resistor: P av = I rms V rms cos φ = (7 . 2 A) (240 V) cos( - 37 ) = 1380 . 04 W , and the power dissipated in the resistor is P av = I 2 rms R, so the resistance is R = P av I 2 rms = 1380 . 04 W (7 . 2 A) 2 = 26 . 6212 Ω . 003 (part 2 of 2) 10.0 points Calculate the reactance X L - X C of the cir- cuit. 1. -14.3257 2. -5.83298 3. -11.356 4. -6.41641 5. -8.47813 6. -17.3734 7. -13.0458 8. -20.0605 9. -10.095 10. -8.17144 Correct answer: - 20 . 0605 Ω. Explanation: For the reactance, tan φ = X L - X C R X L - X C = R tan φ = (26 . 6212 Ω) tan( - 37 ) = - 20 . 0605 Ω . 004 (part 1 of 2) 10.0 points A series RLC circuit in which the 2260 Ω resistor and the 46 nF capacitor is connected to an ac generator whose frequency can be varied. When the current is maximized as a function of frequency the current is 0 . 31 A and the frequency is 57 kHz. Determine the inductance. 1. 0.000103069 2. 0.000403053 3. 0.00016538 4. 0.000630546 5. 0.000206394 6. 0.000169486 7. 0.000158314 8. 0.00013442 9. 0.000496381 10. 0.000563798 Correct answer: 0 . 000169486 H. Explanation: Let : f = 57 kHz = 57000 Hz and C = 46 nF = 4 . 6 × 10 - 8 F . The current in the RLC circuit reaches a max- imum when ω 0 L = 1 ω 0 C L = 1 ω 2 0 C = 1 [2 π f ] 2 C = 1 [2 π (57000 Hz)] 2 (4 . 6 × 10 - 8 F) = 0 . 000169486 H . 005 (part 2 of 2) 10.0 points
Background image of page 2
Version 083 – Midterm 3 – Shih – (58250) 3 Determine the rms value of the generator volt- age at this frequency. 1. 59.7788 2. 495.399 3. 335.84 4. 816.093 5. 92.3764 6. 1028.01 7. 538.052 8. 558.628 9. 545.773 10. 143.514 Correct answer: 495 . 399 V.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/03/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 16

Midterm 3-solutions - Version 083 Midterm 3 Shih(58250 This...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online