solutiontest3_pdf

# solutiontest3_pdf - Version 110 – TEST03 – Tsoi –...

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Unformatted text preview: Version 110 – TEST03 – Tsoi – (59090) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points One application of an RL circuit is the gen- eration of time-varying high-voltage from a low-voltage source, as shown in the figure. 1 H 14 Ω 1223 Ω 12 . 6 V S b a The switch is initially set to position a. Af- ter a long time at position ”a” the switch is quickly thrown to ”b”. Compute the volt- age across the inductor immediately after the switch is thrown from ”a” to ”b”. 1. 1042.08 2. 1113.3 3. 1616.19 4. 1397.25 5. 1234.1 6. 1080.35 7. 1490.76 8. 914.427 9. 1664.64 10. 1063.14 Correct answer: 1113 . 3 V. Explanation: L R 2 R 1 E S b a Let : R 2 = 14 Ω and E = 12 . 6 V . When the switch is at “ a ”, the circuit com- prises the battery, the inductor L , and the resistor R 2 . A long time after the switch has been in position “ a ”, the current is steady. This means that the inductor has no response to the current. In other words, the circuit can be considered as consisting of E and R 2 only, with the inductor reduced to a wire. In this case, the current is simply found by Ohm’s Law I = E R 2 = 12 . 6 V 14 Ω = . 9 A . Let : R 1 = 1223 Ω and I = 0 . 9 A . When the switch is thrown from “ a ” to “ b ”, the current in the circuit is the current passing through R 2 , which was found in Part 1 to be 0 . 9 A . From Kirchhoff’s Loop Law, the initial voltage across the inductor is equal to the initial voltage across R 1 and R 2 . So, we have V L = V R 1 + V R 2 = I R 1 + I R 2 = (0 . 9 A) (1223 Ω) + (0 . 9 A) (14 Ω) = 1113 . 3 V . 002 10.0 points Consider a rotating conducting rod with length a . It is fixed at the end O and is rotating clockwise with angular frequency ω (see figure). The rotation occurs in a region in which there is a constant, uniform magnetic field perpendicular to the plane of rotation (pointed up from the page). The magnitude of this field is B . a ω B O A Version 110 – TEST03 – Tsoi – (59090) 2 If B = 2 . 53 T, ω = 1 . 24 rad / s and a = 2 . 54 m, what is the magnitude of the E induced between O and A ? 1. 14.84 2. 21.2915 3. 31.9495 4. 17.6082 5. 43.5656 6. 10.9654 7. 29.9067 8. 6.79845 9. 11.936 10. 10.12 Correct answer: 10 . 12 V. Explanation: The E induced between the two ends of the rod is given by the work done by the magnetic force in bringing a unit charge from A to O , E ind = integraldisplay O A vector F mag · d vectorr q = integraldisplay a r ω B dr = a 2 ω B 2 = (2 . 54 m) 2 (1 . 24 rad / s) (2 . 53 T) 2 = 10 . 12 V . 003 10.0 points A 16 Ω square loop, whose dimensions are 4 . 7 m × 4 . 7 m, is placed in a uniform 0 . 16 T magnetic field that is directed perpendicular to the plane of the loop (see figure)....
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## This note was uploaded on 06/15/2010 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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solutiontest3_pdf - Version 110 – TEST03 – Tsoi –...

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