homework 7 - bhakta (nnb232) – HW #7 – Antoniewicz –...

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Unformatted text preview: bhakta (nnb232) – HW #7 – Antoniewicz – (56490) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 5.0 points The primary of a step-down transformer has 300 turns turns and is connected to a 130 V rms line. The secondary is to supply 28 A at 6 V. Find the rms current in the primary. Correct answer: 1 . 29231 A. Explanation: Let : V p = 130 V , V s = 6 V , and I p = 28 A . Because of 100 percent efficiency, we have I p V p = I s V s , so the current in the primary is I p = V s V p I s = parenleftbigg 6 V 130 V parenrightbigg (28 A) = 1 . 29231 A . 002 (part 2 of 2) 5.0 points Find the whole number of turns in the sec- ondary, assuming 100 percent efficiency. Correct answer: 14 turns. Explanation: Let : V p = 130 V , V s = 6 V , and N p = 300 turns . Because of 100 percent efficiency, we have V s N p = V p N s , so the number of turns in the secondary is N s = V s V p N p = parenleftbigg 6 V 130 V parenrightbigg 300 turns ≈ 14 turns . 003 10.0 points A solenoid with circular cross section pro- duces a steadily increasing magnetic flux through its cross section. There is a octago- nally shaped circuit surrounding the solenoid as shown. The increasing magnetic flux gives rise to a counterclockwise induced emf E . Initial Case: The circuit consists of two identical light bulbs of equal resistance R con- nected in series, leading to a loop equation E - 2 iR = 0. B B B B X Y i i Figure 1: The corresponding electrical power con- sumed by bulb X and bulb Y are P X and P Y , respectively. Primed ′ Case: Now connect the points C and D with a wire CAD, as in figure 2. B B B B X Y D C A i 2 i 1 i 3 i 3 Figure 2: The corresponding electrical power con- sumed by bulb X and bulb Y are P ′ X and P ′ Y , respectively. What are the ratios P ′ X P X and P ′ Y P Y , respec- tively? bhakta (nnb232) – HW #7 – Antoniewicz – (56490) 2 1. P ′ X P X = 0 and P ′ Y P Y = 1 4 2. P ′ X P X = 2 and P ′ Y P Y = 0 3. P ′ X P X = 0 and P ′ Y P Y = 0 4. P ′ X P X = 4 and P ′ Y P Y = 1 2 5. P ′ X P X = 0 and P ′ Y P Y = 1 2 6. P ′ X P X = 2 and P ′ Y P Y = 1 4 7. P ′ X P X = 1 and P ′ Y P Y = 1 8. P ′ X P X = 4 and P ′ Y P Y = 0 correct 9. P ′ X P X = 2 and P ′ Y P Y = 1 2 10. P ′ X P X = 4 and P ′ Y P Y = 1 4 Explanation: Let E and R be the induced emf and resis- tance of the light bulbs, respectively. For the first case, since the two bulbs are in series, the equivalent resistance is simply R eq = R + R = 2 R and the current through the bulbs is i = E 2 R , so the power consumed by bulb X is P X = parenleftbigg E 2 R parenrightbigg 2 R = E 2 4 R ....
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This note was uploaded on 02/01/2011 for the course PHYSICS 302L taught by Professor Antanioweitz during the Fall '10 term at University of Texas at Austin.

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homework 7 - bhakta (nnb232) – HW #7 – Antoniewicz –...

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