solutions 17 - Patel, Kinal Homework 17 Due: Apr 4 2008,...

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Unformatted text preview: Patel, Kinal Homework 17 Due: Apr 4 2008, 7:00 pm Inst: Weathers 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A transformer has input voltage and current of 12 V and 5 A respectively, and an output current of 0 . 7 A. If there are 1365 turns turns on the sec- ondary side of the transformer, how many turns are on the primary side? Correct answer: 191 . 1 turns. Explanation: Let : n s = 1365 turns , I p = 5 A , and I s = 0 . 7 A . Energy is conserved, so P p = P s I p V p = I s V s V p V s = I s I s For the transformer V n n p n s = V p V s = I s I p n = n s I s I p = (1365 turns) . 7 A 5 A = 191 . 1 turns . keywords: 002 (part 1 of 2) 10 points A step-up transformer is connected to a gen- erator that is delivering 133 V and 110 A. The ratio of the turns on the secondary to the turns on the primary is 1040 to 5. What voltage is across the secondary? Correct answer: 27 . 664 kV. Explanation: Let : n s n p = 1040 5 = 1040 and V p = 133 V . V n so V s V p = n s n p The stepped up voltage in secondary is V s = V p n s n p = (133 V)(1040) 1kV 1000V = 27 . 664 kV . 003 (part 2 of 2) 10 points What current flows in the secondary? Correct answer: 528 . 846 mA. Explanation: Let : I p = 110 A . Energy is conserved, so P p = P s I s V s = I p V p I s = I p V p V s = (110 A)(133 V) 27664 V 10 3 mA 1A = 528 . 846 mA . keywords: 004 (part 1 of 1) 10 points Consider an audio amplifier with an output impedance of 663 and a speaker that has an input impedance of 36 . A circuit replicating this is shown in the figure below....
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This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.

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solutions 17 - Patel, Kinal Homework 17 Due: Apr 4 2008,...

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