Homework_17

# Homework_17 - Nguyen, Thanh – Homework 17 – Due: Apr 4...

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Unformatted text preview: Nguyen, Thanh – 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 11 V and 3 A respectively, and an output current of 0 . 6 A. If there are 1208 turns turns on the sec- ondary side of the transformer, how many turns are on the primary side? Correct answer: 241 . 6 turns. Explanation: Let : n s = 1208 turns , I p = 3 A , and I s = 0 . 6 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 = (1208 turns) . 6 A 3 A = 241 . 6 turns . keywords: 002 (part 1 of 2) 10 points A step-up transformer is connected to a gen- erator that is delivering 130 V and 100 A. The ratio of the turns on the secondary to the turns on the primary is 886 to 2. What voltage is across the secondary? Correct answer: 57 . 59 kV. Explanation: Let : n s n p = 886 2 = 886 and V p = 130 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 = (130 V) (886) · 1kV 1000V = 57 . 59 kV . 003 (part 2 of 2) 10 points What current flows in the secondary? Correct answer: 225 . 734 mA. Explanation: Let : I p = 100 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 = (100 A) (130 V) 57590 V · 10 3 mA 1A = 225 . 734 mA . keywords: 004 (part 1 of 1) 10 points Consider an audio amplifier with an output impedance of 612 Ω and a speaker that has an input impedance of 24 Ω. A circuit replicating this is shown in the figure below....
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## This note was uploaded on 10/24/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Homework_17 - Nguyen, Thanh – Homework 17 – Due: Apr 4...

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