GPII - Homework 4, Circuits - solutions

GPII - Homework 4, Circuits - solutions - honea (cth632)...

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Unformatted text preview: honea (cth632) H4Circuits avram (1956) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points A 47 lamp and a 59 lamp are connected in series and placed across a potential of 47 V. What is the equivalent resistance of the circuit? Correct answer: 106 . Explanation: Let : R 1 = 47 , R 2 = 59 , and V = 47 V . The equivalent resistance of the circuit is R = R 1 + R 2 = 47 + 59 = 106 . 002 (part 2 of 4) 10.0 points What is the current in the circuit? Correct answer: 0 . 443396 A. Explanation: I = V R = 47 V 106 = . 443396 A . 003 (part 3 of 4) 10.0 points What is the voltage drop across the first lamp? Correct answer: 20 . 8396 V. Explanation: This is a series circuit, so the voltage drop across the first lamp is V 1 = I R 1 = (0 . 443396 A) (47 ) = 20 . 8396 V . 004 (part 4 of 4) 10.0 points What is the power dissipated in the first lamp? Correct answer: 9 . 24021 W. Explanation: P 1 = I V 1 = (0 . 443396 A) (20 . 8396 V) = 9 . 24021 W 005 10.0 points Resistances of 2 . 4 , 3 . 6 , and 5 . 1 and a 27 V battery are all in series. Find the potential difference across the first (2 . 4 ) resistor....
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This note was uploaded on 04/12/2011 for the course PHYS 1402 taught by Professor Avram during the Spring '10 term at Austin Community College.

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GPII - Homework 4, Circuits - solutions - honea (cth632)...

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