17. Circuit Review-solutions

# 17. Circuit Review-solutions - horacek(mch2665 – 17...

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Unformatted text preview: horacek (mch2665) – 17 - Circuit Review – gresham – (11223) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. I will be checking for 30 points at noon on Wednesday (3/7)and 60 points noon on Thursday (3/8) 001 (part 1 of 2) 10.0 points A 2 Ω resistor, an 8 Ω resistor, and a 14 Ω resistor are connected in parallel across a 24 V battery. Determine the equivalent resistance for the circuit. Correct answer: 1 . 4359 Ω. Explanation: Let : R 1 = 2 Ω , R 2 = 8 Ω , and R 3 = 14 Ω . The resistors are in parallel, so 1 R eq = 1 R 1 + 1 R 2 + 1 R 3 R eq = parenleftbigg 1 R 1 + 1 R 2 + 1 R 3 parenrightbigg- 1 = parenleftbigg 1 2 Ω + 1 8 Ω + 1 14 Ω parenrightbigg- 1 = 1 . 4359 Ω . 002 (part 2 of 2) 10.0 points b) Determine the current in the circuit. Correct answer: 16 . 7143 A. Explanation: Let : Δ V = 24 V . Δ V = I R I = Δ V R eq = 24 V 1 . 4359 Ω = 16 . 7143 A . 003 10.0 points A 7.0 Ω resistor and a 6.0 Ω resistor are con- nected in series with a battery. The potential difference across the 6.0 Ω resistor is measured as 6 V. Find the potential difference across the bat- tery. Correct answer: 13 V. Explanation: Let : R 1 = 7 . 0 Ω , R 2 = 6 . 0 Ω , and Δ V 2 = 6 V . Basic Concepts: R eq = R 1 + R 2 Δ V = IR I 2 = I Solution: R eq = 7 Ω + 6 Ω = 13 Ω I 2 = Δ V 2 R 2 = 6 V 6 Ω = 1 Ω Δ V = IR eq = (1 Ω)(13 Ω) = 13 V . 004 (part 1 of 2) 10.0 points A hair dryer draws a current of 12 A. a) How long does it take for 6 . 6 × 10 3 C of charge to pass through the hair dryer? Correct answer: 9 . 16667 min. Explanation: Let : I = 12 A and Δ Q = 6 . 6 × 10 3 C . I = Δ Q Δ t Δ t = Δ Q I = 6 . 6 × 10 3 C 12 A · 1 min 60 s = 9 . 16667 min . horacek (mch2665) – 17 - Circuit Review – gresham – (11223) 2 005 (part 2 of 2) 10.0 points b) How many electrons does this amount of charge represent? Correct answer: 4 . 125 × 10 22 . Explanation: Let : q e = 1 . 6 × 10- 19 C . Δ Q = N q e N = Δ Q q e = 6 . 6 × 10 3 C 1 . 6 × 10- 19 C = 4 . 125 × 10 22 electrons ....
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## This note was uploaded on 03/25/2012 for the course PHYSICS 101 taught by Professor Gresham during the Spring '12 term at Florida Memorial.

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17. Circuit Review-solutions - horacek(mch2665 – 17...

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