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solution39 - cherubin(mcc2645 – Homework 3 DC Circuits...

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Unformatted text preview: cherubin (mcc2645) – Homework 3 DC Circuits – Catala – (12054) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the circuit shown in the figure. I 40 V 28 Ω 7 Ω 35 Ω Find its equivalent resistance. Correct answer: 33 . 8333 Ω. Explanation: I E R a R b R c Let : R a = 28 Ω , R b = 7 Ω , and R c = 35 Ω . R c and R b are in parallel, so 1 R bc = 1 R b + 1 R c R bc = parenleftbigg 1 R b + 1 R c parenrightbigg- 1 R bc = parenleftbigg 1 7 Ω + 1 35 Ω parenrightbigg- 1 = 5 . 83333 Ω R bc and R a are in series, so R eq = 28 Ω + 5 . 83333 Ω = 33 . 8333 Ω . 002 (part 1 of 3) 10.0 points A total charge of 18 mC passes through a cross-sectional area of a nichrome wire in 2 . 7 s. a) What is the current in the wire? Correct answer: 0 . 00666667 A. Explanation: Let : Δ Q 1 = 18 mC and Δ t 1 = 2 . 7 s . I = Δ Q Δ t = . 018 C 2 . 7 s = . 00666667 A . 003 (part 2 of 3) 10.0 points b) How many electrons pass through the cross-sectional area in 17.0 s? Correct answer: 7 . 08333 × 10 17 . Explanation: Let : q e = 1 . 6 × 10- 19 C and Δ t 2 = 17 . 0 s . Δ Q = N q e N = Δ Q q e = I Δ t 2 q e = (0 . 00666667 A)(17 s) 1 . 6 × 10- 19 C = 7 . 08333 × 10 17 . 004 (part 3 of 3) 10.0 points c) If the number of charges that pass through the cross-sectional area during the given time interval doubles, what is the resulting cur- rent? Correct answer: 0 . 0133333 A. Explanation: Let : Δ Q 2 = 2 Δ Q 1 . cherubin (mcc2645) – Homework 3 DC Circuits – Catala – (12054) 2 I = 2 Δ Q 1 Δ t 1 = 2 (0 . 018 C) 2 . 7 s = . 0133333 A . 005 (part 1 of 2) 10.0 points The power supplied to the circuit shown in the figure is 7.00 W. E 2 . 0 Ω 14 . 0 Ω 6 . 0 Ω 4 . 0 Ω 2 . 0 Ω a) Find the equivalent resistance of the cir- cuit. Correct answer: 3 . 60784 Ω. Explanation: E R 1 R 2 R 3 R 4 R 5 Let : R 1 = 2 . 0 Ω , R 2 = 14 . 0 Ω , R 3 = 6 . 0 Ω , R 4 = 4 . 0 Ω , and R 5 = 2 . 0 Ω ....
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