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# 1-6 - Create assignment 00224 Homework 6 Feb 22 at 9:30 am...

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Create assignment, 00224, Homework 6, Feb 22 at 9:30 am 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Adding a Resistor 28:03, trigonometry, numeric, > 1 min, nor- mal. 001 A loop circuit has a resistance of R 1 and a cur- rent of 2 A. The current is reduced to 1 . 6 A when an additional 3 Ω resistor is added in series with R 1 . What is the value of R 1 ? Correct answer: 12 Ω. Explanation: Let : I 1 = 2 A , I 2 = 1 . 6 A , and R 2 = 3 Ω . Let the current with R 1 be I 1 and the cur- rent with additional resistance R 2 be I 2 . Then since the emf and the internal resistance of the battery is a constant, we have I 1 R 1 = I 2 ( R 1 + R 2 ) I 1 R 1 = I 2 R 1 + I 2 R 2 R 1 = I 2 I 1 - I 2 R 2 = 1 . 6 A 2 A - 1 . 6 A (3 Ω) = 12 Ω . keywords: Serway CP 18 06 28:04, trigonometry, numeric, > 1 min, nor- mal. 002 Consider the circuit I 30 V S 18 Ω 9 Ω 6 Ω 12 Ω Find the equivalent resistance. Correct answer: 15 Ω. Explanation: I E S R 2 R 3 R 4 R 1 Let : R 1 = 12 Ω , R 2 = 18 Ω , R 3 = 9 Ω , and R 4 = 6 Ω . R 2 , R 3 and R 4 are in parallel, so 1 R 234 = 1 R 2 + 1 R 3 + 1 R 4 = R 3 R 4 + R 2 R 4 + R 2 R 3 R 2 R 3 R 4 R 234 = R 2 R 3 R 4 R 3 R 4 + R 2 R 4 + R 2 R 3 Since R 3 R 4 + R 2 R 4 + R 2 R 3

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1-6 - Create assignment 00224 Homework 6 Feb 22 at 9:30 am...

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