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HW6 - Version One – Homework 6 – Savrasov – 39819 –...

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Unformatted text preview: Version One – Homework 6 – Savrasov – 39819 – May 01, 2006 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Lamp in a Circuit 28:02, trigonometry, numeric, > 1 min, nor- mal. 001 (part 1 of 2) 1 points A lamp having a resistance of 10 Ω is con- nected across 15 V battery. What is the current through the lamp? Correct answer: 1 . 5 A. Explanation: Let : R = 10 Ω and V = 15 V . The current through the lamp is I = V R = 15 V 10 Ω = 1 . 5 A . 002 (part 2 of 2) 1 points What resistance must be connected in series with the lamp to reduce the current to 0 . 5 A? Correct answer: 20 Ω. Explanation: R total = R + R 1 and V = I 1 R total = I 1 R + I 1 R 1 so that R 1 = V- I 1 R I 1 = V I 1- R = 15 V . 5 A- 10 Ω = 20 Ω . Resistors in Parallel 28:02, trigonometry, numeric, > 1 min, nor- mal. 003 (part 1 of 3) 1 points A 16 Ω and a 20 Ω resistor are connected parallel. A difference in potential of 40 V is applied to the combination. What is the equivalent resistance of the parallel circuit? Correct answer: 8 . 88889 Ω. Explanation: Let : R 1 = 16 Ω , R 2 = 20 Ω , and V = 40 V . R 1 and R 2 are in parallel, so 1 R = 1 R 1 + 1 R 2 = R 2 + R 1 R 1 R 2 R = R 1 R 2 R 1 + R 2 = (16 Ω)(20 Ω) 16 Ω + 20 Ω = 8 . 88889 Ω . 004 (part 2 of 3) 1 points What is the current in the circuit? Correct answer: 4 . 5 A....
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HW6 - Version One – Homework 6 – Savrasov – 39819 –...

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