HW6 - Version One Homework 6 Savrasov 39819 May 01, 2006 1...

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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 is con- nected across 15 V battery. What is the current through the lamp? Correct answer: 1 . 5 A. Explanation: Let : R = 10 and V = 15 V . The current through the lamp is I = V R = 15 V 10 = 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 . 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 = 20 . Resistors in Parallel 28:02, trigonometry, numeric, > 1 min, nor- mal. 003 (part 1 of 3) 1 points A 16 and a 20 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 . Explanation: Let : R 1 = 16 , R 2 = 20 , 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 )(20 ) 16 + 20 = 8 . 88889 . 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 May 01, 2006 1...

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