# HW13Sol - Platt, David Homework 13 Due: Oct 7 2005, 4:00 am...

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Unformatted text preview: Platt, David Homework 13 Due: Oct 7 2005, 4:00 am Inst: Ken Shih 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. 001 (part 1 of 4) 10 points Three small lamps are connected to a 12 V battery, as shown in the figure. Assume the battery is ideal (has no internal resistance and given in the figure) and the connecting wires have no resistance. Unlike most real bulbs, the resistances (given in the figure) of the bulbs do not vary. The switch S is closed. 12 V 5 . 5 1 . 4 1 . 6 S a) What is the equivalent resistance of this circuit? Correct answer: 6 . 24667 . Explanation: V R 1 R 2 R 3 S Let : R 1 = 5 . 5 , R 2 = 1 . 4 , and R 3 = 1 . 6 . For resistors in parallel, 1 R eq,p = 1 R a + 1 R b R 23 = 1 R 2 + 1 R 3 - 1 = 1 (1 . 4 ) + 1 (1 . 6 ) - 1 = 0 . 746667 For resistors in series, R eq,s = R a + R b R eq = (5 . 5 ) + (0 . 746667 ) = 6 . 24667 . 002 (part 2 of 4) 10 points b) What is the total current of this circuit? Correct answer: 1 . 92102 A. Explanation: Let : V = 12 V . V = IR I = V R eq = (12 V) (6 . 24667 ) = 1 . 92102 A . 003 (part 3 of 4) 10 points c) What is the current in the 1 . 4 bulb? Correct answer: 1 . 02455 A. Explanation: V 2 = V 3 = V 23 V 23 = IR 23 = (1 . 92102 A) (0 . 746667 ) = 1 . 43436 V ....
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## HW13Sol - Platt, David Homework 13 Due: Oct 7 2005, 4:00 am...

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