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Unformatted text preview: Platt, David Homework 15 Due: Oct 12 2005, 4:00 am Inst: Ken Shih 1 This printout should have 7 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Note: For 003, give your answer in units of muC(E6 C) 001 (part 1 of 2) 10 points Hint: The resistance to the right of A B is the same as the resistance to the right of AB; that is, R AB = R A B . 1 . 9 1 . 9 1 . 9 8 . 75 8 . 75 8 . 75 5 . 4 5 . 4 5 . 4 A A B B What is the resistance R AB between the terminals A and B of this infinitely repeating chain of resistors. Correct answer: 12 . 4362 . Explanation: Let : R 1 = 1 . 9 , R 2 = 8 . 75 , R 3 = 5 . 4 and V AB = 28 . 5 V . Series and Parallel Combinations of Resis tors: R series = R 1 + R 2 + R 3 + 1 R parallel = 1 R 1 + 1 R 2 + 1 R 3 + Since the network is infinitely long, we can redraw it as shown. R 1 R 2 R 3 R AB A A B B Then R AB = R 1 + R 3 + R 2 R AB R 2 + R AB (1) = R 2 AB ( R 1 + R 3 ) R AB R 2 ( R 1 + R 3 ) = 0 . This has roots R AB = 1 2 h ( R 1 + R 3 ) q ( R 1 + R 3 ) 2 + 4 R 2 ( R 1 + R 3 ) i , but since we cant have negative resistances, only the positive root is physical R AB = R 1 + R 3 2 " 1 + r 1 + 4 R 2 R 1 + R 3 # = (1 . 9 ) + (5 . 4 ) 2 " 1 + r 1 + 4(8 . 75 ) 1 . 9 + 5 . 4 # = 12 . 4362 . Note: You could also view this as a collection of n sets of the resistor combination, then take the limit as n becomes infinitely large. Equation (1) then becomes R n = R 1 + R 3 + R 2 R n 1 R 2 + R n 1 , n , which is equivalent....
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 Fall '08
 Turner
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