HW15Sol - Platt David – Homework 15 – Due 4:00 am –...

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Unformatted text preview: Platt, David – Homework 15 – Due: Oct 12 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 7 questions. Multiple-choice 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(E-6 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 Ω 1 . 9 Ω 1 . 9 Ω 8 . 75 Ω 8 . 75 Ω 8 . 75 Ω 5 . 4 Ω 5 . 4 Ω 5 . 4 Ω 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 Ω. Explanation: Let : R 1 = 1 . 9 Ω , R 2 = 8 . 75 Ω , R 3 = 5 . 4 Ω 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 can’t 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 Ω) + (5 . 4 Ω) 2 × " 1 + r 1 + 4(8 . 75 Ω) 1 . 9 Ω + 5 . 4 Ω # = 12 . 4362 Ω . 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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This note was uploaded on 04/10/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

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HW15Sol - Platt David – Homework 15 – Due 4:00 am –...

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