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Unformatted text preview: Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points If the emf of a battery is 6 V and a current of 69 A is measured when the battery is shorted, what is the internal resistance of the battery? Correct answer: 0 . 0869565 . Explanation: Given : E = 6 V and I = 69 A . When the battery with an emf E is shorted, with current I flowing, the internal resistance R i is R i = E I = 6 V 69 A = 0 . 0869565 . 002 (part 1 of 2) 10 points Consider the following two cases. case a: 7 A 7 A = 1 1 R a case b: 7 A 7 A = 1 1 R b Find the ratio of dissipated powers, P a P b . 1. P a P b = 1 8 2. P a P b = 1 2 3. P a P b = 1 4. P a P b = 2 5. P a P b = 1 4 correct 6. P a P b = 4 7. P a P b = 8 Explanation: Given : R 1 = R 2 = R = 1 and I = 7 A . In case a, the two resistors are parallel, so the total resistance is 1 R a = 1 R 1 + 1 R 2 = R 2 + R 1 R 1 R 2 R a = R 1 R 2 R 1 + R 2 = R 2 . In case b, the two resistors are in series, so the total resistance is given by R b = R 1 + R 2 = 2 R. P a = I 2 R a = 1 2 I 2 R P b = I 2 R b = 2 I 2 R = P a P b = 1 2 I 2 R 2 I 2 R = 1 4 . 003 (part 2 of 2) 10 points Find P a . Correct answer: 24 . 5 W. Explanation: P a = 1 2 I 2 R = 1 2 (7 A) 2 (1 ) = 24 . 5 W . 004 (part 1 of 1) 10 points Consider the circuit 22 . 7 45 . 4 4 5 . 4 22 . 7 45 . 4 a b Husain, Zeena Homework 6 Due: Mar 1 2004, 4:00 am Inst: Sonia Paban 2 What is the equivalent resistance between the points a and b ? Correct answer: 39 . 725 . Explanation: Given : R 1 = R = 22 . 7 , R 2 = 2 R = 45 . 4 , R 3 = 2 R = 45 . 4 , R 4 = R = 22 . 7 , and R 5 = 2 R = 45 . 4 . R 1 R 2 R 3 R 4 R 5 a b c d The circuit is redrawn below. R 1 = R R 3 = 2 R R 2 = 2 R R 4 = R R 5 = 2 R a b c d Basic Concepts: R series = R 1 + R 2 + R 3 + 1 R parallel = 1 R 1 + 1 R 2 + 1 R 3 + Solution: This circuit can be analyzed by straightforward series/parallel analysis: R 4 and R 5 are in series, so R 45 = R + 2 R = 3 R . R 2 and R 3 are in parallel with R 45 , so 1 R 2345 = 1 2 R + 1 2 R + 1 3 R = 8 6 R R 2345 = 3 R 4 . R 1 and R 2345 are in series, so R eq = R + 3 R 4 = 7 R 4 = 7 (22 . 7 ) 4 = 39 . 725 . 005 (part 1 of 4) 10 points Twelve resistors, each of equal value R = 6 . 6 , are connected so that each is along one edge of a cube, as shown in the figure. A current of I = 1 . 4 A is introduced at point a. The current will be allowed to exit only along a wire at point b, as drawn....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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