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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 Ω. 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 Ω . 002 (part 1 of 2) 10 points Consider the following two cases. case a: 7 A 7 A = 1 Ω 1 Ω R a case b: 7 A 7 A = 1 Ω 1 Ω 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 Ω 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 Ω) = 24 . 5 W . 004 (part 1 of 1) 10 points Consider the circuit 22 . 7 Ω 45 . 4Ω 4 5 . 4 Ω 22 . 7Ω 45 . 4 Ω 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 Ω. Explanation: Given : R 1 = R = 22 . 7 Ω , R 2 = 2 R = 45 . 4 Ω , R 3 = 2 R = 45 . 4 Ω , R 4 = R = 22 . 7 Ω , and R 5 = 2 R = 45 . 4 Ω . 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 Ω) 4 = 39 . 725 Ω . 005 (part 1 of 4) 10 points Twelve resistors, each of equal value R = 6 . 6 Ω, 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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 Spring '08
 Turner
 Work, Resistor, Correct Answer, Electrical resistance, Series and parallel circuits, Husain, Sonia Paban

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