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# hw6sol - Flores Michael Homework 6 Due Mar 1 2004 4:00 am...

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Flores, Michael – Homework 6 – Due: Mar 1 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 27 questions. Multiple-choice 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 23 V and a current of 39 A is measured when the battery is shorted, what is the internal resistance of the battery? Correct answer: 0 . 589744 Ω. Explanation: Given : E = 23 V and I = 39 A . When the battery with an emf E is shorted, with current I flowing, the internal resistance R i is R i = E I = 23 V 39 A = 0 . 589744 Ω . 002 (part 1 of 2) 10 points Consider the following two cases. case a: 3 A 3 A = 19 Ω 19 Ω R a case b: 3 A 3 A = 19 Ω 19 Ω R b Find the ratio of dissipated powers, P a P b . 1. P a P b = 1 2. P a P b = 1 2 3. P a P b = 1 4 correct 4. P a P b = 1 8 5. P a P b = 2 6. P a P b = 8 7. P a P b = 4 Explanation: Given : R 1 = R 2 = R = 19 Ω and I = 3 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: 85 . 5 W. Explanation: P a = 1 2 I 2 R = 1 2 (3 A) 2 (19 Ω) = 85 . 5 W . 004 (part 1 of 1) 10 points Consider the circuit 21 . 7 Ω 43 . 4 Ω 43 . 4 Ω 21 . 7 Ω 43 . 4 Ω a b

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Flores, Michael – 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: 37 . 975 Ω. Explanation: Given : R 1 = R = 21 . 7 Ω , R 2 = 2 R = 43 . 4 Ω , R 3 = 2 R = 43 . 4 Ω , R 4 = R = 21 . 7 Ω , and R 5 = 2 R = 43 . 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 straight-forward 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 (21 . 7 Ω) 4 = 37 . 975 Ω . 005 (part 1 of 4) 10 points Twelve resistors, each of equal value R = 2 . 9 Ω, are connected so that each is along one edge of a cube, as shown in the figure. A current of I = 9 . 1 A is introduced at point “a”. The current will be allowed to exit only along a wire at point “b”, as drawn. 1 2 3 4 5 6 7 8 9 10 11 12 I a b c d f g e h What is the current through resistor R 2 ? Correct answer: 3 . 03333 A. Explanation: Consider the figure. The current I is in- troduced into node “a”. Three independent resistors of equal value are accessible via this node, namely, resistor R 1 , R 2 , and R 3 . Because these three resistors are equal and the remainder of the circuit is symmetric, the current will split equally between the three paths. Thus, the current in each of the resis- tors R 1 , R 2 , and R 3 is I 1 , 2 , 3 = I 3 = 9 . 1 A 3 = 3 . 03333 A 006 (part 2 of 4) 10 points What is the current through resistor R 7 ?
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