1178864748Xuj - Alvarado Patrick Homework 7 Due Mar 9 2007...

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Alvarado, Patrick – Homework 7 – Due: Mar 9 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The emf of a battery is E = 11 V . When the battery delivers a current of 0 . 8 A to a load, the potential difference between the terminals of the battery is 9 V volts. Find the internal resistance of the battery. Correct answer: 2 . 5 Ω. Explanation: Given : E = 11 V , V load = 9 V , and I = 0 . 8 A . The potential difference across the internal resistance is E - V load , so the internal resis- tance is given by r = E - V load I = 11 V - 9 V 0 . 8 A = 2 . 5 Ω . keywords: 002 (part 1 of 1) 10 points Consider the circuit 23 . 4 Ω 46 . 8 Ω 46 . 8 Ω 23 . 4 Ω 46 . 8 Ω a b What is the equivalent resistance between the points a and b ? Correct answer: 40 . 95 Ω. Explanation: R 1 R 2 R 3 R 4 R 5 a b c d Let : R 1 = R = 23 . 4 Ω , R 2 = 2 R = 46 . 8 Ω , R 3 = 2 R = 46 . 8 Ω , R 4 = R = 23 . 4 Ω , and R 5 = 2 R = 46 . 8 Ω . 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 (23 . 4 Ω) 4 = 40 . 95 Ω .
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Alvarado, Patrick – Homework 7 – Due: Mar 9 2007, 11:00 pm – Inst: Opyrchal, H 2 keywords: 003 (part 1 of 1) 10 points The following diagram shows part of an elec- trical circuit. 11 Ω 3 Ω 6 Ω 7 Ω 2 Ω 4 Ω 5 Ω A B Find the equivalent resistance R eq between points A and B of the resistor network. Correct answer: 4 Ω. Explanation: R 5 R 2 R 6 R 3 R 1 R 4 R 7 A B Let : R 1 = 2 Ω , R 2 = 3 Ω , R 3 = 7 Ω , R 4 = 4 Ω , R 5 = 11 Ω , R 6 = 6 Ω , and R 7 = 5 Ω . Start from the right-hand side in determining the equivalent resistances. Step 1: R 1 , R 2 , and R 3 are in series, so R 123 = R 1 + R 2 + R 3 = 2 Ω + 3 Ω + 7 Ω = 12 Ω . Step 2: R 123 is now parallel with R 4 , so 1 R 1234 = 1 R 4 + 1 R 123 = R 123 + R 4 R 123 R 4 R 1234 = R 123 R 4 R 123 + R 4 = (12 Ω) (4 Ω) 12 Ω + 4 Ω = 3 Ω . Step 3: R 1234 is in series with R 5 and R 6 , so R s = R 1234 + R 5 + R 6 = 3 Ω + 11 Ω + 6 Ω = 20 Ω . Step 4: Finally, R s is parallel with R 7 , so 1 R eq = 1 R s + 1 R 7 = R 7 + R 5 R 5 R 7 R eq = R s R 7 R s + R 7 = (20 Ω) (5 Ω) 20 Ω + 5 Ω = 4 Ω . This problem presents 20 versions all with integer answers. keywords: 004 (part 1 of 2) 10 points Consider the combination of resistors shown in the figure.
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