121homework 7

# 121homework 7 - Felder Jacob Homework 8 Due 9:00 pm Inst...

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Felder, Jacob – Homework 8 – Due: Oct 31 2006, 9:00 pm – Inst: Vitaly 1 This print-out should have 26 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 3) 10 points Three 24 W, 116 V lightbulbs are connected across a 116 V power source. 116 V R 1 R 2 R 3 Find the total power delivered to the three bulbs. Correct answer: 16 W. Explanation: Let : P = 24 W , Δ V l = 116 V , and Δ V s = 116 V . The resistance of each if the three bulbs is given by R = V l ) 2 P = (116 V) 2 24 W = 560 . 667 Ω . As connected the parallel combination of R 2 and R 3 is in series with R 1 . Thus, the equiva- lent resistance of circuit is R eq = R 1 + 1 R 2 + 1 R 3 - 1 = 560 . 667 Ω + 1 560 . 667 Ω + 1 560 . 667 Ω - 1 = 841 Ω . The total power delivered is P total = V s ) 2 R eq = (116 V) 2 841 Ω = 16 W . 002 (part 2 of 3) 10 points Find the potential difference across R 1 . Correct answer: 77 . 3333 V. Explanation: The potential difference across R 1 is Δ V 1 = I R 1 = Δ V s R eq · R = 116 V 841 Ω · (560 . 667 Ω) = 77 . 3333 V . 003 (part 3 of 3) 10 points Find the potential difference across R 2 . Correct answer: 38 . 6667 V. Explanation: The potential difference across R 2 and R 3 is Δ V 2 = Δ V s - Δ V 1 = 116 V - 77 . 3333 V = 38 . 6667 V . keywords: 004 (part 1 of 8) 10 points In the figure below consider the case where switch S 1 is closed and switch S 2 is open. 57 V S 1 S 2 c d a b 13 Ω 27 Ω 38 Ω 49 Ω

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Felder, Jacob – Homework 8 – Due: Oct 31 2006, 9:00 pm – Inst: Vitaly 2 Find the current in the path from a to c . Correct answer: 1 . 11765 A. Explanation: E B S 1 S 2 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 13 Ω , R 2 = 27 Ω , R 3 = 38 Ω , R 4 = 49 Ω , and E B = 57 V . Redrawing the figure, we have E B c d a b R 1 R 2 R 3 R 4 R 1 and R 3 are in series, so R 13 = R 1 + R 3 = 13 Ω + 38 Ω = 51 Ω . R 2 and R 4 are in series, so R 24 = R 2 + R 4 = 27 Ω + 49 Ω = 76 Ω . Simplifying the circuit, we have E B a b R 13 R 24 R 13 and R 24 are parallel, so 1 R ab = 1 R 13 + 1 R 24 = R 24 + R 13 R 13 R 24 R ab = R 13 R 24 R 13 + R 24 = (51 Ω) (76 Ω) 51 Ω + 76 Ω = 30 . 5197 Ω . E B a b R ab R 1 and R 3 are in series, so I 1 = I 13 = E B R 13 = 57 V 51 Ω = 1 . 11765 A . 005 (part 2 of 8) 10 points Find the current in the path from a to d . Correct answer: 0 . 75 A. Explanation: R 2 and R 4 are in series, so I 2 = I 24 = E B R 24 = 57 V 76 Ω = 0 . 75 A . 006 (part 3 of 8) 10 points Find the current in the path from c to b . Correct answer: 1 . 11765 A. Explanation: R 1 and R 3 are in series, so I 3 = I 1 = 1 . 11765 A . 007 (part 4 of 8) 10 points Find the current in the path from d to b . Correct answer: 0 . 75 A. Explanation: R 2 and R 4 are in series, so I 4 = I 2 = 0 . 75 A .
Felder, Jacob – Homework 8 – Due: Oct 31 2006, 9:00 pm – Inst: Vitaly 3 008 (part 5 of 8) 10 points Now consider the case where switch S 2 is also closed, so 57 V S 1 S 2 c d a b 13 Ω 27 Ω 38 Ω 49 Ω Find the current in the path from a to c through the 13 Ω resistor. Correct answer: 1 . 27497 A. Explanation: A good rule of thumb is to eliminate junc- tions connected by zero resistance.

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