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# HW12Sol - Platt David Homework 12 Due Oct 5 2005 4:00 am...

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Platt, David – Homework 12 – Due: Oct 5 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 8 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 Consider the following circuit containing identical bulbs A , B , C , D , and E respec- tively. Denote the potentials across bulbs as V A , V B , V C , V D , and V E . E A B C D E Rank the potentials across the bulbs. 1. V A = V B = V C = V D = V E 2. V A = V B > V A = V D = V E 3. V A = V D = V E > V B > V C 4. V C > V B > V A > V D > V E 5. V A = V B = V C > V D > V E 6. V A = V B = V C > V D = V E correct 7. V E = V D > V A > V B = V C 8. V A = V D = V E > V B = V C 9. V B = V C > V A = V D = V E 10. V A = V C > V B > V D = V E Explanation: The potential across bulbs A , B and C is V . The potential across bulbs D and E is V 2 . Thus i A = i B = i C = V R i D = i E = V 2 R . 002 (part 1 of 2) 10 points Four resistors are connected as shown in the figure. 96 V S 1 c d a b 19 Ω 32 Ω 64 Ω 79 Ω Find the resistance between points a and b . Correct answer: 38 . 7151 Ω. Explanation: E B S 1 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 19 Ω , R 2 = 32 Ω , R 3 = 64 Ω , R 4 = 79 Ω , and E B = 96 V . Ohm’s law is V = I R . A good rule of thumb is to eliminate junc- tions connected by zero resistance.

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Platt, David – Homework 12 – Due: Oct 5 2005, 4:00 am – Inst: Ken Shih 2 E B c a b R 1 R 2 R 3 R 4 The parallel connection of R 1 and R 2 gives the equivalent resistance 1 R 12 = 1 R 1 + 1 R 2 = R 2 + R 1 R 1 R 2 R 12 = R 1 R 2 R 1 + R 2 = (19 Ω) (32 Ω) 19 Ω + 32 Ω = 11 . 9216 Ω . E c a b R 12 R 3 R 4 The series connection of R 12 and R 3 gives the equivalent resistance R 123 = R 12 + R 3 = 11 . 9216 Ω + 64 Ω = 75 . 9216 Ω . E B a b R 123 R 4 The parallel connection of R 123 and R 4 gives the equivalent resistance 1 R ab = 1 R 123 + 1 R 4 = R 4 + R 123 R 123 R 4 R ab = R 123 R 4 R 123 + R 4 = (75 . 9216 Ω) (79 Ω) 75 . 9216 Ω + 79 Ω = 38 . 7151 Ω . or combining the above steps, the equivalent resistance is R ab = R 1 R 2 R 1 + R 2 + R 3 R 4 R 1 R 2 R 1 + R 2 + R 3 + R 4 = (19 Ω) (32 Ω) 19 Ω + 32 Ω + 64 Ω (79 Ω) (19 Ω) (32 Ω) 19 Ω + 32 Ω + 64 Ω + 79 Ω = 38 . 7151 Ω .
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HW12Sol - Platt David Homework 12 Due Oct 5 2005 4:00 am...

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