Practice HW 10 solution

Practice HW 10 solution - song (shs546) Practice HW 10...

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Unformatted text preview: song (shs546) Practice HW 10 Solutions weathers (22202) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 0.0 points In the circuit shown below, the current i in the resistor R doubles its original value when the switch S is closed. 117 . 426 15 15 117 . 426 R E S i Find the value of R . Correct answer: 13 . 0094 . Explanation: For resistors in parallel, 1 R eq,p = 1 R a + 1 R b . For resistors in series, R eq,s = R a + R b . R 1 R 2 R 3 R 4 R E S Let : R 1 = 117 . 426 , R 2 = 15 , R 3 = 15 , R 4 = 117 . 426 , and R eq,c = 1 2 R eq,o . for the current to double when the switch is closed, the equivalent resistance must drop to one-half its value when the switch is open. With the switch open, R 12 = R 1 + R 2 = 117 . 426 + 15 = 132 . 426 R 34 = R 3 + R 4 = 15 + 117 . 426 = 132 . 426 R 1234 = 1 R 12 + 1 R 34 = parenleftbigg 1 R 12 + 1 R 34 parenrightbigg- 1 = parenleftbigg 1 132 . 426 + 1 132 . 426 parenrightbigg- 1 = 66 . 2132 R eq,o = R + R 1234 . When the switch is closed, R 13 = 1 R 1 + 1 R 3 = parenleftbigg 1 R 1 + 1 R 3 parenrightbigg- 1 = parenleftbigg 1 117 . 426 + 1 15 parenrightbigg- 1 = 13 . 3009 R 24 = 1 R 2 + 1 R 4 = parenleftbigg 1 R 2 + 1 R 4 parenrightbigg- 1 = parenleftbigg 1 15 + 1 117 . 426 parenrightbigg- 1 = 13 . 3009 R eq,c = R + R 13 + R 24 . The new resistance is one-half the original, so R + R 13 + R 24 = 1 2 ( R + R 1234 ) 2 R + 2 R 13 + 2 R 24 = R + R 1234 , so R = R 1234- 2 R 13- 2 R 24 R = (66 . 2132 )- 2 (13 . 3009 )- 2 (13 . 3009 = 13 . 0094 . keywords: song (shs546) Practice HW 10 Solutions weathers (22202) 2 002 0.0 points The following diagram shows a closed elec- trical circuit. The ammeter in the center of the resistive network reads zero amperes. E S 1 A 3 5 9 R x Find the electric resistance R x . Correct answer: 15 . Explanation: E S 1 u x y I u R 1 I u R 2 I R 3 I R x I A 0A Let : R 1 = 3 , R 2 = 5 , and R 3 = 9 . If the ammeter reads zero I A = 0 A, the two ends of the ammeter should be equipotentials: V u A = V A . This means that the potential drop from x or y to each side of the ammeter through either the upper u or lower part of the circuit must be the same: V x- V u = I u R 1 (1) V x- V = I R 3 (2) V u- V y = I u R 2 (3) V - V y = I R x (4) Setting the potential across the ammeter...
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Practice HW 10 solution - song (shs546) Practice HW 10...

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