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Practice HW 10 solution

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

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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 0.0points In the circuit shown below, the current i in the resistor R doubles its original value when the switch S is closed. 117 . 426 Ω 15 Ω 15 Ω 117 . 426 Ω R E S i Find the value of R . Correct answer: 13 . 0094 Ω. 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 Ω , R 2 = 15 Ω , R 3 = 15 Ω , R 4 = 117 . 426 Ω , 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 Ω + 15 Ω = 132 . 426 Ω R 34 = R 3 + R 4 = 15 Ω + 117 . 426 Ω = 132 . 426 Ω R 1234 = 1 R 12 + 1 R 34 = parenleftbigg 1 R 12 + 1 R 34 parenrightbigg - 1 = parenleftbigg 1 132 . 426 Ω + 1 132 . 426 Ω parenrightbigg - 1 = 66 . 2132 Ω 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 Ω + 1 15 Ω parenrightbigg - 1 = 13 . 3009 Ω R 24 = 1 R 2 + 1 R 4 = parenleftbigg 1 R 2 + 1 R 4 parenrightbigg - 1 = parenleftbigg 1 15 Ω + 1 117 . 426 Ω parenrightbigg - 1 = 13 . 3009 Ω 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 Ω) - 2 (13 . 3009 Ω) - 2 (13 . 3009 Ω = 13 . 0094 Ω . keywords:
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song (shs546) – Practice HW 10 Solutions – weathers – (22202) 2 002 0.0points 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 Ω 5 Ω 9 Ω R x Find the electric resistance R x . Correct answer: 15 Ω. Explanation: E S 1 u x y I u R 1 I u R 2 I R 3 I R x I A 0 A Let : R 1 = 3 Ω , R 2 = 5 Ω , and R 3 = 9 Ω . 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 equal V u = V ( i.e., Eq. 1 = 2 and Eq. 3 = 4), we have I u R 1 = I R 3 and I u R 2 = I R x , so I u I = R 3 R 1 = R x R 2 R x = R 2 R 3 R 1 = (5 Ω) (9 Ω) 3 Ω = 15 Ω .
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