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Unformatted text preview: toupal (rgt374) – Homework 06 – Chiu – (58295) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A battery with an emf of 14 . 9 V and internal resistance of 0 . 44 Ω is connected across a load resistor R . If the current in the circuit is 1 . 1 A, what is the value of R ? Correct answer: 13 . 1055 Ω. Explanation: Let : E = 14 . 9 V , I = 1 . 1 A , and R i = 0 . 44 Ω . The electromotive force E is given by E = I ( R + R i ) R = E I − R i = 14 . 9 V 1 . 1 A − . 44 Ω = 13 . 1055 Ω . 002 (part 2 of 2) 10.0 points What power is dissipated in the internal re sistance of the battery? Correct answer: 0 . 5324 W. Explanation: The power dissipation due to the internal resistance is P = I 2 R i = (1 . 1 A) 2 (0 . 44 Ω) = . 5324 W . 003 (part 1 of 2) 10.0 points Four resistors are connected as shown in the figure. 95 V S 1 c d a b 1 2 Ω 49Ω 61Ω 8 8 Ω Find the resistance between points a and b . Correct answer: 9 . 63504 Ω. Explanation: E B S 1 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 12 Ω , R 2 = 49 Ω , R 3 = 61 Ω , R 4 = 88 Ω , and E = 95 V . Ohm’s law is V = I R . A good rule of thumb is to eliminate junc tions connected by zero resistance. E B a d b c R 1 R 2 R 3 R 4 toupal (rgt374) – Homework 06 – Chiu – (58295) 2 The series connection of R 2 and R 3 gives the equivalent resistance R 23 = R 2 + R 3 = 49 Ω + 61 Ω = 110 Ω . The total resistance R ab between a and b can be obtained by calculating the resistance in the parallel combination of the resistors R 1 , R 4 , and R 23 ; i.e. , 1 R ab = 1 R 1 + 1 R 2 + R 3 + 1 R 4 = R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) R 1 R 4 ( R 2 + R 3 ) R ab = R 1 R 4 ( R 2 + R 3 ) R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) The denominator is R 4 ( R 2 + R 3 ) + R 1 R 4 + R 1 ( R 2 + R 3 ) = (88 Ω)[49 Ω + 61 Ω] + (12 Ω) (88 Ω) + (12 Ω) [49 Ω + 61 Ω] = 12056 Ω 2 , so the equivalent resistance is R ab = (12 Ω) (88 Ω) [49 Ω + 61 Ω] (12056 Ω 2 ) = 9 . 63504 Ω . 004 (part 2 of 2) 10.0 points What is the current in the 49 Ω resistor? Correct answer: 0 . 863636 A. Explanation: The voltages across R 2 and R 3 , respec tively, (the voltage between a and b ) are V ab = V 23 = 95 V , and we have I 23 = I 3 = I 2 = V ab R 23 = 95 V 110 Ω = . 863636 A . 005 10.0 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 2 Ω 8 Ω 4 Ω R x Find the electric resistance R X . 1. R X = 16 Ω . correct 2. R X = 22 Ω . 3. R X = 8 Ω . 4. R X = 9 Ω . 5. R X = 20 Ω . 6. R X = 4 Ω ....
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This note was uploaded on 04/26/2009 for the course PHY 58310 taught by Professor Chiu during the Spring '09 term at University of Texas.
 Spring '09
 Chiu
 Physics, Work

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