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Unformatted text preview: billing (cab4763) – hw 7 – opyrchal – (121104) 1 This printout should have 15 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 6 . 1 V and internal resistance of 0 . 82 Ω is connected across a load resistor R . If the current in the circuit is 1 . 58 A, what is the value of R ? Correct answer: 3 . 04076 Ω. Explanation: Let : E = 6 . 1 V , I = 1 . 58 A , and R i = 0 . 82 Ω . The electromotive force E is given by E = I ( R + R i ) R = E I R i = 6 . 1 V 1 . 58 A . 82 Ω = 3 . 04076 Ω . 002 (part 2 of 2) 10.0 points What power is dissipated in the internal re sistance of the battery? Correct answer: 2 . 04705 W. Explanation: The power dissipation due to the internal resistance is P = I 2 R i = (1 . 58 A) 2 (0 . 82 Ω) = 2 . 04705 W . 003 10.0 points Four resistors are connected as shown in the figure. 97 V S 1 c d a b 1 8 Ω 48Ω 69Ω 7 6 Ω Find the resistance between points a and b . Correct answer: 12 . 9432 Ω. Explanation: E B S 1 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 18 Ω , R 2 = 48 Ω , R 3 = 69 Ω , R 4 = 76 Ω , and E = 97 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 billing (cab4763) – hw 7 – opyrchal – (121104) 2 The series connection of R 2 and R 3 gives the equivalent resistance R 23 = R 2 + R 3 = 48 Ω + 69 Ω = 117 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 ) = (76 Ω)[48 Ω + 69 Ω] + (18 Ω) (76 Ω) + (18 Ω) [48 Ω + 69 Ω] = 12366 Ω 2 , so the equivalent resistance is R ab = (18 Ω) (76 Ω) [48 Ω + 69 Ω] (12366 Ω 2 ) = 12 . 9432 Ω . 004 (part 1 of 6) 10.0 points Consider the circuit in the figure. 12 . 6 V S 4 . 0 Ω 7 . 0 Ω 5 . 0 Ω 10 Ω 10 Ω 16 Ω a) Find the current in the 5 . 0 Ω resistor....
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 Spring '10
 Fayngold
 Physics, SEPTA Regional Rail, Correct Answer, Jaguar Racing, RAB, Series and parallel circuits

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