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Unformatted text preview: aljabr (faa335) Hw10 Ross (89251) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The emf of a battery is E = 16 V . When the battery delivers a current of 1 A to a load, the potential difference between the terminals of the battery is 14 V volts. Find the internal resistance of the battery. Correct answer: 2 . Explanation: Given : E = 16 V , V load = 14 V , and I = 1 A . The potential difference across the internal resistance is E  V load , so the internal resis tance is given by r = E  V load I = 16 V 14 V 1 A = 2 . 002 (part 1 of 4) 10.0 points A battery has an emf of 12 V and an inter nal resistance of 0 . 15 . Its terminals are connected to a load resistance of 2 . Find the current in the circuit. Correct answer: 5 . 5814 A. Explanation: Let : E = 12 V , R = 2 , and r = 0 . 15 . The total resistance is R + r , so I = E R + r = 12 V 2 + 0 . 15 = 5 . 5814 A . 003 (part 2 of 4) 10.0 points Calculate the terminal voltage of the battery. Correct answer: 11 . 1628 V. Explanation: The terminal voltage V of the battery is equal to V r = E  I r = 12 V (5 . 5814 A) (0 . 15 ) = 11 . 1628 V . 004 (part 3 of 4) 10.0 points Find the power dissipated in the load resis tor. Correct answer: 62 . 3039 W. Explanation: The power dissipated in the load resistor is P R = I 2 R = (5 . 5814 A) 2 (2 ) = 62 . 3039 W . 005 (part 4 of 4) 10.0 points Find the power dissipated in the battery. Correct answer: 4 . 6728 W. Explanation: The power dissipated in the battery is P r = I 2 r = (5 . 5814 A) 2 (0 . 15 ) = 4 . 6728 W . 006 10.0 points The current in a circuit is tripled by connect ing a 747 resistor parallel with the resis tance of the circuit....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue UniversityWest Lafayette.
 Spring '10
 Gavrine

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