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Unformatted text preview: soto (rrs946) – Homework 7 – Hoffmann – (56605) 1 This printout should have 12 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 = 18 V . When the battery delivers a current of 0 . 6 A to a load, the potential difference between the terminals of the battery is 16 V volts. Find the internal resistance of the battery. Correct answer: 3 . 33333 Ω. Explanation: Given : E = 18 V , V load = 16 V , and I = 0 . 6 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 = 18 V 16 V . 6 A = 3 . 33333 Ω . 002 10.0 points Four copper wires of equal length are con nected in series. Their crosssectional areas are 1 . 2 cm 2 , 2 . 2 cm 2 , 4 . 3 cm 2 , and 7 cm 2 . If a voltage of 107 V is applied to the ar rangement, determine the voltage across the 2 . 2 cm 2 wire. Correct answer: 29 . 241 V. Explanation: Let : A 1 = 1 . 2 cm 2 , A 2 = 2 . 2 cm 2 , A 3 = 4 . 3 cm 2 , A 4 = 7 cm 2 , and V tot = 107 V . The resistance of the copper wire is propor tional to the length of the wire and inversely proportional to the cross sectional area of the wire. Since the lengths are the same, and the current i is the same for all resistors connected in series, taking r ∝ 1 A , etc, we have i = V tot r 1 + r 2 + r 3 + r 4 = V 2 r 2 , where V tot is the difference in voltage across all for wires and V 2 is the difference in voltage across the second wire only. Solving for V 2 , we have V 2 = r 2 r 1 + r 2 + r 3 + r 4 V tot = 1 A 2 1 A 1 + 1 A 2 + 1 A 3 + 1 A 4 V tot = 1 2 . 2 cm 2 1 1 . 2 cm 2 + 1 2 . 2 cm 2 + 1 4 . 3 cm 2 + 1 7 cm 2 × (107 V) = 29 . 241 V . 003 10.0 points Four resistors are connected as shown in the figure. 2 1 Ω 5 2 Ω 8 8 Ω 94 V 37Ω S 1 a b c d Find the resistance between points a and b . Correct answer: 37 . 5165 Ω. Explanation: soto (rrs946) – Homework 7 – Hoffmann – (56605) 2 R 1 R 3 R 4 E B R 2 S 1 a b c d Let : R 1 = 21 Ω , R 2 = 37 Ω , R 3 = 52 Ω , R 4 = 88 Ω , and E B = 94 V . Ohm’s law is V = I R . A good rule of thumb is to eliminate junc tions connected by zero resistance....
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This note was uploaded on 11/10/2011 for the course PHYSICS 303L taught by Professor Hoffman during the Fall '11 term at University of Texas.
 Fall '11
 HOFFMAN
 Current, Resistance, Work

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