Homework 7-solutions

# Homework 7-solutions - soto(rrs946 Homework 7 Homann(56605...

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soto (rrs946) – Homework 7 – Hoffmann – (56605) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points 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 0 . 6 A = 3 . 33333 Ω . 002 10.0points Four copper wires of equal length are con- nected in series. Their cross-sectional 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 10.0points Four resistors are connected as shown in the figure. 21 Ω 52 Ω 88 Ω 94 V 37 Ω S 1 a b c d Find the resistance between points a and b . Correct answer: 37 . 5165 Ω. Explanation:

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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. R 2 R 3 R 1 R 4 a b c d The parallel connection of R 1 and R 2 gives the equivalent resistance 1 R 12 = 1 R 1 + 1 R 2 = R 2 + R 1 R 1 R 2 R 12 = R 1 R 2 R 1 + R 2 = (21 Ω) (37 Ω) 21 Ω + 37 Ω = 13 . 3966 Ω .
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