Homework 7-solutions

Homework 7-solutions - gonzales(pag757 – Homework 7 –...

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Unformatted text preview: gonzales (pag757) – Homework 7 – Hoffmann – (56745) 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 10.0 points The emf of a battery is E = 19 V . When the battery delivers a current of 0 . 7 A to a load, the potential difference between the terminals of the battery is 17 V volts. Find the internal resistance of the battery. Correct answer: 2 . 85714 Ω. Explanation: Given : E = 19 V , V load = 17 V , and I = 0 . 7 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 = 19 V- 17 V . 7 A = 2 . 85714 Ω . 002 10.0 points Four copper wires of equal length are con- nected in series. Their cross-sectional areas are 0 . 8 cm 2 , 2 . 5 cm 2 , 1 . 3 cm 2 , and 7 cm 2 . If a voltage of 142 V is applied to the ar- rangement, determine the voltage across the 2 . 5 cm 2 wire. Correct answer: 22 . 1694 V. Explanation: Let : A 1 = 0 . 8 cm 2 , A 2 = 2 . 5 cm 2 , A 3 = 1 . 3 cm 2 , A 4 = 7 cm 2 , and V tot = 142 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 . 5 cm 2 1 . 8 cm 2 + 1 2 . 5 cm 2 + 1 1 . 3 cm 2 + 1 7 cm 2 × (142 V) = 22 . 1694 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: gonzales (pag757) – Homework 7 – Hoffmann – (56745) 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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Homework 7-solutions - gonzales(pag757 – Homework 7 –...

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