hw08a - aldawsari (ma28683) HW08 Betancourt (16856) This...

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aldawsari (ma28683) – HW08 – Betancourt – (16856) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 26.1-26.2 001 10.0 points The emF oF a battery is E = 20 V . When the battery delivers a current oF 0 . 2 A to a load, the potential di±erence between the terminals oF the battery is 18 V volts. ²ind the internal resistance oF the battery. Correct answer: 10 Ω. Explanation: Given : E = 20 V , V load = 18 V , and I = 0 . 2 A . The potential di±erence across the internal resistance is E - V load , so the internal resis- tance is given by r = E - V load I = 20 V - 18 V 0 . 2 A = 10 Ω . 002 (part 1 oF 4) 10.0 points A battery has an emF oF 12 V and an inter- nal resistance oF 0 . 18 Ω. Its terminals are connected to a load resistance oF 1 Ω. ²ind the current in the circuit. Correct answer: 10 . 1695 A. Explanation: Let : E = 12 V , R = 1 Ω , and r = 0 . 18 Ω . The total resistance is R + r , so I = E R + r = 12 V 1 Ω + 0 . 18 Ω = 10 . 1695 A . 003 (part 2 oF 4) 10.0 points Calculate the terminal voltage oF the battery. Correct answer: 10 . 1695 V. Explanation: The terminal voltage V oF the battery is equal to V r = E - I r = 12 V - (10 . 1695 A) (0 . 18 Ω) = 10 . 1695 V . 004 (part 3 oF 4) 10.0 points ²ind the power dissipated in the load resis- tor. Correct answer: 103 . 419 W. Explanation: The power dissipated in the load resistor is P R = I 2 R = (10 . 1695 A) 2 (1 Ω) = 103 . 419 W . 005 (part 4 oF 4) 10.0 points ²ind the power dissipated in the battery. Correct answer: 18
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hw08a - aldawsari (ma28683) HW08 Betancourt (16856) This...

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