HW6-Solutions

# HW6-Solutions - Version PREVIEW HW6 feng (58230) This...

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Version PREVIEW – HW6 – feng – (58230) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. Internal Resistance 07 001 10.0 points After a 4 . 84 Ω resistor is connected across a battery with a 0 . 1 Ω internal resistance, the electric potential between the physical battery terminals is 13 V. What is the rated emf of the battery? Correct answer: 13 . 2686 V. Explanation: Let : R = 4 . 84 Ω , r = 0 . 1 Ω , and V = 13 V . The current drawn by the external resistor is given by I = V R = 13 V 4 . 84 Ω = 2 . 68595 A . The output voltage is reduced by the inter- nal resistance of the battery by V = E − I r , so the electromotive force is E = V + I r = 13 V + (2 . 68595 A) (0 . 1 Ω) = 13 . 2686 V . Internal Resistance 04 002 10.0 points The emf of a battery is E = 10 V . When the battery delivers a current of 0 . 4 A to a load, the potential di±erence between the terminals of the battery is 8 V volts. ²ind the internal resistance of the battery. Correct answer: 5 Ω. Explanation: Given : E = 10 V , V load = 8 V , and I = 0 . 4 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 = 10 V 8 V 0 . 4 A = 5 Ω . Battery with internal resistance 003 (part 1 of 2) 10.0 points The internal resistance r of a battery with emf E is connected to a load resistor with resistance R . I 10 V A B 22 Ω 147 Ω internal resistance ²ind the potential di±erence V BA = V A V B . Correct answer: 8 . 69822 V. Explanation: Let : E = 10 V , R = 147 Ω , and r = 22 Ω . The current through the circuit is i = E r + R ,

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Version PREVIEW – HW6 – feng – (58230) 2 so the potential diFerence V A V B is V A V B = i R = E R r + R = (10 V) (147 Ω) 22 Ω + 147 Ω = 8 . 69822 V . 004 (part 2 of 2) 10.0 points What is the power P dissipated by the load resistor R ? 1. P = E 2 r + R 2. P = R E 2 3. P = r E 2 4. P = ( r + R ) E 2 5. P = E 2 r 6. P = p E r + R P 2 R correct 7. P = p E R P 2 r 8. P = E 2 r R 9. P = E 2 R r 10. P = p E r + R P 2 r Explanation: The power dissipated by the load is P = i 2 R = p E r + R P 2 R . Series Circuit 01 005 (part 1 of 3) 10.0 points The current in the circuit below is 2 am- peres. 0 . 3 Ω 12 V 6 V 0 . 2 Ω R 1 . 5 Ω X Y What is the resistance R ? 1. 1 Ω correct 2. 4 Ω 3. 3 Ω 4. 5 Ω 5. 2 Ω Explanation: Let : R 1 = 0 . 3 Ω , R 2 = 0 . 2 Ω , R 3 = 1 . 5 Ω , E 1 = 12 V , and E 2 = 6 V . R 1 E 1 E 2 R 2 R R 3 X Y ±rom Ohm’s law, the total resistance of the circuit is R total = V I = 12 V 6 V 2 A = 3 Ω . Therefore, the resistance R is R = R total 0 . 3 Ω 1 . 5 Ω 0 . 2 Ω = 1 Ω .
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## This note was uploaded on 04/27/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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HW6-Solutions - Version PREVIEW HW6 feng (58230) This...

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