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Unformatted text preview: Bhakta (nnb232) – HW #4 – Antoniewicz – (56490) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 4.0 points A battery with an internal resistance is con nected to two resistors in series. E X Y 4 Ω 16 Ω 20 Ω . 3 A internal resistance What is the emf E of the battery? 1. E = 6 . 0 V 2. E = 10 . 8 V 3. E = 1 . 2 V 4. E = 12 . 0 V correct 5. E = 13 . 2 V Explanation: E X Y r R 1 R 2 I internal resistance Let : R 1 = 16 Ω , R 2 = 20 Ω , r = 4 Ω , and I = 0 . 3 A . The total resistance of the circuit is R total = r + R 1 + R 2 = 4 Ω + 16 Ω + 20 Ω = 40 Ω , so the emf of the battery is E = I R total = (0 . 3 A) (40 Ω) = 12 V . 002 (part 2 of 3) 3.0 points What is the potential difference across the terminals Y and X of the battery? 1. V Y X = 10 . 8 V correct 2. V Y X = 12 . 0 V 3. V Y X = 13 . 2 V 4. V Y X = 6 . 0 V 5. V Y X = 1 . 2 V Explanation: The potential difference across the termi nals of the battery is V Y X = E  I r = 12 V (0 . 3 A) (4 Ω) = 10 . 8 V , or V Y X = V XY = I ( R 1 + R 2 ) = (0 . 3 A)(16 Ω + 20 Ω) = 10 . 8 V . 003 (part 3 of 3) 3.0 points What power P internal is dissipated by the 4 Ω internal resistance of the battery? 1. P internal = 1 . 2 W 2. P internal = 4 . 8 W 3. P internal = 3 . 6 W 4. P internal = 3 . 2 W 5. P internal = 0 . 36 W correct Explanation: Bhakta (nnb232) – HW #4 – Antoniewicz – (56490) 2 The power dissipated by the r = 4 Ω inter nal resistance is P internal = I 2 r = (0 . 3 A) 2 (4 Ω) = . 36 W . 004 (part 1 of 5) 2.0 points A string of 26 identical Christmas tree lights are connected in series to a 125 V source. The string dissipates 66 W. What is the equivalent resistance of the light string? Correct answer: 236 . 742 Ω. Explanation: Let : n = 26 , V = 125 V , and P = 66 W . Power is defined by P = V 2 R R = V 2 P = (125 V) 2 66 W = 236 . 742 Ω . 005 (part 2 of 5) 2.0 points What is the resistance of a single light? Correct answer: 9 . 10548 Ω. Explanation: The total resistance is n times the resis tance of one bulb, so R = nR 1 R 1 = R n = 236 . 742 Ω 26 = 9 . 10548 Ω . 006 (part 3 of 5) 2.0 points How much power is dissipated in a single light? Correct answer: 2 . 53846 W. Explanation: The total power is n times the power of one bulb, so P = nP 1 P 1 = P n = 66 W 26 = 2 . 53846 W . 007 (part 4 of 5) 2.0 points One of the bulbs quits burning. The string has a wire that shorts out the bulb filament when it quits burning, dropping the resistance of that bulb to zero. All the rest of the bulbs remain burning. What is the resistance of the light string now? Correct answer: 227 . 637 Ω....
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 Fall '10
 Antanioweitz
 Physics, Resistance, Work, Correct Answer, R1 R2, R1 R2 R3

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