This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: zapata (jz883) HW #4 Antoniewicz (56495) 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 16 20 . 3 A internal resistance What is the emf E of the battery? 1. E = 13 . 2 V 2. E = 10 . 8 V 3. E = 6 . 0 V 4. E = 12 . 0 V correct 5. E = 1 . 2 V Explanation: E X Y r R 1 R 2 I internal resistance Let : R 1 = 16 , R 2 = 20 , r = 4 , and I = 0 . 3 A . The total resistance of the circuit is R total = r + R 1 + R 2 = 4 + 16 + 20 = 40 , so the emf of the battery is E = I R total = (0 . 3 A) (40 ) = 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 = 1 . 2 V 3. V Y X = 13 . 2 V 4. V Y X = 6 . 0 V 5. V Y X = 12 . 0 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 ) = 10 . 8 V , or V Y X = V XY = I ( R 1 + R 2 ) = (0 . 3 A)(16 + 20 ) = 10 . 8 V . 003 (part 3 of 3) 3.0 points What power P internal is dissipated by the 4 internal resistance of the battery? 1. P internal = 1 . 2 W 2. P internal = 3 . 6 W 3. P internal = 3 . 2 W 4. P internal = 0 . 36 W correct 5. P internal = 4 . 8 W Explanation: zapata (jz883) HW #4 Antoniewicz (56495) 2 The power dissipated by the r = 4 inter nal resistance is P internal = I 2 r = (0 . 3 A) 2 (4 ) = . 36 W . 004 (part 1 of 5) 2.0 points A string of 25 identical Christmas tree lights are connected in series to a 121 V source. The string dissipates 63 W. What is the equivalent resistance of the light string? Correct answer: 232 . 397 . Explanation: Let : n = 25 , V = 121 V , and P = 63 W . Power is defined by P = V 2 R R = V 2 P = (121 V) 2 63 W = 232 . 397 . 005 (part 2 of 5) 2.0 points What is the resistance of a single light? Correct answer: 9 . 29587 . Explanation: The total resistance is n times the resis tance of one bulb, so R = nR 1 R 1 = R n = 232 . 397 25 = 9 . 29587 . 006 (part 3 of 5) 2.0 points How much power is dissipated in a single light? Correct answer: 2 . 52 W. Explanation: The total power is n times the power of one bulb, so P = nP 1 P 1 = P n = 63 W 25 = 2 . 52 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: 223 . 101 ....
View
Full
Document
 Spring '10
 ANTONIEWICZ
 Physics, Resistance

Click to edit the document details