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HW 19 - terry(ect328 homework 19 Turner(59130 This...

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terry (ect328) – homework 19 – Turner – (59130) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points After a 7 . 76 Ω resistor is connected across a battery with a 0 . 17 Ω internal resistance, the electric potential between the physical battery terminals is 10 V. What is the rated emf of the battery? Correct answer: 10 . 2191 V. Explanation: Let : R = 7 . 76 Ω , r = 0 . 17 Ω , and V = 10 V . The current drawn by the external resistor is given by I = V R = 10 V 7 . 76 Ω = 1 . 28866 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 = 10 V + (1 . 28866 A) (0 . 17 Ω) = 10 . 2191 V . 002 10.0 points A certain coffeepot draws 4 A of current when it is operated on 120 V household lines. If electrical energy costs 10 cents per kilowatt-hour, how much does it cost to oper- ate the coffeepot for 2 hours? 1. cost = 16 cents 2. cost = 8.0 cents 3. cost = 9.6 cents correct 4. cost = 4.8 cents 5. cost = 2.4 cents Explanation: Let : I = 4 A , V = 120 V , r = $0 . 1 / kWh , and t = 2 h . The power that the coffeepot consumes is P = I V , so the cost to operate the coffeepot for 2 hours is C = I V t r = (4 A) (120 V) (2 h) ($0 . 1 / kWh) × parenleftbigg 1 kW 1000 W parenrightbigg = 9.6 cents . 003 10.0 points Four identical light bulbs are connected ei- ther in series (circuit A), or in a parallel-series combination (circuit B), to a constant voltage battery with negligible internal resistance, as shown. E Circuit A E Circuit B
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terry (ect328) – homework 19 – Turner – (59130) 2 Assuming the battery has no internal re- sistance and the resistance of the bulbs is temperature independent, what is the ratio of the total power consumed by circuit A to that consumed by circuit B; i.e. , parenleftbigg P A,T otal P B,T otal parenrightbigg ? 1. P A P B = 1 2 2. P A P B = 1 8 3. P A P B = 8 4. P A P B = 2 5. P A P B = 1 16 6. P A P B = 1 4 correct 7. P A P B = 16 8. P A P B = 1 9. P A P B = 4 Explanation: In circuit A, the equivalent resistance is R A = 4 R , so the electric current through each bulb is i A = V 4 R and the power of each bulb is P A = I 2 R = parenleftbigg V 4 R parenrightbigg 2 R = V 2 16 R .
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