Homework 7-solutions

# Homework 7-solutions - kumar(vk4433 Homework 7...

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kumar (vk4433) – Homework 7 – markert – (55325) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The emf of a battery is E = 19 V . When the battery delivers a current of 0 . 8 A to a load, the potential difference between the terminals of the battery is 17 V volts. Find the internal resistance of the battery. Correct answer: 2 . 5 Ω. Explanation: Given : E = 19 V , V load = 17 V , and I = 0 . 8 A . The potential difference across the internal resistance is E − V load , so the internal resis- tance is given by r = E − V load I = 19 V 17 V 0 . 8 A = 2 . 5 Ω . 002 10.0points Storage batteries are often rated in terms of the amounts of charge they can deliver. How much charge can a 124 A · h battery deliver? Correct answer: 4 . 464 × 10 5 C. Explanation: Let : I = 124 A · h . Since 1 A = 1 C / s, an Ampere-hour is a measure of charge. To find the total charge delivered, we simply convert hours into sec- onds. Q = (124 A · h) parenleftbigg 3600 s h parenrightbigg = 4 . 464 × 10 5 C . 003 10.0points A 13 Ω resistor and a 6.0 Ω resistor are con- nected in series with an emf source. The potential difference across the 6.0 Ω resistor is measured with a voltmeter to be 12 V. Find the potential difference across the emf source. Correct answer: 38 V. Explanation: Let : R 1 = 13 Ω , R 2 = 6 . 0 Ω , and Δ V 2 = 12 V . BasicConcepts: R eq = R 1 + R 2 Δ V = IR I 1 = I 2 = I Solution: I = Δ V 2 R 2 = 12 V 6 Ω = 2 A R eq = 13 Ω + 6 Ω = 19 Ω Δ V = IR eq = (2 A)(19 Ω) = 38 V . 004 10.0points Consider the circuit E 3 . 06 Ω 1 . 81 Ω 4 . 42 A a c b Find the potential difference V a V b . Correct answer: 21 . 5254 V.

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kumar (vk4433) – Homework 7 – markert – (55325) 2 Explanation: Let : I = 4 . 42 A R 1 = 3 . 06 Ω , and R 2 = 1 . 81 Ω . E R 1 R 2 I Currents through resistor elements in a se- ries combination are the same. We can find the potential difference by finding the equiva- lent resistance R = R 1 + R 2 and multiplying the equivalent resistance by the current: V = ( R 1 + R 2 ) I = (3 . 06 Ω + 1 . 81 Ω) (4 . 42 A) = 21 . 5254 V . 005 10.0points The following diagram shows a closed elec- trical circuit. The ammeter in the center of the resistive network reads zero amperes. E S 1 A 14 Ω 38 Ω 7 Ω R x Find the electric resistance R X . 1. R X = 7 Ω . 2. R X = 15 Ω . 3. R X = 11 Ω . 4. R X = 20 Ω . 5. R X = 19 Ω . correct 6. R X = 14 Ω . 7. R X = 8 Ω . 8. R X = 5 Ω . 9. R X = 13 Ω . 10. R X = 6 Ω . Explanation: E S 1 u x y I u R 1 I u R 2 I R 3 I R x I A 0 A Let : R 1 = 14 Ω , R 2 = 38 Ω , and R 3 = 7 Ω . If the ammeter reads zero I A = 0 A, the two ends of the ammeter should be an equipoten- tial: V u A = V A . This means that the potential drop from x or y to each side of the ammeter through either the upper u or lower part of the circuit must be the same. V x V u = I u R 1 (1) V x V = I R 3 (2) V u V y = I u R 2 (3) V V y = I R x (4) Setting the potential across the ammeter equal V u = V ( i.e., Eq. 1 = 2 and Eq. 3 = 4), we have I u R 1 = I R 3 and I u R 2 = I R x , so I u I = R 3 R 1 = R x R 2 R x = R 2 R 3 R 1 = (38 Ω) (7 Ω) 14 Ω = 19 Ω .
kumar (vk4433) – Homework 7 – markert – (55325) 3 006 10.0points Assume: The resistance of each light bulb remains constant.

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• Spring '08
• Turner
• Physics, Work, Resistor, SEPTA Regional Rail, Correct Answer, Electrical resistance, Series and parallel circuits

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