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# solution7_pdf - taherisefat (mt23852) – hw 7 – Opyrchal...

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Unformatted text preview: taherisefat (mt23852) – hw 7 – Opyrchal – (121012) 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 (part 1 of 2) 10.0 points What is the current in a 2 . 35 Ω resistor con- nected to a battery that has a 0 . 2 Ω internal resistance when the potential drop across the terminals of the battery is 14 V? Correct answer: 5 . 95745 A. Explanation: Let : V = 14 V , R = 2 . 35 Ω , and R i = 0 . 2 Ω . Applying Ohm’s law, V = I R I = V R = 14 V 2 . 35 Ω = 5 . 95745 A . 002 (part 2 of 2) 10.0 points What is the emf of the battery? Correct answer: 15 . 1915 V. Explanation: The output voltage is reduced by the inter- nal resistance of the battery by V = E - I R i . Thus the electromotive force is E = V + I R i = 14 V + (5 . 95745 A) (0 . 2 Ω) = 15 . 1915 V / m . 003 (part 1 of 2) 10.0 points The power supplied to the circuit shown in the figure is 10.0 W. E 2 . 0 Ω 14 . 0 Ω 5 . 0 Ω 3 . 6 Ω 2 . 0 Ω a) Find the equivalent resistance of the cir- cuit. Correct answer: 3 . 56916 Ω. Explanation: E R 1 R 2 R 3 R 4 R 5 Let : R 1 = 2 . 0 Ω , R 2 = 14 . 0 Ω , R 3 = 5 . 0 Ω , R 4 = 3 . 6 Ω , and R 5 = 2 . 0 Ω . For resistors in parallel, 1 R eq,p = 1 R a + 1 R b . R 23 = bracketleftbigg 1 R 2 + 1 R 3 bracketrightbigg- 1 = bracketleftbigg 1 14 Ω + 1 5 Ω bracketrightbigg- 1 = 3 . 68421 Ω . For resistors in series, R eq,s = R a + R b . R 234 = R 23 + R 4 = 3 . 68421 Ω + 3 . 6 Ω = 7 . 28421 Ω , and taherisefat (mt23852) – hw 7 – Opyrchal – (121012) 2 R 2345 = parenleftbigg 1 R 234 + 1 R 5 parenrightbigg- 1 = parenleftbigg 1 7 . 28421 Ω + 1 2 Ω parenrightbigg- 1 = 1 . 56916 Ω , so R eq = R 1 + R 2345 = 2 Ω + 1 . 56916 Ω =...
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## This note was uploaded on 01/31/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.

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solution7_pdf - taherisefat (mt23852) – hw 7 – Opyrchal...

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