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solution7_pdf - hussain (wh4892) hw 7 Opyrchal (121104) 1...

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Unformatted text preview: hussain (wh4892) hw 7 Opyrchal (121104) 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 4 . 04 resistor con- nected to a battery that has a 0 . 18 internal resistance when the potential drop across the terminals of the battery is 12 V? Correct answer: 2 . 9703 A. Explanation: Let : V = 12 V , R = 4 . 04 , and R i = 0 . 18 . Applying Ohms law, V = I R I = V R = 12 V 4 . 04 = 2 . 9703 A . 002 (part 2 of 2) 10.0 points What is the emf of the battery? Correct answer: 12 . 5347 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 = 12 V + (2 . 9703 A) (0 . 18 ) = 12 . 5347 V / m . 003 (part 1 of 2) 10.0 points The power supplied to the circuit shown in the figure is 4.00 W. E 2 . 0 12 . 0 8 . 0 4 . 0 2 . 0 a) Find the equivalent resistance of the cir- cuit. Correct answer: 3 . 62963 . Explanation: E R 1 R 2 R 3 R 4 R 5 Let : R 1 = 2 . 0 , R 2 = 12 . 0 , R 3 = 8 . 0 , R 4 = 4 . 0 , and R 5 = 2 . 0 . 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 12 + 1 8 bracketrightbigg- 1 = 4 . 8 . For resistors in series, R eq,s = R a + R b . R 234 = R 23 + R 4 = 4 . 8 + 4 = 8 . 8 , and hussain (wh4892) hw 7 Opyrchal (121104) 2 R 2345 = parenleftbigg 1 R 234 + 1 R 5 parenrightbigg- 1 = parenleftbigg 1 8 . 8 + 1 2 parenrightbigg- 1 = 1 . 62963 , so R eq = R 1 + R 2345 = 2 + 1 . 62963 = 3 . 62963 ....
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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 - hussain (wh4892) hw 7 Opyrchal (121104) 1...

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