Phys121-8sol

# Phys121-8sol - Anderson, Matthew Homework 8 Due: Mar 25...

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Unformatted text preview: Anderson, Matthew Homework 8 Due: Mar 25 2008, noon Inst: Vitaly 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 2) 20 points Four identical light bulbs are connected ei- ther in series (circuit 1) or parallel (circuit 2) to a constant voltage battery with negligible internal resistance, as shown. E Circuit A E Circuit B Compared to the individual bulbs in circuit 1, the individual bulbs in circuit 2 are 1. the same brightness. 2. less than 1 4 as bright. 3. 4 times brighter. 4. more than 4 times brighter. correct 5. 1 4 as bright. Explanation: In circuit 1, the voltage across each light bulb is V = I R = E 4 R R = E 4 , so the power of each bulb in circuit 1 is P 1 = V 2 R = E 2 16 R . In circuit 2, the voltage across each bulb is identical; namely E . Hence the power of each bulb in circuit 2 is P 2 = E 2 R = 1 16 P 1 . We can see that the bulbs in circuit 2 are more than 4 times brighter than the bulbs in circuit 1. 002 (part 2 of 2) 20 points If one of the bulbs in circuit 2 is unscrewed and removed from its socket, the remaining 3 bulbs 1. go out. 2. become dimmer. 3. are unaffected. correct 4. become brighter. Explanation: Since the bulbs are parallel, after one of the bulbs is unscrewed, the voltage across each remaining bulb is unchanged, and the brightness is unaffected. keywords: 003 (part 1 of 3) 20 points A battery with an internal resistance is con- nected to two resistors in series. E x y 2 . 4 10 . 7 29 . 5 . 76 A internal resistance Anderson, Matthew Homework 8 Due: Mar 25 2008, noon Inst: Vitaly 2 What is the emf of the battery? Correct answer: 32 . 376 V. Explanation: E x y r R 1 R 2 I internal resistance Let : R 1 = 10 . 7 , R 2 = 29 . 5 , r = 2 . 4 , and I = 0 . 76 A . The total resistance of the circuit is R total = r + R 1 + R 2 = 2 . 4 + 10 . 7 + 29 . 5 = 42 . 6 , so E = I R total = (0 . 76 A) (42 . 6 ) = 32 . 376 V . 004 (part 2 of 3) 20 points What is the magnitude of the potential differ- ence V Y X across the terminals y and x of the battery? Correct answer: 30 . 552 V. Explanation: The potential difference across the termi- nals of the battery is V Y X = E - I r = 32 . 376 V- (0 . 76 A) (2 . 4 ) = 30 . 552 V , or V Y X = I [ R 1 + R 2 ] = (0 . 76 A) (10 . 7 + 29 . 5 ) = 30 . 552 V ....
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## This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.

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Phys121-8sol - Anderson, Matthew Homework 8 Due: Mar 25...

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