ohw14-solutions

# ohw14-solutions - howard (cah3459) ohw13 turner (56705)...

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howard (cah3459) – ohw13 – turner – (56705) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 10.0 points ±our identical light bulbs are connected ei- ther in series (circuit A) or parallel (circuit B) to a constant voltage battery with negligible internal resistance, as shown. E Circuit A E Circuit B Compared to the individual bulbs in circuit A, the individual bulbs in circuit B are 1. 1 8 as bright. 2. 2 times brighter. 3. 1 16 as bright. 4. the same brightness. 5. not lit up at all. 6. 8 times brighter. 7. 16 times brighter. correct 8. 1 2 as bright. 9. 4 times brighter. 10. 1 4 as bright. Explanation: In circuit A, the voltage across each light bulb is V = IR = E 4 R R = E 4 , so the power oF each bulb in circuit A is P A = V 2 R = E 2 16 R . In circuit B, the voltage across each bulb is identical; namely E . Hence the power oF each bulb in circuit B is P B = E 2 R = 16 P A . We can see that the bulbs in circuit B are 16 times brighter than the bulbs in circuit A. 002 (part 2 of 2) 10.0 points IF one oF the bulbs in circuit B is unscrewed and removed From its socket, the remaining 3 bulbs 1. pop. 2. go out. 3. become brighter. 4. begin to blink in Morse code ”physics sucks”. 5. burn out the battery twice as Fast. 6. begin to ²ash. 7. turn red, white and blue. 8. one become brighter, one became dimmer and one una³ected. 9. become dimmer.

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howard (cah3459) – ohw13 – turner – (56705) 2 10. are unafected. correct 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 unafected. 003 10.0 points 6 . 8V 2 . 3V 4 . 2V I 1 0 . 6 Ω 2 . 4 Ω I 2 6 . I 3 9 . 5 Ω ±ind the power supplied to the 0 . 6 Ω resis- tor at the bottom oF the circuit between the two power supplies. Correct answer: 0 . 553479 W. Explanation: E 1 E 2 E 3 I 1 R A R B I 2 R C I 3 R D Let : R A =0 . , R B =2 . , R C =6 . , R D =9 . , E 1 . , E 2 . , and E 3 =4 . . At a junction (Conservation oF Charge): I 1 + I 2 - I 3 . (1) Kirchhof’s law on the large outside loop: ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 (2) Kirchhof’s law on the right-hand small loop: R C I 2 + R D I 3 = E 3 (3) Using determinants, I 1 = ± ± ± ± ± ± 01 - 1 E 1 + E 2 0 R D E 3 R C R D ± ± ± ± ± ± ± ± ± ± ± ± 11 - 1 R A + R B 0 R D 0 R C R D ± ± ± ± ± ± Expanding along the ²rst row, the numera- tor is D 1 = ± ± ± ± ± ± - 1 E 1 + E 2 0 R D E 3 R C R D ± ± ± ± ± ± - 1 ± ± ± ± E 1 + E 2 R D E 3 R D ± ± ± ± +( - 1) ± ± ± ± E 1 + E 2 0 E 3 R C ± ± ± ± = - [( E 1 + E 2 ) R D -E 3 R D ] - [ R C ( E 1 + E 2 ) - 0] = R D ( E 3 1 2 ) - R C ( E 1 + E 2 ) = (9 . 5 Ω) (4 . - 6 . - 2 . 3 V) - (6 . 6 Ω) (6 . 8 V + 2 . 3 V) = - 106 . 61 V Ω . Expanding along the ²rst column, the de- nominator is D = ± ± ± ± ± ± - 1 R A + R B 0 R D 0 R C R D ± ± ± ± ± ± =1 ± ± ± ± 0 R D R C R D ± ± ± ± - ( R A + R B ) ± ± ± ± 1 - 1 R C R D ± ± ± ± +0 - R C R D - ( R A + R B )( R D + R C ) = (6 . 6 Ω) (9 . 5 Ω) - (0 . 6 Ω + 2 . 4 Ω) (9 . 5 Ω + 6 . 6 Ω) = - 111 Ω 2
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## This note was uploaded on 10/11/2010 for the course PHY 56715 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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ohw14-solutions - howard (cah3459) ohw13 turner (56705)...

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