ohw14-solutions - marout (ssm979) – ohw13 – turner –...

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Unformatted text preview: marout (ssm979) – ohw13 – turner – (56725) 1 This print-out should have 16 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 Four 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 4 as bright. 2. 1 8 as bright. 3. 2 times brighter. 4. 1 16 as bright. 5. 8 times brighter. 6. 1 2 as bright. 7. the same brightness. 8. not lit up at all. 9. 4 times brighter. 10. 16 times brighter. correct Explanation: In circuit A, the voltage across each light bulb is V = I R = 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. one become brighter, one became dimmer and one unaffected. 4. begin to flash. 5. are unaffected. correct 6. turn red, white and blue. 7. become brighter. 8. begin to blink in Morse code ”physics sucks”. 9. burn out the battery twice as fast. marout (ssm979) – ohw13 – turner – (56725) 2 10. become dimmer. 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. 003 10.0 points 9 V 1 . 2 V 3 . 4 V I 1 . 2 Ω 3 Ω I 2 7 . 3 Ω I 3 9 . 9 Ω Find the power supplied to the 0 . 2 Ω resis- tor at the bottom of the circuit between the two power supplies. Correct answer: 0 . 248048 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 . 2 Ω , R B = 3 Ω , R C = 7 . 3 Ω , R D = 9 . 9 Ω , E 1 = 9 V , E 2 = 1 . 2 V , and E 3 = 3 . 4 V . At a junction (Conservation of Charge): I 1 + I 2- I 3 = 0 . (1) Kirchhoff’s law on the large outside loop: ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 (2) Kirchhoff’s law on the right-hand small loop: R C I 2 + R D I 3 = E 3 (3) Using determinants, I 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 E 1 + E 2 R D E 3 R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1- 1 R A + R B R D R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle Expanding along the first row, the numera- tor is D 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 E 1 + E 2 R D E 3 R C R D vextendsingle vextendsingle vextendsingle vextendsingle...
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This note was uploaded on 10/24/2010 for the course PHY 56735 taught by Professor Turner,j during the Fall '10 term at University of Texas.

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ohw14-solutions - marout (ssm979) – ohw13 – turner –...

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