# 8 - kavo(ak22862 – HW08 – Janow –(12121 1 This...

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Unformatted text preview: kavo (ak22862) – HW08 – Janow – (12121) 1 This print-out should have 18 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 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. less than 1 4 as bright. 2. the same brightness. 3. 1 4 as bright. 4. 4 times brighter. 5. more than 4 times brighter. correct 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) 10.0 points If one of the bulbs in circuit 2 is unscrewed and removed from its socket, the remaining 3 bulbs 1. become brighter. 2. are unaffected. correct 3. become dimmer. 4. go out. 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 (part 1 of 3) 10.0 points A battery with an internal resistance is con- nected to two resistors in series. E x y 4 . 5 Ω 13 . 6 Ω 25 . 4 Ω . 47 A internal resistance What is the emf of the battery? Correct answer: 20 . 445 V. Explanation: kavo (ak22862) – HW08 – Janow – (12121) 2 E x y r R 1 R 2 I internal resistance Let : R 1 = 13 . 6 Ω , R 2 = 25 . 4 Ω , r = 4 . 5 Ω , and I = 0 . 47 A . The total resistance of the circuit is R total = r + R 1 + R 2 = 4 . 5 Ω + 13 . 6 Ω + 25 . 4 Ω = 43 . 5 Ω , so E = I R total = (0 . 47 A) (43 . 5 Ω) = 20 . 445 V . 004 (part 2 of 3) 10.0 points What is the magnitude of the potential differ- ence V Y X across the terminals y and x of the battery? Correct answer: 18 . 33 V. Explanation: The potential difference across the termi- nals of the battery is V Y X = E - I r = 20 . 445 V- (0 . 47 A) (4 . 5 Ω) = 18 . 33 V , or V Y X = I [ R 1 + R 2 ] = (0 . 47 A) (13 . 6 Ω + 25 . 4 Ω) = 18 . 33 V . Therefore, the magnitude of V Y X is 18 . 33 V. 005 (part 3 of 3) 10.0 points What power is dissipated by the internal re- sistance of the battery? Correct answer: 0 . 99405 W. Explanation: The power dissipated by the r = 4 . 5 Ω internal resistance is P internal = I 2 r = (0 . 47 A) 2 (4 . 5 Ω) = . 99405 W ....
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8 - kavo(ak22862 – HW08 – Janow –(12121 1 This...

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