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Unformatted text preview: homework 16 – KIM, JI – Due: Mar 10 2008, 4:00 am 1 Question 1, chap 27, sect 2. part 1 of 1 10 points A wire with a resistance R is lengthened to 9 . 4 times its original length by pulling it through a small hole, while the volume of the wire is unchanged. Find the resistance of the wire after it is stretched. Correct answer: 88 . 36 R (tolerance ± 1 %). Explanation: Let : ℓ ′ = 9 . 4 ℓ . The volume V of the wire is given by V = ℓ A, where ℓ is the length of the wire and A is the crosssectional area. When the wire is stretched to a length ℓ ′ , the volume will re main the same. The new crosssectional area is then, ℓ A = ℓ ′ A ′ A ′ = ℓ A ℓ ′ = ℓ A 9 . 4 ℓ = A 9 . 4 . The resistance of a conductor of resistivity ρ is given by R = ρ ℓ A , so R ′ R = ρ ℓ ′ A ′ ρ ℓ A = ℓ ′ A ℓ A ′ = (9 . 4 ℓ ) A ℓ parenleftbigg A 9 . 4 parenrightbigg = (9 . 4) 2 . Thus R ′ = (9 . 4) 2 R = 88 . 36 R . Question 2, chap 27, sect 2. part 1 of 1 10 points A 0 . 97 V potential difference is maintained across a 0 . 8 m length of tungsten wire that has a crosssectional area of 0 . 8 mm 2 and the resistivity of the tungsten is 5 . 6 × 10 − 8 Ω · m. What is the current in the wire? Correct answer: 17 . 3214 A (tolerance ± 1 %). Explanation: Let : V = 0 . 97 V , ℓ = 0 . 8 m , A = 0 . 8 mm 2 = 8 × 10 − 7 m 2 , and ρ = 5 . 6 × 10 − 8 Ω · m . The resistance is R = V I = ρ ℓ A , so the current is I = V A ρ ℓ = (0 . 97 V) (8 × 10 − 7 m 2 ) (5 . 6 × 10 − 8 Ω · m) (0 . 8 m) = 17 . 3214 A . Question 3, chap 28, sect 4. part 1 of 1 10 points Four identical light bulbs are connected ei ther in series (circuit A), or in a parallelseries combination (circuit B), to a constant voltage battery with negligible internal resistance, as shown. E Circuit A E Circuit B homework 16 – KIM, JI – Due: Mar 10 2008, 4:00 am 2 Assuming the battery has no internal re sistance and the resistance of the bulbs is temperature independent, what is the ratio of the total power consumed by circuit A to that consumed by circuit B; i.e. , parenleftbigg P A,Total P B,Total parenrightbigg ? 1. P A P B = 1 2. P A P B = 1 16 3. P A P B = 2 4. P A P B = 1 2 5. P A P B = 4 6. P A P B = 1 √ 8 7. P A P B = 8 8. P A P B = 1 4 correct 9. P A P B = 16 Explanation: In circuit A, the equivalent resistance is R A = 4 R , so the electric current through each bulb is i A = V 4 R and the power of each bulb is P A = I 2 R = parenleftbigg V 4 R parenrightbigg 2 R = V 2 16 R . Thus the total power consumed by all four bulbs in circuit A is P A,Total = 4 P A = V 2 4 R . In circuit B, the equivalent resistance is 1 R B = 1 2 R + 1 2 R = 1 R R B = R , so the electric current through each bulb is i B = V 2 R and the power of each bulb is P B = I 2 R = parenleftbigg V 2 R parenrightbigg 2 R = V 2 4 R ....
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This note was uploaded on 05/05/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Resistance, Work

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