Phys 1112 cp3

# Phys 1112 cp3 - Physics 1112 Spring 2009 University of...

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Unformatted text preview: Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler Solutions toConceptual Practice Problems PHYS 1112 In-Class Exam #2A+2B Thu. Apr. 9, 2009, 11:00am-12:15pm and 2:00pm-3:15pm CP 3.01: If a wire of some length L and a circular cross-section of diameter D has a resistance of 36 kΩ, what will be the resistance of a wire made from the same material, at the same temperature, of length 3 L and the same diameter D ? (A) 4 kΩ (B) 12 kΩ (C) 24 kΩ (D) 72 kΩ (E) 108 kΩ Answer: (E) Using R = ρL/A and A = π ( D/ 2) 2 , we get R ∝ L/D 2 . So, R ∝ L for fixed D . Increasing L → L = 3 L thus increases R → R = 3 R = 3 × 36Ω = 108Ω. CP 3.02: If a wire of some length L and a circular cross-section of diameter D has a resistance of 36 kΩ, what will be the resistance of a wire made from the same material, at the same temperature, of the same length L and diameter 3 D ? (A) 4 kΩ (B) 12 kΩ (C) 24 kΩ (D) 72 kΩ (E) 108 kΩ Answer: (A) Using R = ρL/A and A = π ( D/ 2) 2 , we get R ∝ L/D 2 . So, R ∝ 1 /D 2 for fixed L . Increasing D → D = 3 D thus changes R → R = R/ 3 2 = 36Ω / 9 = 4Ω. CP 3.03: If a wire of some length L and a circular cross-section of diameter D has a resistance of 36 kΩ, what will be the resistance of a wire made from the same material, at the same temperature, of length 18 L and diameter 3 D ? (A) 4 kΩ (B) 12 kΩ (C) 24 kΩ 1 Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler (D) 72 kΩ (E) 108 kΩ Answer: (D) Using R = ρL/A and A = π ( D/ 2) 2 , we get R ∝ L/D 2 . Increasing L → L = 18 L and D → D = 3 D thus changes R → R = 18 × R/ 3 2 = (18 / 9) × 36Ω = 72Ω. CP 3.04: Three different circuits, X , Y and Z , are built with the same three resistors, R 1 > 0, R 2 > 0, and R 3 > 0, and the same battery of battery voltage E , as shown in Fig. 3.04. Compare and rank the magnitude of the currents I 1 through R 1 , observed in the three different circuits. Fig. 3.04 (X) R 1 I 2 R 3 R 2 I 3 I 1 E R 3 I 2 R 2 R 1 I 1 I 3 E I 3 R 3 R 1 I 1 R 2 I 2 E (Z) (Y) I o I o I o (A) Ranking cannot be determined from information given. (B) I 1 ( X ) > I 1 ( Z ) > I 1 ( Y ) (C) I 1 ( Y ) > I 1 ( X ) > I 1 ( Z ) (D) I 1 ( Y ) > I 1 ( Z ) > I 1 ( X ) (E) I 1 ( Z ) > I 1 ( Y ) > I 1 ( X ) Answer: (D) In circuit Y : R 1 is directly connected to battery, hence I 1 ( Y ) = E /R 1 . In circuit Z : R 1 and R 2 in series are connected to battery. Hence I 1 ( Z ) = E / ( R 1 + R 2 ) < I 1 ( Y ), since R 1 + R 2 > R 1 . 2 Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨uttler In circuit X : R 1 , R 2 and R 3 in series are connected to battery. Hence I 1 ( X ) = E / ( R 1 + R 2 + R 3 ) < I 1 ( Z ) < I 1 ( Y ), since R 1 + R 2 + R 3 > R 1 + R 2 ....
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Phys 1112 cp3 - Physics 1112 Spring 2009 University of...

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