t3AsolV2spr2010

t3AsolV2spr2010 - Physics 1112 Spring 2010 University of...

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Physics 1112 Spring 2010 University of Georgia Instructor: HBSch¨uttler Solution to PHYS 1112 In-Class Exam #3A Thu. April 8, 2010, 09:30am-10:45am Conceptual Problems Problem 1: A wire of some length L and with a circular cross-section of diameter D has a resistance of 1600 Ω. A second wire of circlar cross-section and length 5 L , made from the same material, has, at the same temperature, a resistance of 125Ω. What is the diameter of the second wire? (A) 64 D (B) 40 D (C) 8 D (D) 1 8 D (E) 1 64 D Answer: (C) Using R = ρL/A and A = π ( D/ 2) 2 , we get R L/D 2 , so R R = L L D D 2 hence D D 2 = L L R R = 5 L L 125 1600 = 5 × 1600 / 125 = 64 . So, D /D = 64 = 8 or D = 8 D . Problem 2: 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. 3.32. Compare and rank the magnitude of the voltage drops V 2 across R 2 , observed in the three different circuits. Fig. 3.32 (X) R 1 I 3 R 2 R 3 I 2 I 1 E I 1 R 1 R 3 I 3 I 2 E I 1 R 1 R 3 I 3 R 2 I 2 E (Z) (Y) I o I o I o R 2 1
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Physics 1112 Spring 2010 University of Georgia Instructor: HBSch¨uttler (A) V 2 ( X ) > V 2 ( Y ) > V 2 ( Z ) (B) V 2 ( X ) > V 2 ( Z ) > V 2 ( Y ) (C) V 2 ( Z ) > V 2 ( X ) > V 2 ( Y ) (D) V 2 ( Z ) > V 2 ( Y ) > V 2 ( X ) (E) V 2 ( Y ) > V 2 ( Z ) > V 2 ( X ) Answer: (E) In all three circuits, R 2 is in series with another resistance: let’s call it R S . This combination of R 2 and R S thus has an equivalent resistance of R 2 + R S . Also, in all three cicuits, this combination R 2 + R S is connected directly to the battery; so the voltage drop across R 2 + R S is E . Hence current I 2 through R 2 and R S , and corresponding voltage drop V 2 across R 2 are: I 2 = E R 2 + R S and V 2 = R 2 I 2 = R 2 E R 2 + R S . In circuit X , R S consists of R 1 and R 3 connected in series, hence R S ( X ) = R 1 + R 3 and hence R S ( X ) > R 3 . In circuit Y , R S consists of R 1 and R 3 connected in parallel, hence 1 /R S ( Y ) = 1 /R 1 + 1 /R 3 , hence 1 /R S ( Y ) > 1 /R 3 , and hence R S ( Y ) < R 3 . In circuit Z , R S consists of only R 3 , so R S ( Z ) = R 3 . Combining all the foregoing: R S ( Y ) = 1 / [1 /R 1 + 1 /R 3 ] < R S ( Z ) = R 3 < R S ( X ) = R 1 + R 3 , hence I 2 ( Y ) = E [ R 2 + R S ( Y )] > I 2 ( Z ) = E [ R 2 + R S ( Z )] > I 2 ( X ) = E [ R 2 + R S ( X )] , hence V 2 ( Y ) = R 2 I 2 ( Y ) > V 2 ( Z ) = R 2 I 2 ( Z ) > V 2 ( X ) = R 2 I 2 ( X ) . Problem 3: A molecular ion beam containing four different types of ions, called P , Q , R and S here, enters a uniform magnetic field, as shown in Fig. 3.15, with B = 0 above the lower horizontal line, and B perpendicular to, and pointing out of, the plane of the drawing. The incident beam, below the lower horizontal line, is in the plane of the drawing. 2
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Physics 1112 Spring 2010 University of Georgia Instructor: HBSch¨uttler B (out) Fig. 3.15 P S Q R B=0 All ions have the same mass and speed and their semi-circular trajectory radii in the B -field are in a ratio | r P | : | r Q | : | r R | : | r S | = 1 : 2 : 3 : 4 . Given the trajectories as shown in Fig. 3.15, what could be possible charges q P , q Q , q R and q S of the four different ion types in the beam? ( e > 0 is the elementary charge).
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