t3Bsolspr2010

t3Bsolspr2010 - Physics 1112 Spring 2010 University of...

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Unformatted text preview: Physics 1112 Spring 2010 University of Georgia Instructor: HBSchuttler Solution to PHYS 1112 In-Class Exam #3B Thu. April 8, 2010, 11:00am-12:15pm Conceptual Problems Problem 1: A wire of some length 20m and with a circular cross-section of diameter D has a resistance R . A second wire of length 360m and also of circular cross-section, made from the same material, has, at the same temperature, only half the resistance of the first: 1 2 R . What is the diameter of the second wire? (A) 36 D (B) 1 36 D (C) 12 D (D) 6 D (E) 1 6 D Answer: (D) : Using R = L/A and A = ( 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 = (360m) (20m) , 1 2 R R ! = 18 2 = 36 . So, D /D = 36 = 6 or D = 6 D . Problem 2: Three different circuits, Z , Y and X , 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. 3.33 (notice the order!). Compare and rank the magnitude of the voltage drops V 1 across R 1 , observed in the three different circuits. Fig. 3.33 (Z) R 1 I 3 R 2 R 3 I 2 I 1 E I 1 R 2 I 2 I 3 E I 2 R 2 R 3 I 3 R 1 I 1 E (X) (Y) I o I o I o R 3 R 1 1 Physics 1112 Spring 2010 University of Georgia Instructor: HBSchuttler (A) V 1 ( X ) > V 1 ( Y ) > V 1 ( Z ) (B) V 1 ( Y ) > V 1 ( X ) > V 1 ( Z ) (C) V 1 ( Z ) > V 1 ( Y ) > V 1 ( X ) (D) V 1 ( Z ) > V 1 ( X ) > V 1 ( Y ) (E) V 1 ( X ) > V 1 ( Z ) > V 1 ( Y ) Answer: (B) In all three circuits, R 1 is in series with another resistance: lets call it R S . This combination of R 1 and R S thus has an equivalent resistance of R 1 + R S . Also, in all three cicuits, this combination R 1 + R S is connected directly to the battery; so the voltage drop across R 1 + R S is E . Hence current I 1 through R 1 and R S , and corresponding voltage drop V 1 across R 1 are: I 1 = E R 1 + R S and V 1 = R 1 I 1 = R 1 E R 1 + R S . In circuit Z , R S consists of R 2 and R 3 connected in series, hence R S ( Z ) = R 2 + R 3 and hence R S ( Z ) > R 3 . In circuit Y , R S consists of R 2 and R 3 connected in parallel, hence 1 /R S ( Y ) = 1 /R 2 +1 /R 3 , hence 1 /R S ( Y ) > 1 /R 3 , and hence R S ( Y ) < R 3 . In circuit X , R S consists of only R 3 , so R S ( X ) = R 3 . Combining all the foregoing: R S ( Y ) = 1 / [1 /R 2 + 1 /R 3 ] < R S ( X ) = R 3 < R S ( Z ) = R 2 + R 3 , hence I 1 ( Y ) = E [ R 1 + R S ( Y )] > I 1 ( X ) = E [ R 1 + R S ( X )] > I 1 ( Z ) = E [ R 1 + R S ( Z )] , hence V 1 ( Y ) = R 1 I 1 ( Y ) > V 1 ( X ) = R 1 I 1 ( X ) > V 1 ( Z ) = R 1 I 1 ( Z ) . 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. 3.37, with B 6 = 0 above the lower horizontal line, and ~ B perpendicular to, and pointing into, the plane of the drawing....
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t3Bsolspr2010 - Physics 1112 Spring 2010 University of...

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