t3Bsolspr2010

# t3Bsolspr2010 - Physics 1112 Spring 2010 University of...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 1112 Spring 2010 University of Georgia Instructor: HBSch¨uttler 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 = 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 = (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: HBSch¨uttler (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: let’s 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....
View Full Document

## This note was uploaded on 04/23/2010 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at UGA.

### Page1 / 10

t3Bsolspr2010 - Physics 1112 Spring 2010 University of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online