t22306f99 - PHYSICS 2306 TEST #2 5 November 1999 5:00 PM -...

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PHYSICS 2306 TEST #2 5 November 1999 5:00 PM - 6:30 PM 17M01-03 (5:00 PM) and 18M01-03 (6:30 PM) 3”x5” Note Card and Calculator Allowed Solutions _________________________ ____________________ ________________________ (Name - Printed) (Student I.D.) (Signed Pledge - See Below) I pledge that the answers submitted by me represent only my work. Further, I pledge to refrain from communicating any information about this test to anyone until after the results of this test are returned. Please fill in the three entries specified above. Please record your answers to problems #1 through #20 on the op-scan answer sheet. Please remember to darken the entries corresponding to your student I.D on the op-scan sheet and your group number which is based on your Monday recitation time ( 1 for 8 AM, 2 for 9 AM, 3 for 10 AM; 4 for 2 PM, 5 for 3 PM, 6 for 4 PM. Do not put your name on the op-scan sheet . Please return the question sheets to the appropriate stack specified by lecture time; and return the op-scan answer sheet to the other appropriate stack specified by lecture time on the front table. No credit will be given if any question or answer sheets are missing from your exam. Constants and Conversion Factors: charge of electron(magnitude) or proton: e = 1 . 6 × 10 - 19 C mass of electron m e = 9 . 11 × 10 -31 kg mass of proton m p = 1 . 67 × 10 -27 kg Coulomb’s law constant: k e = 1/4 πε o = 8 . 99 × 10 9 (N m 2 /C 2 ) Permittivity of free space: ε o = 8 . 854 × 10 - 12 (C 2 /N m 2 ) Use as the speed of sound in air: v = 340 m/s Permeability of free space: μ o = 4 π × 10 -7 (T m/A)
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1. A copper and an iron wire with the same length, L/2, and cross-sectional area, A, are joined in series to form one long wire of total length L. The resistivity of copper is 1 . 6 × 10 -8 ohm-meter, and the resistivity of iron is 9 . 6 × 10 -8 ohm-meter. If a potential difference of V o = 1 . 4 V is imposed across the opposite ends, then the potential difference between the ends of the copper is: (1) 0.2 V V o (2) 0 . 4 V Copper (Cu) Iron (Fe) A (3) 0 . 5 V L/2 L/2 (4) 0 . 7 V Since area of wires the same, J Cu =J Fe , then E Cu = ρ Cu J Cu and E Fe = ρ Fe J Fe . (5) 1 . 0 V Then E Fe =( ρ Fe / ρ Cu )E Cu . V Cu =E Cu L/2 and V Fe =E Fe L/2 , hence V Fe = ( ρ Fe / ρ Cu ) V Cu . (6) 1 . 2 V V o = 1.4 = V Cu + V Fe = [1 + ( ρ Fe / ρ Cu )] ⋅∆ V Cu = 7 ⋅∆ V Cu V Cu = 1.4/7 = 0.2 V For questions #2 through #5 consider the setup shown here. The currents have reached steady state : R2 R3 R5 a R1 R4 b V1 V3 C R6 3 A V2 R1= 2 , R2= 4 , R3= 8 , R4= 6 , R5= 2 , R6= 1 , V1= 6 V, V2= 12 V,Current thru R1= 3 A 2. The equivalent resistance resistance between points a and b is: (1) 2 Series R23 = R2 + R3 = 12 ; Parallel 1/R234 = 1/R23 + 1/R4 = 1/12 + 1/6 = 1/4 (2) 3 Then R234 = 4 ; which gives series R1234 = R1 + R234 = 2 + 4 = 6 . (3) 4 (4) 6 (5) 8 (6) 12 3. The current through R3 is: (1) 1/2 A R23 I23 = R4 I4 and I23 + I4 = 3 A. Therefore I23 + (R23/R4) I23 = 3 A (2) 1 A Therefore [1 + (12/6)] I23 = 3 A I23 = 1 A and then I4 = 2 A (3) 3/2 A (4) 2 A (5) 5/2 A (6) 3 A 4. The current through R6 is: (1) 1/2 A Assume the current through both R5 and R6 are to the left. Take a clockwise loop (2) 1 A around the parallel branch that contains V1 and V2.
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This note was uploaded on 03/19/2008 for the course PHYS 2306 taught by Professor Ykim during the Fall '06 term at Virginia Tech.

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t22306f99 - PHYSICS 2306 TEST #2 5 November 1999 5:00 PM -...

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