2009S2UnitExam3ASolutions

2009S2UnitExam3ASolutions - ID: A UCSD Physics 2B Answer...

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ID: A 5 UCSD Physics 2B Unit Exam 3A Current, Resistance, Circuits Answer Section MULTIPLE CHOICE 1. ANS: B Each differential area element dA of a circle (cross section) is a ring of thickness dr and length (circumference) 2 π r . The area of this ring is the product, dA = 2 rdr I = J = Jr () 2 0 R = Kr 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2 0 R = 2 Kr 3 dr 0 R = 2 K R 4 4 = 1 2 3.14 122 A m 4 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 166 × 10 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 4 = 1.45 k A TOP: CURRENT DENSITY + INTEGRATION 2. ANS: D This is a series combination. When all resistors are identical R SER = NR = 3 × 33 Ω= 99 Ω TOP: SERIES 3. ANS: D This is basically a unit conversion problem if you think about it: Power = energy time = power × E = P × t = V × I × t = 12.0 J C Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 5.00 C s Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 3.00 min 60 s min Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ = 1.08 × 4 J = 10.8k J TOP: ENERGY 4. ANS: B Important to keep track of direction here. Since the protons and electrons are moving in the same direction, you must subtract the electron contribution to the flux since the charge is negative. Calculate the number of particles times the charge/particle and then divide by time I = Q t = n p e + + n e e t = n p e + n e e + t = n p n e Ê Ë Á Á ˆ ¯ ˜ ˜ e + t = 6.5 × 17 3.5 × Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 1.6 × 19 C Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 1 s = 0.048 A = 48m A TOP: CURRENT
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ID: A 2 5. ANS: B Each parallel pair has resistance R/2, so the total resistance of the circuit is R TOTAL = 3 R + 3 R /2 () = 4.5 R = 225 Ω . Since the resistance between A and B (going around clockwise) is R AB = 2 R + 2 R = 3 R , the ratio of the resistances is 3 R 4.5 R = 2 3 .
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This note was uploaded on 04/04/2010 for the course PHYSICS 2B taught by Professor Lever during the Summer '09 term at UCSD.

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2009S2UnitExam3ASolutions - ID: A UCSD Physics 2B Answer...

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