UnitExam3Asolutions

UnitExam3Asolutions - ID A UCSD Physics 2B Answer Section...

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ID: A 3 UCSD Physics 2B Unit Exam 3A Current, Resistance, Circuits Answer Section MULTIPLE CHOICE 1. ANS: B This is a compound circuit, so we must find the current around the entire loop, and therefore the equivalent resistance of the entire circuit. First, reduce the parallel leg to its equivalent 1 R Parallel = 1 R 1 + 1 R 2 = 1 30.0 Ω + 1 Ω = 1 15.0 Ω so that R = 15.0 Ω This is in series with R 3, so R Series = R 1 + 2 + R 3 = Ω+ Ω= 45.0 Ω The total circuit current is thus I = V R = 75.0 V 45.0 Ω = 1.67 A Since the entire circuit current must pass through R 3 , P 3 = I 2 R 3 = 1.67 A () 2 30 Ω = 83.3 W TOP: NET + POWER 2. ANS: A R = V I 220 V 8.75 A = 25.1 Ω TOP: OHM'S LAW 3. ANS: B The resistors are in parallel, so they’re at the same voltage and independent of each other. P 2 = V 2 R 2 = 50.0 V 2 17.2 Ω = 145 W TOP: POWER
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ID: A 2 4. ANS: D Let's think of this device as being made up of many coaxial shells (layers) of length L , "width" w = 2 π r and thickness dr . Since the current flows directly outward, each layer can be treated as a parallel plate resistor with resistance dR = ρ d A = dr Lw = 2 rL Since each layer acts as one element in a series network (why?), the total resistance is the sum of these layers, that is, an integral: R = = 2 L r a b = 2 L ln b a
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This note was uploaded on 08/23/2008 for the course PHYS 2b taught by Professor Schuller during the Summer '08 term at UCSD.

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UnitExam3Asolutions - ID A UCSD Physics 2B Answer Section...

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