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2009S2UnitExam3BSolutions

# 2009S2UnitExam3BSolutions - ID B UCSD Physics 2B Answer...

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ID: B 4 UCSD Physics 2B Unit Exam 3B Current, Resistance, Circuits Answer Section MULTIPLE CHOICE 1. ANS: A Two ways to do this: 1. 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. Since R 1 = R 2 , R par = R N = 50 Ω 2 = 25 Ω . Now this is in series with R 3, so R Circuit = R 12 + R 3 = 25.0 Ω+ 50.0 Ω= 75.0 Ω . The total circuit current is thus I = V R Series = 120 V 75.0 Ω = 1.60 A Since the entire circuit current must pass through R 3 P 3 = I 2 R 3 = 1.60 A () 2 50.0 Ω = 128 W 2. Think “voltage divider”. Since R 3 = 2 R , V 3 = 2 3 V = 2 3 V = 80 V Now apply the power formula locally to R 3 P 3 = V 3 2 R 3 = V 2 Ω = W TOP: NET + POWER

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ID: B 2 2. ANS: C 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 whose resistance is R = ρ d A . The differential element for this layer is dR = d d A Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ = dr Ar () = 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 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ Plugging in the numers, we get R = 2 L b a Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ = 22 Ω⋅ m 2 1.36m 2.33 2.22 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ = 125 m Ω Note that the diameters a and b are so close that you can get an approximate answer by treating the whole device as a rolled-up plate resistor with d = b a and A = 2 aL . Then R d A = b a 2 = b / a 1 2 L = m 2.33 2.22 1 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 2 = 128 m Ω TOP:
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2009S2UnitExam3BSolutions - ID B UCSD Physics 2B Answer...

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