Exam3_2004Fall_Solutions

Exam3_2004Fall_Solutions - Physics 8.02 TEST 3 MASHUP...

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Physics 8.02 TEST 3 MASHUP SOLUTIONS Fall 2004 PLEASE DETACH THIS SHEET AND USE AT YOUR CONVENIENCE Some (possibly useful) Relations: = surface closed o inside Q ε dA E outside to inside from points dA = = b a a b b to a from moving V V V ds E C Q V U = 1 2 C V 2 = Q 2 2C d A o plate parallel κ = C τ = RC V = i R P joule = i V = i 2 R = V 2 /R 2 ˆ 4 o q r µ π × = vr B 2 ˆ 4 o I r × = ds r dB F = q v × B ext ext I dF ds B τ τ = µ µ × B thru closed path dI ⋅= o Bs G G v = dA B ds E dt d ε = N d φ B dt L = N B i ε = L di dt / L R = U L = 1 2 L i 2 u B = | B | 2 2 0 T = 2 π ω = 1 f Free RLC Circuit No Damping: 0 ( ) sin( ) o Qt Q t ω = , 1 o LC = Driven LRC Circuit: if ε (t) = ε m sin( ω t ), then i (t) = i m sin( ω t - φ ), where i m = ε m R 2 + ω L – 1 ω C 2 1/2 and tan φ = ω L – 1 ω C R 1 LC XLX C == Integrals that may be useful a b dr b a = ) / ln( 1 a b dr r b a = = b a dr r b a 1 1 1 2
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8.02T Fall 2004 Exam 3 Mashup Solutions 1 Problem 1: Four Multiple Choice Questions (20 points) Circle your choice for the correct answer to the four questions given below. A : A parallel plate capacitor with area A and plate separation d is initially charged. We want to discharge it using a conductor of resistivity ρ shaped as a solid cylinder of area A and height d that fits right between the plates of the capacitor. What is the time constant τ that governs the discharge? 1. 0 2 A d ερ = 2. 2 0 d A = 3. 0 = 4. 0 ε = 0 0 ; A d CR R C dA ==⇒ = = B. The circuit shown below contains an AC generator which provides a source of sinusoidally varying emf ( t ) = 0 sin( ω t) , a resistor with resistance R = 1 ohm, and a "black box", which contains either an inductor or a capacitor, or both . We measure the current in the circuit at an angular frequency =30 radians/sec and find that the current is exactly in phase with the driving emf at this frequency. From this, we can determine that the black box must contain a) Only an inductor b) Only a capacitor c) A capacitor and an inductor d) We cannot determine anything about whether there is an inductor or capacitor or both in the black box from this observation alone. Since the current is in phase with the driving EMF we know that we must be in a purely resistive situation, so either there is no C or L in the black box (not an option) or there are both and we are in resonance.
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8.02T Fall 2004 Exam 3 Mashup Solutions 2 C The curve below shows data taken in a driven RLC circuit experiment. The horizontal axis is the time axis. The solid curve represents the power supply voltage. The dotted curve represents the current. This circuit acts 1. inductively 2. capacitively 3. resistively 4. not enough information to determine The solid line (ps voltage) leads the dotted line (current) so we are inductive and driving above the resonant frequency.
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This note was uploaded on 04/07/2008 for the course 8 8.02 taught by Professor Hudson during the Fall '07 term at MIT.

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Exam3_2004Fall_Solutions - Physics 8.02 TEST 3 MASHUP...

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