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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2009 Problem Set 3 Solutions Problem 1: LC Circuit (a) Initially, the capacitor in a series LC circuit is charged. A switch is closed, allowing the capacitor to discharge, and after time T the energy stored in the capacitor is one- quarter its initial value. Determine L if C and T are known. If the energy is at ¼ of it’s initial value then the voltage has fallen in half. That is, () ( ) 2 0 0 2 19 cos 23 3 V T VT V T T L TC LC ππ ωω ω π == = = = = (b) An LC circuit consists of a 90.0-mH inductor and a 1.000- F μ capacitor. If the maximum instantaneous current is 0.100 A, what is the greatest potential difference across the capacitor? By conservation of energy, 22 11 max max LI CV = , so () max max 90 mH 0.1 A 30 V 1.0 μ F L VI C = Problem 2: Filters One of the very useful things that you can do now that you understand AC circuits and impedance is design filters. A filter is a circuit that takes an input and returns an output which, for a certain range of frequencies, is attenuated. There are four basic filter types: low pass (in which low frequencies are let through unattenuated, but high frequencies are blocked), high pass (the opposite), band pass (essentially a combination of a low and high pass filter, letting through signals in a range, or band, of frequencies, and blocking those outside), and stop band (the opposite). There are also notch or trap filters which are similar to stop band filters but try to notch out at a single frequency (such as 60 Hz, which has an annoying habit of creeping into every signal and is frequently eliminated using a notch filter). When trying to analyze filters there are several simple steps to take. First of all, in thinking about what a filter will do, first look at the low and high frequency limits. Recall that at low frequencies inductors look like wires and capacitors like large resistors (or open circuits) while at high frequencies the opposite is true. Problem Set 3 Solutions p. 1 of 8 Fall 2009
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Problem Set 3 Solutions p. 2 of 8 Fall 2009 Problem 2: Filters continued (a) This is a high pass filter. At low frequencies the capacitor will fill up and look like an open circuit, while at high frequencies it will look like a wire. The output voltage is given by: () 0,Out 0 0,in 0,in 22 2 0,Out 0 0,in 0 1 1 11 1 Transfer Func. where is the 3dB point 1 C C R VI R V V RX XR V VR C RC ω ωω == = + + = = ++ To find the slope we go to very low frequencies where our function simplifies: 2 0 0 1 Transfer Func.= 1 + , so clearly the slope is 20 log 2 = 6 dB per octave. Here is a plot of the transfer function: 0.01 0.1 1 10 100 0.01 0.1 1 -40 dB -20 dB 0 dB slope = 6 dB/octave = 20 dB/decade 3dB freq = ω 0 Transfer Func. Reduced Frequency: ω / ω 0 (b) This is a notch filter. At high frequencies the inductor impedance will dominate the circuit and the output will essentially see all of the voltage from that. At low frequencies the capacitor will dominate and again the output will get everything. On resonance the combined impedance of L & C in series will be zero so the output voltage will fall to zero. Let’s calculate: ( ) 0,Out 0 0,in
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