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Constructing a Low-Pass Filter
A series L-R-C circuit is driven with AC voltage of amplitude V_in and frequency omega. Define V_out to be the amplitude of the voltage across the capacitor. The resistance of the resistor is R, the capacitance of the capacitor is C, and the inductance of the inductor is L.
Part A
What is the ratio \frac{V_{\rm out}}{V_{\rm in}}?
Part A.1
Find V_in
Assume that the amplitude of the current in the circuit is I. Write down an expression for the input voltage amplitude V_in.
Hint A.1.a
How to approach the problem

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Hint A.1.b
Combined impedance

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Express your answer in terms of some or all of the variables I, R, the reactance of the capacitor X_C, and the reactance of the inductor X_L.
ANSWER:
  V_in =

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Part A.2
Find V_out

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Express your answer in terms of either R, omega, L, and C or R, X_L=\omega L, and X_C=\frac{1}{\omega C}.
ANSWER:
  \frac {V_{\rm out}} {V_{\rm in}} =  \frac{X_{C}}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}} 

For the following questions it will be useful to write the voltage ratio in the following form:

\frac{V_{\rm out}}{V_{\rm in}} = \frac{1}{\sqrt{(\omega R C)^2+(\omega^2 L C-1)^2}}.

Part B
Which of the following statements is true in the large omega limit (that is, for \omega \gg \frac{1}{RC},\frac{1}{\sqrt{L C}},\frac{R}{L})?
Hint B.1  

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ANSWER:
 \frac {V_{\rm out}} {V_{\rm in}} is proportional to \frac {1} {\omega}.
 \frac {V_{\rm out}} {V_{\rm in}} is proportional to  \frac {1} {\omega^2}.
 \frac {V_{\rm out}} {V_{\rm in}} is proportional to \omega.
 \frac {V_{\rm out}} {V_{\rm in}} is close to 1.
Part C
Which of the following statements is true in the small omega limit (that is, for \omega \ll \frac{1}{RC},\frac{1}{\sqrt{L C}},\frac{R}{L})?
Hint C.1  

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ANSWER:
 \frac {V_{\rm out}} {V_{\rm in}} is proportional to \frac {1} {\omega}.
 \frac {V_{\rm out}} {V_{\rm in}} is proportional to  \frac {1} {\omega^2}.
 \frac {V_{\rm out}} {V_{\rm in}} is proportional to \omega.
 \frac {V_{\rm out}} {V_{\rm in}} is close to 1.
When omega is large, \frac{V_{\rm out}}{V_{\rm in}} \propto \frac{1}{\omega^2}; and when omega is small, \frac{V_{\rm out}}{V_{\rm in}} \approx 1. Therefore, the circuit of this problem has the property that only the high-frequency inputs will be attenuated (reduced in value) at the output, while low-frequency inputs will pass through relatively unchanged. That is why such a circuit is called a low-pass filter.
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