Alternating current : Solving for low-pass filter

  • #1
412
2

Homework Statement



Q] The figure below shows a typical circuit for a low-pass filter. An AC input [itex]V_i~=~10~mV[/itex] is applied at the left end and the output [itex]V_o[/itex] is received at the right end. Find the output voltage as a function of [itex]\nu~(\textrm{frequency})[/itex]

http://img252.imageshack.us/img252/3724/lowpassfilterrb2.jpg [Broken]

Homework Equations



well.. maybe:

[tex]
Z = \sqrt{R^2 + (X^2_C - X^2_L)}
[/tex]

[tex]
i = i_o sin(\omega t + \phi)
[/tex]

[tex]
\phi = tan^{-1}\left(\frac{X_C - X_L}{R}\right)
[/tex]

The Attempt at a Solution



Well.. i'm just thinking that the voltage across the capacitor will be the output voltage. I found out the current in the circuit using the equations and it is coming to be:

[tex]
i = i_o sin\left(\omega t + tan^{-1}\left(\frac{10^6}{2\pi \nu}\right)\right)
[/tex]

This is the current in the first loop. I thought of using Kirchoff's law first.. but i have no idea how to do this in this case as the current will split at the resistor-capacitor junction.

Any help is appreciated. Thanks.
 
Last edited by a moderator:

Answers and Replies

  • #2
454
0
I think you can assume that the output will be connected to a very high resistance (such as an oscilloscope), so there's no current going into the output.

You know how to express the impedance of R, C and the input voltage as complex numbers? then the circuit is just a voltage divider.
 

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