Alternating current : Solving for low-pass filter

[rohanprabhu]

Homework Statement

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

http://img252.imageshack.us/img252/3724/lowpassfilterrb2.jpg

Homework Equations

well.. maybe:

<br /> Z = \sqrt{R^2 + (X^2_C - X^2_L)}<br />

<br /> i = i_o sin(\omega t + \phi)<br />

<br /> \phi = tan^{-1}\left(\frac{X_C - X_L}{R}\right)<br />

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:

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

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.

 

[kamerling]

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.