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:

$$Z = \sqrt{R^2 + (X^2_C - X^2_L)}$$

$$i = i_o sin(\omega t + \phi)$$

$$\phi = tan^{-1}\left(\frac{X_C - X_L}{R}\right)$$

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:

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

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.

 

[Moetasim]

I would approach this problem by first understanding the principles of a low-pass filter and how it works. A low-pass filter is designed to allow low frequency signals to pass through while attenuating high frequency signals. In this circuit, the resistor and capacitor act as a low-pass filter, with the resistor determining the cutoff frequency and the capacitor acting as a frequency-dependent impedance.

To solve for the output voltage as a function of frequency, we can use the equations for impedance and current provided in the homework statement. The impedance of the circuit can be calculated using the equation Z = \sqrt{R^2 + (X^2_C - X^2_L)}, where R is the resistance of the resistor and X_C and X_L are the reactances of the capacitor and inductor, respectively. The current in the circuit can be calculated using the equation i = i_o sin(\omega t + \phi), where i_o is the amplitude of the input current and \phi is the phase angle.

To find the output voltage, we can use Kirchoff's voltage law, which states that the sum of the voltages around a closed loop in a circuit is equal to zero. In this case, the voltage across the capacitor will be the output voltage. Using Kirchoff's law, we can set the voltage across the capacitor equal to the voltage across the resistor, which is equal to the input voltage. This will allow us to solve for the output voltage as a function of frequency.

In summary, to solve for the output voltage as a function of frequency in a low-pass filter, we can use the equations for impedance, current, and Kirchoff's law. By understanding the principles of a low-pass filter and using these equations, we can accurately determine the output voltage for any given frequency.