Question

You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By monitoring the capacitor voltage on an oscilloscope, you see that the voltage decays to half its initial value in 3.40 msms . You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency fcfc

Answer 1

Answer:

The frequency is    $f = 0.221 \ Hz$

Explanation:

From the question we are told that  

     The  time taken for it to decay to half its original size is $t = 3.40 \ ms = 3.40 *10^{-3} \ s$

Let the voltage of the capacitor when it is fully charged be  $V_o$

Then the voltage of the capacitor at time t is  said to be  $V = \frac{V_o}{2}$

   Now  this voltage can be  mathematical represented as

      $V = V_o * e ^{-\frac{t}{RC} }$

Where  RC  is the time constant

   substituting values  

    $\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }$

    $0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }$

    $- \frac{0.5}{RC} = ln (0.5)$

     $-\frac{0.5}{RC} = -0.6931$

     $RC = 0.721$

Generally the cross-over frequency for a low pass filter is mathematically represented as

          $f = \frac{1}{2 \pi * RC }$

substituting values  

           $f = \frac{1}{2* 3.142 * 0.72 }$

           $f = 0.221 \ Hz$