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 a oscilloscope, you see that the voltage decays to half its initial value in $2.5ms$. You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency $f_c$?

Short answer

The crossover frequency is$45Hz$.

Step-by-step solution

Step 1:Given Information

We are given that the time $t_o=2.5\times {10}^{-3}s$.

We have to find out the crossover frequency.

Step 2:Formula Used

Now, as voltage across capacitor is

$V\left(t_o\right)=V_o{e}^{\frac{-t_o}{RC}}$

where $V\left(t_o\right)=\frac{v_o}{2}$(As voltage decays half to its initial value)

Putting the value,

We get ,role="math" localid="1650011861919" $\frac{1}{2}=e^{\frac{-t_o}{RC}}$

On simplifying we get,

$RC=\frac{t_0}{\ln 2}$

Simplify

Now,

using $f_c=\frac{1}{2\mathrm{\pi RC}}$

and putting the value of RC we get,

$f_c=\frac{\ln 2}{2{\mathrm{\pi t}}_{\mathrm{o}}}$

and $f_c=\frac{0.6931}{2\mathrm{\pi }(2.5\times {10}^{-3})}$

On simplifying we get,

$f_c=45Hz$

Hence, the crossover frequency is$45Hz$