Question

A high-pass $RC$filter with a crossover frequency of$1000Hz$ uses a $100\Omega$ resistor. What is the value of the capacitor?

Short answer

The value of the capacitor,$C=1.59\times {10}^{-6}\mathrm{F}$

Step-by-step solution

RC Filter

The opposite is true for an RC high-pass filter, commonly known as an RC Differentiator.

As seen above, the input signal is applied directly to the capacitor via a resistor in parallel with the output.

High-frequency signals can travel through this arrangement, whereas low-frequency impulses are blocked by the capacitor.

Find Capacitance

$f_c=1000\mathrm{Hz}$

$R=100\Omega$

$w=\frac{1}{RC}$

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

Multiply both sides by$R$

$2\pi f_{c}R=\frac{1}{C}$

$C=\frac{1}{2\pi f_{c}R}$

Fill in the blanks in the equation

localid="1650211859861" $C=\frac{1}{2\pi \times \left(1000\right)Hz\times \left(100\right)\Omega }$

$C=1.59\times {10}^{-6}\mathrm{F}$