Question

For an RC filter circuit, show that$V_R=V_C=\frac{{\epsilon }_0}{\sqrt{2}}$at$\omega ={\omega }_c$.

Short answer

The expression is true$V_R=V_C=\frac{{\epsilon }_0}{\sqrt{2}}$.

Step-by-step solution

Given Information

We are given that $\omega ={\omega }_c$and

we have to find out that $V_R=V_C=\frac{{\epsilon }_o}{\sqrt{2}}$.

Simplification

As we know that for an RC filter circuit,

${\omega }_c=\frac{1}{RC}$

Next we will find V_R :

$V_R=\frac{{\epsilon }_{0}R}{\sqrt{R^2+{\displaystyle \frac{R^2{C}^2}{C^2}}}}$

We get ,$V_R=\frac{{\epsilon }_0}{\sqrt{2}}$

Finally, we will find V_C:

$V_c=\frac{{\displaystyle \frac{{\epsilon }_0}{{\omega }_c}}}{\sqrt{R^2+{\displaystyle \frac{1}{{\omega }^2{C}^2}}}}$

$V_C=\frac{{\epsilon }_0}{\sqrt{{\omega }^2{R}^2{C}^2+1}}$

We get,$V_C=\frac{{\epsilon }_0}{\sqrt{2}}$

As $V_R=V_C=\frac{{\epsilon }_o}{\sqrt{2}}$

Hence, it is proved that$V_R=V_C=\frac{{\epsilon }_0}{\sqrt{2}}$.