2
$\begingroup$

I am having a doubt in calculating half power frequency for a given RLC AC circuit. I have attached images of two questions with their solutions. In the first question, the equation for $\cfrac{|V_2|}{|V_1|}$ came out to be:

$\cfrac{|V_2|}{|V_1|} = \cfrac{1}{\sqrt{4+(\omega RC)^2}}$

For calculating half power frequency, it was set equal $\cfrac{1}{\sqrt{2}}$ times the max. value which is $\cfrac{1}{2}$ at $\omega = 0$.

But, in the other problem, the equation came out to be:

$\cfrac{|V_2|}{|V_1|} = \cfrac{\sqrt{1+(\omega RC)^2}}{\sqrt{4 + (\omega RC)^2}}$

For calculating half power frequency, they set it equal to $\cfrac{1}{2}$ (which I think is the max. value at $\omega = 0$.

Can anyone please explain why this difference in solving the problems?

Thanks

Problem 1 Problem 2

$\endgroup$
1
$\begingroup$

The maximum of $$ \left|\cfrac{V_2}{V_1}\right| = \cfrac{\sqrt{1+(\omega RC)^2}}{\sqrt{4 + (\omega RC)^2}} $$ is $1$ at $\omega=\pm \infty$, and you find the half power frequency by solving: $$ \frac{1+(\omega RC)^2}{4+(\omega RC)^2}=\frac{1}{2} $$ which gives $\omega=\pm \sqrt{2}/RC$ enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.