# Calculating half power frequency/cutoff frequency

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1. May 5, 2015

### vishwesh

1. The problem statement, all variables and given/known data

*(Problems attached as images)*

For first problem:

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

For second problem:

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

I have to calculate half power frequency for both cases.

2. Relevant equations

At half power frequency, power absorbed by circuit = max power/2

3. The attempt at a solution

I cannot understand why the problem was solved differently in both cases. In first problem,$\cfrac{V_2}{V_1}$ was set to $\cfrac{1}{2\sqrt{2}}$ and in the second problem it was set to $\cfrac{1}{2}$. While in both problems, max value of $\cfrac{V_2}{V_1}$ occurs at $\omega = 0$ and is equal to $\cfrac{1}{2}$.

Can someone please clear this doubt?

Thanks

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2. May 6, 2015

### Svein

Power = Voltage * Current

3. May 6, 2015

### vishwesh

Thanks for the reply. But, I didn't get you. I am familiar with that formula. I understand that we use that formula to substitute $\cfrac{|V_2|}{|V_1|} = \cfrac{1}{\sqrt{2}}$ but how does it explain the $\cfrac{1}{2\sqrt{2}}$ part ?

Thanks again.

4. May 6, 2015

### Svein

When the impedance is reduced, the current increases. Thus the power ratio is not equal to the voltage ratio.

5. May 6, 2015

### Svein

When the impedance is reduced, the current increases. Thus the power ratio is not equal to the voltage ratio.

6. May 6, 2015

### Staff: Mentor

Check again the frequency and maximum output for the second circuit. How do capacitors behave with frequency?

For the second solution shown, I think the magnitude of the voltage ratio should have been set to $\frac{1}{\sqrt{2}}$ rather than $\frac{1}{2}$

7. May 6, 2015

### vishwesh

Thanks for the reply. I think I got it. The max output should be 1 when $\omega \rightarrow \infty$?

8. May 6, 2015

### Staff: Mentor

Yup. The capacitor starts to look like a short at high frequencies.