Why don't we use the antiderivative factor for Average Voltage

[rtareen]

Below is the really quick derivation for average voltage. However when we do the anti-derivative a factor of $1/\omega$ should come out and the full answer should be $\frac{2V_p}{\pi \omega}$. So why don't we include that? What's going on?

 

[Office_Shredder]

I think there's a typo. The range from $0$ to $\pi$ is not one half a cycle of $\sin(\omega \pi)$. It's probably supposed to be the integral from $0$ to $\pi/\omega$?

 

[rtareen]

I think there's a typo. The range from $0$ to $\pi$ is not one half a cycle of $\sin(\omega \pi)$. It's probably supposed to be the integral from $0$ to $\pi/\omega$?

I wouldn't doubt it. I already found something wrong previously in this book. It doesn't help that there's only one edition. However that doesn't solve the problem of the factor $1/\omega$ not being included?

 

[mfb]

It's a time average, so the factor in front of the integral gets a factor $\omega$, too (we average over a time of $\frac \pi \omega$).

 

[kuruman]

Below is the really quick derivation for average voltage. However when we do the anti-derivative a factor of $1/\omega$ should come out and the full answer should be $\frac{2V_p}{\pi \omega}$. So why don't we include that? What's going on?

Do the integral (anti-derivative). The average of the voltage over half a cycle when time is the variable is the average over half a period T:$$\langle V \rangle = \frac{\int_0^{T/2}V_P\sin(\omega t)~dt}{\int_0^{T/2}dt}=\frac{ V_P\left[1-\cos\left(\frac{\omega T}{2}\right)\right]}{\omega}\times \frac{1}{\frac{T}{2}}.$$ Now substitute $\omega T=2\pi.$ What do you get?

 

[rtareen]

It's a time average, so the factor in front of the integral gets a factor $\omega$, too (we average over a time of $\frac \pi \omega$).

That makes it work out. Thanks!

 

[rtareen]

Do the integral (anti-derivative). The average of the voltage over half a cycle when time is the variable is the average over half a period T:$$\langle V \rangle = \frac{\int_0^{T/2}V_P\sin(\omega t)~dt}{\int_0^{T/2}dt}=\frac{ V_P\left[1-\cos\left(\frac{\omega T}{2}\right)\right]}{\omega}\times \frac{1}{\frac{T}{2}}.$$ Now substitute $\omega T=2\pi.$ What do you get?

I get the answer they got. After a search I realized that was the right answer. But the integral(s) they say would evaluate to that is wrong.

 

[kuruman]

I get the answer they got. After a search I realized that was the right answer. But the integral(s) they say would evaluate to that is wrong.

What integral(s) is (are) that (these) and who are they who say that?

 

[rtareen]

What integral(s) is (are) that (these) and who are they who say that?

If you look at OP you will see they got the answer right but did not set up the integral properly.

 

[kuruman]

If you look at OP you will see they got the answer right but did not set up the integral properly.

Yes, I thought the reference was to some other integral.