- #1

Efeguleroglu

- 24

- 2

We write rms formula as

$$v_{rms}=\sqrt\frac{\int_a^b[f(x)]^2dx}{|b-a|}$$

I know if we take arithmetic mean average voltage will be 0. So we want all voltage values to be positive. Why don't we do that:

$$V_{average}=\frac{\int_a^b\sqrt{[f(x)]^2}dx}{|b-a|}$$

That's first what I did:

$$\phi=ABcos(\omega t)$$

$$d\phi=-AB\omega sin(\omega t) dt$$

$$V=-\frac{d\phi}{dt}=AB\omega sin(\omega t)$$

If we take rms of $$AB\omega sin(\omega t)$$ we find $$\frac{AB\omega}{\sqrt{2}}$$

In my method it's this:

$$f(t)=AB\omega sin(\omega t)$$$$\frac{\int_0^{\frac{\pi}{2}}{f(t)}dt}{\frac{\pi}{2}}=\frac{2AB\omega}{\pi}$$

So please can someone explain why do I have such a confusion?

$$v_{rms}=\sqrt\frac{\int_a^b[f(x)]^2dx}{|b-a|}$$

I know if we take arithmetic mean average voltage will be 0. So we want all voltage values to be positive. Why don't we do that:

$$V_{average}=\frac{\int_a^b\sqrt{[f(x)]^2}dx}{|b-a|}$$

That's first what I did:

$$\phi=ABcos(\omega t)$$

$$d\phi=-AB\omega sin(\omega t) dt$$

$$V=-\frac{d\phi}{dt}=AB\omega sin(\omega t)$$

If we take rms of $$AB\omega sin(\omega t)$$ we find $$\frac{AB\omega}{\sqrt{2}}$$

In my method it's this:

$$f(t)=AB\omega sin(\omega t)$$$$\frac{\int_0^{\frac{\pi}{2}}{f(t)}dt}{\frac{\pi}{2}}=\frac{2AB\omega}{\pi}$$

So please can someone explain why do I have such a confusion?

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