Question

V = 100 sin (200πt + π/4)
using integral calculus, calculate the RMS value of the voltage

Answer 1

So if T is the period of this voltage function, the RMS voltage is
\[ RMS = \sqrt{\frac{1}{T} \int_0^T V(t)^2 dt } \]

So now you just have to figure out T and evaluate this expression

Answer 2

ok i have already worked out that t=5s when the voltage is 70.711v? so could i use these figures into the formular then would i need to differentiate it? sorry i did the assignment 6 months ago and its now a little hazy

Answer 3

You don't need to differentiate it.

But in any case, the answer is \[ \frac{100}{\sqrt{2}} \] You should do the calculation to convince yourself that is right just in case you get asked a question like this and need to do it from first principles.

Answer 4

ok thank you I will give it a go

Answer 5

Actually, I'll show you the trick.

What is
\[ I = \int_0^{2\pi} \sin^2 t \ dt \]?

Well, as cos^2 t has the same period length of 2.pi and should have the same values over that period, just displaced by a pi/2, it must be that
\[ \int_0^{2\pi} \sin^2 t \ dt = \int_0^{2\pi} \cos^2 t \ dt \]
Hence
\[ 2I = \int_0^{2\pi} (\sin^2 t + \cos^2 t) \ dt = \int_0^{2\pi} 1 \ dt = 2\pi \]
That is, \( I = \pi \).

Now that means that the RMS of V(t) = sin(t) must be
\[ RMS = \sqrt{ \frac{1}{2\pi} \pi} = \frac{1}{\sqrt{2}} \]

Answer 6

Hence it must also be the case that the RMS of \( V(t) = V_0 \sin(t) = V_0/\sqrt{2} \) because the \( V_0 \) comes out squared through the integral and then the square root of it taken again.

Finally, if you change the period or displace the function, \( V(t) = V_0 \sin(\omega t + \phi_0) \), it's not hard to convince yourself you get the same result as above for \( I \).

Answer 7

thank you for explaining so indepth for me, I think I know where I am going with it now.

Answer 8

@JamesJ here I was all set to give you a medal for this fine explanation, when I noticed that I already had! I would have given you 2 :-)