If the voltage is AC 220V, Then the maximum value of the voltage (at a specific time) is: 310V or 345V

Please remember these things!

First value 310V = 220V * 1.4142
The second value 345V is obtained through averaging the sin (x) function on the domain 0....Pi


Edit: I put a wrong "l" in the title and I do not know how to change it

 

I remeber the peak voltage is RMS x the square root of two (1.414...).

Would you please demonstrate your second technique? (I don't understand)

 

Would you please demonstrate your second technique? (I don't understand)

I did not know what this 220V AC is. If you say it's RMS then it's clear now.

I thought it was the continuous voltage equivalent to alternative voltage as an average.
To get the average I did the following:
u(t)=Vmax*sin(omega*t)
simply written:
f=sin(x) for x=0....Pi

The integral of sin (x) is - cos (x).
The integral of sin (x) on the 0 ... Pi range (the area of a semi-alternate will be)
cos(0)-cos(Pi)=1+1=2

What constant voltage on the same domain would give the same area?
Vct*Pi=2 => Vct=0.6....... if Vmax=1

But if Vct=220V =>Vmax=345V

 

Sorry, I missed the key word "averaging" in your first post. Now you understand the 220V refers to the RMS value.