B
Bob Eld
Guest
"panfilero" <panfilero@gmail.com> wrote in message
news:f1496677-7937-43e6-9bbd-386d103b675d@p39g2000prm.googlegroups.com...
Yes the RMS current is the RMS voltage devided by R.
No, the "regualr curent" What ever that is, is not the RMS current divided
by the square root of two. With a sign wave, the peak current divided by the
the square root of two is the RMS current, but ONLY for a sign wave.
With any other waveform, the the square root of two relation does NOT hold.
It must be calculated from first principles or measured with a true RMS
meter.
The equation for the RMS current is: Irms = ((int i(t)^2 dt )/ T)^1/2
Where i(t) is the current as a function of time, T is the time over which
the integral is taken, ( 0 to T ), usually one period of the wave.
That is, in words, the RMS current is the square root of the average time
integral of the current squared.
For a square wave of current Ip going plus and minus over a period of T the
RMS value is:
Irms =( ( (Ip^2 *T/2 + (-Ip)^2 * T/2)) / T)^1/2 = ( (Ip ^2)/2 +
(Ip^2)/2)^1/2 = (Ip ^2)^1/2
Irms = Ip, The RMS equals the peak value for a square wave. It very much
more complicated if noise is present.
news:f1496677-7937-43e6-9bbd-386d103b675d@p39g2000prm.googlegroups.com...
I was wondering if I take the RMS voltage of a messy looking square
wave, a noisy square wave, with some measurement device, and I want to
know the regular current of the thing can I take the RMS current and
divide it by the square root of 2 in order to get my regular current?
I know you can do this for sinusoidal ac signals but don't know if the
math still works out the same for other signals...
PS - I got my RMS current value by dividing the RMS voltage value by a
known resistance I have in my circuit
Much Thanks for any responses I get
Yes the RMS current is the RMS voltage devided by R.
No, the "regualr curent" What ever that is, is not the RMS current divided
by the square root of two. With a sign wave, the peak current divided by the
the square root of two is the RMS current, but ONLY for a sign wave.
With any other waveform, the the square root of two relation does NOT hold.
It must be calculated from first principles or measured with a true RMS
meter.
The equation for the RMS current is: Irms = ((int i(t)^2 dt )/ T)^1/2
Where i(t) is the current as a function of time, T is the time over which
the integral is taken, ( 0 to T ), usually one period of the wave.
That is, in words, the RMS current is the square root of the average time
integral of the current squared.
For a square wave of current Ip going plus and minus over a period of T the
RMS value is:
Irms =( ( (Ip^2 *T/2 + (-Ip)^2 * T/2)) / T)^1/2 = ( (Ip ^2)/2 +
(Ip^2)/2)^1/2 = (Ip ^2)^1/2
Irms = Ip, The RMS equals the peak value for a square wave. It very much
more complicated if noise is present.