rms voltage..

sndpgr

Joined Jun 22, 2006
23
Quite an embarrassing question really.
A DC through a resistor R will dissipate V^2/R watts.
when a ac signal v(t) will be applied to same resistor R what will be the power dissipated,I know it will be V(rms)^2*R and I know the value of V(rms).but how do you arrive at that.
kindly help!

recca02

Joined Apr 2, 2007
1,211
see if this helps.
also go thru this if you have the time.
and then its AAC we are talking about.

Audioguru

Joined Dec 20, 2007
11,251
The average power from RMS AC at a certain voltage is exactly the same as power from DC at that same voltage.
The proof is in a heater. Or a big lightbulb that is slow enough not to flicker with AC.

beenthere

Joined Apr 20, 2004
15,808
The whole purpose of using RMS to measure AC is to make AC and DC power calculations exactly equivalent.

thingmaker3

Joined May 16, 2005
5,073
The average power from RMS AC at a certain voltage is exactly the same as power from DC at that same voltage.
I hate to be nit-picky, but there is a difference between "average power" and "RMS power." Just as there is a difference between "average voltage" and "RMS voltage"

But, yes, RMS and DC will yield the same numbers for power, voltage, and current.

Distort10n

Joined Dec 25, 2006
429
I hate to be nit-picky, but there is a difference between "average power" and "RMS power." Just as there is a difference between "average voltage" and "RMS voltage"
To be strictly technical, "average power" implies RMS voltages and currents. "RMS Power" is a misnomer by many engineers:

http://www.eznec.com/Amateur/RMS_Power.pdf
http://www.hifi-writer.com/he/misc/rmspower.htm

It is also fun to read the arguments on the Wikipedia "Talk" page about Audio Power:

http://en.wikipedia.org/wiki/Audio_power

But you are correct, in regards to average or RMS voltage or current waveforms. Using a single cycle sine wave any any given frequency the RMS value is equal to 0.707 of the peak value, while the average is equal to 0.634 (I believe) of the peak value.

sndpgr

Joined Jun 22, 2006
23
thank's for the encouraging replies.
This is what I was I was looking for-
'The heating effect depends on I2R, and so an average of I2 is needed and not an average of I. ' from the website you gave .

Thank you!again!
the average square voltage is then V(t)^2 integrated on a time period and divided by time period.
Am I on right direction.