Here's an interesting question:
If you have a periodic positive voltage feeding a constant current load, do you use the RMS value of the voltage or the average value of the voltage to calculate the power dissipated in the load?
Hi there,
Yes that is interesting
The general form of the voltage would be:
V=A*sin(w*t)+B
and the general form of the current would be a constant:
I=K
and so the instantaneous power would be:
p=V*I=K*(A*sin(w*t)+B)
and the average power is:
P=(1/T)*integral V*I dt
which expanded is:
P=(1/T)*integral [K*(A*sin(w*t)+B)] dt
and since K is a constant we can bring that out of the integral and get:
P=K*(1/T)*integral [(A*sin(w*t)+B)] dt
and since everything to the right of K is just the average voltage, we can write:
P=K*Vavg
and since K was the 'constant' current we can write:
P=I*Vavg
The RMS rule is only good for sine waves or special cases then because what does NOT work here is:
P = Irms*Vrms
so P != Irms*Vrms in this case, where "!=" is the symbol for "not equal to".
One thing that never fails is the integral form of calculating the average power:
P=integral [v*i] dt
and since v and i are more generally functions of time the more accurate writing of that is:
P=integral[v(t)*i(t)] dt
and this should always work as that is the definition of average power.
Just to note, in the past i have gotten good results with that last form even using a numerical integrator where each wave v(t) and i(t) are approximated as piecewise functions with straight lines connecting the measured points, where the points are read off the oscilloscope. With a little care the results are very reasonable.
Many things in electronics are dependent on the situation. For example, the average power is usually important, but in a battery run project the more important thing is the current drain because that affects the run time more than anything else.