Frequency and energy of EM radiation

[jartsa]

If one were to measure power, which is what I started with (i.e. the amount of energy measured in a unit time), then which wave will produce more energy for a fixed amplitude (i.e. they have the SAME average energy per cycle): the one that makes more complete cycle per unit time, or the one that makes less complete cycle per unit time?

Case 1: Let's say we have a 12 V battery, from witch we draw 1 Ampere for one second, zero Amperes for one second, 1 Ampere for one second ... and so on.

Now the RMS voltage of that current is $\frac{12}{ \sqrt 2}$

And the average power is 6 W.Case 2: Let's say we have a 12 V battery, from witch we draw 1 Ampere for two second, zero Amperes for two seconds, 1 Ampere for two seconds ... and so on.

The RMS voltage of that current is still $\frac{12}{ \sqrt 2}$

And the average power is still 6 W.

(The voltage varies between 0 and 12, that is the idea, for some reason I talked about amperage)

 

[ZapperZ]

Case 1: Let's say we have a 12 V battery, from witch we draw 1 Ampere for one second, zero Amperes for one second, 1 Ampere for one second ... and so on.

Now the RMS voltage of that current is $\frac{12}{ \sqrt 2}$

And the average power is 6 W.Case 2: Let's say we have a 12 V battery, from witch we draw 1 Ampere for two second, zero Amperes for two seconds, 1 Ampere for two seconds ... and so on.

The RMS voltage of that current is still $\frac{12}{ \sqrt 2}$

And the average power is still 6 W.

(The voltage varies between 0 and 12, that is the idea, for some reason I talked about amperage)

I don't think this is what we're talking about. The average value doesn't change, because this is the same value averaged over one cycle. But there is a difference if your measurement window measures 10 complete oscillations, or 20 complete oscillations. The device will measure the the total energy within that window. This is what I tried to emphasize in my first post here.

Zz.