3 ms·
Things are a bit more complicated than this. In your toy system a stone that stays still (DC) would have no power (potential energy), which is false. Also in yo
by carlob 9y ago
Things are a bit more complicated than this. In your toy system a stone that stays still (DC) would have no power (potential energy), which is false. Also in your toy system m is constant. In general for AC you have both V and I oscillating.
As a matter of fact the average DC power does not depend on frequency.
You could posit a system where g is alternating between positive and negative values and the h reacts to it (phase-shifted). You'd quickly come the conclusion that the potential energy (averaged over a period of oscillation) does NOT depend on the frequency. (The kinetic energy does, but that's where your analogy breaks).
- arghwhat 9y ago> In your toy system a stone that stays still (DC) would have no power (potential energy). This is false. In the toy system, weight is current, height is voltage. A stone that stays still has constant voltage, not zero voltage. Thus, it would have a 'power'. What the system lacks is to define the stone as a capacitive load. Then it would sorta make sense. It is a hypothetical system, so you can only reason about the aspects the author defined. Tying potential energy in the toy system to real-world potential energy doesn't work. (Btw, potential energy is not power, it is work. Power is work over time.) > As a matter of fact the average DC power does not depend on frequency. Uhm. "DC power" stops existing if the frequency ≠ 0, so in that sense it does depend on frequency. It's true that power itself is not frequency dependent. However, any load is, as reactive losses (parasitic or not) are a function of the frequency. As the power is a function of the load, power ends up being directly tied to the frequency. (A resistive load cannot exist outside of a perfect DC system, so reactive loads will exist).