3 ms·

Well, no, not if the voltage for example doesn't stay constant nor changes with a constant period, because then P=IV becomes an integral that doesn't simplify l

by madez 9y ago

Well, no, not if the voltage for example doesn't stay constant nor changes with a constant period, because then P=IV becomes an integral that doesn't simplify like we are used to.

saagarjha 9y ago
I was under the impression that voltage was mostly constant most of the time, except for brief periods where it was switched to keep up with/scale back to match demand. Assuming constant voltage should be a good approximation in this case, right?
- moccachino 9y ago
  Imagine a stone being lifted. The weight of the stone is the current, and the height it is lifted to is the voltage. Both the weight of the stone and the height have an impact on the energy required to do the lifting. Now imagine you are lifting and lowering it repeatedly. It should be intuitive that doing so 50 times a second requires more power than doing it 25 times a second (One being more lifts than the other, in the same time period).
  - 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).
  - arghwhat 9y ago
    Do you mean weight or mass? Is gravity a thing here? If you define mass/inertia as capacitance, then the resistance to changes make sense. A pure resistive load (which cannot exist, but lets ignore that) is not frequency dependent.
    - moccachino 9y ago
      I'll admit I didn't really think this analogy all the way through.
- carlob 9y ago
  Mmmm, no in alternating current V is alternating as well as I
  - saagarjha 9y ago
    Yes, but they should both be alternating in a periodic fashion, right?
    - madez 9y ago
      The simplification for calculating the power of an alternating current by just using the maximal values and a correction factor is only valid if the signal is alternating in a periodic fashion, which includes constant period. However, the reasons this thread exists is that the electricity power signal does not have a constant period, and is therefore not strictly periodic. Then, the simplification for the integral does not hold anymore, and the integral must be evaluated with some other technique.

by madez 9y ago

Well, no, not if the voltage for example doesn't stay constant nor changes with a constant period, because then P=IV becomes an integral that doesn't simplify like we are used to.

saagarjha 9y ago

I was under the impression that voltage was mostly constant most of the time, except for brief periods where it was switched to keep up with/scale back to match demand. Assuming constant voltage should be a good approximation in this case, right?

moccachino 9y ago
Imagine a stone being lifted. The weight of the stone is the current, and the height it is lifted to is the voltage. Both the weight of the stone and the height have an impact on the energy required to do the lifting. Now imagine you are lifting and lowering it repeatedly. It should be intuitive that doing so 50 times a second requires more power than doing it 25 times a second (One being more lifts than the other, in the same time period).
- 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).
- arghwhat 9y ago
  Do you mean weight or mass? Is gravity a thing here? If you define mass/inertia as capacitance, then the resistance to changes make sense. A pure resistive load (which cannot exist, but lets ignore that) is not frequency dependent.
  - moccachino 9y ago
    I'll admit I didn't really think this analogy all the way through.
carlob 9y ago
Mmmm, no in alternating current V is alternating as well as I
- saagarjha 9y ago
  Yes, but they should both be alternating in a periodic fashion, right?
  - madez 9y ago
    The simplification for calculating the power of an alternating current by just using the maximal values and a correction factor is only valid if the signal is alternating in a periodic fashion, which includes constant period. However, the reasons this thread exists is that the electricity power signal does not have a constant period, and is therefore not strictly periodic. Then, the simplification for the integral does not hold anymore, and the integral must be evaluated with some other technique.