In what relation is the energy input in an alternating current circuit to its frequency?

I'd guess I have to compute something like

$$E=\int P(\omega,t) dt=\int U(\omega,t) I(\omega,t) dt, $$

but if say

$$U(\omega,t)\propto\sin{(\omega t)},$$

then it seems part of the integral is $\propto\frac{1}{\omega}$, while I would expect the energy to grow with $\omega$.


Note that your formula

$$E=\int P(\omega,t) dt=\int U(\omega,t) I(\omega,t) dt$$ can be rewritten as

$$E=\int U(\omega,t) I(\omega,t) dt=\int\frac{U^2(w,t)}{Z}dt$$ Now, let $U=U_0\sin(wt)$ and $Z=const$ which is reasonable during a short period of time $t$.Thus:

$$E=\frac{1}{Z}\int_0^{t} U^2(w,t)dt=\frac{U_0^2}{Z}\int_0^{t} \sin^2(w,t)dt$$ Next:

$$\int_0^{t} \sin^2(w,t)dt=\frac{t}{2}-\frac{\sin(2wt)}{4w}$$ So, if $w\rightarrow\infty$, then $E$ does not depend of $w$.

Edit: Additions:

To be more specific, the circuit's impedance $Z$ depends on the frequency as

$$Z=\sqrt{r(w)^2+\left ( wL-\frac{1}{wC}\right)^2}$$ where $L$ is the circuit's impedance and $C$ is the circuit's capacitance and $r$ is the resistance which depends also on $w$ due to the skin effect.

That means $Z \rightarrow wL$ as $w\rightarrow\infty$ if $L\neq 0$

So an answer, closer to reality is that


  • $\begingroup$ Do you know how I can regulate the $\omega$? Is there an upper bouont for the $\omega$ I can make the thing working with? And where does that bound come from? $\endgroup$ – Nikolaj-K Apr 24 '12 at 7:18
  • $\begingroup$ @NickKidman en.wikipedia.org/wiki/Frequency_changer $\endgroup$ – Martin Gales Apr 24 '12 at 14:54

Consider just the power delivered: $$ P(t) = U(\omega t') I(\omega t') $$ Consider the simple case $ U(t')=U_0\sin(\omega t')$ and $ I(t')=I_0\sin(\omega t')$. Then the power delivered is $$P(t) = U_0 I_0\sin^2(\omega t)$$ $$={ U_0 I_0\over 2}\{1-cos(2\omega t)\}$$

We can break this into two terms: $$P_{DC}={U_0 I_0\over 2}$$ and $$P_{2\omega} = { U_0 I_0\over 2}cos(2\omega t)$$

$P_{DC}$ represents the average power flowing out of our power supply. It is $\propto$ voltage and current. Since it is constant in time, the energy delivered is rising linearly in time. That is, the longer we keep our device plugged in to our power supply, the more energy the power supply has delivered to it.

$P_{AC}$ delivers no net power. For half of the $2\omega$ cycle, it is dumping some extra power in, for the other half, it extracts the extra power dumped in. It is a reflection of the fact that current and voltage are not constant, but in fact are varying sinusoidally and only delivering $P_{DC}$ on average.

And yes, if you integrate $P(t)$ to get $E(t)$ you will see a $1/\omega$ term in the answer. I'll leave it to you to think about why. Hint: higher frequency peak powers don't last as long as lower frequency ones.

  • $\begingroup$ I be damned, you understood Nick's question. I had no idea what was the problem, I always think in terms of one-cycle averaged power. I guess I am ready for physics.SE retirement, and I am serous about that. $\endgroup$ – Pygmalion Apr 23 '12 at 19:00
  • $\begingroup$ @Pygmalion: Well, I wrote a formula for the energy in terms of power and the question was literally "In what relation is the energy input in an alternating current circuit to its frequency?". I wonder how can one interpret the question in a way, which would lead to an answer which does not involve the symbol E. But that's no complaint in any way. I just wonder why you wonder. The question was very direct and straight forward, no? Of course it often happens that one answers a question, which one had oneself at one point or a question one had been asked several times before. $\endgroup$ – Nikolaj-K Apr 23 '12 at 20:43
  • $\begingroup$ But anyway, I did the calculations before the question myself. It was only when I found the result that I wondered about it. My question is motivated by the following conceptial problem, which after the two posted answers I still don't get: If I consider the energy input, which I have to provide in one second interval (whatever the time $T$ may be doesn't really matter, here I consider the energy I need in one second of time), why do higher $\omega$ seem to mean less energy? Yes the sharp peaks imply that the integral will be small, but I find it counterintuitive that faster change = less $E$. $\endgroup$ – Nikolaj-K Apr 23 '12 at 20:43
  • $\begingroup$ The DC part of the energy is independent of $\omega$. It is only this "oscillating" energy which seems to go positve AND THEN NEGATIVE which is $\propto 1/\omega$. We know from the power equation that we have constant POWER at each peak, and that power is $P_{PEAK} = U_0 I_0$. But if that peak POWER is only flowing energy for a time $dt \propto 1/\omega$ then the amount of energy flowed in each peak is $\propto 1/\omega$. That is, if we turn the power +/-/+/- really fast, our + and - peaks of energy are smaller than if we toggle it really slowly. Hope this helps! $\endgroup$ – mwengler Apr 23 '12 at 23:09
  • $\begingroup$ I have more mathematical reasoning. If you are integrating energy as above, you eventually end up with a cyclical (trigonometrical) function with period $T$ in the end. Now, in physics only finite integrals make sense, so you must limit your integration within limits of that period, otherwise you will get the same result for upper limit $t$ and upper limit $t+T$. If frequency is larger, then the limiting period T is smaller and you get smaller energy... $\endgroup$ – Pygmalion Apr 24 '12 at 6:10

If you have a circuit with static elements (e.g. combination of resistors, capacitors and inductors), then for

$$U(t) = U_0 \sin(\omega t)$$

you have

$$I(t) = I_0 \sin(\omega t + \phi)$$.

You get $I_0$ and $\phi$ from complex impedance $Z(\omega)$:

$$I_0 = \frac{U_0}{|Z|}, \tan \phi = \left(\frac{\text{Im}(Z)}{\text{Re}(Z)}\right).$$

You can also observe voltage and current as complex numbers:

$$U(t) = U_0 e^{i\omega t}, I(t) = I_0 e^{i\omega t}, U_0 = Z I_0$$

EDIT: If you are only concerned by term $\omega^{-1}$, use $T_0$ instead of $\omega$:

$$U(t) = U_0 \sin(\frac{2 \pi t}{T_0})$$

you have

$$I(t) = I_0 \sin(\frac{2 \pi t}{T_0} + \phi)$$.

You will end up with the expression that energy is proportional to $T_0$. This only tells you that energy is smaller, because time in which it was transferred is smaller.

  • $\begingroup$ Mhm, that doesn't really that answer the question, does it? And you ask "why would part of the integral be proportional to $\omega^{-1}$" ... isn't this what you get when you integrate the phases? $\endgroup$ – Nikolaj-K Apr 23 '12 at 18:10
  • $\begingroup$ I am not quite sure what do you mean by energy as an integral of time. Usually you would do average power over one cycle, that is integrate energy for one voltage cycle and divide it by time of the cycle $1/T_0 = \omega/2\pi$, so you lose $1/\omega$. $\endgroup$ – Pygmalion Apr 23 '12 at 18:18
  • $\begingroup$ My last guess is that you must know that you are integrating cyclical function and they are defined only over one cycle (e.g. $\sin$ from 0 to $2 \pi$). Larger frequency, shorter period, smaller energy. $\endgroup$ – Pygmalion Apr 23 '12 at 18:47

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