为什么在计算平均功率时使用均方根，而不是简单地计算电压/电流的均方根？

28

P = I_{eff}^{2} \times R

$P = I_{\text{eff}}^2 \times R$ 其中是有效电流。为了使功率平均，必须是平均电流，因此我推测有效电流是平均电流。

I_{eff}

$I_{\text{eff}}$

I

$I$

在那种情况下，为什么不只是 $I_{\text{eff}}$

I_{eff} = \frac{1}{t} \int_{0}^{t} | i | d t

$I_{\text{eff}} = \frac{1}{t}\int_{0}^{t} |i|dt$

相反，它的定义如下：

I_{eff} = \sqrt{\frac{1}{t} \int_{0}^{t} i^{2} d t}

$I_{\text{eff}} = \sqrt{\frac{1}{t}\int_{0}^{t} i^2dt}$

因此，使用这两个表达式来计算得出不同的答案。 $P$

为什么会这样呢？对我来说完全是无稽之谈。我只能猜测我误解了有效电流是平均电流。但是，如果不是这种情况，那么当不是平均电流时，我将看不到如何成为平均功率。 $P$ $I_{\text{eff}}$

power-supply power circuit-analysis

— 金名
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50

对于交流电，平均电压/电流为零。

— 罗杰·罗兰德

9

功率与电流的平方成正比，而不与电流的大小成正比。

— 储朱

26

因为如果您想要平均功率，则必须计算功率并对其平均，而不是不是功率的东西。

— Neil_UK

4

“要使功率平均$ I $必须是平均电流”-这就是您错的地方。

— user253751 '16

6

@drobertson "Root mean square" = root of mean of square, which is not the same as mean of root of square, and therefore not the same as mean of absolute value.

— user253751

56

Take a simple example where the sums are trivial. I have a voltage that is on 50% of the time and off 50% of the time. It is 10V when it is on. The average voltage is thus 5V. If I connect a resistor of 1 ohm across it, it will dissipate 100W when it is on and 0W when it is off. The average power is thus 50W.

Now leave the voltage on all the time but make it 5V. Average voltage is still 5V, but the average power is only 25W. Oops.

Or suppose I have the voltage only on 10% of the time, but it is 50V. The average voltage is 5V again, but the power is 2500W when on, and 0W when off, so 250W average.

In reality to calculate power in general you have to integrate (instantaneous voltage) * (instantaneous current) over a period of the waveform to get the average (or from 0 to some time t as in your example to find the power over some interval).

If (and it's a big if) the load is a fixed resistor R you can say that v= i*R, so instantaneous power is i^2 * R and so then you can integrate i^2 over the period to get the "RMS current", and multiply by R later (since it's fixed it doesn't enter into the integral).

RMS current is not particularly useful if the load is something nonlinear like a diode. It can be useful in analyzing losses in something like a capacitor with a given ESR. The losses (and resulting heating effect which shortens the capacitor life) will be proportional to the RMS current, not the average.

— Spehro Pefhany
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34

For power to be average i must be average current, so I am surmising that the effective current is the average current.

In short, average voltage x average current only equals average power when the voltage and current are DC quantities. Think about the following example: -

If you applied 230 V AC from your utility power outlet to a heating element, it would get warm or even hot. It is taking power that you can be billed for. 230 V AC is a sine wave and all sine waves have an average value of zero. The resulting current flowing through the heating element is also a sine wave with an average value of zero.

So, using average voltage x average current produces zero average power and clearly that is wrong. It is RMS voltage x RMS current that is going to give a meaningful answer (irrespective of whether it's DC or AC).

You have to go back to basics and ask yourself what power is - it is voltage x current and these are instantaneous values multiplied together. This results in a power waveform like this: -

Because of the act of multiplication, the power waveform now has an average value that is non-zero. Taking this a step further, if the load resistor were 1 ohm then, the amplitude of current will equal the amplitude of the applied voltage so, the power becomes the average of $v^2$ .

这导致我们说功率是the mean of the square of voltage（或电流），并且在本示例中假设我们选择1欧姆，我们也可以说产生该功率的有效电压是square root of the mean of the voltage squared或“ RMS”值。

$v_{pk}$ $v^2_{pk}$

$\dfrac{v^2_{pk}}{2}$ . Then taking the square root to get the effective voltage we get $\sqrt{\dfrac{v^2_{pk}}{2}}$ or $\dfrac{v_{pk}}{\sqrt{2}}$

In effect the RMS value of an AC voltage (or current) is the equivalent value of a DC voltage (or current) that produces the same heating effect in a resistive load.

So no, average voltage or average current is irrelevant but average power is king.

— Andy aka
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Good explanation

— crowie

Note that average power equals RMS voltage times RMS current if and only if voltage and current are proportional.

— Peter Green

Does this multiplication mean that non-resistive loads have a power curve that is sometimes negative? Does this mean that the naive average of the power is different from VRMS*IRMS? Is the difference related to the power factor?

— Random832

1

@Random832 - it seems your comment should have come after mine but yes, I was careful with the words not to imply any power factor in order to avoid unnecessary complication in the answer. Power only equals Vrms x I rms in an ac circuit for loads that have a PF of 1.

— Andy aka

1

@anhnha yes, the general case is always the product of instantaneous v and i. In fact power factor is never (a brave word to use) used to sensibly calculate power. I have left plenty of other answers on this subject that you may have seen.

— Andy aka

16

The devil is in the details when you work out the math.

Given that instantaneous power $P_\text{inst} = i^2 \cdot R$ , then the average power is:

P_{avg} = \bar{P_{inst}} = \bar{i^{2} \cdot R} = \bar{i^{2}} \cdot R = \frac{1}{T} \int_{0}^{T} i^{2} d t \cdot R

$P_\text{avg}= \overline{P_\text{inst}} = \overline{i^2 \cdot R} =\overline{i^2} \cdot R = \frac{1}{T}\int_{0}^{T}i^2dt \cdot R$

The effective DC current is that which dissipates the same average power

P_{avg} = I_{eff}^{2} \cdot R

$P_\text{avg}=I_\text{eff}^2 \cdot R$ then it follows:

I_{eff}^{2} = \frac{1}{T} \int_{0}^{T} i^{2} d t

$I_\text{eff}^2 = \frac{1}{T}\int_{0}^{T}i^2 \ dt$

I_{eff} = \sqrt{\frac{1}{T} \int_{0}^{T} i^{2} d t}

$I_\text{eff} = \sqrt{\frac{1}{T}\int_{0}^{T}i^2 \ dt}$

If you look at average voltage/current and RMS voltage/current, they are different because of the properties of integrals. In other words,

\int_{a}^{b} i^{2} d t \neq [\int_{a}^{b} i d t]^{2}

$\int_{a}^{b}i^2 \ dt \neq\Big[\int_{a}^{b}i \ dt\Big]^2$ If this property were true, then the squared could be pulled out of the integral and cancel with the square root.

Additionally, there is the issue of the $\frac{1}{T}$ underneath the square root which would also cause issues.

In summary, it is because the math does not work out that way.

— Addison
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This is the more precise and correct answer, IMO.

— hcabral

4

Average power is just the integral of work, over some finite time period, divided by that time period. For your case, each instant of work is:

d U = P_{t} \cdot d t = R_{t} \cdot I_{t}^{2} \cdot d t

$\textrm{d}U = P_t\cdot \textrm{d}t =R_t\cdot I_t^2\cdot \textrm{d}t$

So, you integrate that to get total work for some finite period and then, to convert that into an average power value, you just divide it by the finite period. Or:

\bar{P} = \frac{1}{t_{1} - t_{0}} \int_{t_{0}}^{t_{1}} R_{t} \cdot I_{t}^{2} \cdot d t

$\overline{P}=\frac{1}{t_1-t_0}\int_{t_0}^{t_1} R_t\cdot I_t^2\cdot \textrm{d}t$

If $R_t$ is a constant over time, then:

\bar{P} = R \cdot \frac{1}{t_{1} - t_{0}} \int_{t_{0}}^{t_{1}} I_{t}^{2} \cdot d t

$\overline{P}=R\cdot\frac{1}{t_1-t_0}\int_{t_0}^{t_1} I_t^2\cdot \textrm{d}t$

But if you want to now construct some kind of fictional effective current that fits the $R\cdot I_{eff}^2$ model, then by simple inspection of the above equation it must be the case that:

\begin{aligned} \bar{P} = R \cdot I_{e f f}^{2} = R \cdot \frac{1}{t_{1} - t_{0}} \int_{t_{0}}^{t_{1}} I_{t}^{2} \cdot d t \\ ∴ I_{e f f}^{2} & = \frac{1}{t_{1} - t_{0}} \int_{t_{0}}^{t_{1}} I_{t}^{2} \cdot d t \end{aligned}

$\begin{align*} \overline{P}=R\cdot I_{eff}^2=R\cdot\frac{1}{t_1-t_0}\int_{t_0}^{t_1} I_t^2\cdot \textrm{d}t \\ \therefore~~~~~~~~~~~~~ I_{eff}^2 &=\frac{1}{t_1-t_0}\int_{t_0}^{t_1} I_t^2\cdot \textrm{d}t \end{align*}$

It's just an equivalent substitution, right?

And then obviously:

\begin{aligned} I_{e f f} & = \sqrt{\frac{1}{t_{1} - t_{0}} \int_{t_{0}}^{t_{1}} I_{t}^{2} \cdot d t} \end{aligned}

$\begin{align*} I_{eff} &=\sqrt{\frac{1}{t_1-t_0}\int_{t_0}^{t_1} I_t^2\cdot \textrm{d}t} \end{align*}$

If you start things so that $t_0=0$ and set $t_1=t$ then you get your own equation. It's that easy, really.

— jonk
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Nice clean answer. I am sure you would appreciate some digression into 2-norm of Hilbert's spaces too...

— carloc

3

Imagine two currents flow simultaneously through your load:

DC current of 1A
AC current with 1A amplitude

The total current will look something like this:

Now, if we apply your formula for $I_{eff}$ , we will get 1A, as if the AC component produced zero power. I hope you agree that this makes even less sense than the original formula.

— Dmitry Grigoryev
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2

Consider $R=1\Omega$ and and a current of 1A for one second and 10A for another second. What's the average power?

Obviously, it is

\bar{P} = \frac{1 s \cdot 1 A^{2} \cdot 1 Ω + 1 s \cdot 10 A^{2} \cdot 1 Ω}{2 s} = 50.5 W

$\bar{P}=\frac{1s\cdot 1A^2\cdot 1\Omega+1s\cdot 10A^2\cdot 1\Omega}{2s}=50.5W$

Let's rewrite this:

\bar{P} = 1 Ω * \underset{= I_{e f f}^{2}}{\underset{⏟}{(\frac{1 s \cdot 1 A^{2} + 1 s \cdot 10 A^{2}}{2 s})}}

$\bar{P}=1\Omega*\underbrace{\left(\frac{1s\cdot 1A^2+1s\cdot 10A^2}{2s}\right)}_{=I_{eff}^2}$

On the other hand, the average current is 5.5A, which gives an "average power" of 30.25W.

The point is, the power formula contains the square of the current, so the effective current is higher than just the average of the (absolute value of) the current.

— sweber
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2

Let me put this in more general terms: Instant power P(t) dissipated over a load is a product (in mathematical sense as multiplication) of V(t) and I(t). Or I(t)*I(t)/R for that matter. Average power is therefore an average[I(t)*I(t)]/R. The paradox is in the well-known mathematical theorem that an average of a product of variable functions is not equal to product of their averages,

[(V(t)I(t)] != [V(t)]*[I(t)];

equivalently,

[I(t)^2] != [I(t)]*[I(t)]

To illustrate this basic calculus problem to some extreme, assume that you have a resistor load of 1 Ohm, and the voltage is pulsed as 10V for 10% duty cycle, 10% up, 90% no voltage. The real dissipated power is 10V*10A = 100W for 10% of the duty cycle, and zero for the rest of duty cycle. So the average power dissipated by this resistor is 10W.

Now, if you take (or even measure!) the averages separately using separate meters, the average [V] of this pulsed waveform will come up as 1V, and the average of I will come as 1A. Multiplying the measured results one might come to a conclusion that the power consumed by this "device" is only 1W, which will be totally wrong by a factor of 10!!!.

This is a typical mistake in many disciplines and applications. For example this mistake is in the basis of many bogus claims of some magical water heaters that produce more output than the "consumed electricity" usually explained by "cold fusion", or some other BS. There are even patents granted on these "pulsed heaters".

— Ale..chenski
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