Efficiency= DC power output/Ac power input. DC power= I²(av.)*R. why average current?

AC power= I²(RMS)*R. Why RMS current and not average this time?


The reason is that in AC, the average current is zero (or close to zero).

You could take the absolute value but if you notice the formula for power you use

$$P= I_{rms}^2 R$$

So, by squaring AC current you get a positive value that you can get the average.

enter image description here

If you think about it, Power is Energy divided by time. So especially in the case of AC, where $I^2$ continuously fluctuates you want to obtain an average value over a cycle or a multiple of cycles.

However the energy, is the area under the curve in the following graph

enter image description here

It turns out, that if AC current is a pure sine (most of the time it is), the energy in a cycle is equal to $\frac{1}{2}I^2R\Delta T$, so when you do the math you obtain that the $I_{rms}=\frac{1}{\sqrt{2}} I_0 = \frac{\sqrt{2}}{2} I_0$

To be more verbose mathematically, the energy for a cycle with duration $T_{cycle}$ is equal to :

$$E = \int_0^{T_{cycle}}I^2(t)R dt \xrightarrow{I(t)=I_0 sin(\frac{2\pi}{T_{cycle}}t)}$$

Keep in mind that, if the period of the cycle is $T_{cycle}$, then the angular frequency is $\omega=\frac{2\pi}{T_{cycle}}$.

$$E = \int_0^{T_{cycle}}I_0^2 sin^2(\frac{2\pi}{T_{cycle}}t)R dt $$ $$E = I_0^2 R\int_0^{T_{cycle}} sin^2(\frac{2\pi}{T_{cycle}}t)dt $$

but because $$\int_0^{T_{cycle}} sin^2(\frac{2\pi}{T_{cycle}}t)dt = \left[\frac{t}{2} -\frac{1}{4}\sin\left(2\frac{2\pi}{T_{cycle}}t\right)\right]_{0}^{T_{cycle}}$$

Because $\sin\left(2\frac{2\pi}{T_{cycle}}t\right)$ for $t=0$ and $T_{cycle}$ then

$$\int_0^{T_{cycle}} sin^2(\frac{2\pi}{T_{cycle}}t)dt = \frac{T_{cycle}}{2}$$

So finally you get

$$E = I_0^2 R\frac{T_{cycle}}{2}$$

In order to obtain the average Power you divide by the duration of the cycle (i.e. $T_{cycle}$)

$$P_{av} = \frac{I_0^2 R\frac{T_{cycle}}{2}}{T_{cycle}} = \frac{1}{2}I_0^2 R$$

Because the idea is to use a "mean" value that gives the same result (let's call that $I_{rms}$, you need to have

$$P_{av} = = \frac{1}{2}I_0^2 R=I_{rms}^2 R $$

From this you can get (assuming current follows a sinusoidal distribution) that:

$$I_{rms} =\frac{ \sqrt{2}}{2} I_0$$


It may help to recall that the rms value of the sinusoidal waveshape produces the equivalent heating in a resistor as does a straight d.c. voltage of the same numerical value.


The equation DC power= I²(av.) *R is only an approximation at best.

If we take a periodic current trough a resistor R we can use Fourier series to calculate dissipated power. The point is that each harmonic is orthogonal to all the others and hence each power can be individually calculated and later on added up.

$$i(t)=I_0+I_1\cos(\omega t + \varphi_1)+ I_2\cos(2\omega t + \varphi_2)+\ldots =\sum_{k=0}^\infty I_k \cos(k\omega t+ \varphi_k)$$

$$P=R\left(I_0^2+I_1^2+\ldots \right)=R\sum_{k=0}^\infty I_k^2$$

We may now assume that the DC component, the zero-order harmonic, is giving the most of the total power and hence ignore the higher order ones.

$$P\approx R I_0^2$$

I don't think such an assumption is good enough for a half wave rectifier though.

On the other hand being rigorous is pretty simple: what we all are after is average power and this is defined as $$ P=\frac{1}{T}\int_Tv(t)i(t)\mathrm{d}t $$ over a T timeslot. This is always true but a few semplificative hypothesys may be taken.

It may come the case when one of the two quantitie is constant, like $v(t)=V$ or $i(t)=I$ at least over the integration period. The above simplifies into $$P=\frac{1}{T}\int_T V\; i(t)\mathrm{d}t=V\;\underbrace{\frac{1}{T}\int_T i(t)\mathrm{d}t}_{I_\mathrm{AVG}}=V\;I_\mathrm{AVG}$$ or in the other case $$P=\frac{1}{T}\int_T I\; v(t)\mathrm{d}t=I\;\underbrace{\frac{1}{T}\int_T v(t)\mathrm{d}t}_{V_\mathrm{AVG}}=I\;V_\mathrm{AVG}$$

These are the cases where say a constant voltage source supplies a variable current, average one is what's needed to calcurate the power

A slightly different case is when a variable quantity $v(t)$ or $i(t)$ is forced on a resistive load

$$P=\frac{1}{T}\int_T R\; i^2(t)\mathrm{d}t= R\;\frac{1}{T}\int_T i^2(t)\mathrm{d}t= R\;\underbrace{\frac{1}{T}\int_T i^2(t)\mathrm{d}t}_{I_\mathrm{RMS}^2}= R\;I_\mathrm{RMS}^2$$

given we define $I_\mathrm{RMS}=\sqrt{\frac{1}{T}\int_T i^2(t)\mathrm{d}t}$

This is the case you wish to calculate say the power lost in a resistive distribution line supplying a given current.

The same can be done with voltage leading to the other well known $V_\mathrm{RMS}=\sqrt{\frac{1}{T}\int_T v^2(t)\mathrm{d}t}$

  • $\begingroup$ While calculating power or heat generated while using a pulsating DC current active for half cycle(eg. half wave rectifier) we should find average of I² (i.e; I²rms) but why everywhere instead of I²rms*R,,,,, they use I²avg.*R to calculate power? $\endgroup$ Jan 31 at 6:35
  • $\begingroup$ images.app.goo.gl/52tJbVr8rNpE1jC5A Sir pls look at this image at "Output DC power" and pls tell me why they use average current to calculate efficiency of half wave rectifier? $\endgroup$ Jan 31 at 6:46

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