# why we say that in three phase system total power delivered to the balanced three-phase circuit is the same at any instant

My question is: why we say that in three phase system total power delivered to the balanced three-phase circuit is the same at any instant?. This is how I understood that one phase at one point of time gives all the power then the second and third and so again from the beginning I do not know if my assumption is correct and I would like to hear your opinion.

thanks.

It comes down to the fact that in an ideal system, each phase has a sine waveform, and the well known relationship

$$sin(x)^2 + cos(x)^2 = 1$$ for all X, and power is proportional to V^2.

Now that relationship obviously applies to 2 phases 90 degrees apart. How can you apply it to 3 phases 120 degrees apart?

Take the first phase as Sin(X). The other two are $$sin(x + 120 degrees)$$ and $$sin(x - 120 degrees)$$

A bit of vector analysis (draw the three out to illustrate) lets you resolve each of the latter two phases into sine and cosine components. So $$sin(x + 120) = A * sin(x) + B * cos(x)$$ I'll let you figure out the values of A and B.

Repeat for the other phase $$sin(x + 120) = C * sin(x) + D * cos(x)$$

Now square each of the three phases and sum over the squares... this is a bit of an exercise, but you should get

$$sin(x)^2 + cos(x)^2$$

which is of course 1.

No, the assumption isn't correct.

At any given time, the voltage on at least two of the phases will be non-zero. The total instantaneous power will be the sum of all the powers delivered by each of the three phases, and is near enough constant over one cycle.

See the answer by Munadil Fahad to Why do we need 3 phase power supply? for some pretty graphs.

You assumption is incorrect. At each instant the load power is shared amongst the three phases.

Imagine for simplicity the voltages and currents are in phase. It works out even if they are not, but is more difficult to see.

Consider phase voltages:

$$V_a = V\cos(\omega t), V_b = V\cos(\omega t - 120°), V_c = V\cos(\omega t + 120°)$$

Then at $t=0$, power in phase A is zero, but the other two are non-zero. A little while later all three phases will be non-zero. After a third of a cycle, power in phase B will reach zero while the other two are non-zero.

So it can be seen that at least two of the individual instantaneous phase powers are usually non-zero.

To see that the total power is nonetheless constant, use the cosine multiplication trig identity $\cos\alpha \cos\beta = \frac{1}{2}(\cos(\alpha + \beta)+\cos(\alpha - \beta))$ to multiply voltage and current:

$$P_a = V\cos(\omega t) \cdot I\cos(\omega t) = \frac{VI}{2}[\cos(2\omega t)+\cos(0°)]$$ $$P_b = V\cos(\omega t- 120°) \cdot I\cos(\omega t- 120°) = \frac{VI}{2}[\cos(2\omega t+120°)+\cos(0°)]$$ $$P_c = V\cos(\omega t+ 120°) \cdot I\cos(\omega t+ 120°) = \frac{VI}{2}[\cos(2\omega t-120°)+\cos(0°)]$$

Since $\cos(0°) = 1$ the sum of powers is:

$$P_a+P_b+P_c = \frac{VI}{2}[\cos(2\omega t)+\cos(2\omega t+120°)+\cos(2\omega t-120°)]+\frac{3VI}{2}$$

Clearly $\frac{3VI}{2}$ is constant, so we need only concentrate on the other part. Applying the cosine multiplication trig identity in reverse this time, we have:

$$\cos(2\omega t)+\cos(2\omega t+120°)+\cos(2\omega t-120°) = \cos(2\omega t)+2\cos(2\omega t)\cos(120°)$$

Since $\cos(120°) = -\frac{1}{2}$ this reduces to zero, leaving just the constant part. Thus, the instantaneous combined power of the three phases does not depend on $t$. In other words, it "is the same at any instant".

A graphical representation of these equations is shown in this graph. $P_a$ is red, $P_b$ is yellow, $P_c$ is blue and the sum is orange.