The total power of a three-phase AC circuit is equal to three times the single phase power.
So if the power in a single phase of a three-phase system is ‘P’, then the total power of the three-phase system would by 3P (provided the three-phase system is perfectly balanced).
But if the three-phase system is not exactly balanced, then the total power of the system would be the sum of the power of individual phases.
Suppose, in a three phase system, the power at R phase is PR , at Y phase is PY and at B phase is PB, then total power of the system would be
This is simple scalar sum, since power is a scalar quantity. This is the season, if we consider only single phase during calculating and analyzing of three phase power, it is enough.
Let us consider, network A is electrically connected with network B as shown in the figure below:
Let us consider the expression of the voltage waveform of a single phase system is:
Where V is the amplitude of the waveform, ω is the angular velocity of propagation of the wave.
Now, consider the current of the system is i(t) and this current has a phase difference from the voltage by an angle φ. That means current wave propagates with φ radiant lag in respect of the voltage. The voltage and current waveform can be represented graphically as shown below:
The current waveform in this case can be represented as:
Now, the expression of the instantaneous power,
[where Vrms and Irms is the root mean square value of voltage and current waveform]
Now, let us plot the term P versus time,
It is seen from the graph that, the term P does not have any negative value. So, it will have a nonzero average value. It is sinusoidal with a frequency twice of system frequency. Let us now plot second term of the power equation, i.e. Q.
This is purely sinusoidal and has a zero average value. So from of these two graphs, it is clear that P is the component of power in an AC circuit, which actually transported from network A to network B. This power is consumed in network B as electric power.
Q on the other hand does not really flow from network A to network B. Rather it oscillate between network A and B. This is also component of power, actually flowing into and out of the inductor, capacitor like energy storage elements of the network.
Here, P is known as the real or active part of the power and Q is known as imaginary or reactive part of the power.
Hence, P is called real power or active power, and Q is called imaginary or active power. The unit of active power is Watt, whereas the unit of reactive power is Voltage Ampere Reactive or VAR.
We have already considered,
where, S is the product of root mean value of voltage and current i.e.
This product of RMS value of voltage and current of a system is referred as apparent power is Voltage Ampere or VA. So,
This can be represented in complex form as
Again, the expression of the real power is
where ɸ is the angle between voltage and current phasor. So,
So, here in the expression P, cos ɸ is the factor which determines the real power component of an apparent power S.
This is why the term cos ɸ in the expression of real power is called power factor. For both positive and negative value of ɸ, cos ɸ is always positive.
This implies, regardless of the sign of ɸ (which is dependent on whether the current is lagging or leading the voltage) real power is always positive.
That means it flows from the sending end (Network A) to receiving end (Network B). We have also shown the same earlier when looking at the waveform for real power.
Now if the current is leading the voltage, then the angle between voltage and current phasor is negative, taking the voltage phasor as reference:
In that case the reactive component of the power is negative,
The relation between apparent power to active power and reactive power can be represented in trigonometric form as shown below.
Now, if current is lagging the voltage the angle between voltage and current phasor is positive, taking the voltage phasor as reference.
In this case, the reactive component of power is positive. Since,
The power triangle is represented as shown below.
If the impedance of the network is capacitive, the current leads the voltage, and in case of inductive network the current lags voltage. So we can conclude, the reactive power is negative in the case of capacitive reactance and it is positive and in the case of inductive reactance.
If the network is purely resistive, there would not be any angular difference between current and voltage. Hence,
So the reactive power in this case would be,
Thus, there is no reactive power generated or consumed in the network.