AC circuit is generally three-phase for electrical distribution and transmission. Single-phase circuits are commonly used in our domestic supply network.
The total power of a three-phase circuit is three times the single-phase power.
If the power in one phase of a three-phase system is ‘P’, the total power of the three-phase system would be 3P (assuming the three-phase system is perfectly balanced).
However, if the three-phase system is not exactly balanced, its total power is the sum of the power of each phase.
Assuming that in a three-phase system the power is in phase R PR, in phase Y PY, and in phase B PB, then the total power of the system would be
P = PR + PY + PB
It’s a simple scalar sum since the power is a scalar quantity. This is the season, considering that a single phase is sufficient in the calculation and analysis of the three-phase power.
Note that network A is electrically connected to network B, as shown in the following figure:
Consider the expression of the voltage waveform of a single-phase system:
Where V is the amplitude of the waveform, is the angular velocity of the wave. AC circuit
It is now assumed that the current of the system is i (t) and that this current has a phase difference with respect to the voltage of an
This means that a current wave with a radiation delay propagates from the voltage. The waveform of voltage and current can be graphically represented as follows:
The current waveform can be represented in this case by:
Well, the expression of instantaneous power, AC circuit
[whereVrms andIrms are the root mean square of the voltage and current waveform]
Now let’s draw the term P in terms of time,
The graph shows that the term P has no negative value. So he will have an average nonzero. It is sinusoidal with a frequency twice that of the system. Now let’s drill up the second term of the power equation, i.e. Q. AC circuit
second power equation
This is purely sinusoidal and has a mean of zero. It thus becomes clear from these two diagrams that P is the component of the supply in an AC circuit which is actually transported from the network A to the network B. This energy is consumed in the network B as energy electric.
On the other hand, Q does not really go from network A to network B. It tends to shuttle between networks A and B. It is also a component of the food entering and leaving the inductor an energy storage element of the network.
Here P is known as the real or active part of the power and Q as the imaginary or reactive part of the power.
Actiue Power (P) = VrmsIrmscosØ and Reactive Power (Q) = VrmsIrmssinØ
Therefore, P is called active power or active power and Q is imaginary power or active power. The active power part is watts, while the reactive power part is the blind-amp voltage or VAR.
We have already thought about
where S is the product of the root mean value of the voltage and the current, i.e.
This RMS voltage and current product of a system are called the
apparent power of Ampere Voltage or VA. then,
This can be represented in complex form as
Again, the expression of real power
where ɸ is the angle between the voltage and current phasors. then,
In the expression P, therefore, cos ɸ is the factor which determines the actual power component of the apparent power S.
For this reason, the term cos ɸ in the expression of active power is called the power factor. For the positive and negative values of ɸ, cos ɸ is always positive.
This implies that regardless of the sign of ɸ (depending on whether the current is too late or the voltage ahead) the active power is always positive.
This means that it goes from the sender (network A) to the sender (network B). We’ve also shown the same thing once when we looked at the waveform for real power.
When the current is ahead of the voltage, the angle between the voltage and current phasors is negative, taking the voltage vector as a reference:
In this case, the reactive component of the power is negative,
Energy Triangle AC circuit
The relationship between apparent power in real power and reactive power can be represented in trigonometric form, as shown below.
If the current is too late at the voltage, the angle between the voltage and current phasors is positive, taking the voltage phasor as a reference.
In this case, the reactive component of the power is positive. there,
The power triangle is shown below.
When the impedance of the network is capacitive, the current causes the voltage, and in the case of an inductive network,
the current is backward. We can, therefore, conclude that reactive power is negative for capacitive reactance, positive for inductive
If the network is purely resistive, there would be no difference in the angle between current and voltage. Therefore
So the reactive power, in this case, would be
Thus, no reactive power is generated or consumed in the network.