What is apparent power?
Apparent power is the product of root-mean-square voltage and the root-mean-square (RMS) current.
Different types of power dissipate in AC power systems i.e. Active, Reactive, Apparent, and complex powers.
The measure of alternating current (AC) power is obtained from the multiplication of the root-mean-square voltage and the root-mean-square (RMS) current. In DC circuits there is only resistance, the Voltage and currents are in phase. While in AC circuit resistance is the combination of resistance and reactance. The current and Voltages are not in Phase in AC Circuits.
We can denote this Power using (S) and the formula is given below
S = Erms * Irms
S is the apparent power in Volt Ampere.
Erms is the root-mean-square (rms) voltage in volts.
Irms is the rms current in amperes.
Formula and unit
The basic formula for the calculation of this type of power for any circuit is given below.
- S = VI for single phase loads
- S = √3VI for three phase loads
The Unit is Volt Ampere. But this is very small and people normally use it in terms of “KVA”.
If the power (Apparent) in the electric circuit is supplied from a power supplier to the grid. It includes both real & reactive power consumption in the loads. By the help of the power factor triangle which shows the relationship between apparent, active, reactive power and power factor we cane find another formula as follows
From the triangle above we find S = √ (True power2 + Reactive Power2)
S = √(Q2 + P2)
Q is reactive power consumption in load (VAR)
P is active power consumption in load (W)
Example of apparent power
Assume we have a three phase 10 KW load working on a 400 Volt power source. This Load has 0.85 power factor. We need to calculate the apparent power of this load.
In this example we have the load active power which is 10 KW, And the power factor of the load is 0.85 we calculate current first as follows
P = √3 V * I * PF ⇒ I = P / (√3 *V * PF ) = 10,000/(1.73*400*0.85)
I = 17 A
Then S =√3 V * I
S = 1.73*400 * 17 = 11.7 KVA
Note that the active power is greater than the apparent one. This means that this load draws reactive power. This is clear as the PF is 0.85
Then the question is: Could S(KVA) be less than P(KW)?
S is either less than or equal to P. In case of unit PF the two powers are equals.
Points to Be Noted
- When the impedance is only pure resistance, the value of S(VA) will be the same as that of true power. (We will discuss true Power Later)
- In case, if the reactance exists. The value of S(VA) will be greater than the true power.