# What are active power, reactive power and apparent power?

In electrical research, there are three basic forms of power, namely: active power, reactive power and apparent power.
Active power (P or PTRUE) is the power consumed by the resistive element in watts. For example, the energy converted into heat or power is generated by active power.

Reactive power (PVAR) is the power used to generate a magnetic or electrostatic field. The energy is first stored in the circuit and then returned to the circuit. Reactive power is generated by capacitors or inductors, and the unit is “Var (VAR), that is, volt-ampere reactive power.

The sum of active power and reactive power is called apparent power (PVA), and the unit is volt-ampere (VA). The rated power of many electrical products is expressed in volt-amperes. Apparent power reflects the relationship between total current, total voltage and total impedance. The power company sends out the apparent power. When the power is marked by volt-amperes, the relationship is PVA=total volts×total amperes. If the power supply voltage is 120v, then I=PVA/120; if the power supply voltage is 240V, then I=PVA/240. For example, the rated power of the motor is 2400VA, and it can work under dual voltage (120V/240V). If the motor is connected to a 120V power supply, I=VA/E =2400/120A =20A, if the motor is connected to a 240V power supply, then I=VA/E=2400/240A = 10A, please note that if the power remains the same When the voltage is doubled, the current will be halved.

1. R-C circuit power
In a purely capacitive AC circuit, all energy is stored in the capacitor during a part of the cycle, and then returned to the circuit during the other part of the cycle. If there are both R and C in the circuit, part of the energy is stored in the capacitor, and part of the energy is consumed by the resistance.

1). Power of R-C series circuit
example 1
Figure 1 shows three different series circuits: ① pure resistance; ② pure resistance; ③ resistance and resistance coexist. This example illustrates the power calculation method for each circuit.
(1) Pure resistive series circuit (see Figure 1a)
IR=E/R=120/100A=1.2A
PTRUE =E×IR=120×1.2W=144W
PVAR=E×IX=0W (no resistance component)
PVA=ET×IT=120×.2W=144W

(2) Pure resistance series circuit (see Figure 1b)
Ic=E/Xc=120/100A=1.2A
PTRUE=E×IR=0W (no resistive element)
PVAR =E×Ic=120 ×1.2W=144W
PVA=ET×IT=120 ×1.2W=144W (resistance)
(3) Resistance and resistance series circuit (see Figure 1c)
IT=ET/Z=120/141.42A =0.85A
PTRUE=I²×R=0.85²× 100w =72.25W
PVAR=I²×Xc=0.85²×100w=72.25W
PvA=ET×IT=120×0.85 W= 102W (combination of active power and reactive power)
or
PvA=I²×Z=0.85²×141.42 W= 102.2W (The deviation from the above formula is caused by rounding error)
As mentioned earlier, the total power is the sum of the power of various forms, and the calculation method is:
PT=√(P²TRUE + P²VAR) =√(72.25²+72.25²)W=√(5220 + 5220)W= 102.1W

Please note that the apparent power is equal to the total power.
Example 1 shows the relationship between active power, reactive power and apparent power. If the reactive power is zero, the apparent power is equal to the active power; if the active power is zero, the apparent power is equal to the reactive power. The apparent power is equal to the total power of the circuit, and the energy comes from the power company.

①Power triangle
The relationship between active power, reactive power and apparent power can be described by a right triangle, which is called a power triangle. Reactive power is the vertical side of this triangle, active power is the horizontal side, and apparent power or total power is the hypotenuse. In Figure 2, the vertical edge is located below the horizontal edge. Here it is assumed that the capacitive reactive power is negative and the inductive reactive power is positive. The circuit represented by this figure has active power and capacitive reactive power. The angle φ is the phase angle between the applied voltage and the total current.

In a circuit containing only one capacitor, the phase difference between the voltage and the current is 90°, and the current leads the voltage. Energy is stored in the capacitor for part of an AC cycle and returned to the circuit at another time in the same cycle. If the product of voltage and current is added in the entire cycle, the negative reactive power and the positive reactive power are equal in number, so the total power is zero.

②Power curve
Figure 3 shows the charge and discharge power curve. The power curve is the product of E×I in a complete cycle. When the power curve is positive, the energy is stored in the capacitor, when the power curve is negative, the energy is returned to the circuit. When resistance is added to the circuit, the phase difference between voltage and current is less than 90°. The larger the resistance, the smaller the phase difference. When the power supply voltage and current are in phase, the circuit is purely resistive. When the product of E×I produces active power, the voltage and current must be both negative or positive at the same time. In Figure 3, no net power is produced. Because when the voltage and current differ by 90°, the polarity is the same 50% of the time, and the polarity is opposite 50% of the time. (See the comparison of power curve and power factor at the back of this chapter.)

Example 2
A resistor and a capacitor are connected in series, as shown in Figure 4. The resistor is 60Ω, Xc is 30Ω, and the power supply is AC120V. Calculate Z, IT, P, PVAR and PVA, and draw a schematic diagram of the power triangle.

Z= √(R²+X²c)=√(60² +30²)=√(3600 +900)Ω=67Ω
IT=E/Z=120/67A=1.79A
P=I²×R=1.79²×60W= 192.24W
PVAR=I²×Xc=1.79²×30W=96W
PVA=I²×Z=1.79²×67W=214.7W

(As mentioned earlier, the apparent power can also be calculated by √(P²TURE+P²VAR).)
The power triangle is shown in Figure 5.
Example 2 moves the power curve in Figure 3 to the positive direction, reducing the part of the curve below the zero axis. This is because the angle between voltage and current is not 90°, so the active power increases. The power curve is very useful when counting electricity production or electricity consumption over a period of time.

③The power of R-C parallel circuit
Example 3 In Figure 6, there is a resistor and a capacitor in parallel. R is 60Ω, Xc is 30Ω, and the power supply is AC120V. Calculate Z, IR, IC, IT, PVAR and PVA, and draw a schematic diagram of the power triangle.
(1) Calculate Z:
Z=R×Xc/√(R²+X²)
=60×30/√(60²+30²)Ω
=1800/√(3600 +900)Ω=26.86Ω
(2) Calculate IR and IC (remember that the voltage drop on each parallel branch is equal):
IR=E/R=120/60A=2A
IC=E/Xc=120/30A =4A
(3) Calculate IT:
IT=E/Z= 120/26.86A =4.47A
(4) Calculate the active power:
P=I²×R=2²×60W=240W
(1) Calculate reactive power:
PVAR=I²c×Xc=4²×30W=480W
(2) Calculate apparent power:
PVA=I²T×Z=4.47²×26.86W=536.6W

(7) The power triangle is shown in Figure 7.

2. R-L circuit power
In a purely inductive AC circuit, the energy output by the power supply is stored in a magnetic field for part of a cycle and returned to the circuit during another part of the cycle. When the circuit has both R and L, part of the energy is stored in the magnetic field and part of the energy is consumed by the resistance.

①The power of R-L series circuit
When resistance and inductance are present in the circuit, the phase difference angle between voltage and current is less than 90°. Therefore, similar to the R-C circuit, the active power is dissipated at this time, and the apparent power is increased due to the addition of resistance.
Example 4
A resistor and inductor are connected in series, as shown in Figure 8. The resistor is 3Ω and the inductor is 60Ω. The power supply voltage is AC480V. Z, IT, P, PVAR and PVA, and draw a power triangle.
(1) Calculate Z:
Z=√(R²+X²L)=√(30²+60²)Ω= √(900+3600)Ω=67Ω
(2) Calculate IT:
IT=E/Z=480/67A=7.16A
(3) Calculate the active power:
P=I²R×R=7.16²×30W=1538W
(4) Calculate reactive power:
PXL=I²L×XL=7.16²×67W=3080W
(5) Calculate apparent power:
PVA=I²T×Z=7.16²×67W=3434.7W

(7) The power triangle is shown in Figure 9.

②The power of R-L parallel circuit
Example 5
A resistor and an inductor are connected in parallel, as shown in Figure 10. Calculate active power, reactive power, and apparent power, and draw a power triangle.
Active power is
P=I²×R=16²×30W=7680W
Reactive power is
PVAR=I²XL×XL=8²×60W=3840w
Apparent power is
PVA=√(P²TRUE+P²VAR)=√(7680²+3840²)W=8606W
The power triangle is shown in Figure 11.