Introduction to Electrical Watts, Vars, VoltAmps and Power Factor.

1. INTRODUCTION TO
ELECTRICAL POWER and POWER FACTOR
By Dan Gray, B.A.Sc., P. Eng.
Electrical loads such as heating elements and incandesent lights consume
Active Power. They convert electrical energy into heat or light. When an electrical
load is consuming only Active Power the current and voltage sine waves will be in
phase as in figure 1. The symbol for this type of power is (P) and is measured in
Watts. The power waveform obtained by multiplying the voltage and current
waveforms is always positive shown by the green wave. The current waveform is
shown in red and the voltage waveform in blue. The frequency of the current and
voltage waves on the North American grid is 60 Hertz. The frequency of the power
wave is 120 Hertz. The power waveform is producing 2 pulses for every cycle of
voltage and current. Electrical Power is calculated using RMS values of current
and voltage. The term RMS means the square Rootof the Mean value of the
Squared function. The rms value of a sine wave is the amplitude x 0.707. The
amplitude of house voltage is approximately 170 volts so its rms value is 0.707 x
170 = 120 volts. For example if the rms house voltage is 120 volts and the rms
current to a heater is 5 amperes the electrical power to the heater would be 120 x 5
= 600 Watts. A direct current battery producing 120 volts and 5 amps to a heater
would be supplying 600 Watts of constant power. The pulsating power wave has a
peak value of 1200 Watts and minimum value of zero, but its average value is 600
Watts which is equivalent to the battery example.
Figure 1: Ideal Resistor current and voltage in phase

2. When a coil of wire (inductor) is connected to a sinusoidal voltage the
current that flows will lag the voltage waveform as in Figure 2. Forthe ideal
inductor with zero resistance the current lags the voltage wave by 90 degrees. The
power waveform is the productof the voltage and current waves and has equal
negative and positive areas under the power wave. The positive areas are times
when the inductor absorbs energy and the negative areas are times when the
inductor returns energy back to the energy sourcesuch as a generator. This type of
power is called Inductive Reactive Power (symbol is QL) and is measured in Vars
which is determined by multiplying the rms voltage and current. For example if the
voltage is 120 volts and the current is 4 Amps the reactive power QL = 120 x 4 =
480Vars lagging. VAR is an acronym meaning Volt Amps Reactive. This type of
power does not do any useful work but does require extra current to supply it to
motor coils.
Figure 2: Ideal Inductor current lags voltage by 90 degrees
When a capacitor is connected to a sinusoidal voltage the current that flows
will lead the voltage waveform as in Figure 3.Forthe ideal capacitor with zero
resistance the current leads the voltage wave by 90 degrees. The power waveform
is the productof the voltage and current waves and has equal negative and positive
areas under the power wave. The positive areas are times when the capacitor
absorbs energy and the negative areas are times when the capacitor returns energy
back to the energy source. This type of power is called Capacitive Reactive Power
(symbol is QC) and is measured in Vars and is determined by multiplying the rms
voltage and current. For example if the voltage is 120 volts and the current is 3
Amps the reactive power QC = 120 x 3 = 360Vars leading. This type of power does
not do any useful work but can be used to supply reactive power to a motor coil.

3. Figure 3: Ideal Capacitor current leads voltage by 90 degrees
A motor requires reactive power because of the magnetic field produced by
the stator coils. The positive portion of the power wave is equal to the negative
portion of the power wave so no active power is supplied to the ideal coil. This
reactive power is measured in Vars and is an exchange of energy that is
transmitted from the alternator to the motor load and then returns back to alternator
and does no useful work but does require extra current flow in the transmission
line. The current in the transmission line produces heat in the resistance of the line.
The extra current required to supply reactive power increases transmission line
power loss and also increases the voltage drop on the line which reduces the
voltage at the load.
The power flow to a motor consists of Active Power and Reactive Power.
The electrical Active power supplied is converted to mechanical power to turn the
motor shaft. Some power is lost due to rotor and stator heating, bearing and
windage friction. The Reactive power oscillates between the motor and the source
as it produces the magnetic field for the motor. Figure 4 is an example of motor
current lagging the voltage by 45 degrees or π/4 radians.

4. Figure 4: Motor current lagging voltage by 45 degrees.
The Apparent power has units of Volt Amps (VA) and is determined using a
right-angle triangle. The hypotenuse represents the Apparent power(S) and the
Active(P) and Reactive(Q) powers are the other two sides of the triangle so that
S2 = P2 + Q2. The angle Ɵ between S and P is the power factor angle and the ratio
P/S = Cosine(Ɵ) and is the Power Factor.TheActive power P = S x cos(Ɵ)Watts
and the Reactive power is Q = S x sin(Ɵ) Vars. Ɵ is also the phase angle between
the current and voltage waves.

5. A capacitor of the correctcapacitance value to make the Power factor equal
to 1.0 can be put in parallel with the motor terminals and it will supply reactive
power to the motor and the motor will supply reactive power to the capacitor. The
capacitor and motor exchange reactive power with each other and the transmission
line only has to supply true power current. The transmission line current is reduced
and the power loss on the line is reduced. The voltage at the motor terminals will
increase since the voltage drop on the line is reduced.
The following diagram shows the effect of connecting a capacitor in parallel
with an induction motor. The motor on the left is receiving 100 Amps from the 440
volt system. The Active power supplied is 35.2 KW which requires 80 Amps of in
phase current. The Reactive power supplied is 26.4 KVars which requires 60 Amps
of lagging current. The middle diagram shows the same motor with a 361.7 micro
Farad capacitor connected in parallel with the motor terminals. This value of
capacitance supplies 26.4 KVars of Reactive power to the motor and the 440 volt
sourcedoes not have to supply any Vars to the motor. The source only has to
supply the Active power of 35.2 KW with current of 80 Amps of in phase current.
The capacitor supplies 60 Amps of lagging current to supply the required vars for
the motor. Since the 440 volt source is supplying only Active power the Power
Factor= 1.0 and the sourcecurrent and voltage are in phase. The right diagram
shows the PF corrected to 0.9. The capacitor supplies 9.4 KVars and 21.4 Amps of
lagging current to the motor and the source supplies 35.2 KW, 17.0 KVars, and
88.9 Amps lagging the voltage by 25.84 degrees. This current consists of 80 Amps
of true power current and 38.6 Amps of reactive power current to make the total
reactive current supplied to the motor equal to 60 Amps. Power factor correction
reduces the current from the 440 volt supply, power loss and voltage drop in the
resistance of the cable supplying the motor will be reduced which will increase
voltage at the motor terminals and increase motor torque.