Theory of Power in AC Circuits III and Power factor

Power Factor:

In common, an ac circuit will have a combination of resistive and reactive components and the reactive elements might be inductive or capacitive as shown in the circuit of figure below. The overall impedance of such a network, as seen by a current or voltage source driving it, has a phase and magnitude and it is this phase angle that determines the Power Factor of the network.

Figure: A General AC Circuit having Reactive Elements

Z = R ± jX and Φ = tan-1 (± X/R)

Here the net reactance might be either capacitive or inductive.

Power factor = cos Φ

In order to maximize the active power we need:

cos Φ =1 that implies Φ = 0

This signifies that, as seen by source, there must be no phase shift between the voltage and current, that is, the network must ideally look purely resistive.

The total impedance of an ac network can be expressed in terms of constituent components that comprise an equivalent resistance, an equivalent inductance providing a positive reactive component and an equivalent capacitance providing a negative reactive component. Then:

Impedance = Resistance + Inductive Component + Capacitive Component

Z = REQ + j XLEQ – j XCEQ

When XLEQ = XCEQ, then the total reactive component of the impedance is zero and hence:


Φ = tan-1 (± X/R) = 0

Power factor = cos Φ = 1

If this is the case then, 

Sin Φ = sin 0 = 0 and

Apparent Power = Active Power = VRMS IRMS = (Vm Im)/2

Reactive Power = j VRMS IRMS sin Φ = j [(Vm Im)/2] sin Φ = 0

This signifies that, in this case, all of the power delivered to network by the source is dissipated in the resistive elements. That is, all the power delivered is consumed and becomes helpful or “active” power.

Thus, in order to accurate the power factor of any ac network to unity, the total effective reactance should be made equivalent to zero. This is attained by making the capacitive reactive component of the impedance equivalent and opposite to the inductive reactive component. When the network has only one kind of reactance then a reactance of the opposite type should be added in order to neutralize it.

Power Factor Correction:

Consider comprising a capacitor to neutralise the consequence of the motor inductance and correct the Power Factor as shown in figure below.


Figure: Correction of the Power Factor of the Circuit

Z = [(R + jωL) (1/jωC)]/ [R + jωL + (1/jωC)]


Multiplying the numerator and denominator by jωC:

Z = (R + jωL)/(jωCR – ω2LC + 1)

Z = (R + jωL)/[(1 - ω2LC) + jωCR]



By breaking this into its equivalent components provides:

REQ = R/[(1 – ω2LC)2 + (ωCR)2]

jXLEQ = (jωL)/ [(1 – ω2LC)2 + (ωCR)2]

-jXCEQ = [(-jωC) (ω2L2 + R2)]/ [(1 – ω2LC)2 + (ωCR)2]

From this the corresponding resistive, inductive and capacitive components can be recognized. It is apparent that cancellation of the inductive and capacitive components needs the values of positive and negative imaginary terms which should be equivalent but opposite. This needs:

(ωL)/ [(1 – ω2LC)2 + (ωCR)2] = [(ωC) (ω2L2 + R2)]/ [(1 – ω2LC)2 + (ωCR)2]

This gives:

ωL = ωC (ω2L2 + R2)

L = C (ω2L2 + R2)

And hence,

C = L/(ω2L2 + R2)

For the component values given L = 200mH, R = 50? at a frequency of 50Hz

C = (200 x 10-3)/(62.82 + 502) = (200 x 10-3)/ (3943.8 + 2500) = 0.2/6443.8 = (0.2 x 106)/ 6443.8 μF

C = 31 μF

Then with the net reactance removed:

Z = REQ = R/[(1 – ω2LC)2 + (ωCR)2]

For the given values of R = 50?, L = 200mH and C = 31µF:

REQ = 50/[(1 - 3142 x 0.2 x 31 x 10-6)]2+ [314 x 31 x 10-6 x 50]2

REQ = 50/ [(1- 0.61)2 + (0.49)2] = 50/(0.15 + 0.24) = 50/0.39 = 128.2 ?

In this situation, Im = Vm/|Z| = Vm/REQ = 310/128.2 = 2.42 A

Then for the modified circuit with Power Factor Cos Φ = 1

PAPP = PAVE = Vm Im/2 = (310 x 2.42)/2 = 375 W

And with Sin Φ = 0

Reactive Power = PIMAG = j (Vm Im/2) sin Φ = j [(310 x 3.86)/2] x 0 = 0 W

Note that the active or average power has not been raised but in fact remains similar. The apparent power though has been reduced and therefore the power that must be delivered by the source is as well decreased. However, the inactive, reactive or imaginary power has been removed from the viewpoint of the source that no longer has to generate this power. This signifies that all the power drawn from the source is now just helpful active power and no surplus power are demanded that is not used.

In practice, some reactive power is drawn from the source throughout initial transient conditions then this power oscillates between the capacitance and inductance present beneath steady state conditions however is not dissipated.

Latest technology based Electrical Engineering Online Tutoring Assistance

Tutors, at the, take pledge to provide full satisfaction and assurance in Electrical Engineering help via online tutoring. Students are getting 100% satisfaction by online tutors across the globe. Here you can get homework help for Electrical Engineering, project ideas and tutorials. We provide email based Electrical Engineering help. You can join us to ask queries 24x7 with live, experienced and qualified online tutors specialized in Electrical Engineering. Through Online Tutoring, you would be able to complete your homework or assignments at your home. Tutors at the TutorsGlobe are committed to provide the best quality online tutoring assistance for Electrical Engineering Homework help and assignment help services. They use their experience, as they have solved thousands of the Electrical Engineering assignments, which may help you to solve your complex issues of Electrical Engineering. TutorsGlobe assure for the best quality compliance to your homework. Compromise with quality is not in our dictionary. If we feel that we are not able to provide the homework help as per the deadline or given instruction by the student, we refund the money of the student without any delay.