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## 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 ComponentZ = R

_{EQ }+ j X_{LEQ }– j X_{CEQ}When X

_{LEQ }= X_{CEQ}, then the total reactive component of the impedance is zero and hence:Z = R

_{EQ}Φ = tan

^{-1}(± X/R) = 0Power factor = cos Φ = 1

If this is the case then,

Sin Φ = sin 0 = 0 and

Apparent Power = Active Power = V

_{RMS}I_{RMS}= (V_{m}I_{m})/2Reactive Power = j V

_{RMS}I_{RMS}sin Φ = j [(V_{m}I_{m})/2] sin Φ = 0This 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 CircuitZ = [(R + jωL) (1/jωC)]/ [R + jωL + (1/jωC)]

Z = R

_{EQ}+ jX_{LEQ}- jX_{CEQ}Multiplying the numerator and denominator by jωC:

Z = (R + jωL)/(jωCR – ω

^{2}LC + 1)Z = (R + jωL)/[(1 - ω

^{2}LC) + jωCR]Rationalising:

By breaking this into its equivalent components provides:

R

_{EQ}= R/[(1 – ω^{2}LC)^{2 }+ (ωCR)^{2}]jX

_{LEQ }= (jωL)/ [(1 – ω^{2}LC)^{2}+ (ωCR)^{2}]-jX

_{CEQ}= [(-jωC) (ω^{2}L^{2}+ R^{2})]/ [(1 – ω^{2}LC)^{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 – ω

^{2}LC)^{2}+ (ωCR)^{2}] = [(ωC) (ω^{2}L^{2}+ R^{2})]/ [(1 – ω^{2}LC)^{2}+ (ωCR)^{2}]This gives:

ωL = ωC (ω

^{2}L^{2}+ R^{2})L = C (ω

^{2}L^{2}+ R^{2})And hence,

C = L/(ω

^{2}L^{2}+ R^{2})For the component values given L = 200mH, R = 50? at a frequency of 50Hz

C = (200 x 10

^{-3})/(62.82 + 50^{2}) = (200 x 10^{-3})/ (3943.8 + 2500) = 0.2/6443.8 = (0.2 x 106)/ 6443.8 μFC = 31 μF

Then with the net reactance removed:

Z = R

_{EQ}= R/[(1 – ω^{2}LC)^{2}+ (ωCR)^{2}]For the given values of R = 50?, L = 200mH and C = 31µF:

R

_{EQ}= 50/[(1 - 314^{2}x 0.2 x 31 x 10^{-6})]^{2}+ [314 x 31 x 10^{-6}x 50]^{2}R

_{EQ }= 50/ [(1- 0.61)^{2}+ (0.49)^{2}] = 50/(0.15 + 0.24) = 50/0.39 = 128.2 ?In this situation, I

_{m}= V_{m}/|Z| = V_{m}/R_{EQ}= 310/128.2 = 2.42 AThen for the modified circuit with Power Factor Cos Φ = 1

P

_{APP}= P_{AVE}= V_{m}I_{m}/2 = (310 x 2.42)/2 = 375 WAnd with Sin Φ = 0

Reactive Power = P

_{IMAG }= j (V_{m}I_{m}/2) sin Φ = j [(310 x 3.86)/2] x 0 = 0 WNote 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.

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