Power and Power Factor

POWER AND POWER FACTOR

Average power:

Average power is the product of the effective values of both voltage and current multiplied by cosine of the phase angle between voltage and the current.

P_αν = V_eff I_eff cos Ɵ

If we consider a purely resistive circuit the phase angle between voltage and current is zero. Hence, the average power is

If we consider a purely reactive circuit (ie, purely capacitive or purely inductive), the phase angle between voltage and current is 90°. The average power is zero, P _αν  = 0.

The average power is the power dissipated in the resistive part only.

Example: 12

A voltage of V(t) = 100 sin ωt is applied to a circuit. The current flowing through the circuit is i(t) = 15 sin (ωt - 30°). Determine the average power delivered to the circuit.

Effective value of the voltage V_eff = 100/√2

Effective value of current I_eff = 15/√2

Average power = P_αν = V_eff I_eff cos Ɵ

= 100/√2 × 15/√2 cos 30°

= 649.5 W

Apparent Power and Power Factor

In case of sinusoidal voltage applied to the circuit, the product of voltage and current is apparent power. The apparent power is expressed in volt amperes or VA.

Apparent power = V_eff I_eff

Power factor:

Power factor is defined as the ratio of average power to the apparent power.

Power factor (Pf) = cos Ɵ = P_av / V_eff I_eff

In case of sinusoidal sources the power factor is the cosine of the phase angle between voltage and current.

Pf= cos Ɵ

As the phase angle between voltage and current increases the power factor decreases. The power factor varies from 0 to 1. For purely resistive circuits, the phase angle between voltage and current is zero, the power factor is unity. For purely reactive circuits the phase angle between voltage and current is 90°, the power factor is zero. In a RC circuits power factor is known as leading power factor. In an RL circuit the power factor is known as lagging power factor.

Example: 13

A sinusoidal voltage V = 50 sin wt is applied to a series RL circuit. The current in the circuit is given by i = 25 sin(ωt - 53). Determine (a) apparent power (b) power factor and (c) average power.

Apparent power = P = V_eff I_eff

V _eff = V_m/√2 = 50/√2

P = 625 VA

Power factor = cos Ɵ

Ɵ = 53°

cosƟ = cos 53°

= 0.6

Average power P _αν = V_eff I_eff cos Ɵ

= 625 × 0.6 = 375 W

Reactive power.

Average power dissipated is

P_av = V_eff I_eff cos Ɵ

From the impedance triangle

cos Ɵ = R/|Z|

V_eff = I_eff Z

Substitute cosƟ and V_eff in average power equator

The above equation gives the average power dissipated in a resistive circuit.

If we consider pure inductor,

P_r = iV_L

= iL di/dt

i = I_m sin (ωt+0)

P_r = I_m² sin (ωt+Ɵ) Lω cos (ωt+Ɵ) = I_m² /2 ωLsin 2 (ωt+Ɵ)

P_r = I_eff² ωLsin 2 (ωL+Ɵ)

P_r = I_eff² WL

P_r = I_eff² X_L

where X_L = z sin Ɵ

 P_r = I_eff² z sin Ɵ

= (I_eff z) I_eff sin Ɵ

P_r = V_eff I_eff  sin Ɵ VAR

Power Triangle

The power triangle is shown in fig 1.30

Average power, P_αν = P_a cos Ɵ

Reactive power, P_r = P_a sin Ɵ

where

P_a= Apparent power

Example: 14

In a certain RC circuit, the true power is 300 W and the reactive power is 1000 W what is the apparent power?

True power, P_av  = VI cos Ɵ = 300 W

Reactive power, P_r = VI sin Ɵ = 1000 W

tan Ɵ = 3.33

Ꮎ = tan^-1 3.33 = 73.3.

Apparent power, P_a = 

=1043.9 VA