## LESSON 19. Reactive power and power triangle

19.1. Reactive Power

We know that the average power dissipated is

${P_{av}}={V_{eff}}\left[ {{I_{eff}}\,\,\cos \,\theta } \right]..................................................\left( {19.1} \right)$

From the impedance triangle shown in fig. 19.1

$\cos \;\theta \,=\,{R \over {\left[ Z \right]}}..................................................\left( {19.2} \right)$

$and\,\,{V_{eff}}\,\,=\,\,{I_{eff}}\,\,Z..................................................\left( {19.3} \right)$

If we substitute Eqs (19.2) and (19.3) in Eq. (19.1), we get

Fig.19.1

${P_{av}}`={I_{eff}}\,\,Z\,\left[ {{I_{eff}}\,{R \over Z}} \right]$

$=\,\,\,I_{eff}^2\,\,R\,\,watts..................................................\left( {19.4} \right)$

This gives the average power dissipated in a resistive circuit.If we consider a circuit consisting of a pure inductor, the power in the inductor.

${P_r}=i{v_L}..................................................\left( {19.5} \right)$

$=iL\,{{di} \over {dt}}$

Consider $i={I_m}\,\sin \,(\omega t\, + \,\theta )$

Then ${P_r}=I_m^2\,\sin \,\left( {\omega t + \theta } \right)L\omega \,\,\cos \,\,\left( {\omega t + \theta } \right)$

$={{I_m^2} \over 2}\,\left( {\omega L} \right)\sin \,\,2\,\,\left( {\omega t + \theta } \right)$

$\Pr=I_{eff}^2\left( {\omega L} \right)\,\sin \,2\,\left( {\omega t + \theta } \right)..................................................\left( {19.6} \right)$

From the above equation, we can say that the average power delivered to the circuit is zero.  This is called reactive power.  It is expressed in volt-amperes reactive (VAR).

${P_r}=I_{eff}^2\,{X_L}\;\,VAR..................................................\left( {19.7} \right)$

From Fig.19.1, we have

${X_L}=Z\sin \,\theta ..................................................\left( {19.8} \right)$

Substituting Eq. 19.8 in Eq. 19.7, we get

${P_r}=I_{eff}^2\,Z\,\sin \,\theta \;$

$=\left( {{I_{eff}}Z} \right){I_{eff\,}}\sin \,\theta$

$=\,{V_{eff}}\,{I_{eff}}\,\,\sin \,\theta \,\,VAR$

19.2. The Power Triangle

A generalized impedance phase diagram is shown in Fig.19.2.  A phasor relation for power can also be represented by a similar diagram because of the fact that true power Pav and reactive power Pr differ from R and X by a factor I2eff as shown in Fig. 19.2.

At any resultant in time, Pa is the total power that appears to be transferred between the course and reactive circuit.  Part of the apparent power is true power and part of it is reactive power.

${P_a}=I_{eff}^2\,Z$

The power triangle is shown in Fig19.2

From Fig. 19.3, we can write

${P_{true}}={P_a}\cos \,\theta$

or average power Pav = Pa cos θ

and reactive power Pr = Pa sin θ

Fig.19.2                                Fig. 19.3

Last modified: Thursday, 26 September 2013, 10:33 AM