Question 10.4: An electrical load operates at 240 V rms. The load absorbs a...

a) The power factor is described as lagging, so we know that the load is inductive and that the algebraic sign of the reactive power is positive. From the power triangle shown in Fig. 10.10,

P = \left|S\right| \cos \theta ,

Q = \left|S\right| \sin \theta .

Now, because \cos \theta = 0.8, \sin \theta = 0.6.

Therefore

\left|S\right|=\frac{P }{ \cos \theta }=\frac{8 kW}{0.8} = 10 kVA,

Q = 10 \sin \theta =6kVAR,

and

S = 8 + j6 kVA,

b) From the computation of the complex power of the load, we see that Using Eq. 10.21,

= 8000 W.

Solving for I_{eff},

I_{eff}= 41.67 A.

We already know the angle of the load impedance, because it is the power factor angle

\theta =\cos^{-1}\left(0.8\right)= 36.87^{\circ } .

We also know that θ is positive because the power factor is lagging, indicating an inductive load. We compute the magnitude of the load impedance from its definition as the ratio of the magnitude of the voltage to the magnitude of the current:

\left|Z\right| =\frac{ \left|V_{eff}\right|}{\left|I_{eff}\right|}=\frac{240}{41.67} = 5.76.

Hence,

Z = 5.76 \angle 36.87^{\circ }\Omega = 4.608 + j3.456 \Omega.