a) All voltage and current phasors in this problem are assumed to represent effective values. Note from the circuit diagram in Fig. 10.14 that\pmb{I}_{s}=\pmb{I}_{1}+\pmb{I}_{2}.The total complex power absorbed by the two loads is
S = (250)\pmb{I}^{*}_{s}
= (250)\left(\pmb{I}_{1}+\pmb{I}_{2}\right)^{*}
= (250)\pmb{I}^{*}_{1}+(250)\pmb{I}^{*}_{2}
= S_{1} + S_{2}.
We can sum the complex powers geometrically, using the power triangles for each load, as shown in Fig. 10.15. By hypothesis,
S_{1}= 8000 – j\frac{8000(.6)}{(.8)}
= 8000 – j6000 VA,
S_{2}= 20,000 (.6)+j 20,000(.8)
= 12,000 + j16,000 VA.
It follows that
S = 20,000 + j10,000 VA,
and
\pmb{I}^{*}_{s}=\frac{20,000 + j10,000}{250} = 80 + j40 A.
Therefore
\pmb{I}_{s} = 80 – j40 = 89.44 \angle -26.57° A.
Thus the power factor of the combined load is
pf = \cos(0 + 26.57°) = 0.8944 lagging.
The power factor of the two loads in parallel is lagging because the net reactive power is positive.
b) The apparent power which must be supplied to these loads is
\left|S\right| = \left|20 + j10\right| = 22.36 kVA.
The magnitude of the current that supplies this apparent power is
\left|\pmb{I}_{s}\right|= \left|80 – j40\right| = 89.44 A.
The average power lost in the line results from the current flowing through the line resistance:
P_{line} = \left|\pmb{I}_{s}\right|^{2}R = \left(89.44\right)^{2}\left(0.05\right) = 400 W
Note that the power supplied totals20,000+400=20,400 W, even though the loads require a total of only 20,000 W.
c) As we can see from the power triangle in Fig. 10.15(c), we can correct the power factor to 1 if we place a capacitor in parallel with the existing loads such that the capacitor supplies 10 kVAR of magnetizing reactive power. The value of the capacitor is calculated as follows. First, find the capacitive reactance from Eq. 10.37:
X =\frac{\left|V_{eff}\right|^{2}}{Q}
=\frac{\left(250\right)^{2}}{-10,000}
= -6.25 \Omega.
Recall that the reactive impedance of a capacitor is -1/ωC, and \omega=2\pi\left(60\right) = 376.99 rad/s, if the source frequency is 60 Hz. Thus
C =\frac{-1}{\omega X}=\frac{-1}{\left(376.99\right)\left(-6.25\right)} = 424.4 \mu F.
The addition of the capacitor as the third load is represented in geometric form as the sum of the two power triangles shown in Fig. 10.16. When
the power factor is 1, the apparent power and the average power are the same, as seen from the power triangle in Fig. 10.16(c). Therefore, the apparent power once the power factor has been corrected is
|S| = P = 20 kVA.
The magnitude of the current that supplies this apparent power is
\left|\pmb{I}_{s}\right|=\frac{20,000}{250} = 80 A.
The average power lost in the line is thus reduced to
P_{line} = \left|\pmb{I}_{s}\right|^{2}R = \left(80\right)^{2}\left(0.05\right) =320 W.
Now, the power supplied totals 20,000 + 320=20,320W. Note that the addition of the capacitor has reduced the line loss from 400 W to 320 W.
