Question 16.6: In Figure 16–6 the load ZL is a 100-Ω resistor in series wit......
In Figure 16–6 the load ZL is a 100-Ω resistor in series with a capacitor whose reactance is −60 Ω. The source voltage is 880 V(rms). Find the complex power delivered to the load and the load power factor.
By inspection the node-voltage equation at node A is
\frac{\textbf{V}_{L} − \textbf{V}_{S}}{50} + \frac{\textbf{V}_{L}}{j40} + \frac{\textbf{V}_{L}}{Z_{L}} = 0
Solving for VL with VS = 880 ∠ 0° V and ZL = 100 − j60 Ω yields VL = 411 + j309 V (rms). The magnitude of the load current is
|\textbf{I}_{L}| = \frac{|\textbf{V}_{L}|}{|Z_{L}|} = \frac{|411 + j309|}{|100 − j60|} = 4.41 A (rms)
The complex power delivered to the load and the load power factor are
SL = |IL|² ZL = 4.41²(100 − j60)
= (1.95 − j1.17) × 10³ VA
pf = \frac{P_{L}}{|S_{L}|} = \frac{1.95 × 10^{3}}{|1.95 × 10^{3} − j1.17 × 10^{3}|} = 0.857
The power factor is leading since QL is negative.