Question 16.DE.11: The inductive load ZL in Figure 16–10 draws an apparent powe......
The inductive load ZL in Figure 16–10 draws an apparent power of 2 kVA at a lagging power factor of 0.8 when the rms load voltage is 880 V(rms) at 60 Hz. Find the value of the capacitance C needed to raise the power factor of the combination to 0.95 lagging.
For |SL| = 2 kVA at a lagging power factor of 0.8, the complex power delivered to ZL is
S_{L} = 2000 × (0.8 + j\sqrt{1 – 0.8^{2}})
= 1600 + j1200 VA
Since PL = 1600 W, the parallel combination must draw an average power of 1600 W because the capacitor does not draw any average power. To deliver 1600 W at a power factor of 0.95, the complex power delivered to the parallel combination must be
S_{P} = \frac{1600}{0.95} × (0.95 + j\sqrt{1 – 0.95^{2}})
= 1600 + j526 VA
The capacitive reactive power needed to produce this result is the difference between the reactive power in SP and SL. That is,
QC = Im(SP) − Im(SL)
= 526 − 1200 = −674 VAR
Since QC = −2πfC|VL|², the required capacitance is found to be
C = \frac{−Q_{C}}{2πf |\textbf{V}_{L}|^{2}}
= \frac{−(−674)}{2π60(880)^{2}} = 2.31 × 10^{−6} F