a) When the capacitor is absent, what is the current through the source?
b) What is the voltage across the motor?
c) What is the complex power of the motor?
d) What capacitance should be added to correct the power factor at the motor?
e) What is the current through the source after that capacitance is added?
Zeq = (10.0 + j) ohm, so
= vs / Zeq
= 450 V / (10.0 ohm \/ 5.7 deg)
= 44.8 A \/ -5.7 deg
= (7 + j) ohm
= 7.07 ohm \/ 8.1 deg
= (44.8 A \/ -5.7)(7.07 ohm \/ 8.1)
= 317 V \/ 2.4 deg
= vm . im*
= (317 V \/ 2.4)(44.8 A \/ 5.7)
= 14.2 kVA \/ 8.1 deg
= 14.0 kW + j2.0 kVAR
XC should be added such that 317 V across it will contribute
a reactive power of -2.0 kVAR, thereby eliminating that from the inductive reactance.
Since Q = V2 / XC,
= V2 / Q
= (317 V)2 / 2000 VAR
= 52 ohm
A capacitance of 51 microfarad yields a reactive capacitance of 52 ohm at 60 Hz.
In this part, we check that we accomplished what we wanted in part d).
There are numerous ways to proceed; perhaps the easiest is to evaluated the equivalent
impedance of the network after the capacitance is added. The capacitor in parallel with
the motor yields an equivalent impedance of
= ZmZc / (Zm + Zc)
= (7+j)(-j52) / (7-j51)
= (7.07 \/ 8.1)(52 \/ -90) / (51.5 \/ -82.2)
= 7.1 ohm
Note that the equivalent impedance of the parallel combination has become real, with no net reactance. In general, this is the goal in power factor correction.
The total impedance of the simple circuit is 10.1 ohm; the current through the source is now (450 V) / 10.1 ohm) = 44.4 A.
So the current is now in phase with the voltage source; the magnitude of the current is the same as before (within the round off error of the calculations), but the phase shift has been eliminated, and thus the power factor corrected!