Question 23.5: For the Y-connected load in Fig. 23.23: a. Find the average ...
For the Y-connected load in Fig. 23.23:
a. Find the average power to each phase and the total load.
b. Determine the reactive power to each phase and the total reactive power.
c. Find the apparent power to each phase and the total apparent power.
d. Find the power factor of the load.
a. The average power is
P_{\phi}=V_{\phi} I_{\phi} \cos \theta_{I_{\phi}}^{V_{\phi}}=(100 V )(20 A ) \cos 53.13^{\circ}=(2000)(0.6).
= 1200 W.
P_{\phi}=I_{\phi}^{2} R_{\phi}=(20 A )^{2}(3 \Omega)=(400)(3)=1200 W.
P_{\phi}=\frac{V_{R}^{2}}{R_{\phi}}=\frac{(60 V )^{2}}{3 \Omega}=\frac{3600}{3}=1200 W.
P_{T}=3 P_{\phi}=(3)(1200 W )=3600W.
or
P_{T}=\sqrt{3} E_{L} I_{L} \cos \theta_{I_{\phi}} ^{V_{\phi}}=(1.732)(173.2 V )(20 A )(0.6)= 3600 W .
b. The reactive power is
Q_{\phi}=V_{\phi} I_{\phi} \sin \theta_{I_{\phi}}^{V_{\phi}}=(100 V )(20 A ) \sin 53.13^{\circ}=(2000)(0.8).
= 1600 VAR.
or Q_{\phi}=I_{\phi}^{2} X_{\phi}=(20 A )^{2}(4 \Omega)=(400)(4)=1600 VAR.
Q_{T}=3 Q_{\phi}=(3)(1600 VAR )= 4800 VAR.
or
Q_{T}=\sqrt{3} E_{L} I_{L} \sin \theta_{I_{\phi}}^{V_{\phi}}=(1.732)(173.2 V )(20 A )(0.8)= 4800 V A R.
c. The apparent power is
S_{\phi}=V_{\phi} I_{\phi}=(100 V )(20 A )=2000 VA.
S_{T}=3 S_{\phi}=(3)(2000 VA )= 6000 V A.
\text { or } S_{T}=\sqrt{3} E_{L} I_{L}=(1.732)(173.2 V )(20 A )= 6000 VA.
d. The power factor is
F_{p}=\frac{P_{T}}{S_{T}}=\frac{3600 W }{6000 VA }= 0.6 \text { lagging }.