An electrical load operates at 240 V rms. The load absorbs an average of 5 KW at lagging power factor of 0.8 The reactive power Q is:
Select one
a 3.75 KVAR
b 4.50 KVAR
c 6.00 KVAR
d 6.75 KVAR
e 5.25 KVAR
Average power, P =5KW= 5KW=5KW
But P=/S/cosθP=/S/ cos \thetaP=/S/cosθ
Given power factor cosθ=0.8cos \theta=0.8cosθ=0.8
=sinθ=1−cos2θ=sin \theta=\sqrt{1-cos^2\theta}=sinθ=1−cos2θ
=sinθ=1−0.82=0.36=0.6=sin \theta=\sqrt{1-0.8^2}=\sqrt{0.36=0.6}=sinθ=1−0.82=0.36=0.6
P=/S/cosθP=/S/ cos \thetaP=/S/cosθ
5KW=/S/∗0.85KW=/S/*0.85KW=/S/∗0.8
/S/=50000.8=6250/S/=\frac{5000}{0.8}=6250/S/=0.85000=6250
Reactive power, Q =/S/sinθ=/S/ sin\theta=/S/sinθ VAR
Q=6250(0.6)=3750Q=6250(0.6)=3750Q=6250(0.6)=3750 VAR
S=P+jQS=P+jQS=P+jQ
S=5000+j3750S=5000+j3750S=5000+j3750
S=5+j3.75S=5+j3.75S=5+j3.75 KVA
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