Given : The load voltage shown in the above circuit is an rms phasor voltage. Re

Question

Given: The load voltage shown in the above circuit is an rms phasor voltage.

Required:
a. Calculate the rms magnitude and angle of the voltage source, Vs.
b. Calculate the rms magnitude and angle of the phasor current, Ib.
c. Calculate the real and reactive power delivered by the voltage source, Vs.
d. Calculate the apparent power and power factor (in percent) of the voltage source, Vs.

Given: The load voltage shown in the above circuit is an rms phasor voltage. Required: a. Calculate the rms magnitude and angle of the voltage source, Vs. b. Calculate the rms magnitude and angle of the phasor current, Ib. c. Calculate the real and reactive power delivered by the voltage source, Vs. d. Calculate the apparent power and power factor (in percent) of the voltage source, Vs.

Explanation / Answer

the total impedence of the parallel branches is

(R+jXl)||(1/jXc)= 45+15j

total impedence= 1.8+j4+45+15j=46.8+60j

current from 40+j20 branch is

I=250/(40+20j)=5-2.5j A

Ic=250/(-300j)=.833j

Ib=Ic+I=.833j+5-2.5j= 5-1.67j

Ib=5.27<-18.43

b) Vs=250<0+Ib(1.8+4j)

Vs=266.22<3.66

=14.77

cos=.967

real power= P=VICos= 1356.7 Watt

reactive power= Q=VI sin=357.67 VAR

Apparent power=|S|=(P^2+Q^2)^1/2= 1403.1 VA

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