Respuesta :
Given Information:
Vab = 208 V
ZY = Z1 = 10 < -25° Ω
Zdelta = Z2 = 5 < 40° Ω
Required Information:
Line current = Ia = ?
Active Power = P = ?
Reactive Power = Q = ?
Answer:
[tex]I_{a} = 51.02<-30.235[/tex] ° A
[tex]P = 15.88[/tex] [tex]kW[/tex]
[tex]Q = 9.25[/tex] [tex]kVAR[/tex]
Explanation:
First, convert the wye connected load into delta
[tex]Z_{delta} = 3*ZY[/tex][tex]= 3*(5 < 40)[/tex] [tex]= 15 < 40[/tex]° Ω
[tex]Z2 = 15 < 40[/tex]° Ω
Now both of the loads and also the source are delta connected, now we can proceed further.
Find the equivalent load impedance, since the loads are connected in parallel;
[tex]Z_{eq} = Z1*Z2/(Z1 + Z2)[/tex]
[tex]Z_{eq} = (10 < -25)*(15 < 40)/(10 < -25 + 15 < 40)[/tex]
[tex]Z_{eq} = 7.06 + j0.029[/tex] Ω
[tex]Z_{eq} = 7.06 < 0.235[/tex]° Ω
The phase current can be found by
[tex]I_{ab} = V_{ab} /Z_{eq}[/tex]
[tex]I_{ab} = 208/(7.06 < 0.235)[/tex]
[tex]I_{ab} = 29.46 - j0.1208[/tex] A
[tex]I_{ab} = 29.46 < -0.235[/tex]° A
The relation between phase current [tex]I_{ab}[/tex] and line current [tex]I_{a}[/tex] is
[tex]I_{a} = \sqrt{3} I_{ab}<-30[/tex]
[tex]I_{a} = \sqrt{3}*29.46<-30-0.235[/tex]
[tex]I_{a} = 51.02<-30.235[/tex] ° A
The Active Power is
[tex]P = \sqrt{3} V_{ab}I_{a} cos(30.235)[/tex]
[tex]P = \sqrt{3}*208*51.02*cos(30.235)[/tex]
[tex]P = 15.88[/tex] [tex]kW[/tex]
The Reactive Power is
[tex]Q = \sqrt{3} V_{ab}I_{a} sin(30.235)[/tex]
[tex]Q = \sqrt{3}*208*51.02*sin(30.235)[/tex]
[tex]Q = 9.25[/tex] [tex]kVAR[/tex]
