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qpzm77gg
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R = 100Ω, L=0.5H, supply voltage vs(t) = 12sin(377t) in Figure 1a
XL = wL =377(0.5) = 188.5Ω
vrms = 8.485∠0
Z=213.38∠62.05 Ω
I = 8.485∠0/ 213.38∠62.05 = 0.0398∠-62.05 A
VR = 3.98∠-62.05 V
VL = 7.502∠27.95 V
PT = 0.03982(100) = 0.158W
QT = QL = 0.299 VAR
ST = 0.338∠62.05 VA
Fp = 0.467a capacitor is added to the circuit in series in figure 1b, and Fp = 1
S=0.158∠0
Q=0
PT=0.0346W
I=0.0186A
Z=455.7Ω
C=14.1μF
If Fp in figure 1b is equal to 0.8
Q=0.02595VAR(ind)
I=0.0051∠36.87 A
Z=1662.82∠-36.87 Ω
For the capacitance of here, I'm so interrogative.
If I used the method of QT =QL - Qc
Qc = 0.0049-0.02595 = -0.02105VAR
Xc=809.3Ω
But I used the method of Vs=Vc+VR+VL
Vc = 8.485-0.0051(100)-0.0051(188.5)=7.01V
Xc=1374.5Ω
Why the answers are different when I used these two methods, and it's correct all of my above answers? If some are wrong, please feel free to indicated the mistake. The Figure 1a and 1b are uploaded the files.
Thanks
XL = wL =377(0.5) = 188.5Ω
vrms = 8.485∠0
Z=213.38∠62.05 Ω
I = 8.485∠0/ 213.38∠62.05 = 0.0398∠-62.05 A
VR = 3.98∠-62.05 V
VL = 7.502∠27.95 V
PT = 0.03982(100) = 0.158W
QT = QL = 0.299 VAR
ST = 0.338∠62.05 VA
Fp = 0.467a capacitor is added to the circuit in series in figure 1b, and Fp = 1
S=0.158∠0
Q=0
PT=0.0346W
I=0.0186A
Z=455.7Ω
C=14.1μF
If Fp in figure 1b is equal to 0.8
Q=0.02595VAR(ind)
I=0.0051∠36.87 A
Z=1662.82∠-36.87 Ω
For the capacitance of here, I'm so interrogative.
If I used the method of QT =QL - Qc
Qc = 0.0049-0.02595 = -0.02105VAR
Xc=809.3Ω
But I used the method of Vs=Vc+VR+VL
Vc = 8.485-0.0051(100)-0.0051(188.5)=7.01V
Xc=1374.5Ω
Why the answers are different when I used these two methods, and it's correct all of my above answers? If some are wrong, please feel free to indicated the mistake. The Figure 1a and 1b are uploaded the files.
Thanks