The problems about power factor changed in RCL circuit, help~

[qpzm77gg]

R = 100Ω, L=0.5H, supply voltage v_s(t) = 12sin(377t) in Figure 1a

X_L = wL =377(0.5) = 188.5Ω
v_rms = 8.485∠0
Z=213.38∠62.05 Ω
I = 8.485∠0/ 213.38∠62.05 = 0.0398∠-62.05 A
V_R = 3.98∠-62.05 V
V_L = 7.502∠27.95 V
P_T = 0.0398<sup>2(100) = 0.158W
Q_T = Q_L = 0.299 VAR
S_T = 0.338∠62.05 VA
F_p = 0.467a capacitor is added to the circuit in series in figure 1b, and F_p = 1
S=0.158∠0
Q=0
P_T=0.0346W
I=0.0186A
Z=455.7Ω
C=14.1μF

If F_p 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 Q_T =Q_L - Q_c
Q_c = 0.0049-0.02595 = -0.02105VAR
X_c=809.3Ω
But I used the method of V_s=V_c+V_R+V_L
V_c = 8.485-0.0051(100)-0.0051(188.5)=7.01V
X_c=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</sup>

 

[gneill]

Your calculations for the required capacitor in Figure (a) appear to be fine.

For Figure (b), what are the given component values and conditions? Are the voltage source, resistance, and inductance the same as before?

 

[qpzm77gg]

Your calculations for the required capacitor in Figure (a) appear to be fine.

For Figure (b), what are the given component values and conditions? Are the voltage source, resistance, and inductance the same as before?

Yes, They are same as before, and then step and step add the capacitor in the circuit and change the power factor from 1 to 0.8.
Thanks

 

[gneill]

Yes, They are same as before, and then step and step add the capacitor in the circuit and change the power factor from 1 to 0.8.
Thanks

Just to be clear, the components start out as Vs, R, L, and C from part (a) where the power factor is 1, and you wish to add another capacitor to bring the power factor down to 0.8?

 

[qpzm77gg]

Just to be clear, the components start out as Vs, R, L, and C from part (a) where the power factor is 1, and you wish to add another capacitor to bring the power factor down to 0.8?

The complete question is uploaded in the file.
Thanks a lot.

 

[gneill]

Okay. Having the complete question helps.

For part (c)(iv), finding the capacitance, if you know the power factor and the resistance (the real part of the impedance), then you should be able to find the corresponding reactance. After all,
$$ pf = cos(\theta) = \frac{R}{\sqrt{R^2 + X^2}}$$
X will be positive or negative depending upon whether the current is said to be lagging or leading. You can then find XC from X = XL + XC.

I haven't been able to follow how you've approached parts (i) through (iii). I think if it were me I would have been tempted to do the parts in reverse order

 

[qpzm77gg]

Q_T = Q_L-Q_c
Because all I are same in the in series circuit.
X_T=X_L - X_c
Q_T/I²=188.5-X_c
X_c = 1186.19Ω

another method,
|Z|=1662.82
|Z|² = R²+X_c2²
1662.82² = 100²+X_c2²
X_c2=1659.8Ω
Q_T=I²(1659.8)
Q_T=0.0432VAR(cap)
Q_c=Q_L-Q_T
Q_c=0.0049+0.0432=0.0481VAR(cap)
X_c=0.0481/0.0051²=1849.3Ω


There are also difference between the answers, why?

 

[gneill]

Q_T = Q_L-Q_c
Because all I are same in the in series circuit.
X_T=X_L - X_c
Q_T/I²=188.5-X_c
X_c = 1186.19Ω

Is Q a positive or negative value if the pf is leading?

another method,
|Z|=1662.82
|Z|² = R²+X_c2²
1662.82² = 100²+X_c2²
X_c2=1659.8Ω
Q_T=I²(1659.8)
Q_T=0.0432VAR(cap)
Q_c=Q_L-Q_T
Q_c=0.0049+0.0432=0.0481VAR(cap)
X_c=0.0481/0.0051²=1849.3Ω


There are also difference between the answers, why?

I have suspicions about your value for the impedance. Since you know the real part R is 100 Ohms, you should be able to find the magnitude of the impedance given the power factor. You should also be able to find the magnitude of the reactive part.