Finding voltages and power in an RLC circuit with 200μF capacitor

[builder_user]

I see.Thanks
I've found
Uc(t)=7.7328-16.0384j
Ul(t)=0.0014+0.003j
Ur(t)=0.224+0.108j
e(t)=7.9554-15.92746j
Are the result correct? 'cause I'm not sure...

 

[gneill]

I'll have to re-work the problem; I've lost track of the calculations that I did previously. It may take a while, as I've got several things on the go at the moment. I hope that's okay.

To summarize, you are given a series RLC circuit and an expression for the voltage waveform that exists across the inductor and capacitor combination. The inductor is L = 3mH, the resistor is R = 4 Ω, and the capacitor is C = 200μF. The order of the components seems to change depending upon what voltages are to be calculated.

The current across the series connected LC pair is given to us as:

Ulc = 17.89V*sin(1000(degrees/s)*t - 64 degrees)

You are looking to find the values for the voltage supply e(t), and the voltages across the individual components as well as several (shuffled) component pairs, and the reactive and complex power used by the circuit. Does that about sum it up?

 

[builder_user]

yes.

I've found P
p=e(t)*I*=e(t)*0.027j=0.43+0.216j
P-reactive - - - - - - 0.216 var?
P-active - - - - - - - - - - 0.43 W?

 

[gneill]

Values that I'm seeing (To compare with the values you've calculated):

For:
ω = 1000 deg/sec; φ = -64° ; B = 17.89V; Ulc(t) = B*sin(ωt + φ)
R = 4Ω ; L = 3mH ; C = 200μF

Impedances:
ZL = 52.36 mΩ {milli Ohms}
ZC = -286.48Ω
ZLC = -286.48Ω {ZL + ZC}
Z = 4 - 286.48Ω {Total impedance of series RLC}

Phasors:
Ulc = B(cos(φ) + jsin(φ)) = 7.842 - j16.079 V ; |Ulc| = 17.89V ; Angle: -64°

I = Ulc/ZLC = 0.056 + j0.027 A ; |I| = 62.46 mA ; Angle: 26°

e = I*Z = 8.067 - j15.97 V; |e| = 17.89 V ; Angle: -63.2°

Ul = I*ZL = -1.434 + j2.939 mV ; |Il| = 3.27 mV ; Angle: 116°

Uc = I*ZC = 7.844 - j16.082 V; |Uc| = 17.89 V ; Angle: -64°

Ur = I*R = 0.225 + j0.11 V ; |Ur| = 250 mV ; Angle: 26°

Power:
P = e * conjugate(I) = 0.0156 - j1.117 W ;
|P| = 1.12 VA {Apparent power}
Re(P) = 0.0156 W = ; 15.6 mW {Real power dissipated}
Im(P) = -1.12 VAR {Reactive power -- negative means it's "capacitive" looking - current is leading voltage}

 

[builder_user]

Some of your results are similar to mine but some of them are completely different.
I think your results are correct so I'll use them.

P = e * conjugate(I) = 0.0156 - j1.117 W ;

I can't get this result

 

[gneill]

Some of your results are similar to mine but some of them are completely different.
I think your results are correct so I'll use them.

You should check your math to see if you can't arrive at the same results; it's important to be able to work these sorts of problems before things get even more complicated! Besides, you never know, I might have mucked up somewhere!

P = e * conjugate(I) = 0.0156 - j1.117 W ;

I can't get this result

You may need to hang on to more decimal places in your intermediate results.

e = 8.0670 - j15.9699 V
I = 0.0561 + j0.0274 A

e*conj(I) = (8.0670 - j15.9699)*(0.05614 - j0.02738) W

= [8.0670 x 0.05614 - (-15.9699 x -0.02738)] + j[-15.9699 x 0.05614 + 8.0670 x -0.02738]

= [0.4529 - 0.4373] + j[-0.2209 + -0.8966]

= 0.0156 - j1.117

 

[builder_user]

I've already checked it. The core of the problem was accuracy.