# How to calculate the voltage of the capacitor in this circuit?

i have a problem with this circuit.
I am trying to calculate all the voltages and currents but i think i did something wrong.

I think the voltages $U_L$ and $U_C$ are wrong because they seem a bit high and i am also not sure if the formulas used are the right ones in this case.

$$Z = j\omega L + \dfrac{1}{\dfrac{1}{R} + j\omega C} =11.57 e^{j62.16°}$$ $$I = 19.02 e^{-j112.16°}A$$

corrected: $$U_L = I\omega L = 646.66e^{-j22.16°}V$$ $$U_C = U_0 - U_L = 463.6 e^{-j189.36°}V$$

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Some may feel otherwise, but currently we accept homework questions. If you think it is a poor one please feel free to ask the user to improve it. – Kortuk May 12 '11 at 20:40
Agree. Bad homework questions (showing no effort except cutting and pasting the original question) will get no or poor answers, clever homework questions that specify one or two things that are not clear should have the chance to get good answers. – zebonaut May 13 '11 at 19:10
Well, i think i showed some effort. The question is just not really wide-ranging. – madmax May 13 '11 at 20:47

Your $\mathbf{I_L}$ is correct, but your phase angle for $\mathbf{U_L}$ is wrong. An inductor's voltage leads the current by 90 degrees; you subtracted. The voltages are larger than the input because the circuit is reactive, but total power is conserved and you can check that $\mathbf{U_q} = \mathbf{U_L} + \mathbf{U_C}$ by cross-verifying $\mathbf{U_C}=\mathbf{I_L}(\mathbf{Z_C} \parallel R)$.

Edit: Here are the values I calculated with Python:

from math import pi, e
w = 2 * pi * 636.62
R = 0.11e3
ZL =  1j * w * 8.5e-3
ZC = -1j / w / 10e-6
Z = ZL + (R * ZC) / (R + ZC)
U = 220 * e ** (1j * pi * -50 / 180)
I = U / Z
UL = I * ZL
UC = U - UL
IC = UC / ZC

def polar_str(x):
from math import atan2
angle = atan2(x.imag, x.real) / pi * 180
return "({:7.3f}, {:8.3f})".format(abs(x), angle)

print('\n'.join(v + ' = ' + polar_str(eval(v)) for v in
[' R','ZL', 'ZC', ' Z', ' U', ' I', 'UL', 'UC', 'IC']))

output: (mag, phase)

R = (110.000,    0.000)
ZL = ( 34.000,   90.000)
ZC = ( 25.000,  -90.000)
Z = ( 11.567,   62.155)
U = (220.000,  -50.000)
I = ( 19.019, -112.155)
UL = (646.657,  -22.155)
UC = (463.659,  170.649)
IC = ( 18.546,  -99.351)

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Ok, so i just have to swap the phase angle for $U_L$ with the one from $U_C$. Thought that $\phi_(Uc) = \phi_(ic) - 90°$ – madmax May 12 '11 at 21:14
@madmax, $\angle U_C = \angle I_C - 90^\circ$ and $\angle U_L = \angle I_L + 90^\circ$. However, $I_C \not= I_L$ since the capacitor is in parallel with a resistor. The phase of $U_C$ is determined by the parallel impedance $Z_C \parallel R$. If you need $I_C$ you can use the current divider rule or just calculate $\frac{U_C}{Z_C}$. – eryksun May 12 '11 at 22:32
When i calculate $\frac{U_C}{Z_C}$, i get the same value and phase as $I_L$. Although i use $U_C$ and Z without $\omega L$. $I_C$ should be 18.5 A. – madmax May 13 '11 at 8:38
Thank you for the calculated values. The problem is my $Z_C$. My $Z_C = 24.38 e^{-j77.2°}$ $Z_C$ is just Z without $\omega L$, isn't it ? – madmax May 13 '11 at 21:00
@madmax, that's the value of $R \parallel Z_C$. The phase shift of an ideal capacitor is $-90^\circ$. – eryksun May 13 '11 at 21:16