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Here is the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

\$ V_i = \frac{8}{\sqrt 2} \angle0° V\\ V_c = \frac{6}{\sqrt 2} \angle45° V\\ R = 1000\space\Omega \$

I'd like to know how to find the capacitance with these given values. What I've tried:

\$ V_c = V_i \times \frac{X_c}{R\space\space+\space\space X_c} \\ X_c = \frac{V_c\space R}{V_i\space\space - \space\space V_c} \$

Here is the problem, just solving for Xc, I got a complex number with a angle different than -90°. So I forced Xc to be -90°, then when I solve for the capacitance c, I get a complex number:

\$ X_c = 1058.7141\angle 93.4716182° \\ \frac{1}{j\frac{500}{2\pi}c} = 1058.7141\angle 93.4716182° \\ c = -0.00001184768784 + j\space 7.18682779 \times 10^{-7} \$

What am I possibly missing?

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It's a malformed question because with these components, the phase angle of \$V_C\$ must be \$-90^\circ\$. The only way it could be \$45^\circ\$ would be if there were also an inductance in the circuit.

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  • \$\begingroup\$ Makes a lot of sense! thanks! But the calculation is right? \$\endgroup\$ Dec 13, 2014 at 17:47
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    \$\begingroup\$ @JoãoPauloOliveiraFernandes: The equations are correct up to point at which you get a value for the complex impedence, but it's not really \$X_C\$ but rather \$X_N\$ where \$N\$ is some strange compound part with the impedance you calculated. A complex capacitance has no real physical sense. \$\endgroup\$
    – Edward
    Dec 13, 2014 at 18:24
  • \$\begingroup\$ Ok. Since the given values are right, I should have found a Xc with -90°, right? After that would be just match the complex amplitudes to find c, right? \$\endgroup\$ Dec 13, 2014 at 18:28

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