The circuit operates in sinusoidal regime.

\$V_e\$ operates at 50 Hz frequency.

The eff. voltages (rms) of \$V_e\$ and \$V_a\$ are, respectively, \$100\$ V and \$50\$ V.


simulate this circuit – Schematic created using CircuitLab

I tried to solve it with 2 methods:

First method - phasors diagrams

\$I=\frac{V_A}{100}=\frac{1}{2}\$ A


simulate this circuit

So \$|V_C|=\sqrt{V_e^2-V_A^2}=\sqrt{100^2-50^2}=\sqrt{7500}=50\sqrt{3}\$

Now: \$|V_C|=|\frac{1}{jwC} I|=\frac{1}{wC}|I|=\frac{1}{wC}\frac{1}{2}\implies C=\frac{1}{2\omega|V_C|}=\frac{1}{2\cdot50^2 \sqrt3}\approx115.47 \mu F\$

Second method

\$V_C=V_e-V_A=50\$ V

\$I_R=\frac{V_A}{100}=\frac{1}{2}\$ A

\$V_C=I\frac{1}{j\omega C}\implies C=\frac{I}{V_C\cdot j\omega}=\frac{\frac{1}{2}}{j 50\cdot 50}=0.0002j\$ F

Why is in this case \$C\$ a complex number?

However, in both cases, the result appears to be wrong, since the solution is \$18.38 \mu F\$.

Could anybody explain-me where I am wrong with the two ways, please?


1 Answer 1


Your first method is almost correct. You went wrong at the very last step, when you plugged in numbers. Hint: look at the units for every variable.

Your second method failed because you're not accounting for phase. \$V_C\$ is not in phase with \$V_e\$, \$V_A\$, or \$I\$, so you can't just add or subtract their magnitudes and get a physically meaningful value. In your first approach, you already found that:

$$|V_C| + |V_A| > |V_e|$$

This looks like it violates Kirchoff's Voltage Law, but it really doesn't. KVL says that:

$$v_c(t) + v_a(t) = v_e(t)$$

In terms of complex exponentials (aka cosines), that gives:

$$|V_C|e^{j(\omega t + \phi_C)} + |V_A|e^{j(\omega t + \phi_A)} = |V_e|e^{j(\omega t + \phi_e)}$$

which in turn means that:

$$|V_C|e^{j\phi_C} + |V_A|e^{j\phi_A} = |V_e|e^{j\phi_e}$$

or, in phasor notation:

$$V_C \angle \phi_C + V_A \angle \phi_A = V_e \angle \phi_e$$

Try again with the phases included and you should be able to get the right answer.

EDIT: The equation you've come up with is:

$$100 \mathrm V \angle \phi_e = V_C \angle {- \frac \pi 2} + 50 \mathrm V \angle 0$$

This is one equation in two unknowns. You need another equation. There are two options:

$$\phi_C = \tan^{-1} \frac{-V_C}{V_A} = \tan^{-1} \frac {-V_C}{50 \mathrm V}$$

$$|V_C|^2 = |V_e|^2 - |V_A|^2 = (100 \mathrm V)^2 - (50 \mathrm V)^2$$

Inverse tangents are awkward to deal with, since they tend to confuse computer algebra systems. The second equation is much easier, can be solved by itself, and leads directly to the answer you want.

  • \$\begingroup\$ I understood the error in the 1st method. For as concern the 2nd method, to apply the KVL I have to know the three phases. Now, assuming the phase of \$V_A\$ as \$0\$, the phase of \$V_C\$ would be \$-\frac{\pi}{2}\$. But what about the phase of \$V_e\$? \$\endgroup\$
    – sl34x
    Commented May 23, 2015 at 11:20
  • 1
    \$\begingroup\$ I've updated my answer. \$\endgroup\$
    – Adam Haun
    Commented May 23, 2015 at 17:48
  • \$\begingroup\$ Just one question: if you treat it with the \$arctan\$, isn't \$\phi_C=tan^{-1} \frac {V_C}{V_A}\$? You see it from the phasor diagram, right? \$\endgroup\$
    – sl34x
    Commented May 23, 2015 at 18:27
  • 1
    \$\begingroup\$ No. In terms of a right triangle on the unit circle, the tangent of the angle equals the length of the opposite side divided by the length of the adjacent side. (Remember SOHCAHTOA!) You can also express it as y/x, as you may remember from basic physics. Or imaginary/real. Here, \$V_e\$ is the hypotenuse, \$V_C\$ is the (pure imaginary) opposite side, and \$V_A\$ is the (pure real) adjacent side. \$\endgroup\$
    – Adam Haun
    Commented May 23, 2015 at 19:17
  • \$\begingroup\$ I don't understand that minus sign though \$\endgroup\$
    – sl34x
    Commented May 23, 2015 at 19:31

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