I have a problem understanding how phasors work and I'll use a problem from a recent exam to illustrate the misunderstanding.

Note: Underlined variables such as \$\underline{V}\$ are considered here to be complex numbers, while non-underlined are considered to be magnitudes of the complex numbers.

Here's the text:

We have voltage generator \$E=2\sqrt{7} \mbox{ } V\$ with angular frequency \$\omega=10^6 \mbox{ } s^{-1}\$ and internal resistance \$R_g=0.5\sqrt{3} \mbox{ } k\Omega\$ connected to parallel connection of impedance \$Z\$ and coil \$L\$. Current is \$I=I_1=I_2=4 \mbox{ } mA\$.

Calculate complex value of \$\underline{Z}\$ and inductivity of \$L\$.

Here's the picture of the circuit:


So from phasor theory I know that they represent complex numbers in the form of \$ \underline{I}=I_e e^{j\psi_0}\$, where \$I_e\$ is effective value of the current and \$\psi_0\$ if the initial phase of the current. So on phasor, the line would be of length \$I_e\$ and have an angle \$\psi_0\$ to the phase axis.

Next, I know that on a line which only has a resistor the voltage and current are in phase and on the phasor, they would be on a same line. If we have a line which has an inductor, the current will lag with respect to voltage by \$ \frac{\Pi}{2}\$. If we have a capacitor, then the current will be in front of the voltage by \$ \frac{\Pi}{2}\$.

Next on a phasor currents on a circuit should form a closed figure.

If we don't have any starting phases in a circuit, we can set the phase of one element to

zero, proclaim it the reference phase and calculate the rest of the phases with respect to it.

So according to my reasoning, I can set the phase of the generator to zero and get \$ \underline{E}=2\sqrt{7} e^{j0}\mbox{ } V\$ and now I have the complex voltage.

Next, I know the effective current \$I\$ and the resistance of \$ R_g\$, so I can calculate the voltage drop across the resistor and this way get the voltage which the impedance \$Z\$ and the inductance \$L\$ see. So \$R_gI=2\sqrt{3}\mbox{ }V\$ and the impedance and the inductance see the voltage of \$U_1=\sqrt{7} \mbox{ }V\$. Next since the resistor is in this case ideal, the \$U_1\$ is in phase with \$E\$.

Next, I know that the current \$I_1\$ can be obtained by following equation \$ \underline{I}_1=\frac {\underline{U_1}}{\underline{Z}}=\frac{U_1 e^{j0}}{Ze^{j\phi}}=\frac{U_1}{Z}e^{0-\phi}\$ From this I can get the \$Z\$.

Next, I know that the impedance \$\underline{Z_l}\$ of the line on which the inductor is is \$\underline{Z_l}=j\omega L\$ and I know that \$\underline{I_2}=\frac{\underline{U_1}}{\underline{Z_l}}=\frac{U_1 e^{j0}}{\omega L e^{\frac{\Pi}{2}}}=\frac{U_1} {\omega L} e^{0- \frac{\Pi}{2}}= \frac{U_1} {\omega L} e^{- \frac{\Pi}{2}}\$ From this I can get the L.

For currents \$ I_1\$ and \$I_2\$ to have same effective value the \$\phi\$ needs to be \$ \frac{Pi}{2}\$ and the impedance \$ Z\$ needs to be mostly capacitive. In this case the current \$I\$ needs to be in phase with the voltage \$ E\$ but it isn't because if it were, the effective value would be determined by the resistance \$R_g\$ and it would be different.

So since the two currents going out of the current \$I\$ have same effective value, I concluded that the three currents need to form a triangle with sides of the same length, like on this picture:

phasor diagram, yes I know it's pathetic

In that case the angle on one of the currents needs to be \$\frac {\Pi} {3} \$ and on the other current it needs to be \$-\frac {\Pi} {3} \$. In that case the current \$\underline{I}\$ is in phase with the voltage \$\underline{U_1}\$ and \$\underline{I_2}\$ is lagging by \$ \frac {\Pi}{3}\$ while the \$\underline{I_1}\$ moved forward by\$ \frac {\Pi}{3}\$. However in that case the current \$ \underline{I_2}\$ isn't lagging by \$ \frac {\Pi}{2}\$ which it should because the only component on its branch is an inductor.

So my problem is that I have a bunch of rules and I can't determine when whey should be applied, as I have shown.

Can anyone explain to me where my reasoning is wrong? For example in the inductor branch, in which cases will current lag behind voltage by \$ \frac {\Pi}{2}\$ and in which by some other number? When can I rely on voltage and current on a purely resistive branch to be in phase? According to Kirchhoff's law the sum of currents in a node should be zero, so on a phasor currents for that node should form a closed figure but in this case they don't.

In my books this isn't clearly explained and most problems we have in problem collections don't have detailed solutions.

Correct solutions for this specific problem are \$\underline{Z}=250(\sqrt{3}-j) \mbox{ } \Omega\$ and \$L=0.5 \mbox{ } mH\$

  • \$\begingroup\$ @Marco Ceppi Are there any news on the homework tag? Latest thing I heard was that we're still using it here. \$\endgroup\$
    – AndrejaKo
    Jul 27, 2011 at 16:57
  • \$\begingroup\$ Looking through the meta I don't see any reference to tagging a post with homework or not. To be honest it seems like a meta tag that doesn't add any context to the post other than to just be there \$\endgroup\$ Jul 27, 2011 at 17:50
  • \$\begingroup\$ meta.electronics.stackexchange.com/questions/115/… The homework tag should be going away, but our mods don't have the time to mess with tags. \$\endgroup\$
    – Kellenjb
    Jul 27, 2011 at 17:54
  • \$\begingroup\$ Are you sure I = I1 = I2 = 4mA? If I = I1 + I2 then I must be 8mA. \$\endgroup\$
    – Thomas O
    Jul 29, 2011 at 19:38
  • \$\begingroup\$ @Thomas O The problem says: Effective (as in not complex) \$ I=I_1=I_2= 4 \mbox{ } mA\$. I'm 100% sure. I failed the exam because of that problem :). \$\endgroup\$
    – AndrejaKo
    Jul 29, 2011 at 20:02

1 Answer 1


I give up. I can't solve the problem given, I think more information is needed beyond what is in the problem statement, and I wouldn't be saying that if I had not hacked away at it and wound up at this point. To begin with, the problem is as follows.

We have voltage generator \$E=2\sqrt{7} \mbox{ } V\$ with angular frequency \$\omega=10^6 \mbox{ } s^{-1}\$ and internal resistance \$R_g=0.5\sqrt{3} \mbox{ } k\Omega\$ connected to parallel connection of impedance \$Z\$ and coil \$L\$. Current is \$I=I_1=I_2=4 \mbox{ } mA\$. Calculate complex value of \$\underline{Z}\$ and inductivity of \$L\$.

My claim is that this is unsolvable. I owe a little explanation for for my claim before I change the problem and solve something different. Basically, the fact that \$\underline{Z}\$ and \$L\$ are unknown gives 3 unknowns. Combined with the power factor of the circuit, this gives 4 real unknowns. You can do mesh analysis or node analysis and find that you will have 2 complex equations, minus one reference. You're one short.

Here is what I would add:

Assume that the magnitude of \$I_1\$ and \$I_2\$ are equal.

The only way I know to do this is to use the answer given in the problem, so now that I have that out of the way I'll hack away at this. I'll introduce only \$Z_{e}\$, which is the combined impedance of the 2 parallel components. I might also forget some of the vector bars, forgive me please. Start at the voltage source and note the following, using the general \$|V|=|I| |Z|\$ property.

$$|E| = |I| |Z_g+Z_e|$$

$$|Z_g+Z_e| = \frac{ |E| }{|I|} = 500 \sqrt{7}$$

Now I'll define my reference and follow through the voltage a bit. The notation I use is \$U_1\$ for that obvious voltage point after the resistor. I'm using \$-\psi\$ for the current angle because I already know it's a net inductive circuit, which is just from knowledge of the solution.

$$ E = 2 \sqrt{7} \angle 0 $$ $$ I = \frac{1}{250} \angle -\psi$$ $$ U_1 = E - R I = 2 \sqrt{7} - 2 \sqrt{3} \angle -\psi$$

I need to write the equation for the equivalent inductance.

$$ Z_e = \frac{1}{ \frac{1}{Z} + \frac{1}{j \omega L} } $$

Anyway, I'll just skip some steps and write the values. I hope to come back and put more in later. Sorry about the lack of actual circuit analysis in this answer.

$$ \psi = arctan( \frac{1}{3 \sqrt{3} } )$$ $$ Z = 250 \angle -\frac{\pi}{3} $$ $$ Z_e = 250 \angle \frac{\pi}{3} $$ $$ I_1 = \frac{1}{250} \angle arctan( \frac{2}{\sqrt{3}} )$$ $$ I_1 = \frac{1}{250} \angle -arctan( \frac{5 \sqrt{7}}{\sqrt{21}} )$$

It's already redundant to say this, but these numbers give the \$Z=250(\sqrt{3}-j)\$ and \$L=0.5 mH\$. It would also work to say that Z is a resistor of \$250 \sqrt{3} \Omega \$ in series with a \$ 4 nF\$ capacitor.

I think this was a bad question, and I hope I've given enough breadcrumbs of a consistent answer for your to prove this to someone else. Maybe I'm wrong, but if my current analysis is right, I would hate to have for anyone to be given this on a test.

  • \$\begingroup\$ I spent about an hour playing with this as well, also came to the conclusion that not enough information was provided. I also could certainly be wrong, its been a long time since I did any steady state AC analysis. \$\endgroup\$
    – Mark
    Jul 30, 2011 at 3:30
  • \$\begingroup\$ At least I got the \$U_1\$ correctly on the exam... Anyway, +1 for the effort. \$\endgroup\$
    – AndrejaKo
    Jul 30, 2011 at 7:01

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