Impedance in a multi-frequency AC signal

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.

In short: no.

In "long": it doesn't mean that and, while it may be possible to be so, don't forget that the current capability is directly related to the cross-wire section (area), which means making it generate more than it was planned to will most surely heat it up and, possibly, permanently deteriorate the transformer.

However, seeing that you have 3x8V, it may be possible to connect the windings in parallel, thus granting you the ability to deliver 3 x current, but for that you need to separate the windings so that they are completely separate, then connect them properly in parallel (hot end to hot end), i.e. not in anti-parallel.

Even so, I would discourage this because the windings themselves may not be equal, that is, the 0-6 one may have the average turn length less than the 18-24 one would, meaning that the internal secondary impedances will differ, resulting in other possible cases of deterioration. If, by any chance, you get here, equality may be brought by inserting series resistances on each winding (two at least), at the cost of more losses.