Circuit with coupled coils

All i need to do here is to find equivalent thenenin U12 generator with it's impedance.Known values are: \$\ E, L_1 , L_2 , k, \omega\$ This is what i have so far:

\$\ \frac{E}{j\omega L1}j\omega\ L_{12}=U_{12}=E_T \Rightarrow \frac{E}{L1}k\sqrt{L1L2} = E_T\$

i believe that this is correct since i don't know what else could i do here.

But i don't know how to find equivalent impedance. Any ideas?

  • \$\begingroup\$ Can you find the open-circuit output voltage and the short-circuit output current (the current that would be delivered if the output port were shorted)? \$\endgroup\$ – The Photon Apr 16 '16 at 14:39
  • \$\begingroup\$ @ThePhoton as you can see i found Thevenin voltage (correctly i hope) where i considered that circuit is opened, so there's no current in that branch since opened circuit is like having infinitely huge resistance in series, so when i am looking for Thevenin voltage there's only voltage coming from the inductor that is "coupled" with L2 (correct me if my reasoning is wrong). Why do i need short-circuit output current? I am supposed to find impedance between the two points, which would be Thevenin generator's inpednace. \$\endgroup\$ – cdummie Apr 16 '16 at 16:44
  • \$\begingroup\$ Example: I have a box containing a battery and resistor (unknown values). I measure its voltage at the terminals of the box with no load and I get a value of 10V. This is the Thevenin (open circuit) voltage). Now if I put a short circuit across its terminals and measured the current I get a value of 2A. The equivalent Thevenin resistance would be 10/2 ohms or 5 ohms. In your case its an impedance but the principle is the same. \$\endgroup\$ – JIm Dearden Apr 16 '16 at 17:03

As with any examples like this ONLY the leakage inductances produce an output impedance. For instance, if coupling is perfect (k=1) then the output terminals appear as a perfect voltage source with zero impedance.

Start with primary leakage inductance - this is \$L_1(1-k)\$. This will be "transformed" (or referred) to the secondary side by the turns ratio squared or \${L_2/L_1}\$. The rest of the primary (the coupled part) no longer plays any role in producing an impedance in series with the output terminals.

The referred_to_secondary primary leakage inductance will add to the actual leakage inductance of the secondary \$L_2(1-k)\$.

So looking into the terminals 1 and 2 you will see an inductance of: -

\$L_2(1-k) + L_1(1-k)\cdot{L_2/L_1}\$.

This looks like it reduces to an inductance of \$2\cdot L_2(1-k)\$

The above assumes that "E" is a perfect voltage source i.e. has zero source impedance.

For the open circuit output voltage you have to calculate what is seen on the coupled part of the primary. Basically E is attenuated by the potential divider formed by leakage and coupled inductance.

This is simply \$\dfrac{k L_1}{kL_1 + (1-k)L_1}\$ or \$k\$.

So, the open circuit output voltage is turns ratio x k x E where turns ratio is \$\sqrt{L_2/L_1}\$.


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