This is my problem enter image description here

Now I know that ideally $$V_1/V_2 = 1:α$$ and $$(J_1-J_3) = α (J_2-J_3)$$

where J1 is the current of the left mesh, J2 is the current of the right mesh and J3 is the current of the mesh in the middle(the one that passes through Zx)

If V0 = 0 then that means that ZL is short-circuited thus $$V_2 = 0$$ as well. Can I use equation of currents for the ideal transformer if V1/V2 can't be used?

  • \$\begingroup\$ How will you use the currents if they are infinite? \$\endgroup\$ – Eugene Sh. Sep 14 '17 at 19:55
  • \$\begingroup\$ Then what do you propose? Is there another way to write Zx in correlation with jωL, α, Vin and V1? \$\endgroup\$ – user3601507 Sep 14 '17 at 20:20
  • \$\begingroup\$ lim notation could help. \$\endgroup\$ – Eugene Sh. Sep 14 '17 at 20:21
  • \$\begingroup\$ We can't use those. We can only use the ideal transformer equations and Kirchoff's Laws. This question was in a past exam and nobody knows the answer, plus the professor won't reply to our mails about it. So nobody really knows what to do aside from use the current equation and writing J2 = 0 since ZL * J2 = V2 = V0 = 0. \$\endgroup\$ – user3601507 Sep 14 '17 at 20:23
  • \$\begingroup\$ Why is V2 = 0 a problem? Can you think of the transformer as a current transformer? \$\endgroup\$ – Vladimir Cravero Sep 14 '17 at 20:31

The component jwL is transferred to the secondary and its new impedance is: -

\$\dfrac{j\omega L}{N^2}\$ where N is the transformer ratio primary to secondary.

This then simplifies the circuit and allows you to remove the transformer because once jwL transfers to the secondary, you can replace the transformer with a voltage source of Vin/N feeding into jwL/\$N^2\$.

The problem then boils down to solving a potential divider formed by Zx and jwL/\$N^2\$. At the top of the potential divider is Vin and at the bottom of the potential divider is Vin/N and, at the centre point is 0 volts. ZL plays no part in this analysis because it connects to the centre point and the centre point produces 0 volts.

Here's how V2 becomes the new input voltage and jwL transfers to the secondary: -

enter image description here

"ratio" = N in the picture.

And the next step is to ignore the transformer completely and just treat the problem as a potential divider with one voltage being Vin and the other voltage being V2 aka Vin/N.

The unknown impedance will be capacitive BTW.

  • \$\begingroup\$ Yup so basically ZL is short-circuited as I mentioned. My question is, can you use the transformer equations since V0 = 0 and V1/V0 does not make any mathematical sense? \$\endgroup\$ – user3601507 Sep 16 '17 at 17:42
  • \$\begingroup\$ You are missing my point in my answer. ZL doesn't have to be short-circuited to get 0 volts across it. Putting 0 volts on it gives 0 volts on it and that is exactly what my answer is telling you. You use the turns ratio to move jwL to the secondary and then you can throw away the transformer and substitute it with a voltage source. \$\endgroup\$ – Andy aka Sep 16 '17 at 19:21
  • \$\begingroup\$ I see. My Sadiku textbook says that you can't transfer to secondary because primary+secondary are connected via Zx. Unless it works in this case because the transformer's primary/secondary current doesn't flow through Zx. \$\endgroup\$ – user3601507 Sep 16 '17 at 19:30
  • \$\begingroup\$ I'll add a picture. Do you see what I have done and why it is now solvable? \$\endgroup\$ – Andy aka Sep 16 '17 at 19:36
  • \$\begingroup\$ Yup, I know what you meant because I understand how to transfer the primary to secondary and vice versa but I wasn't 100% sure that it could be done here because of this:i.imgur.com/S27np9I.png \$\endgroup\$ – user3601507 Sep 16 '17 at 20:37

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