# Ideal inductor in series with a resistor:

The question is to find the thevenin equivalent circuit across a and b. The inductors are ideal transformers. I stretched the inductors 180 degrees to make the circuit easier to analyze. From there, I can see that the left hand inductor is parrallel with the voltage source and is in series with the 2 ohms resistance. An ideal transformers has infinite inductance and zero resistance. In my book, it is said that the LHS inductor has a voltage drop of $1\angle 0$. But, how can that happen it would leave no voltage for the resistance??

• if it is an idealized transformer and in the absence of knowledge of leakage inductance, there will be zero current taken through any winding - this leaves exactly one volt on the left inductor. – Andy aka Nov 21 '13 at 20:36
• @Andyaka: Inductors seem to have special rules-they don't seem to stop. For example, in this circuit(i.stack.imgur.com/Jpstf.png), $I_x$ is zero. So, the dependent votlage source is zero, so its a short. Doesn't that mean that $V'_{Th}$ is zero. – user29568 Nov 21 '13 at 20:45
• There's a dependent current source and this will be zero BUT this doesn't appear to be anything to do with your question. – Andy aka Nov 21 '13 at 21:12
• Well, you are quite right. I should have opened a seperate question for it. But, my real question is how is it that inductors work in a odd way. How is it a dependent current source?It is CCVS. – user29568 Nov 21 '13 at 21:14
• As such will V'Th be zero? No; the controlled voltage source behaves as a wire when the secondary is open (there is zero volts across the controlled source since the primary current is zero). But this doesn't imply that the secondary voltage is zero. In fact, by inspection, the secondary voltage is 160V and so: $V'_{Th} = -160V$ – Alfred Centauri Nov 21 '13 at 22:12

But, how can that happen it would leave no voltage for the resistance??

Andy's given the correct answer to this question in the comments. This is the complete analysis.

Call the left winding primary and the right winding secondary. See that the secondary winding is open and thus there is zero secondary current.

Since this is an ideal transformer, the primary current must also be zero and thus, by Ohm's Law, the voltage across the $2 \Omega$ resistor, in series with the primary, is zero.

By KVL, the voltage across the primary is $-1V$ so the voltage across the secondary is $-4V$.

Finally, the open circuit voltage is $V_{ab} =V_{th} = 1 - 4 = -3V = 3\angle 180 V$

To find the Thevenin impedance directly, zero the independent voltage source and see that the $2 \Omega$ resistor is across the primary which, reflected to the secondary, appears as $2 \cdot 4^2 = 32 \Omega$. Thus, the Thevenin Impedance is $Z_{th} = 32 \Omega$.