# Non-Ideal Transformer Equations

I saw the following non-ideal transformer in my textbook as an exercise. Find the equations for VA and VB. Then, find the required frequency of V1 with all elements such that the transformer is resonant if k = 0.

I tried to find the voltage equations for both points A and B:

$$v_A (jw) = jwi_A(jw) L_1 + jwi_B(jw) k \sqrt{L_1 L_2}$$ $$v_B (jw) = jwi_A(jw) k \sqrt{L_1 L_2} + jwi_B(jw) L_2$$

Are my equations correct? Also, I believe that resonance is reached when the impedances all cancel out but I am not sure how to apply it here.

• If k=0 then there is no mutual inductance, right? – relayman357 May 10 '20 at 3:14
• I think so? Would that eliminate a term from each equation then? – bth May 10 '20 at 3:18
• $\color{red}{\text{STOP RIGHT NOW}}$ - go back and withdraw the amendment you have just made to your question. You originally asked a question that NEVER SAID "not to use the capacitors & resistors for Part A" and I have answered that original question in good faith and with respect. Your late amendment makes my answer seem wrong so go back and undo the change. – Andy aka May 10 '20 at 15:41
• OK @bth thanks for rolling it back. With capacitors bypassed and resistors open circuit there is no secondary current hence $i_B$ is zero. But $v_B$ is still present but you still need to decide how you apply a reference node to the secondary or the question is incomplete. But we can say that $v_A$ = V1 – Andy aka May 10 '20 at 16:02
• So, you need to calculate what the voltage across kL1 (my diagram) is (fairly easy) then, if the reference node is also the lower node as per the primary side), $v_B$ is that voltage x square root of L2/L1. – Andy aka May 10 '20 at 16:08

For Part A you should redraw the circuit like this: -

The extra inductances I've added are: -

• (1 - k)L1
• (1 - k)L2

These are the leakage inductances that will produce resonance. The 100% coupled parts serve only as an impedance transformer such that the secondary impedances (these)....

• (1 - k)L2
• C1
• R4

... can be transferred to the primary via the 100% coupling. But you need to know how to transform those impedances and therefore, you need to know the turns ratio. The turns ratio is: -

$$\sqrt{\dfrac{kL2}{kL1}} = \sqrt{\dfrac{L2}{L1}}$$

Can you take it from here?

Part B (k = 0) is trivial because it just involves L1 and C2.

• Thanks! So would my equations for Part A change to this? $$v_A (jw) = (1-k) L_1 jw*i_A(jw) + jw \sqrt{\frac{L_2}{L_1}}*i_B(jw)$$ $$v_B (jw) = jw \sqrt{\frac{L_1}{L_2}} * i_A(jw) + i_B \sqrt{\frac{L_1}{L_2}}((1-k)L_2 + R_{C1} || R_4)$$ I am not sure if I fully understand coupling here – bth May 10 '20 at 15:17
• I don't understand your initial equations in your question hence I didn't bother try to cover that ground. Given also that single point node equations for a voltage with no reference node make no sense. Equations that seem to include currents ($i_A$ for example) do require that the circuit indicates where these currents are. I see no equation that uses the capacitors hence your formula are probably incorrect (if i could decipher them). – Andy aka May 10 '20 at 15:28
• My apologies, the question states not to use the capacitors & resistors for Part A but to use them for Part B hence why I tried to avoid using them. – bth May 10 '20 at 15:36