# Ideal Transformer

Discussion in 'Homework Help' started by Hitman6267, May 26, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
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This is the first time I attempt solving a problem like this and I don't know from where to start.

Do I need to move the R, L, and C to the last circuit on the right ?
What happens to Vs ?
What is the factor that makes |Zth| minimum ?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Yes - using successive impedance transformation for the two transformers

It will be multiplied by the combined transformer ratio a1xa2

When L & C are at series resonance condition

3. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
To move an impedance to the right I have to multiply it by (a1)^2 (from first circuit to the second) and if want to move it to the left I multiply it by 1/[(a1)^2]

Correct ?

How did you come to that conclusion ? Is it necessary to move Vs ?

Is the condition that w = wzero = 1/ radical(LC) ?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Yes

Keep in mind that the impedance from the left-hand transformer primary side will be modified twice in the transition to the right-hand secondary.

Well, the first (left hand) transformer multiplies the voltage Vs by a1 and the second (right hand) by a2.

You want the equivalent relative to the nodes a & b - so yes you will need to define the equivalent source voltage at a & b.

That's right

$\omega_{0}=\frac{1}{\sqrt{LC}}$

5. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
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I don't see where Vs fits in this equation.

I moved L and C to the last circuit on the right by multiplying them by (a2)^2 and applied the formula I just quoted. (didn't work)

Where do R and Vs play a role ?

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Impedances are transformed according to the rule you gave - you can't just multiply the capacitance by a1^2 if that's what you mean. Xc=1/(ωC).

Why should Vs fit the equation for ω0?

7. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
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I transformed the L and C to impedances and then multiplied them by a2^2.

The questions is : Find the value of ω in krad/s for which |ZTh| is minimum.

You told me to me that for |ZTh|to be minimum, ω = ω0

We have $\omega_{0}=\frac{1}{\sqrt{LC}}$

How should I relate this equation to the newly calculated impedances and Vs ?

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Do you think the resonant frequency will change if you transform the elements across the network to their equivalent values on the far right? In terms of terminal conditions an equivalent circuit has to behave in the same manner as the original circuit - otherwise it isn't equivalent.

9. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
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So you're saying I should apply the formula $\omega_{0}=\frac{1}{\sqrt{LC}}$ for the values of L and C I was given ? If so, it didn't work.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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What do you mean by it didn't work? Do you have an answer from the text that differs from yours?

11. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
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I submit it to an online form (set up by my professor) that tells me if my answers are correct or not.

This is what I did,
Question: Find the value of ω in krad/s for which |ZTh| is minimum.
Formula: $\omega_{0}=\frac{1}{\sqrt{LC}}$

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Hitman6267 likes this.
13. ### Hitman6267 Thread Starter Member

Apr 6, 2010
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worked, thank you for your help

Edit:

In the circuit same circuit using the same values and Vs=5.8ej0. We short circuit at a-b, i.e. ZL=0. Find the maximum value of |I3| in mA for variable ω.

I was able to do it by using I1N1=I2N2 relationships. Why is that value (for my values it is (I1)/6) considered to be the maximum value of I3 ?

Last edited: May 27, 2010
14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
At ω=ω0 the inductive and capacitive reactances cancel - it is a series resonant condition.

The two ideal transformers under this resonant condition can be replaced by a single equivalent ideal transformer with a 1:6 turns ratio. An ideal 1:6 turns ratio transformer with a short-circuited secondary will see a maximum secondary current of 1/6th of the maximum primary short circuit current - in the ideal equivalent. The maximum possible primary equivalent short-circuit current is Vs/R. So the maximum possible equivalent secondary short-circuit current (i.e. effectively at location a-b shorted) is Vs/(6R)