Show equivalence: Z = i*omega (L_1 - M) + (i*omega*M...

.olli

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May 28, 2017
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Hi.

I need to show the following equivalence



\(\displaystyle Z\, =\, i\omega\, (L_1\, -\, M)\, +\, \dfrac{i\omega M\, \left(i\omega\, (L_2\, -\, M)\, +\, X\right)}{i\omega M\, +\, i\omega\, (L_2\, -\, M)\, +\, X}\)

. . . . .\(\displaystyle =\, ...\, \dfrac{i\omega L_1\, \left(i\omega\, (L_2 n^2\, -\, L_1)\, +\, Xn^2\right)}{i\omega L_1\, +\, i\omega\, \left(L_2 n^2\, -\, L_1\right)\, +\, Xn^2}\)

\(\displaystyle \mbox{with }\, n\, =\, \dfrac{L_1}{M}\, \mbox{ and }\, i\, \mbox{ the imaginary unit.}\)



Here is what I tried:

- bring everything to one fraction
- I see that the denominator can be simplified by dropping the iwM term, however I am not sure that it's helpful
- different approaches to combine terms

I am not getting there. Thanks for help.

Best,
Olli
 

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Last edited by a moderator:
Hi.

I need to show the following equivalence



\(\displaystyle Z\, =\, i\omega\, (L_1\, -\, M)\, +\, \dfrac{i\omega M\, \left(i\omega\, (L_2\, -\, M)\, +\, X\right)}{i\omega M\, +\, i\omega\, (L_2\, -\, M)\, +\, X}\)

. . . . .\(\displaystyle =\, ...\, \dfrac{i\omega L_1\, \left(i\omega\, (L_2 n^2\, -\, L_1)\, +\, Xn^2\right)}{i\omega L_1\, +\, i\omega\, \left(L_2 n^2\, -\, L_1\right)\, +\, Xn^2}\)

\(\displaystyle \mbox{with }\, n\, =\, \dfrac{L_1}{M}\, \mbox{ and }\, i\, \mbox{ the imaginary unit.}\)



Here is what I tried:

- bring everything to one fraction
- I see that the denominator can be simplified by dropping the iwM term, however I am not sure that it's helpful
- different approaches to combine terms

I am not getting there. Thanks for help.

Best,
Olli
Did you try multiply&divide the second term by the conjugate of its denominator?
 
Last edited by a moderator:
Hi.
I looked into that now. However, I think it was leading me further away from the result I need (as far as I can tell)...
 
Hey.
Anyone has a suggestion for another approach to this? I still don't see it.
Thanks for any hint!
Olli
 
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