- #1

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Unless my rearrangement is wrong, z=R+iwl, so Re(z) should definitely be R!

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- Thread starter whatisreality
- Start date

- #1

- 290

- 1

Unless my rearrangement is wrong, z=R+iwl, so Re(z) should definitely be R!

- #2

HallsofIvy

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From [tex]\frac{1}{z}= \frac{1}{R}+ \frac{1}{iWL}[/tex], multiply on both sides by all denominators, RiWLz, to get RiWL= iWLz+ Rz= (R+ iWL)z. Solve that for z to find the real and imaginary part.

- #3

wabbit

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So you're saying 1/(R+iwL)=1/R+1/(iwL) ?

- #4

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- #5

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z = -R^{2}ω^{2}L^{2}/(iωL+R) is my second attempt!

Edit: maybe not. L squared too. And that makes life more difficult! What do I do about i on the denominator? Multiply by the conjugate?

Edit: maybe not. L squared too. And that makes life more difficult! What do I do about i on the denominator? Multiply by the conjugate?

Last edited:

- #6

HallsofIvy

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[tex]\frac{a+ bi}{c+ di}= \frac{a+ bi}{c+ di}\frac{c- di}{c- di}[/tex][tex]= \frac{ac+ bd+ (bc- ad)i}{c^2+ d^2}=[/tex][tex]\frac{ac+ bd}{c^2+ d^2}+ \frac{bc- ad}{c^2+ d^2} i[/tex]

Here you want to multiply numerator and denominator by [tex]R- i\omega L[/tex].

- #7

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z = RiwL/(R+iwL)

Realise the denominator:

(iR^{2}wL+Rw^{2}L^{2})/(R^{2}+w^{2}L^{2})

Realise the denominator:

(iR

- #8

wabbit

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Try this as a warm up : forget the exercise, just compute [tex]\frac{1}{a+ib}[/tex]. Then go back to z.

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