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anonymous

  • one year ago

Find the magnitude of

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  1. anonymous
    • one year ago
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    j=i imaginary part

  2. ganeshie8
    • one year ago
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    magnitudes get divided when you divide two complex numbers : \[\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}\]

  3. ganeshie8
    • one year ago
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    so the magnitude of \(\large \eta^2 \) is given by : \[\large |\eta^2| = \dfrac{\omega\mu}{\sqrt{\sigma^2+\omega^2\epsilon^2}}\]

  4. ganeshie8
    • one year ago
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    simply take the square root

  5. anonymous
    • one year ago
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    Why did you take off the square root at the numerator only?

  6. anonymous
    • one year ago
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    and how did the denominator turn to that form?

  7. ganeshie8
    • one year ago
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    the magnitude of numerator, \(j\omega \mu\), is just \(|\omega \mu|\) right

  8. ganeshie8
    • one year ago
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    its the same formula that you know magnitude of \(a+jb\) is given by \(\sqrt{a^2+b^2}\)

  9. ganeshie8
    • one year ago
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    its just that mathematicians use \(i\) and physicists use \(j\) they both are same

  10. anonymous
    • one year ago
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    I need time to digest it. Thanks for explanation :)

  11. anonymous
    • one year ago
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    One more question: They ask for magnitude of \(\eta\) , not \(\eta^2\), why did you turn it to \(\eta^2\)?

  12. ganeshie8
    • one year ago
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    \(|\eta| = \sqrt{|\eta^2|}\)

  13. anonymous
    • one year ago
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    And after going around, you get back to \(|\eta|\), right? Thanks again. :)

  14. ganeshie8
    • one year ago
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    Exactly, that wasn't supposed to be hard haha!

  15. ganeshie8
    • one year ago
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    technically the answer is simply \( \sqrt{\dfrac{\omega\mu}{\sqrt{\sigma^2+\omega^2\epsilon^2}}}\)

  16. anonymous
    • one year ago
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    :)

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