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Impedance ?

  1. Jan 17, 2008 #1
    I have trouble figuring out how my textbook came up with the totals and am looking for step by step help. Here is what the text shows.

    Z = R + j0 = R = 56 Ohm (in rectangular form [XL = 0])
    Z = R < 0 degrees = 56 < 0 degrees Ohm (in polar form)

    Z = 0 + jXL = j100 Ohm (in rectangular form [R = 0])
    Z = XL < 90 degrees = 100 < 90 degrees Ohm (in polar form)

    Z = R + jXL = 56 Ohm + j100 Ohm

    Z = square root(R^2 + X^2L)<tan^-1(100 Ohm/56 Ohm) = 115<60.8 degrees Ohm

    I think I figured out how to get the first number 115 but I'm having trouble on how the 60.8 degrees was determined. But a step by step explanation on how to get both numbers would be really helpful.
  2. jcsd
  3. Jan 18, 2008 #2


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    R and jXL form the orthogonal sides of a rectangle triangle. Z is the hypotenuse. It's modulus is [tex]\sqrt{R^2 + X_L^2}[/tex] and the phase is the angle between the hypotenuse and the side R: [tex]tan^{-1}\frac{X_L}{R}[/tex]
  4. Jan 18, 2008 #3
    I think I figured out where I went wrong. I know I need to divide XL/R and then multiply it by tan^-1. Only problem is I don't know what tan^-1 is. What does tan^-1 equal?
  5. Jan 18, 2008 #4


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    You don't have to multiply for anything. [tex]tan^{-1}[/tex] is the trigonometric function inverse of the tangent. It means the arc whose tangent is...
  6. Jan 18, 2008 #5
    I'm sorry I don't understand. Can you give a step by step example?
  7. Jan 18, 2008 #6


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    Have you ever studied trigonometry? Are you familiar with the functions sine, cosine and tangent?
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