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What is V(K) of inductor given coupling coefficent?

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    how to do this!! help me. my caluclator can't solve 3 rectangular form mesh currents equations so i cannot solve for individual currents for all 3 loops!! i just guessed on this answer. i put 400j(I1)-(1600+400)j(I2) or D.

    A. -j400 I1+j1600 I2 (A)
    B. j400 I1+j400 I2 (A)
    C. -j400 I1-j400 I2 (A)
    D. j400 I1-j1200 I2 (A)
    Last edited: Jun 12, 2013
  2. jcsd
  3. Jun 12, 2013 #2

    rude man

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    Not my choice.

    Write down the equation for voltage drop across L1 (jX = j400) due to one or more of i1, i2 and i3.
  4. Jun 12, 2013 #3
    ok using mesh equations?
    (3+j400)I1-(j400)I2=20 (for mesh 1)

    don't know if i'm doing this right. I know usually V1=M*di2/dt for mutual inductors especially with this arrangement of dots but not sure how to apply that. voltage drop normally across the inductor would be Vk=j400(I1-I2) but not sure with this arrangement.
    Last edited: Jun 12, 2013
  5. Jun 12, 2013 #4

    rude man

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    Without mutual inductance Vk=j400(I1-I2) would be correct, but there is mutual inductance M.

    The voltage drop across L1 is due to the current in L1 and the current in L2. What is the current in L2 and how does it affect the voltage across L1?

    The voltage sources will not appear in your equation. You are dealing with the currents i1, i2 and/or i3 only.
  6. Jun 12, 2013 #5
    I am guessing its jw(L1(I1-I2)+M(I2)) based on a similar equation in my book. Would this be right? M I got as k*sqrt(L1*L2)=8. So j400I1+j400I2?
  7. Jun 12, 2013 #6

    rude man

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    Ah, yes, straight A!
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