Analysis of Circuit containing coupled coils

Discussion in 'Homework Help' started by eleceng16, Jun 2, 2012.

  1. eleceng16

    Thread Starter New Member

    Mar 26, 2012
    6
    0
    Hi,

    I'm trying to work out the loop equations of the circuit attached ... I actually have the answer but I am have troubling understanding how it came about.

    The second loop equation (the right one) is :

    0 = (4 - j2 + j6 +j8 + (2xj4))I2 - (4+j8)I1 -j4I1

    The bold is the part which I cannot understand ... Why is the mutually induced voltage term multiplied by 2?

    Thanks in advance!!:)
     
  2. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    How do you account for the mutual inductance in the j6 ohm coil, in relation to I2? How about in the j8 ohm coil? Which coil does I2 go through?
     
  3. eleceng16

    Thread Starter New Member

    Mar 26, 2012
    6
    0
    Ok I got it :) I didn't realise the self induced voltage term has the same polarity as the mutual induced term in the second loop!!
     
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