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Kirchhoff's law - Simultanious Equations - Question

  1. Apr 25, 2007 #1
    When Kirchhoff's law are applied to a particular electrical circuit the current I^1 & I^2 are connected by the equations;

    27 = 1.5 I^1 + 8 (I^1 - I^2)
    -26 = 2 I^2 - 8 (I^1 - I^2)

    Please show working out =)
  2. jcsd
  3. Apr 25, 2007 #2


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    Why was this posted under "differential equations"?

    Taking I^1= x and I^2= y because your notation was confusing me, you have
    27= 1.5x+ 8(x-y)= 1.5x+ 8x- 8y= 9.5x- 8y and
    -26= 2y- 8(x- y)= 2y- 8x+ 8y= 10y- 8x.

    That is you want to solve the equations 8x- 10y= 26 and 9.5x- 8y= 27.

    You might try multiplying the first equation by 4 (so you get -40y) and the second equation by -5 (so you get +40y) and adding those equations to eliminate y.

    No, I'm not going to "show working out". I am going to assume that you are capable of doing the basic algebra here yourself! The practice will do you good.
  4. Apr 25, 2007 #3
    sorry about the place of posting i was unable to work out which of the mathimatics sections it should go under.

    and sorry about the "I to the power of 1 and I to the power of 2"

    dont worry about the working out =) i will be able to carry on from there

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