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Nice and tricky question on electric circuits

  1. Feb 9, 2006 #1
    Hi, this is a very nice but (at least for me) quite confusing problem on electric circuits:

    Before you read this, it will be helpful to have had a look at the attached picture (sorry - the quality is quite nasty)


    By considering each half of the circuit on the left below as a potential divider, one can show that

    Z1/Z2 = Z3/Z4

    The bridge circuit on the right of the picture is said to be balanced when the detector D registers no voltage difference between its terminals. Use the above equation to find formulae for R and L in terms of the other components when the circuit is balanced.

    OK, so this is what I tried:

    Z1= R + XL

    Z2 = R2

    Z3 = R3

    [tex]Z4 = (\frac {1} {R4} + iwC4)^{-1} [/tex]

    Equation 1

    as derived from Z1/Z2 = Z3/Z4


    [tex] R + iwL = R2*R3*(\frac {1} {R4} + iwC4) [/tex]

    Equation 2

    Now, regard the series connection on the respective sides of the potential divider.

    given: U(Z2) = U(Z4) (A)

    left hand side:

    [tex] U(left) = ( Z1 + Z2)*I = \frac {U(0)} {2} [/tex]

    solve for I to calculate

    [tex] U (Z2) = \frac {Z(2)*U(0)} {2* (Z3 + Z4)} [/tex]

    right hand side:

    like lhs

    [tex] U(Z4) = \frac {Z(4)*U(i)} {2(Z(1)*Z(2)}[/tex]

    So now we put that in eq. (A)

    to get:

    [tex] R + iwL = R3/R2*(\frac {1} {R4} + iwC4)^{-2} [/tex]


    But now I don't know how to solve for L and R as w is not given and I don't know how to deal with those complex numbers to find L and R.

    Can anyone help?? That would be absoluetly awesome!!! :smile:

    Attached Files:

    Last edited: Feb 9, 2006
  2. jcsd
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