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    I will address the descriptive parts of this posting by means of an update in the next few hours. I wanted to get the calculative solution off to you quickly as I noticed the posting due to expire within the next hour

    The above can be redrawn more formally as two parallel networks connected together by the mutual capacitance as shown below

    As no source resistance is specified we set R_S=0Ω so that all the source voltage appears across the input network. Also since no secondary circuit source voltage is specified we set V_S2=0 as shown by the modified circuit below

    We now find the parallel combination of R=R_2 ||R_L2 in the secondary network from
    R=R_2 ||R_L2=(1/R_2 +1/R_L2 )^(-1)
    R=R_2 ||R_L2=(1/1000+1/2000)^(-1)=667Ω
    Then we can replace both R_2 and R_L2 by a single equivalent resistor Ras shown below

    The above circuit is simply a voltage divider circuit with two complex impedances denoted by Z_12 and Z_2 as shown

    We use circuit theory to derive the magnitude of the impedances at a angular frequency ω as shown below
    Z_12=1/(jωC_12 )=-j 1/(ωRC_12 )
    Which has magnitude
    |Z_12 |=1/(ωRC_12 )
    Z_2=(1/R+1/(1/jωC_22 ))^(-1)=((1+jωRC_22)/R)^(-1)

    Z_2=R/(1+jωRC_22 )=(R(1-jωRC_22))/(1+(ωRC_22 )^2 )
    Z_2+Z_12=(R(1-jωRC_22))/(1+(ωRC_22 )^2 )-j 1/(ωRC_12 )
    Z_2+Z_12=R/(1+(ωRC_22 )^2 )-j{(ωR^2 C_22.ωRC_12+1+(ωRC_22 )^2)/({1+(ωRC_22 )^2}ωRC_12 )}
    Z_2+Z_12=(ωR^2 C_12-j{ω^2 R^3 C_22.C_12+1+(ωRC_22 )^2})/({1+(ωRC_22 )^2}ωRC_12 )
    Which has ...

    Solution Summary

    EMC calculation of crosstalk voltage induced in victim conductor due to capacitive coupling between conductors are examined.