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Voltage at high frequency AC with resistors and capacitors

  1. Jan 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Given the circuit below we are asked to calculate [itex]C_x[/itex] and [itex]R_x[/itex]. We are also given a diagram with logarithmic scale showing the voltage [itex]U_x[/itex] as a function of frequency. In the diagram we can see that [itex]U_x[/itex] increases at lower frequencies, but as we reach higher frequencies the [itex]U_x[/itex] eventually becomes a constant [itex]75[/itex]mV. Now, the suggested solution to this problem states that because [itex]U_x[/itex] becomes constant at high frequencies, it indicates that all of the voltage drops over the capacitances only.

    According to the solution we have

    [tex]
    U_x = 10\frac{Z_2}{Z_2 + 50 + Z_1},
    [/tex]

    where [itex]Z_1[/itex] and [itex]Z_2[/itex] are the impedances to the right. And then at high frequencies we may neglect the [itex]50\Omega[/itex] in the denominator, thus resulting in the high frequency limit

    [tex]
    U_x = 10\frac{C_x}{C+C_x}.
    [/tex]

    circuit.png
    2. Relevant equations


    3. The attempt at a solution

    The reason I find this suggestion surprising is because I find the impedance to the right
    [tex]
    Z_1 = \frac{1}{j\omega C_x},
    [/tex]
    [tex]
    Z_2 = \frac{R_y}{j\omega R_y C+1},
    [/tex]

    where [itex]R_y[/itex] is the resistors in parallel and [itex]C[/itex] is the capacitors in parallel. But if [itex]\omega\to\inf[/itex] then [itex]Z_1\to 0[/itex] and [itex]Z_2\to R_y[/itex]. This would suggest that we can neglect voltage drop over the capacitors. But this could not be correct. Voltage division would then show that in principle all voltage would fall over [itex]R_y[/itex] which is contradictory to the fact that [itex]U_x=75[/itex]mV, since the source is at [itex]10[/itex]V. So what am I missing in the above reasoning?


    Thank you.
     
  2. jcsd
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