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LCR circuits - phase difference between C and L

  1. Sep 11, 2010 #1
    I'm writing up a prac that involved LCR circuits.

    One part of the prac involved measuring the phase difference between the voltage across the capacitor (V_C) and the inductor (V_L), while the frequency of the input signal was varied.

    This was done by measuring the distance between peaks for (V_C) and (V_L) on an oscilloscope.

    I got varying values for the phase difference, with a phase difference of zero when the input frequency matched the resonant frequency.

    V_C may be expressed as (-1/Cw)I cos wt.

    V_L may be expressed as Lw I cos wt.


    When looking at these equations, I don't understand why the phase difference depended on the signal's frequency. The way I interpret the equations is that V_C and V_L are permanently 180 degrees out of phase (both being cos of wt, but one negative, the other positive). I can understand that their amplitudes won't always be the same, but it seems to me they should always be 180 degrees out of phase.

    Where have I gone wrong in my understanding?

    Thanks
     
  2. jcsd
  3. Sep 11, 2010 #2
    Your equations for the voltages across the capacitor and inductor are over-simplified. The proper equations, in the engineering notation, are (using I(ω) = I0cos ωt)

    VC = (+1/jωC) I(ω) = (-j/ωC) I(ω) = (+1/ωC)I0sin ωt

    and VL = jωL I(ω) = -ωL I0sin ωt

    [Added - Note the 180-degree phase difference between VC and VL]

    J is a shorthand operator for the derivative operator j that changes the phase of the current by 90 degrees. so

    jsinωt --> +cosωt, and jcos(ωt) --> -sin(ωt).

    These are both derived from the basic equations VL = L dIL/dt, and IC = C dVC/dt

    Bob S
     
    Last edited: Sep 11, 2010
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