Lecture 4a: Transient Recovery Voltage Theory - Power System Transients Fall 2020 - Lubkeman

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  • Опубликовано: 22 ноя 2024

Комментарии • 7

  • @maxelmar3200
    @maxelmar3200 7 месяцев назад

    Dr. Lubkeman, why does the voltage across the capacitor Jump upon clearing the fault? The voltage across the Capacitor won't change instantaneously Just like the current into the inductor. Please elaborate on this point if possible. Thank you

    • @dr.davidlubkeman284
      @dr.davidlubkeman284  7 месяцев назад

      The voltage is still zero right after the fault clears, according to the boundary condition for the capacitor. However voltage quickly jumps to a magnitude of 2x peak voltage at a very high frequency given by w0=sqrt(1/LC). Frequency is high since that stray capacitance is very small. In a computer simulation it may appear to change instantaneously, but there is still a small time duration associated with that jump.

  • @joshim9065
    @joshim9065 Год назад

    Great work!!!

  • @joshim9065
    @joshim9065 Год назад

    @22:02 sir, how a fault condition makes Vc(0)=0?, If a voltage source is shorted , does it mean voltage becomes zero? Capacitor is charged already to source voltage level at the point of zero current crossing. How Vc(0)=0, when there is a bolted fault?

    • @dr.davidlubkeman284
      @dr.davidlubkeman284  Год назад

      For a bolted fault at the circuit breaker, the fault current will be essentially
      If = sqrt(2)*Vm*cos(wt)/(jwL) by the time the signal would be sent to the breaker to clear the fault. A bolted fault with Rf = 0 will give us a voltage of zero (Vc = Rf*If) across that stray capacitance regardless of when the fault is cleared. The boundary conditions are not the same as what we saw in the power factor correction switching case.

    • @joshim9065
      @joshim9065 Год назад

      @@dr.davidlubkeman284 okay sir thank you