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Discussion in 'Electronic Basics' started by Peter, Jan 2, 2004.

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  1. Peter

    Peter Guest

    In a circuit with an inductor, the voltage leads the current by 90 degrees.
    Why 90 degrees. Does this change as inductance changes to more or less than

  2. Greg Neill

    Greg Neill Guest

    The voltage across an ideal conductor is given by
    the expression:

    V = L*dI/dt

    If the current is a sine wave, the voltage will
    be a cosine wave (derivative of sine is cosine).
    A cosine is equivalent to a sine shifted by 90
    degrees. But which way is the shift?

    Since the current cannot change instantaneously
    in an inductor, we know that the voltage change
    will precede the current change.
  3. For sine waves and pure inductance, the voltage always leads the
    current by 90 degrees. If you change the value of inductance, the
    magnitude of the current changes (assuming you apply the same voltage
    and frequency) but the phase relationship stays 90 degrees. This all
    comes out of the property of inductance where the voltage and current
    relate instantaneously by V=L*(di/dt) which says in words that the
    voltage across an inductance of L henries is L times the rate of
    change of the current through the inductance in amperes per second.
    If the voltage is sinusoidal, this differential equation indicates
    that the current is also a sine type wave but it has to be shifted by
    90 degrees for the peak voltage to occur where the current is going
    through zero, but changing the most rapidly (having the highest
    magnitude of di/dt).
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