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inductor impedance - why no minus sign?

Discussion in 'Electronic Basics' started by Sean McIlroy, May 21, 2007.

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  1. Sean McIlroy

    Sean McIlroy Guest


    if an inductor has inductance L then the voltage V across it and the
    current I through it satisfy

    V = (-1) * L * dI/dt

    and yet its impedance to a sinusoidal voltage with angular frequency w
    is NOT the expected

    (-1) * L * j * w

    but rather

    (+1) * L * j * w

    how come? or rather: what am i missing THIS time?

  2. JeffM

    JeffM Guest

  3. John Larkin

    John Larkin Guest

    No. If the current is measured flowing into the inductor, as for a
    resistor, there's no -1 term. Increasing positive I produces a
    positive V.

  4. Eeyore

    Eeyore Guest

    The minus term applies to induction in a coil from a magnetic field does it not

  5. Sean McIlroy

    Sean McIlroy Guest

    ok, but how does that square with the verbal formulation that the
    voltage is in the direction that "opposes" the change in current? if
    it were a resistor instead of an inductor then a positive voltage drop
    would be associated with a positive current flow, which seems to me to
    "oppose" a DECREASE in current, not an increase.

  6. Z = V/I and V = -L*dI/dt

    if I = A*sin(w*t) then you can compute Z. If you use phasors then you'll
    see... if you just plug in directly then its not easy but you can get the

    An easy way is to use complex numbers(well, this is essentially phasors),

    I = A*sin(w*t) = A*Re(-j*e^(j*w*t))= A<-90


    V = -L*dI/dt = -L*A*w*cos(w*t) = -L*A*w*Re(e^(j*w*t)) = L*A*w<180

    Hence V/I = [L*A*w<180]/[A<-90] = wL<[180-90] = wL<[90] = jwL

    Hopefully that makes sense. You can get it by just dividing the
    quanity -L*A*w*cos(w*t) with A*sin(w*t) and then computing the magnitude and

    Note that V/I is phasor division... You cannot do
    A*sin(w*t)/[-L*A*w*cos(w*t)] because that is not what the impedance is(its
    not defined as the instantaneous voltage divided by the instantanous current
    but by phasor division).
  7. John Larkin

    John Larkin Guest

    I can't deal with that verbal formulation. For a 2-terminal device,
    with one end grounded, voltage is usually measured at the ungrounded
    terminal and current is measured flowing into that terminal.

    For a resistor, positive current makes positive voltage.

    E = I * R

    For a capacitor, positive current ramps up positive voltage.

    E = integral(I/C)

    For an inductor, increasing positive current causes the terminal to
    have a positive voltage.

    E = L di/dt

    all positive. A negative sign would violate conservation of energy.

  8. Guest

    1) It's the induction voltage accross the coil, not quite the voltage
    accross it unless R = 0
    2) Is it V(t) = -L dI/dt ?
    3) It depends on how you mark the (-) and (+) for V(t) on your
    drawing. If you swap the (-) and (+) then the negative sign infront
    of the L disapears
    4) Why the current flows from (+) to (-) while electrons going the
    other way?
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