# inductor impedance - why no minus sign?

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

1. ### Sean McIlroyGuest

hola

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?

peace
stm

3. ### John LarkinGuest

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.

John

4. ### EeyoreGuest

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

Graham

5. ### Sean McIlroyGuest

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.

peace
stm

6. ### Jon SlaughterGuest

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
value.

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

then

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
phase:

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 LarkinGuest

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.

John

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?