inductor impedance - why no minus sign?

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

Reply to
Sean McIlroy
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You forgot the ^0.5

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Reply to
JeffM

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

Reply to
John Larkin

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

Graham

Reply to
Eeyore

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

Reply to
Sean McIlroy

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

Reply to
Jon Slaughter

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

Reply to
John Larkin
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 electr> hola
Reply to
ccon67

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