inductor impedance - why no minus sign?

You forgot the ^0.5

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

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

Graham

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

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

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