DC voltage across the inductor?

[php]

what happens when u apply DC voltage across the inductor? i.e is the voltage drop only across the series internal resistance of the coil? can we just determine this resistance by simply saying: V_supply = I x r(series)? or do we have to include the inductance?

 

[Ouabache]

hint: Do you know the impedance of an inductor?

 

[php]

hint: Do you know the impedance of an inductor?

whats an impedance?
do you mean X_L=wL?

 

[cabraham]

V = Ri + L*di/dt

i = (V/R)*(1 - exp(-Rt/L))

If R=0 (superconducting), then

i = Vt/L

Claude

 

[Ouabache]

Before we get too far along, I hope you have more in your circuit besides the inductor, otherwise what do you think would happen when you apply a voltage across a coil of wire?

whats an impedance?
do you mean X_L=wL?

You're on the right track, however you've given only the inductive reactance.
The impedance Z of an inductor includes both real (pure resistance) and imaginary terms (reactance).
$Z = R + jX_L = R+ j \omega L$
So the voltage across it would be? $V = IZ = I (R + j \omega L)$

Since you are applying only a DC voltage (no frequency component), what happens to the reactance term?

 

[fizz_it]

Most likely the wire will get hot. You are right, at DC the resistance is only the series resistance of the wire.

If you were to apply an AC signal then it is different. As you increase the frequency the perceived resistance to the AC signal would increase. Many small signal inductors are rated at 100MHz. The rating will be something like 300 ohms at 100MHz meaning that to the source, the line the signal is traveling on will look like a 300 ohm resistor. This is in magnitude impedance format, which is the perceived resistance. At slightly over DC the perceived resistance will be near zero ohms, as the frequency increases, so will the perceived resistance. There is a point at which the resistance will reach a maximum and then it will go back down

There are a few simplifications here but the concept is sound.