Why is the voltage across an inductor negative?

[Drakkith]

I think I must have missed something when going over inductors for my electric circuits class.

The voltage across an inductor is described by the equation: $V(t) = L\frac{di(t)}{dt}$
For the case where current is flowing through the inductor until $t=0$, at which point the voltage/current supply is switched off, the voltage then takes the form: $V(t) = L\frac{de^{\frac{-t}{\tau}}}{dt} = -L{\tau}^{-1}e^{\frac{-t}{\tau}}$

My question is, why is this voltage negative? Wouldn't that oppose the current flow? Obviously I'm not understanding something, as this appears to contradict the fact that an inductor opposes changes in current. I thought the voltage would be a positive number, so as to continue current flow in its current direction.

 

[timthereaper]

As dumb an answer as this is, I think it's just a consequence of Lenz' law:

From Wikipedia:

If the current is decreasing, the induced voltage will be negative at the terminal through which the current enters and positive at the terminal through which it leaves, tending to maintain the current. Energy from the magnetic field is being returned to the circuit; the inductor is said to be "discharging".​

 

[Drakkith]

As dumb an answer as this is, I think it's just a consequence of Lenz' law:

From Wikipedia:

If the current is decreasing, the induced voltage will be negative at the terminal through which the current enters and positive at the terminal through which it leaves, tending to maintain the current. Energy from the magnetic field is being returned to the circuit; the inductor is said to be "discharging".​

Hmmm. Well, I just now realized that since the inductor is providing voltage to the circuit, it acts like a voltage source. As such, the current would be flowing from the positive terminal to the negative terminal of the inductor, which is exactly how it works for a voltage source.

 

[berkeman]

The voltage across an inductor is described by the equation: V(t)=Ldi(t)dtV(t) = L\frac{di(t)}{dt}
For the case where current is flowing through the inductor until t=0t=0, at which point the voltage/current supply is switched off, the voltage then takes the form: V(t)=Lde−tτdt=−Lτ−1e−tτ

Can you show a circuit where you see the voltage supply being "switched off"? Often in basic circuit analysis of inductor circuits, a switch will be shown to open circuit the connection to the source supply. In those cases, you can get very large voltage transients, limited only by the parasitic capacitance of the circuit. (Or the explicit capacitance in circuits like flyback transformers for generating high voltages).

http://www.global.tdk.com/techmag/electronics_primer/img/img_electronics_primer_inductor_vol2_1.gif

 

[Drakkith]

Can you show a circuit where you see the voltage supply being "switched off"? Often in basic circuit analysis of inductor circuits, a switch will be shown to open circuit the connection to the source supply.

Sorry Berk, that's what I meant. I didn't know quite how to word it though.

 

[Averagesupernova]

If you do a search you will find this has been discussed already on this forum with some members giving some pretty good ways to get your mind wrapped around what is happening. If you think that the inductor 'does what it has to' in order to maintain the status quo concerning current flow it is very helpful.

 

[cabraham]

Hmmm. Well, I just now realized that since the inductor is providing voltage to the circuit, it acts like a voltage source. As such, the current would be flowing from the positive terminal to the negative terminal of the inductor, which is exactly how it works for a voltage source.

Inductors behave as current sources, not voltage sources.

 

[Drakkith]

Inductors behave as current sources, not voltage sources.

Indeed it does.