# Inductor behavior connected directly to DC (no resistor)?

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1. Oct 10, 2015

### ME77

If an ideal inductor is connected directly (no resistor present) to a DC voltage source, why does it behave like a short-circuit at time t=0? If voltage across an inductor is related to the change in current (i.e. di/dt), then shouldn't the voltage across the inductor be very large at time t=0 because the current goes from zero to some value instantaneously (assume no resistor is series). This would imply that the inductor should act like an open-circuit at time t=0. Wouldn't the inductor oppose the initial current spike and prevent the initial current from flowing? I know that this is not true; however, I am trying to determine why not.

2. Oct 10, 2015

### Staff: Mentor

It is neither a short nor an open. Can you write the equation that relates the current through the inductor to the voltage across it?

3. Oct 10, 2015

### ME77

berkeman - The equation is I(t) = Vc/R * (1-e^(-t*L/R)). Using the equation, I understand that the current becomes infinity (or indeterminate) when R=0. However, I am trying to understand what physically is happening. If R=0, does a magnetic field still form when current flows through an ideal conductor? Thanks.

4. Oct 10, 2015

### Staff: Mentor

I had a more fundamental equation in mind:

$$v(t) = L \frac{di(t)}{dt}$$

That's the way I think about inductors. What happens when you apply a step voltage across an inductor?

BTW -- the current through the ideal inductor only becomes infinite in infinite time. It is a linear ramp from t=0 to whenever. And with real inductors, the current is limited by the inductor's series resistance (check the datasheets for this number) and the source voltage.

5. Oct 11, 2015

### meBigGuy

It is true and your thoughts to the contrary are incorrect. Why do you think the ideal inductor conducts at t=0? It starts with 0 current and ramps linearly while building a magnetic field.