# What is the value of the opposing voltage created by the inductor?

Assume I've connected an inductor to a DC or AC source. The voltage across the inductor due to the source current would be L di/dt. But we know that the inductor induces/creates a voltage to oppose the source voltage. My question is, what is the value of this opposing voltage. Surely it can't be L di/dt because then the opposing voltage and source voltage would be equal thus making current to be zero, right?

• Instantaneously, yes, it is equal and opposing. Then the opposing voltage decays exponentially, allowing an exponentially increasing current flow. – TonyM Aug 20 '19 at 11:11
• So intially the opposing voltage is the same as the applied voltage right? What then causes the opposing voltage to decay? – noorav Aug 20 '19 at 13:05
• As I often politely post when welcoming newcomers to the site, this is not an on-line technical encyclopedia, copied out to you on demand. The information you're asking for is freely available and well detailed on the interweb. It's all there waiting for you :-) People will help you take the next step if your question shows that you've done as much as you possibly could on your own. – TonyM Aug 20 '19 at 18:48

The current is zero but only instantaneously. Inductors give graphs of voltage that upward slope as DC voltage surges and then gradually reaches its set level (which it technically never reaches but gets continually closer). The graph starts at 0V and then ascends. It doesn't immediately leap up to max voltage like a square wave

If you connect an ideal inductor across an ideal DC source, the current is initially zero, and rises linearly with time. The voltage, of course, is the applied DC voltage, so the current rises as di/dt = V/L amperes/second.

With an inductor that has some DC resistance, the current does not increase indefinitely but exponentially approaches the applied voltage divided by the resistance with time constant $$\\tau = L/R \$$. If it's a superconducting coil that has zero DC resistance, it increases linearly until it hits the critical current and then resistance appears.

If sinusoidal AC is applied, the steady-state current is the voltage divided by the reactance, which is $$\X_L = \omega L\$$.

• So intially the opposing voltage is the same as the applied voltage right? What then causes the opposing voltage to decay? – noorav Aug 20 '19 at 13:05
• It stays the same (by definition because you've fixed the voltage with your external source) and the current increases. – Spehro Pefhany Aug 20 '19 at 14:00
• Oh so if the opposing voltage remains same and the current increases, does it mean that the resistance or impedance is decreasing? – noorav Aug 20 '19 at 14:38
• There's no resistance in the first case, and I don't think that "impedance" is the best way to think of it, unless the applied voltage is sinusoidal and the steady state (after things have settled down) is what you're after. The frequency components of a unit step are a bit unpleasant and not very intuitive- at least to me (a couple Dirac delta/unit impulse functions). – Spehro Pefhany Aug 20 '19 at 16:00