# 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 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 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 at 18:48

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\$$.