Voltage drop across an ideal inductor

Question

I was afraid to type the title of the question though there are lot of answers to similar questions. I spent half a day going reading these questions and trying to understand the answers posted and other sources on the internet and remaining half of the day trying to understand the difference between EMF and potential drop which branched off from the earlier questions as I felt it is crucial to understand the difference. All I was able to understand is "EMF is produced by anything other than electrostatic field while potential drop is produced due to electrostatic field". These are the list of questions which I found a bit useful:

1. Voltage drop across an inductor
2. How can we derive the polarity of the induced voltage in an inductor?
3. How to determine the polarities of inductor?

But, I feel each of these questions have different explanations and I feel hard to conclude anything from these. Finally, I decided to understand only the ideal inductor (as most problems deals with only ideal inductors) and I just want to know how to find voltage produced across an inductor in positive current direction and how to apply it in simple circuit analysis problems. I would be thankful If I could get an answer to the question, "Why voltage drop is negative of induced EMF for an ideal inductor".

Answer 1

I just want to know how to find voltage produced across an inductor in positive current direction and how to apply it in simple circuit analysis problems.

In circuit analysis, you can simply view an ideal inductor as a component with this behavior:

$$\frac{dI}{dt} = \frac{V}{L}$$

There is no need to worry about what magnetic field is found within the inductor, which direction the coil is wound, or even what physical mechanism produces the emf across the device.

You combine this behavioral rule (also called a "constitutive relationship") with Kirchhoff's Current and Voltage Laws and the constitutive relationships for the other components in the circuit to find the voltage across or current through the inductor and the other devices in the circuit.

Why voltage drop is negative of induced EMF for an ideal inductor

I'm not clear on where you got this claim, or how you are interpreting the sign of the induced emf.

If it helps, you can consider that the inductor is an energy storage device.

If you have an inductor with initial current of 0 A, then you know it is discharged and not storing any energy in its magnetic field. If you apply a voltage that is positive between its left terminal and its right terminal, you know that it will start to store energy in the inductor. That means that you must be delivering electrical energy to the inductor. In order to do that, (a short time after you apply the voltage) current must be flowing from left to right through the inductor (from the higher-voltage terminal to the lower-voltage terminal).

If you were to apply positive voltage to the left terminal and get current from right to left, that would imply the inductor is providing energy to the rest of the circuit. This does happen when the inductor is previously charged and is discharging into the circuit. But as this happens the magnitude of the current will fall. The derivative of the current will still be positive if current is taken as a negative value, with reference direction from the higher-voltage terminal to the lower voltage terminal.