Inductor Voltage Drop

Thread Starter


Joined Feb 13, 2009
I feel like a complete idiot. I'm a first year grad student and I still don't understand inductors. My question is about the sign of the voltage drop across an inductor.

Consider the gorgeous ascii art circuit below, where the C's are meant to be the inductor and the > indicates the direction of current.

Va --->---CCCCC------ Vb

Now everyone knows that the voltage drop across an inductor is

Va - Vb = L di/dt.

So according to this equation, if the current is increasing in the direction shown, the voltage drop (Va-Vb) will be positive. But I thought that an inductor was supposed to oppose a changing current. By that logic, when the current increases in the direction shown, shouldn't the inductor induce an opposite voltage (Vb > Va), in which case Va - Vb would be negative.

I'm sure there is a simple answer, but I've been pulling my hair out trying to resolve this.



Joined Nov 9, 2007
Does this sketch help?

If you start at point A and work your way around the loop you will see that what you have called the voltage drop across the inductor is in fact a voltage rise so all the signs are correct as are the polarities.


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I think, the confusing part is that when current through inductor is increasing, the voltage drop across inductor is positive (Va > Vb). It makes a feeling that this positive voltage, in turn, will try to increase the current, so that inductor is not opposing further current rise.

In fact, you have to look at this voltage from the point of view of external current source. The higher positive voltage across inductor, the more it opposes external current source to provide more current to the inductor. In this sense, inductor is trying to maintain current constant by increasing the voltage across the inductor, thus making more "resistance" for external source.

This is just another explanation of what was said in previous replies.

Thank you.


Joined Feb 4, 2008
Imagine that the inductor is a battery with its polarity the same as the voltage across the inductor. Then think in which way has the battery to be placed as to oppose the current flow. I hope you understand batteries. ;)
Note that an inductor is not a battery but this will help you understand.

Thread Starter


Joined Feb 13, 2009
What you have all said makes sense. But I thought that current flowed from higher to lower potential. Thus if the voltage (Va-Vb) increases, wouldn't that be inducing a current in the same direction not in the opposing direction?
Current flows from higher to lower potential in the resistor. In the inductor it can flow in any direction even if voltage = 0. Higher and lower potentials define only rate of change of the current.
By the way, similar situation with capacitors, just change V->I and I->V. The current defines rate of change of the voltage, not the voltage itself. But capacitor is easier to understand, since it has direct analogy with, for instance, tank which is filling in with water. For inductor, there is no such analogy (may be, inertia of moving water?).


Joined Apr 9, 2009
I thought an inductor induces a voltage that counteracts the source voltage, and as a result, induces a current that opposes the source current. So you can think of the overall voltage drop across the inductor as the sum of two separate voltages: the voltage applied by the source and the counter voltage (smaller) induced by the magnetic field. When you add these two together, you get the overall inductor voltage.
Although the induced current may seem to flow in the "wrong" direction, as compared to the polarity of the overall inductor voltage, it does flow from + to - of the induced voltage (which is a counter voltage).
You can think of the total current flowing through the inductor as the sum of the source current and the induced current (smaller). The induced current opposes the source current, so it slows down the overall inductor current. Hence the di/dt.
Does that make sense? (I've been wrestling with inductors for days.)