# Why doesn't the potential difference across an inductor increase over time?

For school, I was trying to solve a question that features a circuit with a resistor, an inductor, and a battery all connected in series. It then shows an increasing concave down graph of something vs. t - the graph started at the origin and had a horizontal asymptote at some positive value of x. The "something" could be:

A. The potential difference across the resistor

B. The potential difference across the inductor

and/or C. The current in the circuit.

I had reasoned that it would be all 3, because when an inductor is first connected to a battery at time t=0, it doesn't allow electricity to flow, but as time goes on, it allows more and more electricity to flow until it is essentially acting like a wire. Therefore, current will increase, and voltage will also increase across the entire circuit. However, when this problem was graded, it turned out that B was incorrect, and I was not told why. Why is this the case?

• "voltage will also increase across the entire circuit" This is a curious statement. What do you mean by "across the whole circuit"? I ask this because it's directly relevant to the question you're asking. May 12 '18 at 13:59
• @PhilipWood: as you may know, many engineering undergraduates in the U.S. must take one or sometimes two courses in electricity even if their major is not electrical engineering. As it happens, I have instructed such courses a few times, to about 500 undergraduates in total. In my experience, not a few such undergraduates will speak of voltage or tension in such words: "voltage across the entire circuit." They tend to speak thus just after I have corrected them to say "voltage across" rather than "voltage through," so I believe that a common mode of misunderstanding is implicated here.
– thb
May 12 '18 at 17:51
• @PhilipWood: the only reliable cure I know for such misunderstanding is to assign homework exercises in Kirchhoff's voltage law (KVL). In my experience, KVL exercises don't actually take hold on the first go, so one must occasionally return to them over a period of months. All I can say is that, eventually, the confusion dissipates. OP is likely in that middle stage of misunderstanding now. To OP: keep after it; heed the answers here; they're good ones.
– thb
May 12 '18 at 17:56
• @thb Yes indeed. I taught Physics for a long time in the UK, and am the first to acknowledge that 'voltage' is a very difficult idea, and understanding of it develops gradually. May 12 '18 at 18:10
• @PhilipWood: I see. You were ahead of my comments all along.
– thb
May 12 '18 at 18:57

We know that an inductor resists change in current and does so by creating a potential in opposition to current change; the quantity known as inductance is the ratio between voltage across the inductor's terminals and rate of change of current through it.

So, beginning with a steady state open circuit (zero current flow), when your circuit is closed there will be a voltage applied by the battery across the series combination of inductor and resistor. Since the inductor will oppose any change in current flow, the battery voltage will appear across the inductor, the voltage across the resistor will be zero and current through the circuit will remain zero... but only for an instant.

Current will begin to flow through the circuit, rising as a function of voltage and inductance; since it is flowing through the resistor, the voltage across the resistor will increase in proportion to the current through it. This will reduce the voltage across the inductor, which will slow the rate at which current will increase through the circuit, slowing the rate at which the voltage across the resistor increases and the voltage across the inductor decreases. This process continues along an exponential curve, asymptotically approaching a steady state where the voltage across the inductor has fallen to zero, all of the battery voltage appears across the resistor, and circuit current is battery voltage divided by resistance.

To summarize a few things:

• Circuit current rises exponentially from zero, asymptotically approaching a constant value defined by battery voltage and resistance
• Voltage across the resistor remains in direct proportion to current through it, so rises exponentially from zero, asymptotically approaching battery voltage
• Voltage across the inductor starts at battery voltage and falls exponentially, asymptotically approaching zero

The above is true for an ideal circuit (inductor has zero resistance). A real-world inductor will have internal resistance, and would be modeled for the purposes of analysis as an ideal inductor in series with an ideal resistor.

Therefore, current will increase, and voltage will also increase across the entire circuit.

It's not clear exactly what you mean by "across the entire circuit", but think about your model of a battery.

For a simple analysis that is usually used for circuits like this, the battery is considered as a constant voltage source. Therefore, by Kirchoff's Voltage Law,

$$V_b = V_l + V_r$$

where $V_b$, $V_l$ and $V_r$ are the voltages across the battery, the inductor, and the resistor respectively (and when you choose the same sign convention I did for each of them, but since you didn't bother to include a circuit diagram in your question, I don't feel obligated to provide one to indicate the sign conventions in my answer).

So if the voltage across the resistor increases, the voltage across the inductor must decrease.

The voltage across the inductor is proportional to the rate of change of current which is a maximum at the start when the current and the voltage across the resistor are zero.

As time progresses the current increases at a slower rate as does the voltage across the resistor but the voltage across the inductor decreases.

After a long period of time the current is constant and so the voltage across the inductor is zero.