# How does current flow in a purely inductive circuit if the net voltage is zero?

Considering the equation,

$$E=−L\frac{di}{dt}$$

The negative sign in the above equation indicates that the induced emf opposes the battery's emf.

If we're talking about a purely inductive circuit, the induced emf is equal and opposite to applied emf. Isn't it just like two identical batteries in opposition?

If that's the case, how does the current flow?

• No. An inductor is akin to a current source, a battery is a voltage source. Jun 12, 2022 at 21:43
• Does this answer your question? If induced voltage (back-emf) is equal and opposite to applied voltage, what drives the current? Jun 12, 2022 at 22:12
• @hyportnex changing magnetic field produces an electric field and electric field has some relation with potential difference i.e $\delta V=E.r$. So it won't be incorrect to say that inductor is a secondary voltage source. Jun 12, 2022 at 22:43
• > "the induced emf opposes the battery's emf" Yes but only if the inductor is connected to the battery with no other devices. But the formula is general and the sign says induced emf in the inductor opposes the change of current in the inductor, irrespective of where it comes from, whether from battery, or from capacitor, or other source or energy, or even from disconnecting the inductor. Jun 12, 2022 at 23:20

How does current flow in a purely inductive circuit if the net voltage is zero?

The problem in this question is that it is based on a completely wrong assumption. This concept of “net voltage” isn’t really a thing. In fact, by Kirchoff’s voltage law your “net voltage” is guaranteed to be zero. So the net voltage being zero does not imply anything about the current.

Isn't it just like two identical batteries in opposition?

No, an inductor is not like a battery. A battery has a voltage that is independent of the current. An inductor has a voltage that is proportional to the change in the current. (A capacitor has a voltage that is proportional to the integral of the current) They are not the same, and having them with opposite voltages does not imply any cancellation of current.

• the same thing applies even when only a capacitor is used,we get applied voltage = voltage across the capacitor as they are connected by a wire,the curent still flows in that case,how do we clarify that situation,thanks. Jun 13, 2022 at 16:41
• I added a sentence that covers the capacitor case. Since it isn't part of the original question I put it in parentheses. The two main points remain: (1) your "net voltage" idea is always zero per KVL, regardless of the devices used (2) net voltage being zero does not imply any cancellation of current since different devices have different voltage-current relationships
– Dale
Jun 13, 2022 at 16:47
• so if we use kirchhoff's laws,e - v = ir,here i is finite but r is 0,is this assumption correct or something is still there here. Jun 13, 2022 at 20:44

When the source voltage is suddenly made zero then the current will be decreasing at some rate and if an ideal inductor is present in a circuit then that decaying current will cause the magnetic flux to decrease with time through the loop of that inductor.

And In accordance with faradays law if the magnetic flux through a conducting loop is changing with time then there should be an induced EMF in that inductor and that EMF will be induced such that it will support the decaying current (Lenz Law) and this induced EMF causes the current not to decay quickly so, that is why you find the current is still present in the circuit even after the removal of source voltage.

• in the books,its written e=v at every point of time as the inductor and source are connected by a wire,how that is explained in simple explanations not by mathematical equations,thanks. Jun 12, 2022 at 20:21
• @sachin It will be very helpful if you clarify in your book what are these $e$ and $v$ used for. Jun 12, 2022 at 20:29
• e stands for source voltage and v the back emf. Jun 12, 2022 at 20:31
• Some statements are bit confusing in your comment. But still If you are saying that the condition $e=v$ holds for every point of time for a constant source voltage then maybe they are trying to indicate that the current is now flowing steadily without variation and hence the source voltage will be equal to the potential difference between the ends of that inductor. It might be the case that I interpreted your question in comment incorrectly due to lack of clearance so that's why I'm saying to elaborate it in depth. Jun 12, 2022 at 20:47
• @sachin I got something for you and I think this might help you to understand it better if you are trying to ask this. here is that question following with answer to that physics.stackexchange.com/questions/184541/… Jun 12, 2022 at 21:11

If we're talking about a purely inductive circuit, the induced emf is equal and opposite to applied emf. Isn't it just like two identical batteries in opposition?

If that's the case, how does the current flow?

It isn't, because two real batteries of same emfs acting against each other produce equals emfs whether there is current changing or not changing, and the fact they are real means they have non-zero internal resistance $$R$$. Then second circuital law from Mr. Kirchhoff states

$$\mathscr{E} + (-\mathscr{E}) = 2RI$$

which implies zero current, $$I=0$$.

When a real inductor is connected to a battery, current will flow through the circuit, because there is no static equilibrium like above; the current has to change in time in order for the induced EMF to exist, so initial change of current after the connection is made, is further maintained as the current increases. Because the real inductor also has some internal resistance $$R_c$$ (let's ignore real inductor's capacitance for now), the second circuital law from Mr. Kirchhoff states

$$\mathscr{E} - L\frac{dI}{dt} = (R + R_c)I$$ which does not imply $$I=0$$. In this case, induced EMF does not completely cancel battery's emf, because there needs to be some emf remaining to push the increasing current against the resistance forces in the circuit.

In the case of a perfect inductor and perfect battery, both resistances would be zero and we would get

$$\mathscr{E} - L\frac{dI}{dt} = 0.$$ which still would not imply $$I=0$$; instead, current would keep increasing indefinitely.

• in the overall question,the cases are taken to be perfectly ideal that is without any resistance,your approach takes the question from purely inductive to purely LR circuit,so things are not geting clear,thanks. Jun 13, 2022 at 16:57
• @sachin We cannot just assume that something certainly would happen with ideal batteries of zero resistance, or ideal inductors. We have no experience with such devices. Instead, we have to derive from experience (which is always with non-zero resistance), what is the expected behaviour as the resistance decreases towards the zero. This is what my answer points to. We can see that in the case of the inductor with internal resistance connected to a battery, there is a well-behaved unique non-zero current due to the battery, which increases in time, and this is the case even if $R=0$. Jun 13, 2022 at 18:43
• If you do not understand or like this answer, I recommend studying Kirchhoff's circuital laws. Then re-read this answer. Jun 13, 2022 at 18:45

There is a perfect mechanical analogy. If we push an idle car in a flat ground, we feel a force $$F = -m\frac{dv}{dt}$$ where $$m$$ is the mass of the car and $$v$$ its velocity. Of course, at $$t=0$$, the velocity is zero, but nevertheless its derivative is not zero.

The situation of two opposing bateries is static, like two sumô fighters pushing each other, while staying at the same spot.

• analogies are like study of history through stories,will be grateful if we stick to scientific approaches,thanks. Jun 13, 2022 at 17:00
• I agree. But in this case, the math (the differential equation) is the same. Jun 13, 2022 at 20:51