simulate this circuit – Schematic created using CircuitLab

OK, I always have a hard time to understand inductors. Now, I do know that inductors will not let current through them to change instantaneously. And the voltage induced across them is given by the formula: $$ V_{ind} = L \frac{di}{dt}. $$ Suppose, the circuit was at steady state. So the current through the inductor is 1 A at t = 0-. At t = 0, I move the SW1 to position B. Now, using the above formula, the magnitude of the induced voltage across the inductor should be 1 V. And the polarity is such that $$ V_{node\, C} = GND - 1 V = 0 - 1 = -1 V. $$ So, node C is at -1 V now, I know that the inductor will try to keep current flowing from C to GND, but the polarity forces me to think the other way around, that is current should be flowing from GND to C from both sides(also from B to C). Then I become getting confused, node C looks like a new ground, a sink to current.

So, it is clear that I am having some hard time on this concept, any help would be appreciated.


  • 3
    \$\begingroup\$ electronics.stackexchange.com/questions/470171/… \$\endgroup\$ – G36 Jun 3 at 16:35
  • 1
    \$\begingroup\$ The current will flow in the same direction as before. From C --->L---->GND---->R1--->C \$\endgroup\$ – G36 Jun 3 at 16:40
  • 1
    \$\begingroup\$ allaboutcircuits.com/textbook/direct-current/chpt-15/… and the next chapter \$\endgroup\$ – G36 Jun 3 at 16:41
  • \$\begingroup\$ @G36 I'll have a look at them, thanks. \$\endgroup\$ – muyustan Jun 3 at 16:47
  • \$\begingroup\$ node C looks like a new ground ... the ground is a chosen reference point from which to measure voltages in the circuit ... you can use node C as a reference right from the start if you want to \$\endgroup\$ – jsotola Jun 3 at 17:33

When you throw the switch, the inductor circuit is changing from "being a motor" to "being a generator" and it tries to keep the +1 amp flowing by altering the node C voltage (the only node that it can alter) to ensure that +1 amp still circulates at that instant following switch-over. The only viable voltage at node C that ensures this is -1 volt.

This forces 1 amp (at that instant) to flow through both resistor and inductor in the same direction prior to the switch changing position. The voltage clearly has to be -1 volts (node C) across the resistor to satisfy ohms law for the resistor for 1 amp flowing. This is because one side of the resistor has been connected to 0 volts by the switch changing position.

At the instant the switch changes over, you can assume the inductor to be equivalent to a constant current source of 1 amp and that means that whatever load impedance is connected across it (\$Z_{EXT}\$ = 1 ohm in your example), the voltage produced is 1 amp x \$Z_{EXT}\$. But only for that instant.

There is also one more thing that can be said at that instant; because we know that Faraday's equation is true at all times for an inductor (\$V = L\frac{di}{dt}\$) AND, because the inductor voltage has to be -1 volt, the rate of change of current is now -V/L or -1 volt / inductance. So we know the terminal voltage expressed by the inductor, the current and, the rate of change (fall) of current that will happen at that instant.

What happens hereafter is an exponentially decaying current best described by this picture: -

enter image description here

Picture taken from this slide show (Physics 121 - Electricity and Magnetism, Lecture 12 - Inductance, RL circuits)

| improve this answer | |
  • \$\begingroup\$ Ok, I think it is more clear now. Now, please assume inductor has released all its stored energy and circuit is now at SS again with zero current. I then toggle the switch at t=T let's say. For finding di/dt, I know current at T- is 0 but what it is at T+? Will I consider coil as a short circuit at T- ? \$\endgroup\$ – muyustan Jun 3 at 18:31
  • \$\begingroup\$ At the instant the switch changes over the current is still zero but initially rising at a rate of V/L (as per \$V = L\frac{di}{dt}\$). That rate will reduce over time because of the presence of R1 and instead of rising at V/L (as per a pure inductance across a fixed supply voltage) it will rise as per the well know exponential formula depicted in this picture. \$\endgroup\$ – Andy aka Jun 3 at 18:37
  • \$\begingroup\$ I think I mislead you mistakenly. The problem is how to find V using the formula L*di/dt? I need di/dt to find V induced across the inductor. di/dt is I(at t=T+) - I(at t=T-). Initial current(at T-) s 0, but for the current at T+, will I calculate it by replacing inductor with a short circuit? \$\endgroup\$ – muyustan Jun 3 at 18:46
  • \$\begingroup\$ Di/dt isn’t that formula because that formula is just centred on current with no measure of time. Are you trying to understand why the voltage reverses at time = T? I’m guessing what you might be asking because realistically, I don’t understand what your incorrect di/dt formula is affecting your thinking. If you want to know why voltage reverses, it’s because the rate of increase of current at T- becomes the same rate but negative at T+ but, I’m guessing that is what you want. \$\endgroup\$ – Andy aka Jun 3 at 20:22
  • \$\begingroup\$ why did you say the formula has no measure of time? Isn't time between T+ and T- considered as dt ? I don't have any problem with the sign/direction of di/dt, thus voltage induced. The problem is, for the first case(at t=0), the current just before toggling the switch was easy, 1 A, and just after toggling it was 0(the current pushed by rhe battery, I know the current won't decreasw to 0 because of the magnetic energy stored on the inductor, but when we want to calculate V across inductor, we used di/dt as 1A/s). So I just get confused for second case(t=T), about the di/dt ....... \$\endgroup\$ – muyustan Jun 3 at 20:37

Well both are correct. Current needs a loop to flow in. The loop contains the ground wire, resistor and inductor. 1A flows into ground symbol at one place, so 1A must flow out of ground symbol at another place. If you replace the grounds with just a wire you will see it better. Also as 1A flows via the resistor, there is 1V drop over it, and since one end of the resistor is at 0V, and given which way current flows, there must be -1V at the other end of the resistor.

| improve this answer | |
  • \$\begingroup\$ I think where I mostly have difficulty is that the inductor voltage and current are not following passive sign convention. Should I consider the inductor like a voltage source? Becuse sources also are not following passive sign convention, at least I think that way. \$\endgroup\$ – muyustan Jun 3 at 16:49
  • 2
    \$\begingroup\$ @muyustan The passive sign convention doesn't say anything about the actual direction of current flow or the actual polarity of voltage...it tells us what we must assume the direction and polarity to be if we want to calculate power. Sometimes the power for an inductor is positive, meaning that it is storing energy from the circuit and sometimes the power is negative, meaning that the inductor is returning energy to the circuit. \$\endgroup\$ – Elliot Alderson Jun 3 at 16:54
  • \$\begingroup\$ @ElliotAlderson ok, let me put it in this way, so, at t=0+ the bottom handside(according my schematic) of the inductor is + and upper handside of it is -. However, the current flows from - to + which is unintuitive to me, because it is from + to - other times. This is exactly the thing on my mind, I might be misused passive sign convention thing. \$\endgroup\$ – muyustan Jun 3 at 16:58
  • 3
    \$\begingroup\$ @muyustan You need to get used to the idea that current can flow in either direction through an inductor, a capacitor, or even an ideal voltage source. Sometimes these elements absorb power and sometimes they provide power. If you assume current and voltage that follow the P.S.C. but find that the actual value of one of these is negative then the calculated power will also be negative, meaning that the element is supplying power to the circuit at that instant in time. \$\endgroup\$ – Elliot Alderson Jun 3 at 17:03
  • \$\begingroup\$ @ElliotAlderson thanks, it helped. \$\endgroup\$ – muyustan Jun 3 at 17:10

but the polarity forces me to think the other way around, that is current should be flowing from GND to C from both sides(also from B to C).

The inductor is not a resistor, so it has no reason to have current and voltage have the same sign. In fact, since an ideal inductor does not dissipate any power, current and voltage are required to have different signs part of its duty cycle.

That point C has a voltage of -1V momentarily is not a mistake: in fact, this is how voltage inverters deriving a negative voltage from a positive one via switched inductors work.

| improve this answer | |

It sounds paradoxical but electrical circuits are best understood through non-electrical analogies. Thus, the intuitive idea of ​​inductors and capacitors is like containers that can store kinetic and potential energy.


For example, we can intuitively understand the behavior of an inductor through the mechanical property of inertia.

Imagine that you are pushing a car that can not start (you are the source, the car is the load). At the beginning there is a pressure caused by the counteraction of the car... but it gradually decreases to zero with the acceleration of the car. You let go of the car and it moves on its own.

Then you stand in front of the car and now it starts to put pressure on you (now the car is the source, you are the load). In this moment there is the same pressure as in the beginning above; then it gradually decreases to zero.

I hope this mechanical analogy will help you to understand why "+" appears at the bottom end of the inductor.

Regarding the magnitude of the self induced voltage at the moment of switching, it would be equal to the exciting voltage - I.R1 = V. So, first a voltage source V passes current I = V/R1 through the inductor; then, the inductor passes the same current I through the resistor R1 thus creating the same initial voltage V = I.R1.


In a similar way, we can intuitively understand the behavior of the dual electrical storing element - the capacitor, through the mechanical property of springiness.

Imagine that you are pushing a spring (you are the source, the spring is the load). At the beginning there is no pressure caused by the counteraction of the spring... but it gradually increases to maximum with the compression of the spring. You tie the spring and it stands in a tense state.

Then you release the spring and now it starts to put pressure on you (now the spring is the source, you are the load). At this moment, there is the same movement as in the beginning above; then it gradually decreases to zero.

Regarding the magnitude of the current at the moment of switching, it would be equal to the initial current created by the exciting voltage V by the help of the resistor R1 (I = V/R1). So, in the very beginning, the voltage source V passes current I = V/R1 through the capacitor; then, at the moment of switching, the capacitor applies the same initial voltage across the resistor R1 thus creating the same initial current I = V/R1.

Inductor vs Capacitor

Finally, let's compare the behaviour of the two dual electrical storing elements after removing (zeroing) the source:

The inductor continues passing the current in the same direction while the capacitor passes the current in the opposite direction through the "source".

As if the inductor is "grateful" to the source and continues acting in the same direction while the capacitor is not "grateful" and "cheats" on it by opposing:)

| improve this answer | |
  • \$\begingroup\$ I think what confuses me is that I think in such a way that the current has to be entering the inductor from its positive terminal all the time. This makes the contradiction in my head. I assume I have to understand that at the switching moment, when inductor starts to act as a source, I should evaluate sign convention as I do for batteries, i.e, the other sources well known. So, current leaves from + terminal of a battery, so it is normal to leave not enter from the + terminal of the inductor when it is sourcing energy to the circuit. \$\endgroup\$ – muyustan Jun 9 at 17:32
  • \$\begingroup\$ This was my final evaluation about this question. \$\endgroup\$ – muyustan Jun 9 at 17:32
  • \$\begingroup\$ Exactly... At this moment, the inductor transmutes from a load to a source because it has accumulated "kinetic energy" and now begins supplying the load... \$\endgroup\$ – Circuit fantasist Jun 9 at 17:38
  • 1
    \$\begingroup\$ My strength is not in the calculation but in the intuitive explanation:) \$\endgroup\$ – Circuit fantasist Jun 9 at 18:10
  • 1
    \$\begingroup\$ Thanks for your efforts \$\endgroup\$ – muyustan Jun 9 at 18:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.