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When an AC source is connected with an inductor we have a continuous change in voltage and current across the inductor.

According to Farady and Lenz law we will get an equal and opposite voltage against the source voltage for each change of voltage or current across the inductor.

Now since we always get an equal and opposite voltage across the inductor for each change in source voltage so it means that the current will never be able to flow in the inductor but yet current is always flowing in each text book. I am confused about it. Please guide me. I shall be grateful to you. Note that it is an ideal case and we have no resistance in the circuit.

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    \$\begingroup\$ I explain it here electronics.stackexchange.com/questions/470171/… \$\endgroup\$ – G36 Mar 23 at 13:49
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    \$\begingroup\$ Also, try to read this allaboutcircuits.com/textbook/direct-current/chpt-15/… and the next "chapter" Inductors and Calculus. \$\endgroup\$ – G36 Mar 23 at 14:51
  • \$\begingroup\$ @G36 please read my answer below and also commeny on that. Thanks for your guidance. \$\endgroup\$ – Alex Mar 27 at 19:32
  • \$\begingroup\$ @G36 copy paste your answer below that you have given in the link. i was to accept this answer. \$\endgroup\$ – Alex Apr 25 at 9:32
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There are two issues with AC, how the steady state works, and how we get to the steady state from from initial zero starting conditions.

You are concerned that current will never flow. Let's look at the starting conditions of no initial current in the inductor. When we connect it to the supply, there will typically be a finite voltage across the terminals. With a voltage across the inductor, we know the current will build, at a rate given by dI/dt = V/L. So now we have a current flowing.

Let's address your ideal circuit. You have a voltage source, with no internal resistance. You have an inductor, with no internal resistance. That means you have a loop with no resistance, so the current is undefined. If you input this circuit into a simulator like SPICE, it would object, some would say 'matrix not invertible', the smarter ones would say 'zero impedance loop'.

The lack of resistance means that the current that started flowing when you first connected the inductor, the initial transient, would continue forever, without dying out. In the real world, there's always resistance (except for superconductors where the initial transient continues forever without dying out).

This means that if you want to solve for current in this particular circuit, you will have the superposition of any everlasting initial transient, and the long term steady state result. This will complicate your mental picture of what's going on.

The easiest way to resolve this is to connect the inductor at a time which gives you no initial transient. Paradoxically, this is at the peak of the AC voltage.

Connect the inductor. AC voltage is at its peak, inductor current is zero. Current starts to build, and continues to build until the input voltage has dropped to zero. As the voltage dips negative, now current starts to fall, and for a nice symmetrical waveform, current reaches zero by the time the voltage waveform is at its negative peak. That's the first half cycle done, run through the same argument swapping the signs for the second half cycle. This is one cycle of the steady state.

There are two ways of interpreting this current change with voltage behaviour that's summed up by the equation dI/dt = V/L. We could say that the current changes because of the applied voltage. We could say that the change in current generates a voltage exactly equal to that applied. Actually what happens is that both things happen at the same time. We cannot meaningfully say that one causes the other, in the way that kicking a ball causes it to move, because the moving ball certainly does not cause the kick, this cause-effect is not reversible.

For resistive inductors, we can switch on at any point in the waveform, and the initial transient will gradually be damped out by the resistance, leaving only the steady state. The higher the resistance, the faster we'll settle to the steady state. It was only in this zero resistance case that we had to start at peak voltage, to start already in the steady state.

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  • \$\begingroup\$ You said that "when you first connected the inductor, the initial transient, would continue forever, without dying out.". What would be the intial transiet condition in this circuit. Can you please explain that transient condition? \$\endgroup\$ – Alex Mar 23 at 10:23
  • \$\begingroup\$ @Alex By choosing to switch on at peak voltage, there is no initial transient. If instead I switched on at zero volts, then there would be a large DC current flowing, on average. This is why zero-volt switching ICs are so bad with transformer loads, they switch at the worst possible time. The DC dies out, but not before causing a large inrush current that often blows fuses. Don't confuse 'initial transient' with 'what happens when you switch on. The total current = steady state + initial transient. My answer chose to switch when total current was already steady state. \$\endgroup\$ – Neil_UK Mar 23 at 13:49
  • \$\begingroup\$ i am confused about the transient and steady state current in this circuit. Can you give me a link or can you explain your self about the topic little further? Because in ac all looks steady state. \$\endgroup\$ – Alex Mar 23 at 14:40
  • \$\begingroup\$ @Alex It's a linear system, so if there are mulitple solutions, a weighted sum of those solutions will also be a solution. There are two easily obtainable solutions, the steady state solution which is periodic at the AC frequency, and a transient solution which decays with a characteristic time constant which depends on the resistance. When R=0, so it does not decay. Maybe you should do a transient simulation of this circuit with a SPICE simulator, with R=1e-12 (can't be zero) and several reasonable small values for R, starting at peak and zero volts, and compare the traces. \$\endgroup\$ – Neil_UK Mar 23 at 14:57
  • \$\begingroup\$ Neil_UK check my answer. I found it good aswell besides yours \$\endgroup\$ – Alex Mar 27 at 19:19
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You have missed one thing: The induced voltage is the result of a change in the current. The current must be changing, otherwise no induced voltage exists. The current changes in a way that the voltage which is caused by the induction between the terminals of the inductor is just equal with the fed voltage (assuming resistance=0)

You seemingly think the induced voltage is there at first and then "no voltage difference, no current can be born" That's reversing the cause and the consequence. As well one could think a freely falling object cannot fall because the air resistance has grown as high as the gravity. The error is the same. Air resistance needs the motion to exist.

The induction law in ideal coils (=no resistance, no capacitance, no radiation as radiowaves) has the following quantitative form: The current changes at rate = fed voltage divided by the inductance. That's the formula for the time derivative of the current.

If you connect A volts constant DC to a coil which is ideal and has inductance = B Henrys the current grows infinitely at rate A/B. If A=1V and B=1H, the current grows one ampere per second.

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    \$\begingroup\$ You are just restating what is known by the OP without addressing the problem of how current can flow when the back emf exactly equals the applied voltage. \$\endgroup\$ – Andy aka Mar 23 at 8:43
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    \$\begingroup\$ Hmm interesting.. I think the OP is seeing two equal emf sources connected in parallel. Then the entire wire connecting the positive terminals of the voltage sources floats to constant potential. Hence he/she's arguing that no current can flow. But this is not an issue because the wire resistance is assumed to be \$0\$, so any amount of current can flow as \$\frac{0}{0}\$ can take any finite value. I'm waiting to see few more answers XD \$\endgroup\$ – beccaboo Mar 23 at 9:11
  • \$\begingroup\$ "If you connect A volts constant DC to a coil which is ideal and has inductance = B Henrys the current grows infinitely at rate A/B. If A=1V and B=1H, the current grows one ampere per second." I dont understand this. The rate of change of current will not induce again an equal and opposite voltage? Which will again stop the rate of change of current? \$\endgroup\$ – Alex Mar 24 at 6:02
  • \$\begingroup\$ "again" is a wrong idea. At the very start after connecting the wires the current growth rate is A/B and that stays until something is changed, The induced voltage is =A. Practical parts have resistance which slows the growth down. Check this electronics.stackexchange.com/questions/282053/… There's a device which has persuaded many people to change their opinions about induction. \$\endgroup\$ – user287001 Mar 24 at 8:05
  • \$\begingroup\$ Check my answer. \$\endgroup\$ – Alex Mar 27 at 19:21
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Think about what happens when you have 0 volts applied across an inductor. You might say that there isn't any current and that would be one truth from many other truths. The reality is that if there was 1 amp flowing through an inductor and the driving voltage was instantly brought to 0 volts, that 1 amp would continue to flow forever (assuming a perfect inductor).

What I'm trying to say is that you can't define the current through an inductor by the voltage across it even if the back-emf appears to make the "true" or "real" voltage zero at any particular instant in time. The current in an inductor is dependent on previous events in time.


Another way of looking at it: the back emf being equal to the driving voltage makes the effective voltage across the inductor equal to zero. But we're not talking about the external terminals of the inductor any more: -

schematic

simulate this circuit – Schematic created using CircuitLab

We're talking about an invisible untouchable node. Given there is 0 volts across this "internal inductor" the impedance has to be zero (because it is also DC hence, current is 0/0 i.e. indefinable at any instant.

This means that there is nothing useful to know about current flow in a perfect inductor based on saying the back-emf equals the applied voltage

We are left with the old tried and tested relationship, that being: -

$$V = L\dfrac{di}{dt}$$

And back-emf tells us nothing about the current that flows.

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  • \$\begingroup\$ check my answer. I found it a good one as well. \$\endgroup\$ – Alex Mar 27 at 19:17
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    \$\begingroup\$ @Alex in your answer, you said inductance is flux multiplied by current. Actually, inductance (per turn) is flux divided by current. If you have one turn then inductance is flux divided by current. \$\endgroup\$ – Andy aka Mar 27 at 20:04
  • \$\begingroup\$ What about rest of the answer? \$\endgroup\$ – Alex Mar 28 at 4:33
  • \$\begingroup\$ @Alex fix your equations - they are glaringly wrong. \$\endgroup\$ – Andy aka Mar 28 at 8:41
  • \$\begingroup\$ Thats the copy paste mistake \$\endgroup\$ – Alex Mar 28 at 10:44
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An inductor stores energy in the form of a magnetic field.

As you stated in the question: When DC voltage is connected over an inductor the rising current will be opposed by the induced back-EMF. This will only last so long as the field is increasing to its full strenght (until saturation).

So when voltage is applied, the inductor will start storing energy. In order to store energy, current will flow. When AC voltage is applied, the inductor will be charging and discharging the energy which is stored in the magnetic field.

During the positive half cycle of the source voltage, the inductor will store energy and during the negative half cycle, it will release the same energy that it previously stored.

Only when the frequency of the applied voltage is to high, the pulses are to brief to overcome the back-EMF, so the inductor will block the current. This is why inductors can be used in analog filters.

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  • \$\begingroup\$ So you mean that if the frequency is low then there will be no problem for the current to flow in the inductor? Voltage and current will have the same frequency in this case. Every time voltage switches direction current will also switch its direction. And for every change in voltage and current from the source there will be equal an back emf in the inductor to oppose the change in current. \$\endgroup\$ – Alex Mar 24 at 4:33
  • \$\begingroup\$ As the frequency increases the current will decrease. This because the impedance of the inductor is proportional to the source frequency. In this circuit the voltage over the inductor will be equeal to the source voltage. The current will have the same frequency but will lag the voltage by 90 degrees. This phase shift /delay is there because the inductor opposes every change in current. The amplitude of the current will depend on the source frequency. \$\endgroup\$ – ArjanP Mar 24 at 8:20
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For the ideal case which you are referring to there is no resistance and hence the current will be able to flow even though there is no potential difference between the voltage source and inductor.
Of course, in reality there is a finite (interconnect) resistance between the inductor and the voltage source. Because of this, the potential drop across the inductor (in sinusoidal steady state) will always be lower than the voltage source allowing the current to flow through the resistor.

In response to the comment, consider a wire with a resistance R. For a current I to flow through it, the resistor needs a potential difference V across it, which is given by, $$V = IR$$ This is because of Ohm's Law. Now, smaller the resistance, the less is the voltage required to maintain the same current.
In the limiting case of extremely small R, an extremely small voltage is required for current I.

As an alternative explanation, if R = 0, V = 0 and the equation becomes: $$0 = I * 0$$. Clearly, all values of current will satisfy this equation. So now, the wire does not decide (or bother) what current will flow through it. It is the inductor which decides the current.

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  • \$\begingroup\$ The question said: Note that it is an ideal case and we have no resistance in the circuit \$\endgroup\$ – Andy aka Mar 23 at 10:17
  • \$\begingroup\$ I have explained both the cases, ideal and non-ideal \$\endgroup\$ – sarthak Mar 23 at 10:18
  • \$\begingroup\$ "For the ideal case which you are referring to there is no resistance and hence the current will be able to flow even though there is no potential difference between the voltage source and inductor." whats meant by that? Can you please explain. How there can be current without potential difference? \$\endgroup\$ – Alex Mar 23 at 10:47
  • \$\begingroup\$ @Alex See my edits \$\endgroup\$ – sarthak Mar 23 at 11:38
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Adding to Andy's awesome answer:

velocity = displacement/time

Given the instantaneous value of velocity, can you find the displacement? No, because the displacement depends on the initial values and the history:
\$s(t) = s(0) + \int\limits_0^t v(t)\,dt\$

At any particular time, does it make sense to have some displacement but \$0\$ velocity? Yes, after driving all day you rest.


Suppose you're at rest and some giant kicks on your ass with some force \$F\$. By newton's third law you will kick back at the giant with the same but opposite force \$-F\$. And that giant continues kicking you with variable force \$F(t)\$. To honor newton, you will also continue kicking back at the giant with force \$-F(t)\$. Here at any time both forces are equal and opposite. Do you stop accelerating because you're generating your own opposite force? No because you cannot feel the effect of your force on your own. You can feel only the external forces on you, so you accelrate at a rate \$\frac{F}{m}\$.


voltage = velocity
current = displacement
time = time

The voltage source kicks the inductor. Inductor kicks back with the same but opposite emf. However the inductor cannot feel its own emf, so it allows the current to change at a rate \$\frac{V}{L}\$.

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    \$\begingroup\$ @Alex yes good question, but it's the same question. Think of the circuit as two object system: 1) voltage source and, 2) inductor. Forward emf acts on the inductor and inductor offers inertia \$\omega L\$, so a current of \$\frac{V}{\omega L}\$ flows through the inductor. Back-emf acts on the voltage source, but voltage source offers infinite inertia so no current flows in the reverse direction. In the circuit, the clockwise current is \$\frac{V}{\omega L}\$ and the counter-clockwise current is \$0\$. The total current is \$\frac{V}{\omega L} - 0 = \frac{V}{\omega L}\$ clockwise. \$\endgroup\$ – beccaboo Mar 26 at 13:19
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    \$\begingroup\$ Key thing is the back-emf doesn't act on the inductor. Similarly the forward emf doesn't act on the voltage source. Back-emf acts on the voltage source. Forward emf acts on the inductor. They're equal and opposite. But they DON'T cancel each other out because they act at different places. \$\endgroup\$ – beccaboo Mar 26 at 13:25
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    \$\begingroup\$ I think you feel that current cannot flow because potential difference is \$0\$ between the wire ends connecting the voltage source and the inductor. However this is not an issue because we're assuming the resistance of wire is \$0\$. Current = \$\frac{\Delta V}{R} = \frac{0}{0}=\text{can be finite nonzero value}\$. On snow you can have a finite velocity even when the ground is flat(height difference is \$0\$) because snow offers \$0\$ friction. \$\endgroup\$ – beccaboo Mar 26 at 13:33
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    \$\begingroup\$ Disregard the downvotes and read my answer again with confidence paying attention to where each force is acting. I'm sure you will get more clarity. Good luck:) \$\endgroup\$ – beccaboo Mar 26 at 13:40
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    \$\begingroup\$ If you want more analogies, consider me slapping on your face. Your face reacts back on my palm with the same force. Both forces are equal and opposite, yet you will feel the pain and your face moves because the reaction force offered by your face is not on your face. It is on my hand. Your face doesn't know anything about the reaction force it produced. \$\endgroup\$ – beccaboo Mar 26 at 13:56
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"The answer to your question lies in the fact that you are dealing with two different types of electric field (conservative and non-conservative) and that the non-conservative electric field owes its existence to a changing magnetic flux produced by a changing current.

The definition of self-inductance is \$L=\dfrac {\Phi}{I}\$ where \$\Phi\$ is the magnetic flux and \$I\$ is the current.

Differentiating the defining equation with respect to time and then rearranging the equation gives $$\dfrac{d\Phi}{dt} = L\dfrac{dI}{dt} \Rightarrow \mathcal E_{\rm L} = - L\dfrac{dI}{dt} $$ after applying Faraday's law where \$\mathcal E_{\rm L}\$ is the induced emf produced by a changing current.

The electric field associated with the changing magnetic flux is non-conservative.

Consider a circuit consisting of an ideal cell of emf Vs, a switch and an ideal inductor all in series with one another.

At time t=0 the switch is closed. The initial current must be zero which you can understand with an appreciation of the fact that mobile charge carriers have inertia and thus cannot undergo an infinite acceleration.

The conservative field produced by the cell is trying to increase the current from zero but the non-conservative field produced by the inductor is trying to stop the current changing. Which field wins? At t=0 there is no current so it would appear that it is a draw between the two fields, but the non-conservative field can only stop a current flowing at t=0 on condition that the current changes. So the current has to increase despite the opposition of the non-conservative field and so it continues with the current increasing due to the conservative field despite the opposition of the non-conservative field. All that the non-conservative field can do is slow down the rate at which the current changes; it can never stop the current changing as then it (the non-conservative field) would no longer exist"

By Farcher

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    \$\begingroup\$ Fix your equations, for example, \$L = \frac{Φ} {I}\$ Also, if this explanation suits you and you say "Aha I finally got it", I have no problem with that. We can view it also in this way: From Ampere's law, we know that the current in the conductor product the Magnetic field. And this change in the Magnetic field "induced voltage" in the inductor (Faraday's law). And this voltage, by definition (Lenz Law), opposes any external effort to change the existing flux (or current) in an inductor. \$\endgroup\$ – G36 Mar 28 at 15:41
  • \$\begingroup\$ I see you have finally accepted your own answer but this answer, although accurate as a piece of text, does not address the fundamental question you initially raised and that question is basically: how can current flow when the back emf exactly opposes the applied voltage. I also have issue with your math: Inductance per turn (not inductance) is flux divided by current. Also you are quoting Farcher - can you provide a better link to exactly who Farcher is or where the text came from. This is an important question so, any accepted answer MUST be scrutinized. \$\endgroup\$ – Andy aka Apr 18 at 8:34
  • \$\begingroup\$ Is it this that you copy and pasted? \$\endgroup\$ – Andy aka Apr 18 at 11:20
  • \$\begingroup\$ I undo it @Andy ake. Respect. \$\endgroup\$ – Alex Apr 19 at 22:52
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Noooooooooooooo. Of course you will have resistance ( more like impendance ), which will limit the average current ,equal to L2pf and will cause a phase shift of 90 degrees between voltage and current limiting even more power.

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    \$\begingroup\$ I didn't downvote, but the OP explicitly said this is an ideal case, which means both the voltage and the inductance have zero internal resistance. \$\endgroup\$ – a concerned citizen Mar 23 at 9:59
  • \$\begingroup\$ he said about resistance , not impedance that's completely different. \$\endgroup\$ – Jelly Strawberry Mar 23 at 10:00
  • \$\begingroup\$ no, he asked about current flowing, and sad the circuit was ideal, so no resistance was there. @JellyStrawberry again, you're lacking the basic terminology skills here. \$\endgroup\$ – Marcus Müller Mar 23 at 10:02
  • \$\begingroup\$ The question said Note that it is an ideal case and we have no resistance in the circuit. \$\endgroup\$ – Andy aka Mar 23 at 10:02
  • \$\begingroup\$ ideal inductors have impendance in an ac circuit \$\endgroup\$ – Jelly Strawberry Mar 23 at 10:04

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