# Questions about how an inductor works

Hi all I am confused with the inductor working, I want to understand it in very much detail. Pl see the following image, First Thing : Here we have an initially uncharged inductor L in fig 1 with a switch S and a const. voltage source of voltage V.
Now in fig 2 let the switch is closed at t=0, current flows and a magnetic field sets-up, this magnetic field produces back emf and so current stops at t=0+ and the inductor behaves as open circuit

NOW THE CONFUSIONS

the texts say that inductor doesn't allow sudden change of current, OK, then if the inductor was uncharged its i(0-) = 0 and so i(0) = i(0+) = 0 so how the magnetic field sets -up ? and how the back emf produced ? and who stopped the current ?

and texts say at t=0 voltage across inductor (back emf) will be V but from equation V(inductor) = L(di/dt) now if there was back emf, there must be flux change so there must be a sudden current change so di/dt = infinity so V(inductor) = infinity not V

I am deeply confused

Pls help me

#### Attachments

scottdave
Homework Helper
Your image did not show up for me. Does your textbook have a chart showing the voltage or current waveforms for an inductive circuit? That would provide some answers that I think you are looking for.

Baluncore
When connected, current starts to flow and increase. The magnetic field is proportional to the current.
The fundamental equation is; V = L ⋅ di / dt.
The energy stored in the magnetic field will be E = ½ ⋅ L ⋅ I2

The maximum current that might be reached over time will be limited by the series resistance in the circuit.
You have not shown that as a resistance in series with the inductor.

When you finally open the switch, where will the current and it's magnetic energy go?
It will produce a high voltage spike that will destroy the switch.
You will need a flyback diode to prevent the infinite voltage.

Your confusion arises, I think, from your attempts to model an ideal inductor with real-world values.

A) An ideal inductor has inductance only. If you go on this path, you have to accept things like infinite voltages and rates of change of current. But it’s a better starting place for understanding.

B) A real inductor has series winding resistance and parallel winding capacitance (and the connecting wires have resistance, the battery internal resistance...etc). This is far more complex to model, but the values calculated are more digestible.

You are trying to reconcile an inductor from A with the values from B.

• jrmichler
jim hardy
Gold Member
Dearly Missed
Now in fig 2 let the switch is closed at t=0, current flows and a magnetic field sets-up, this magnetic field produces back emf and so current stops at t=0+ and the inductor behaves as open circuit
That's just plain wrong.
Back EMF is in proportion to rate of change of current,
So as soon as voltage gets applied, current commences to change.
Its rate of change is in proportion to applied voltage.
Constant voltage results in constant rate of change of current.
Draw a graph. Current is neither zero nor constant, it rises from zero with constant rate of change (aka slope)

the texts say that inductor doesn't allow sudden change of current, OK, then if the inductor was uncharged its i(0-) = 0 and so i(0) = i(0+) = 0 so how the magnetic field sets -up ? and how the back emf produced ? and who stopped the current ?
Go back to your basic calculus - what's the time integral of a constant ?

V(inductor) = L(di/dt)

so iinductor = ∫Vdt X 1/L

you need to make that integral-derivative relation between volts and flux&current intuitive. Work on that . You'll need it when you get to AC circuits.

old jim

• Merlin3189, Fisherman199, Paul Colby and 1 other person
This is what you said
First Thing : Here we have an initially uncharged inductor L in fig 1 with a switch S and a const. voltage source of voltage V.
Now in fig 2 let the switch is closed at t=0, current flows and a magnetic field sets-up, this magnetic field produces back emf and so current stops at t=0+ and the inductor behaves as open circuit

What you should have said is At t=0 current begins to flow and builds up a magnetic field. At maxium current the field is static and the current continues to flow.
The problem is here there is no maxium current so it becomes infinite as you have drawn the circuit. The current never stops flowing ( perhaps you were thinking of a capacitor)

CWatters
Homework Helper
Gold Member
You can summarise what happens as follows.

At t <0 the current is zero.
At t=0 the current is zero.
At t=0+ the current increase from zero...

This is because...

Vapplied causes a changing current (di/dt = V/L).
A changing current (di/dt) causes a changing magnetic field (dB/dt)

You don't really need to think about the back emf but if you insist..

A changing magnetic field (dB/dt) produces a Voltage Vback = Ldi/dt

It's similar to the way a DC motor produces a back emf. See also Lenz's law.

• Fisherman199 and jim hardy
Hi all yours fruitful comments are greatly appreciated thanks a lot, but my confusion is that as CWatters and jim hardy said that "At t=0+ the current increase from zero..." and "So as soon as voltage gets applied, current commences to change." seems to violate the sentence that inductor doesn't support sudden change of current"

also

if the Vapplied causes a changing current (di/dt = V/L), and Back EMF is in proportion to rate of change of current so Vback = Ldi/dt = L x (V/L) = V so by KVL / Ohms law => iinductor should be = (Vapplied - Vback)/R = (V - V)/R = 0

and

you are saying "So as soon as voltage gets applied, current commences to change."

????

pls explain me like a new born baby from ground up even the finest thing pls in terms of magnetic field, flux, charge etc in detail step-by-step
pls

Hi all yours fruitful comments are greatly appreciated thanks a lot, but my confusion is that as CWatters and jim hardy said that "At t=0+ the current increase from zero..." and "So as soon as voltage gets applied, current commences to change." seems to violate the sentence that inductor doesn't support sudden change of current"

A changing current, and a "sudden change in current" are two different things. Current in an inductor will obviously change, but it does so with a defined slope di/dt. An inductor will NOT perform a step change in current, like it won't go from 0 to 10A in zero time, much like a capacitor will not do a step change in voltage, but you can change the voltage on a capacitor with a defined slope dv/dt.

An inductor will however do a step change in voltage, much like a capacitor will do a step change in current.

• Merlin3189 and jim hardy
Yes, it immediately begins to change from zero at T0. The inductor stores energy in a magnetic field, proportional to the current in the inductor, think of energy as a "quantity of something". Analogy to matter : When we have an empty beaker, and we begin to pour water into it, at T0, there is zero water, but the RATE of change of water in the beaker is a positive number. You can not instantly make the beaker have 1 liter, it can "instantly" see a fill rate of 1 L/ Second. The rate can change "instantly", not the quantity.

In an ideal inductor model (electrical circuit analysis) , we only consider the Inductance (L) and essentially nothing else ( but the necessary non-ideal element we typically use is resistance of the windings which can be exceptionally low). We do not consider flux or the M field, and as such there are a wide variety of ways to make an inductor with the same inductance.

So Vemf = L * dI/dt

For physics / mechanics energy related analogy, we can look at kinetic energy and a Mass. We can not instantly change the Ek of a Mass. For example T0 Velocity of M is 0 , and T+ it is 100 x/t.... we must accelerate the mass from speed zero. However - at the instant we apply a force, the mass begins to move.

we have the same mathematical relationship:

Force = Mass * dv/dt

CWatters
Homework Helper
Gold Member
Hi all yours fruitful comments are greatly appreciated thanks a lot, but my confusion is that as CWatters and jim hardy said that "At t=0+ the current increase from zero..." and "So as soon as voltage gets applied, current commences to change." seems to violate the sentence that inductor doesn't support sudden change of current"

Ok so I know it's not quite the same but...

Cars have inertia so they cant manage an instantaneous step change in velocity but clearly they can mange to change velocity.

jim hardy
Gold Member
Dearly Missed
"So as soon as voltage gets applied, current commences to change." seems to violate the sentence that inductor doesn't support sudden change of current" you cannot cause an instantaneous change of current through an inductor. It would require infinite volts.

again go back to your basics of derivative and integral .

#### Attachments

• CWatters
CWatters
Homework Helper
Gold Member
if the Vapplied causes a changing current (di/dt = V/L), and Back EMF is in proportion to rate of change of current so Vback = Ldi/dt = L x (V/L) = V so by KVL / Ohms law => iinductor should be = (Vapplied - Vback)/R = (V - V)/R = 0

What is R in your equation? It's not shown in your circuit.

Meanwhile consider a simple circuit with just a battery (Vbat), wire, switch and a resistor R instead of the inductor. The resistor will also generate a back emf when you pass a current through it... Vback = IR

Now I could write the equation I = (Vbat - Vback)/R however in that case R wouldn't be the value of the resistor it would be the resistance of the wire and switch connecting the two together. In an ideal circuit the resistance of the wire is 0 so that equation isn't going to work for ideal wire.

Most people forget the negative sign.

V= - L di/dt

You connect a voltage to an inductor, you have now determined its terminal voltage, the ensuing di/dt is a result of that forced terminal voltage and the resulting di/dt causes a d(flux)/dt that induces a voltage opposite to the applied terminal voltage satisfying KVL.

• CWatters and jim hardy
jim hardy
Gold Member
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back to basics of derivative...

time derivative of position is velocity. Likewise, integral of velocity is position.
No mass can achieve an instantaneous change in position, for that would require it to move at speed of light.
It must acquire some velocity and its position will be the time integral of its velocity.

Time derivative of velocity is acceleration.
Any mass can achieve an instantaneous change in acceleration because F=MA and there's no derivative or integral in that formula.

You need to make current and inductance as intuitive as velocity and mass. It's training your brain to think in derivatives.
Working the math should help.

The simple resistor does not integrate or differentiate, its voltage E = R X I not R X dI/dt

inductors and capacitors are derivative/integral devices.
Use the position/velocity/force analogies to make your brain work derivatives right. It's a vital thinking tool for budding EE's.

• Fisherman199
Hi all thaaaaaaaaaaanks a loooooooooooot now i feel i am getting but what i understood is below,

first let the circuit has R (practical inductor, wires etc...)
The moment/instant AT WHICH V is applied current i begins instantly to increase from i=0 A, hence at this moment can't make any drop across R and therefore all V appears across inductor terminals (to satisfy KVL) = L di/dt, with a rate(slope di/dt) of V/L A/sec (The rate can change "instantly", not the quantity.)

[The reason of this rate is that at the beginning of current increase i=0 (because it begins from 0), so can't make any drop across R and therefore all V appears across inductor terminals (to satisfy KVL) = L di/dt, so di/dt = V/L]

AM i correct till here ?? pls check and every word

• jim hardy and Tom.G
CWatters
Homework Helper
Gold Member
Yes. I think you are correct.

All of the applied voltage appears across the inductor because the initial current is zero (and that means no voltage drop in the resistance of the wires).

The voltage across the coil causes the current to increase from zero at rate di/dt = V/L.

what i am going to say may be silly but.....
as we have seen till now that current begins from zero because electrons (or any object that has some mass) on the application of force (V) can start to move from rest (i.e., from speed 0) but if so, then why the current in a resistor (R only, no inductor now) connected accross a voltage source V suddenly increase to its final value = V/R ???????????? electrons in the resistor should also start from zero and speed up with a rate.

pls tell me all these in terms of electrons, force, electric-magnetic fields i.e., very basic

i want to understand it from the root

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jim hardy
Gold Member
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What's the rest mass of electric charge?
In absence of an inductor's magnetic field, what force besides its inertia opposes its acceleration?
Charge can accelerate so quickly that for everyday circuit analysis we ignore the infinitesimal delay.

what i am going to say may be silly but.....
as we have seen till now that current begins from zero because electrons (or any object that has some mass) on the application of force (V) can start to move from rest (i.e., from speed 0) but if so, then why the current in a resistor (R only, no inductor now) connected accross a voltage source V suddenly increase to its final value = V/R ???????????? electrons in the resistor should also start from zero and speed up with a rate.

pls tell me all these in terms of electrons, force, electric-magnetic fields i.e., very basic

i want to understand it from the root
pls pls

Ok so inductance doesn't come from kinetic energy of the moving electrons. Inductance is energy stored in a magnetic field, only moving charge can create a magnetic field, but for us that's generally just electrons.

So as soon as charge moves, a magnetic field is created, and that stores energy. It is not possible to have current (ie flow rate of charge) without creating any magnetic field, so a pure resistance is a hypothetical construct for the very reasons you cite. Since you cannot isolate the fields created by a moving charge, and the field due to a charge difference (due dc resistance or proximity) you always have some energy stored in a magnetic and in the electric field therefore any time charge moves you cannot avoid inductance and capacitance.

Basically what we call a resistor is a thing that has far more dc resistance than inductance or capacitance, if you look at any RF engineering you quickly realise a resistor does actually have both those.

• jim hardy
jim hardy
Gold Member
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A straight wire has inductance. That's why a long wire radio antenna works.

In circuit analysis we start out with DC and ignore stray inductance and capacitance because they only come into play when frequency becomes substantial as in radio work.

I think OP is still forming his basic thought steps.
Best way to do that is work lots of problems.
Learn "What" before "Why".

old jim

CWatters
Homework Helper
Gold Member
what i am going to say may be silly but.....
as we have seen till now that current begins from zero because electrons (or any object that has some mass) on the application of force (V) can start to move from rest (i.e., from speed 0) but if so, then why the current in a resistor (R only, no inductor now) connected accross a voltage source V suddenly increase to its final value = V/R ???????????? electrons in the resistor should also start from zero and speed up with a rate.

pls tell me all these in terms of electrons, force, electric-magnetic fields i.e., very basic

i want to understand it from the root
pls pls
As others have said, that's wrong. The not due to the mass or inertia of electrons.

If you put a magnet near a coil you only get a voltage on the coil while the magnet is moving. That's because the voltage is determined by the rate of change of flux in the coil. If it's stationary the flux through the coil is constant so no voltage. The faster you spin a dynamo the faster the magnets move past the coil, the faster the flux changes and the higher the voltage. Ok with this?

The flux in a coil also depends on the current flowing through it. To make a stronger electromagnet you put more current through it right? So another way to change the flux in a coil is to change the current flowing through it. So the voltage is proportional to the rate of change of current . We can write this as an equation...

V= L * dI/dt

or you can rearrange it to..

dI/dt= V/L

..if you prefer.

Notice that nothing in this equation says the current has to "start" at zero. It depends what happened to the inductor in the past. You might have had a constant 1000A flowing in an inductor for 100 years but if the voltage right now is V (and V isn't zero) then the current will be changing away from 1000A at a rate equal to..

dI/dt = V/L

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jim hardy
Gold Member
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as we have seen till now that current begins from zero because electrons (or any object that has some mass) on the application of force (V) can start to move from rest (i.e., from speed 0) but if so, then why the current in a resistor (R only, no inductor now) connected accross a voltage source V suddenly increase to its final value = V/R ???????????? electrons in the resistor should also start from zero and speed up with a rate.

In circuit analysis,,,
Charge ,
be it electrons or the mysterious positive charge we call Coulombs,
we consider massless.
If we didn't we'd have to put terms in Kirchoff's laws to account for gravity .
and a circuit probably wouldn't work if you turned it upside down.

That's a problem with the "water analogy" . Water has lots of mass.

So train your brain that a change in voltage can propagate down EDIT: or UP a wire at nearly the speed of light.

Charge is very nimble.

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• Fisherman199