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I've been studying about inductors lately, so I'm going to tell you how inductors work based on what I understood, please correct me if I'm wrong.

As soon as the switch is closed (in a DC circuit with an inductor), there is an induced voltage (lets say E) which is produced to oppose the flow of current which is produced due to the voltage (V) of the battery. At this instant, the value of induced voltage is equal to the applied voltage i.e V=E (the direction of E is opposite of V). Now, since the net voltage in the circuit is 0 (because V=E and are in opposite directions) there will be no flow of current (so the current now, just after closing the switch is 0). We know that back EMF is produced only when there is a change in current/flux through the inductor. Now that there is no change in current/flux through inductor (as no net voltage) there is no need for the back EMF to be present, therefore it starts wearing off. Let’s just say dE amount of back EMF wore off / reduced in time dt. Now there is a very small potential difference dE between the applied voltage V and the reduced induced EMF (E-dE). This produces a current di. Now as soon as the current is produced, the inductor tries to oppose it by getting back up the induced EMF. BUT note that the new induced EMF won't be equal to the initial induced EMF i.e E, this time the induced EMF produced by inductor will be equal to E-dE Because since its the property of inductor to try and not change the current (i.e it neither tries to increase nor decrease current) in the circuit, and right now the current in the circuit as discussed before is di. So the new induced EMF produced such that it won't change this current di, to do so it has to maintain a constant potential difference equal to dE. So it brings back up the induced EMF to E-dE, so the potential difference between applied voltage and induced voltage is dE and current is di.

NOW that the current is constant (di), (and back EMF is produced ONLY when there is a change in current), the back EMF starts wearing off and this whole process repeats until back EMF value reaches 0 and the current in the circuit is purely due to applied voltage V and the inductor starts acting as a wire, where as in the beginning it was acting as an open switch as no current was passing (due to no net voltage).

Can any one please confirm what I understood about inductors is correct or not? If not, please help me understand in simple words and preferably not using complex mathematics and complex theories. (I'm just a teenager and just started learning about inductors.)

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    \$\begingroup\$ No, it's wrong right from the 1st sentence. The back emf does not oppose any current change in the inductor (just as the voltage developed across a resistor due to current flow does not oppose the flow of current in that resistor). It's just a back emf that's all; in your simple circuit the back emf does not play a role at all. \$\endgroup\$
    – Andy aka
    Sep 30, 2023 at 18:53

2 Answers 2

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There's a problem with trying to argue around causation in inductor circuits. The formula for voltage and rate of change of current is \$V=L\frac{di}{dt}\$. This is an equation, it asserts the two sides are equal. It says that there is simply a constant of proportionality between voltage across an inductor, and rate of change of current through an inductor, called the inductance. It says nothing about causation. It is just as true to say the voltage causes the current to change, as it is to say that the current change causes the voltage.

When we choose to say one causes the other, we usually pick that which suits our narrative best.

So , as soon as the switch is closed (in a DC circuit with an inductor), there is an induced voltage (lets say E) which is produced to oppose the flow of current which is produced due to the voltage (V) of the battery .

We close the switch, so the story is that we apply voltage to the inductor. The current through the inductor will begin to ramp, at a rate given by V/L. When the current starts ramping, that generates the countering back EMF, which in the absence of resistive voltage drops, will be equal to the applied voltage.

at this instant , the value of induced voltage is equal to the applied voltage i.e V=E (the direction of E is opposite of V). Now,since the net voltage in the circuit is 0(because V=E and are in opposite directions) there will be no flow of current (so the current now, just after closing the switch is 0).

There is no flow of current at t=0 because a) there was no flow of current before t=0 and b) a finite voltage can only ramp the current finitely fast. It has not yet had time to increase the current to a finite value.

Net voltage in the circuit being 0 does not mean no flow of current. In this case, it means the circuit has no resistance (because we are either considering an ideal circuit, or it's built using superconductor with zero resistance).

we know that back emf is produced only when there is a change in current/flux through the inductor.now that there is no change in current/flux through inductor (as no net voltage) there is no need for the back emf to be present ,therefore it starts wearing off.

No. At and after t=0, when there is an applied voltage, and so when the current is increasing, there is a back EMF equal to the applied voltage. The back EMF does not 'wear off'.

lets just say dE amount of back emf weared off / reduced in time dt .

No.

now there is a very small potential difference dE between the applied voltage V and the reduced induced emf (E-dE) . this produces a current di.

No.

now as soon as the current is produced , the inductor tries to oppose it by getting back up the induced emf .BUT note that the new induced emf wont be equal to the initial induced emf i.e E ,

There is only one value of EMF produced, and that is equal to the applied voltage. Thinking of multiple EMFs is wrong. It appears that you are complicating the picture in your struggle to understand. Relax, and have another go at the simple picture.

this time the induced emf produced by inductor will be equal to E-dE

This almost sounds like you're trying to reason about another voltage in the circuit, which causes the well known change in rate of current rise when we connect a normal, resistive, coil to a voltage source. This voltage is that across the circuit resistance. Current flow through this resistance causes an additional voltage that causes the back EMF to be different from the applied voltage. In the ideal superconducting case, the back EMF is always, always, equal to the applied voltage.

Because since its the property of inductor to try and not change the current(i.e it neither tries to increase nor decrease current) in the circuit

I don't like 'try', it anthropomorphizes the component. It simply produces a back EMF which has the direction to oppose what is causing the current change.

which , and right now the current in the circuit as discussed before is di.so the new induced emf produced such that it wont change this current di , to do so it has to maintain a constant potential difference equal to dE. so it brings back up the induced emf to E-dE , so the potential difference between applied voltage and induced voltage is dE and current is di .

schematic

simulate this circuit – Schematic created using CircuitLab

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It may be easier to use an analogy that corresponds more to our physical world around us.

When you apply a force to accelerate a mass, it's inertia immediately counters that force, exactly and oppositely. The inertial force resists any change in the velocity of a mass and is always equal to and exactly opposite to the applied force. Meanwhile, the applied force does cause the velocity of the mass to change. If the force is constant then the change in velocity is also constant. In this case, we can say that \$\frac{\text{d}}{\text{d}t}v=\frac{F}{m}\$.

Similarly, when you apply a force (here, a voltage) to an inductor there is an immediate counter-force (back-emf) to that force that is exactly opposite to it. Meanwhile, this applied voltage does induce a rate of change in the inductor's current. So in this case we say that \$\frac{\text{d}}{\text{d}t}I_L=\frac{V_L}{L}\$.

Inductance is analogous to mass. The current in the inductor is analogous to the velocity of that mass.

You can continue to find more analogous ideas between these two fields, as well, all the way up to energy (which must be conserved.) So, where you have momentum, \$m\cdot v\$, you will find Webers as \$L\cdot I_L\$. But also where you have impulse, \$F\cdot \Delta t\$, which has the same units as momentum, you also have the applied voltage times time, \$V_L\cdot \Delta t\$, which also has the same units, Webers. So when you apply a voltage across an inductor you are providing an impulse that changes its momentum over time.

Just think in terms of this kind of analogy. It helps.

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