Is self induced emf in an inductor will equal and opposite to the applied voltage

[mobinmk]

hii frnds,

my elementary doubt is that

if i applied 10v in an inductor, by faradys law back emf is produced .
is it equal and opposite??

if it is equal and opposite back emf is -10v

then no magnetic flux, cancel each other, no current flow thru it

pls help me..

am confused..

 

[CataM]

In a circuit formed with a power supply and an inductor (ideally with no resistance), the emf would equal the power supply and opposed. This is correct regarding Kirchhoff voltage law too. In DC that would happen for a short period of time, in AC would happen as long as the supply is connected.

 

[BradtheRad]

It may be easier to picture what happens when no voltage is applied, but a flux field is present. We add a resistive load.

The flux field collapses, inducing current. A voltage is created at each end of the coil (positive at one end, negative at the other).

Current goes down, voltage goes down, until both are at zero.

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The voltage created by the inductor is counter-EMF. It is easier to measure when no other voltage is present. That is why I offered the scenario above.

Counter-EMF is not easy to measure when we hook up (or theorize hooking up) a supply sending an external voltage.

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I have a Youtube video which portrays inductor behavior... A custom animated simulation which shows flux fields growing and collapsing, counter-emf, electrons (current bundles) travelling through wires.

The counter-emf is conceptualized as a lug, or plug, or slug, moving back and forth inside the inductor. This seems fitting because it acts immediately, in reaction to any change in external voltage.

 

[mobinmk]

In a circuit formed with a power supply and an inductor (ideally with no resistance), the emf would equal the power supply and opposed. This is correct regarding Kirchhoff voltage law too. In DC that would happen for a short period of time, in AC would happen as long as the supply is connected.

Thankz CataM,

IN dc ,
what happen after a short period of time ??
eg

when i applied 10v battery to inductor, inductor resistance is 1 ohm

then current is i = 10/1 = 10A,

their is current rise from 0 (at t=0) to 10A (at t= 0++)

in this transient period ( t= 0 to 0++)

intresting thing happen.

am trying to understand , pls help me

 

[c_mitra]

am confused..

You are not alone. Take heart.

Let me begin at the beginning: an inductor has an inductance that is given by L=v(t)/(di/dt); this comes from the definition: v(t)=L*(di/dt)

It is important to understand the key parameter called (di/dt); this suggests that the current needs time to build up in an inductor and this cannot happen instantaneously. This is very different from a resistor where the current does build up in no time and this is the root cause of all phase delay. Similar stuff happens in capacitors but that is with respect to the voltage and we are not talking about capacitors right now.

Now we come to your question. You apply 10V to an inductor that has an inductance of L. The back emf produced will be L*(di/dt) but how do we get this quantity (di/dt)?

When you were in school, you were taught charging and discharging of an inductor and capacitor but they were always in series with some resistor. Apparently we cannot get rid of the resistance problem.

Perhaps it was not explained this way, but the resistance is essential in defining another quantity called time constant. Without the resistance the capacitor charges instantly and the inductor (di/dt) becomes infinity. The reality is that all capacitors and inductors have some series resistances in real life (cf. superconducting magnets).

Although (di/dt) is very large, L*(di/dt) will be exactly -10V and the current will circulate at 0V (because no resistance) and there will be considerable magnetic flux.

In real life, the back emf is always less than the applied voltage. That allows the current to build up and decay- again with a phase or delay that is determined by the series resistance.

 

[mobinmk]

It may be easier to picture what happens when no voltage is applied, but a flux field is present. We add a resistive load.

The flux field collapses, inducing current. A voltage is created at each end of the coil (positive at one end, negative at the other).

Thanks BratheRad

what u mean flux field ??

i know about flux, field,

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.

also gravitational field , magnetic field etc

Flux is the amount of “something” (electric field or whatever you want) passing through a surface.
The total flux depends on strength of the field, the size of the surface it passes through, and their orientation.


inductor is a set of coils,

how field is present, when no voltage applied ??

again am confused

pls help me

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You are not alone. Take heart.

Let me begin at the beginning: an inductor has an inductance that is given by L=v(t)/(di/dt); this comes from the definition: v(t)=L*(di/dt)

It is important to understand the key parameter called (di/dt); this suggests that the current needs time to build up in an inductor and this cannot happen instantaneously. This is very different from a resistor where the current does build up in no time and this is the root cause of all phase delay. Similar stuff happens in capacitors but that is with respect to the voltage and we are not talking about capacitors right now.

Now we come to your question. You apply 10V to an inductor that has an inductance of L. The back emf produced will be L*(di/dt) but how do we get this quantity (di/dt)?

When you were in school, you were taught charging and discharging of an inductor and capacitor but they were always in series with some resistor. Apparently we cannot get rid of the resistance problem.

Perhaps it was not explained this way, but the resistance is essential in defining another quantity called time constant. Without the resistance the capacitor charges instantly and the inductor (di/dt) becomes infinity. The reality is that all capacitors and inductors have some series resistances in real life (cf. superconducting magnets).

Although (di/dt) is very large, L*(di/dt) will be exactly -10V and the current will circulate at 0V (because no resistance) and there will be considerable magnetic flux.

In real life, the back emf is always less than the applied voltage. That allows the current to build up and decay- again with a phase or delay that is determined by the series resistance.

Thankz c mitra,

Really i got energy from ur reply.

am trying to learn inductor deeply on the electromagnetism domian( electromagnetic field theory)

then only i can visulaize whats happen in inductor,

you explain in ( network theory)

Although (di/dt) is very large, L*(di/dt) will be exactly -10V and the current will circulate at 0V (because no resistance) and there will be considerable magnetic flux.

In real life, the back emf is always less than the applied voltage. That allows the current to build up and decay- again with a phase or delay that is determined by the series resistance.

How the current will circulate ??

applied voltage is 10v
back emf is -10v

these 2 voltage oppose each other then how current circulate?

eg 2 men push a mass of body 100kg each other in opposite direction

2 forces will be equal and opposite.(10N &-10N)

then mass of body ALWAYS stationary.


also other ques

in real life back emf is always less than applied voltage,

why?? any losses.

by lenzs law ( conservation of energy & newtons third law)

is it violates??

one thing am asked in the working principle of transfomer

eg

i applied 11kv voltage at the primary side of transfomer

winding resistance is 10 ohm

is current 11kv/10 = 1.1kA

IS this inrush current??

at first energization of transfomer their is humming noise

is this to bulid up magnetization of core ??

why current lag 90 deg??

if the self induced emf is -11kv

then it oppose the applied voltage

no self induction, no mutual induction....

wow...

is it miracle happen here

am totaly confused.....

pls help me

 

[c_mitra]

How the current will circulate ??

applied voltage is 10v; back emf is -10v

these 2 voltage oppose each other then how current circulate?

eg 2 men push a mass of body 100kg each other in opposite direction

2 forces will be equal and opposite.(10N &-10N)

then mass of body ALWAYS stationary.

I shall reply in your style: just like Newton's law says that without an external force, an object can be either stationary or will move in a straight line with a constant velocity.

A stationary object is equivalent to an object moving in a straight line with const velocity; that is called an inertial frame- both are equivalent.

Consider a metal ring with zero resistance. (somehow) you produce a constant current in this ring. This current will be associated with a magnetic field. Because there is zero resistance, the current will continue to move in the ring indefinitely and the magnetic field will be stable. This is like (not same or even similar) the inertial frame in the Newton's first law.

(how to produce the initial current? Imagine you cut the ring, insert a battery and then when the desired current has been obtained, you bypass the battery (now the ring is closed). In real life, we use the same magnetic induction to induce a current in the ring- there is no need to cut or close the ring)

The applied voltage is 10V; this voltage will cause an infinite current because the resistance is zero. The induced voltage will be -10V and they will cancel exactly leaving zero volt and const current in the ring. That the current is flowing can be seen by the measurement on the magnetic field produced. You will be needing energy to create the magnetic field but no energy is needed to maintain it.

Still puzzled?

in real life back emf is always less than applied voltage, why?? any losses.
by lenzs law ( conservation of energy & newtons third law)

You are right; the losses are due to resistance.

You have studied electric motors. When you apply the power, the current flows because of the potential. The potential causes the rotor to, you guessed it right, rotate. That process produces a back emf and this back emf is dependent on the motor rotation. The current that flows in the windings of the motor is due to the net potential (applied voltage - back emf produced).

If the motor stops, the back emf is no more (it vanishes) and the motor coils sees the full applied external voltage and the current becomes very large and it gets damaged.

If the external voltage is removed, the back emf is there is the opposite direction but the circuit is open and the current cannot flow and the motor comes to a stop slowly.

If the external voltage is removed and the motor power input is shorted, back emf now forces current in the opposite direction and the motor comes to a stop very rapidly.

By the way, Lenz's law does not claim that the back emf produced must be equal- it just says that they must be in opposite direction.

Leave transformer for another day.

 

[CataM]

By the way, Lenz's law does not claim that the back emf produced must be equal- it just says that they must be in opposite direction.

I think that is the biggest point.
In ideal case it is equal, in real life not.

 

[BradtheRad]

how field is present, when no voltage applied ??

It is the situation immediately after voltage is removed. The flux field is due to current flowing. (I guess I left out the fact current is flowing.)

It's natural for us to picture the scenario which starts with everything=0. However there is the extremely useful characteristic of an inductor, which generates high voltage as a behavior of counter-emf. (Example, boost converter.) It is a result of Amperes and Henries and Ohms. It obeys the formulas, and the L*R time constant.

We learn from the normal scenarios, of course, but our information is not complete unless we evaluate the inductor's generator action. It is a phenomenon which is pretty much unique among electronic components. We can find counterpart roles for the inductor and capacitor in many ways, but the capacitor does not quite do the same thing as an inductor can do.

The inductor is current operated. It resists a change in current flow. (It is the definition of an inductor.) Therefore counter-emf is a secondary behavior. It derives from the primary behavior (which has to do with current). It adds up to a lot of things going on. Magnetic and electrical forces come into play. How to explain the process in all its parts? It's certainly not easy to explain it, nor get a grasp on it.

 

[c_mitra]

I now understand your source of confusion. You are using the ideas of electrostatics and magnetostatics when you should be considering the ideas from electrodynamics which covers both changing electric and magnetic fields.

1. Consider charging a capacitor: current flows to the capacitor plates which builds up a potential and the current stops when the potential becomes equal and opposite to the external applied potential. Work is done in this process and the energy is stored in the dielectric as electric field.

2. Consider charging an inductor: as current flows, a magnetic field is produced and this magnetic field opposes the applied potential and the current and an ideal inductor can support infinite current. The energy is stored in the magnetic field of the core material.

3. When a current flows, it is associated with a magnetic field. A changing magnetic field is capable of inducing an electric field. This electric field is the cause of the back emf. A steady current does not induce any electric field. A steady electric field (no current is flowing) does not cause any magnetic field.

4. Capacitors are intended to be charged from a voltage source. Inductors are supposed to be charged from a current source. Both are otherwise very similar. Capacitors cannot stand a DC current. Inductors cannot stand a DC voltage.

5. When you charge a capacitor via an inductor, you get a resonant tank circuit. That is not having any stable state.

6. A LC combination also radiates electromagnetic radiation and energy is also lost via that path (even without any resistance).

 

[mobinmk]

I shall reply in your style: just like Newton's law says that without an external force, an object can be either stationary or will move in a straight line with a constant velocity.

A stationary object is equivalent to an object moving in a straight line with const velocity; that is called an inertial frame- both are equivalent.

Consider a metal ring with zero resistance. (somehow) you produce a constant current in this ring. This current will be associated with a magnetic field. Because there is zero resistance, the current will continue to move in the ring indefinitely and the magnetic field will be stable. This is like (not same or even similar) the inertial frame in the Newton's first law.

(how to produce the initial current? Imagine you cut the ring, insert a battery and then when the desired current has been obtained, you bypass the battery (now the ring is closed). In real life, we use the same magnetic induction to induce a current in the ring- there is no need to cut or close the ring)

The applied voltage is 10V; this voltage will cause an infinite current because the resistance is zero. The induced voltage will be -10V and they will cancel exactly leaving zero volt and const current in the ring. That the current is flowing can be seen by the measurement on the magnetic field produced. You will be needing energy to create the magnetic field but no energy is needed to maintain it.

Thnks C_MITRA


Very nice explaination,

now i clear the concept,

at first i think

if we apply a voltage across inductor , suddenly oppose it , and no current flow thru it.

can u explain

why the time delay di/dt this results to phase delay >> power factor..

on the basis of electromagnetic field theory.

if the applied emf and back emf will be qual, then no time delay no phase shift . is it ?

also how the field collaspe?? then back emf decrease, current be flow constant

is lenzs law offers 90 deg time displacement?

 

[CataM]

is lenzs law offers 90 deg time displacement?

Faraday found out that an induced emf appears when changes flux/current which is why 90 deg phase shift appears.
Lenz only said that the emf opposes that change in flux/current and set the negative sign. Together form Faraday-Lenz law.

 

[c_mitra]

if we apply a voltage across inductor , suddenly oppose it , and no current flow thru it.

1. You need to understand the reaction to sine waves first. An application of a high voltage *suddenly* is equivalent to a sine wave with very short period or very high frequency. If you study datasheets carefully, you will notice that they rarely draw square pulses, they invariably make trapezoidal waves, and they often show the rise and fall times of the pulses. This is important because zero rise and fall times are practically impossible.

2. So we apply a sine wave voltage source to an inductor. Strictly speaking, we should apply a current source to an inductor because an ideal inductor cannot stand a DC voltage. But what happens if you apply a sine wave AC voltage to an inductor?

3. A current begins to pass and this creates a magnetic field. This magnetic field is *not* constant and is increasing in the same way as the current (forget about the voltage right now). A changing magnetic field is producing an electric field (here the rate of change of magnetic field comes into play; this is important). This electric field is now proportional to the rate of change of the magnetic field (or the current). As the inductor is now immersed in this electric field, it produces a voltage (not a current; note the difference).

4. Applied voltage was a sine wave; induced voltage is a cosine wave (derivative). It is equivalent to say that a 90deg phase shift has taken place.

5. If no current flows through the inductor, there will be no induced voltage. This is very important to understand.

why the time delay di/dt this results to phase delay >> power factor..

on the basis of electromagnetic field theory.

You have not understood electromagnetic theory well. It is tricky to explain in a simple forum but you can study and come back with specific questions and I will be happy to reply (as far as possible).

if the applied emf and back emf will be qual, then no time delay no phase shift . is it ?

Not quite correct. Consider a thin circular ring with *zero* resistance. You place this on the top of an electromagnet. You feed the electromagnet with a linear current ramp (0 to I over time T). A linear current ramp means dI/dt is constant and the voltage induced in the ring will be constant.

Because the ring has zero resistance, any voltage (howsoever small) induced will produce a large current and this current will be producing a magnetic field and this magnetic field will oppose the primary magnetic field and the ring will jump off. In reality, the ring will produce a magnetic field that will just cancel the external magnetic field.

also how the field collaspe?? then back emf decrease, current be flow constant

Constant current in an inductor does not produce any back emf.