Inductor in D.C Circuits: Exploring Self-Inductance and Induced E.M.F

In summary, according to Faraday's law, the voltage through an inductor is proportional to the derivative of the magnetic flux. When a ferromagnetic core is inserted into a coil, a voltage appears at the coil's terminals. If the variation is big enough the induced emf may be greater than the voltage of the battery.
  • #1
Amith2006
427
2
1) A bulb is connected in series with an inductor and a 6 volt D.C supply. A soft iron core is inserted into the coil quickly. It is said that there is a momentary increase in the intensity of the bulb because at that particular instant the induced e.m.f is greater than the e.m.f of the source because self inductance increases and hence induced e.m.f increases. Is it true?
 
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  • #2
Hmmm. My first reaction is that it is not true, since the inductance does not matter in a DC circuit, and since a moving non-magnetized core does not induce an EMF in a coil that is stand-alone. But you should write out whatever EMF term they are referring to in order to be sure...
 
  • #3
Now, if the d.c source is replaced by an a.c source then will it be true?
 
  • #4
Depends on the value of the final inductor and the AC frequency. Let's say that the AC reactive impedance of the inductor matches the real resistance of the light bulb at the AC frequency of the source. What will be the effects of the inductor? Remember that current lags the voltage drop across the inductor by some amount here... What does that do to the power available to the light bulb? What about the extreme case where the inductive reactance far exceeds the resistance of the light bulb?
 
  • #5
berkeman said:
Depends on the value of the final inductor and the AC frequency. Let's say that the AC reactive impedance of the inductor matches the real resistance of the light bulb at the AC frequency of the source. What will be the effects of the inductor? Remember that current lags the voltage drop across the inductor by some amount here... What does that do to the power available to the light bulb? What about the extreme case where the inductive reactance far exceeds the resistance of the light bulb?
Suppose the a.c source has a rms voltage of 6 volts and frequency of 100 Hz, what would be the effect on the intensity of the bulb? The self inductance of the inductor is not given. Is it possible to answer this question without it?
 
  • #6
berkeman said:
Hmmm. My first reaction is that it is not true, since the inductance does not matter in a DC circuit, and since a moving non-magnetized core does not induce an EMF in a coil that is stand-alone. But you should write out whatever EMF term they are referring to in order to be sure...
Hmmm
That is certainly true for a stand alone coil.
And inductance is unimportant in a steady state DC circuit.

However, if you instantaneously created an iron core in the operating coil then it will increase inductance and cause a momentary dimming of the bulb, just like switching an inductor into the circuit.

OTOH, when inserting a core into an operating coil it's going to get magnetized as it aproaches the coil.
The energy used to move the core has to go somewhere.

Since the coil should pull the core in like any solenoid then I would think there ought to be a momentary dimming followed by brightening as the core is stoped.

If the core is pushed, exceeding the acceleration the coil might impart to the core, then I might think you would just get brightening.

The exact answer isn't clear to me.
 
  • #7
Amith2006 said:
Suppose the a.c source has a rms voltage of 6 volts and frequency of 100 Hz, what would be the effect on the intensity of the bulb? The self inductance of the inductor is not given. Is it possible to answer this question without it?
Sure, the exact value of the inductace is not important.
Just pick some values and plug them into the equation and see what happens to current, thus brightness.
 
  • #8
That was nice discussion.Thanks.
 
  • #9
The answers are wrong! According to Faraday´s law, the voltage through an inductor is proportional to the derivative of the magnectic flux.
[tex]v(t)=\frac{d\phi}{dt}[/tex]
In a linear, time invariant inductor [tex]\phi(t) = Li(t)[/tex], so the only way to make a voltage to appear is to make the current time varying. [tex]v(t) = L\frac{di}{dt}[/tex]
When you introduce a ferromagnectic core into the coil, the flux is varied, so a voltage appears at the terminals of the inductor. If the variation is big enough the induced emf may be greater than the voltage of the battery.
Of course, after the nucleus is inserted there is no more field variation and the coil will behave as a small resistor.

edited to add: the answer from NoTime is correct.
 
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  • #10
NoTime said:
Since the coil should pull the core in like any solenoid then I would think there ought to be a momentary dimming followed by brightening as the core is stoped.
Hey, good point. I didn't think of the solenoid effect on the core. Thanks for that!
 
  • #11
NoTime said:
Hmmm


Since the coil should pull the core in like any solenoid then I would think there ought to be a momentary dimming followed by brightening as the core is stoped.
No, If you insert the core quick enough, there will be an alteration in the magnectic field (see my previous post). This alteration in the magnectic field will give origin to a voltage that will oppose the applied dc voltage. So, you are right in saying that there will be a momentary dimming. But the increasing induced voltage may overcome the dc and produce a net negative voltage in excess of that supplied by the source.
Since a bulb has no polarity, it will glow more strongly with this negative voltage.
But when the core is stopped, there will be no more variation in the field, so no induced voltage and the lamp will glow as before the insertion of the core.
 
  • #12
SGT said:
No, If you insert the core quick enough, there will be an alteration in the magnectic field (see my previous post). This alteration in the magnectic field will give origin to a voltage that will oppose the applied dc voltage. So, you are right in saying that there will be a momentary dimming. But the increasing induced voltage may overcome the dc and produce a net negative voltage in excess of that supplied by the source.
Since a bulb has no polarity, it will glow more strongly with this negative voltage.
But when the core is stopped, there will be no more variation in the field, so no induced voltage and the lamp will glow as before the insertion of the core.
In this particular statement, I was considering the coil pulling the core in only and stoping it (the core will oscilate a bit here). No external force application.
Now the core entering the coil should increase the coils inductance thus lowering the current temporarily, a back EMF effect as you note, dimming the bulb.
However, as the now moving core tries to leave the coil
that back EMF should now become a forward EMF.
 
  • #13
Quiz question -- what happens to the bulb current if you yank the core out of the inductor?
 
  • #14
Frankly, I'm not entirely certain, either for pushing it in or yanking it out.
I suspect it depends on magnetic field diffusion into the core.
Looking on the web, small steel core might have a diffusion time of 100ms or so.
Practically speaking I think the core might act like a diamagnetic material if moved fast enough.

So I'll say the bulb will dim.
 
  • #15
NoTime said:
Frankly, I'm not entirely certain, either for pushing it in or yanking it out.
I suspect it depends on magnetic field diffusion into the core.
Looking on the web, small steel core might have a diffusion time of 100ms or so.
Practically speaking I think the core might act like a diamagnetic material if moved fast enough.

So I'll say the bulb will dim.
In any case, pushing or pulling, the field will oppose the movement, so in a first moment the lamp will dim. After that, the polarity of the induced voltage will reverse and the lamp willgo brighter if the movement is fast enough.
 

What is an inductor?

An inductor is a passive electronic component that stores energy in the form of a magnetic field when a current flows through it. It consists of a coil of wire wrapped around a core material, typically made of iron or a ferrite compound.

How does an inductor work in a d.c circuit?

In a d.c circuit, an inductor resists changes in current flow. When a voltage is applied to an inductor, the current starts to flow, and the inductor creates a magnetic field that opposes the change in current. As a result, the inductor initially acts as a short circuit and then gradually increases its resistance as the current stabilizes.

What is the role of an inductor in a d.c circuit?

The main role of an inductor in a d.c circuit is to store energy in its magnetic field. It also helps to smooth out the current flow and reduce voltage spikes, making the circuit more stable.

What are the characteristics of an ideal inductor in a d.c circuit?

An ideal inductor in a d.c circuit has zero resistance and can store an infinite amount of energy in its magnetic field. It also has a constant inductance value regardless of the current or voltage applied to it.

What happens to an inductor in a d.c circuit when the current is turned off suddenly?

When the current in an inductor is turned off suddenly, the magnetic field collapses, creating a high voltage spike. This phenomenon is known as inductive kickback, and it can damage electronic components in the circuit. To prevent this, a diode is often placed in parallel with the inductor to provide a path for the current to flow and dissipate the energy from the collapsing magnetic field.

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