# Inductor in d.c circuits

1. Oct 2, 2006

### Amith2006

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?

2. Oct 2, 2006

### Staff: Mentor

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. Oct 3, 2006

### Amith2006

Now, if the d.c source is replaced by an a.c source then will it be true?

4. Oct 3, 2006

### Staff: Mentor

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. Oct 8, 2006

### Amith2006

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. Oct 8, 2006

### NoTime

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. Oct 8, 2006

### NoTime

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. Oct 9, 2006

### Amith2006

That was nice discussion.Thanks.

9. Oct 10, 2006

### SGT

The answers are wrong! According to Faraday´s law, the voltage through an inductor is proportional to the derivative of the magnectic flux.
$$v(t)=\frac{d\phi}{dt}$$
In a linear, time invariant inductor $$\phi(t) = Li(t)$$, so the only way to make a voltage to appear is to make the current time varying. $$v(t) = L\frac{di}{dt}$$
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.

Last edited by a moderator: Oct 10, 2006
10. Oct 10, 2006

### Staff: Mentor

Hey, good point. I didn't think of the solenoid effect on the core. Thanks for that!

11. Oct 11, 2006

### SGT

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. Oct 11, 2006

### NoTime

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. Oct 11, 2006

### Staff: Mentor

Quiz question -- what happens to the bulb current if you yank the core out of the inductor?

14. Oct 12, 2006

### NoTime

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. Oct 13, 2006

### SGT

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.