# Inductor and lamp in parallel

1. Mar 10, 2016

### Steve F

Trying to figure out why an inductor in parallel with a lamp and battery will cause the lamp to GRADUALLY increase in brightness to a steady glow. I understand the concept of a coil of wire, producing a changing magnetic field when the current running through it changes then the changing magnetic field induces an EMF in the wire such that it opposes the change in current. But I can't figure out how that causes a lamp to gradually increase to a steady glow. Any help appreciated. See link below for the circuit...

Cheers!

Last edited by a moderator: May 7, 2017
2. Mar 10, 2016

### Hesch

When the battery is connected, the lamp will be lit immediately.
Then the lamp will gradually decrease in brightness, because the current through the inductor will be increased, until the inductor short-circuits the battery.

If the lamp and the inductor were connected in series, the lamp would behave as described in the question.

3. Mar 10, 2016

### Staff: Mentor

Hi Steve F.

You haven't got the idea right; it would have said in series. What is the reference where you read this?

4. Mar 10, 2016

### alw34

Where did you get a description of the bulb increasing in brightness??

What do you see as the final steady state circuit condition? I see the light out and all the current flowing through the inductor, a near short.

As you posted, inductance opposes any CHANGE in current flow. But in a steady state dc state, the inductor offers no inductance v[t] = L di/dt and is zero.

For some details, check out:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indcur.html#c1

and click on TRANSIENT BEHAVIOR OF THE COIL

5. Mar 29, 2016

### Steve F

Hi thanks for the replies. As I am new to this I am only seeing them now, sorry about that!

Anyway, my understanding too was that the bulb would decrease in brightness. It would need to be connected to an inductor in series to increase in brightness.

I will try to find the question and take a photo.

Steve

6. Mar 29, 2016

### Steve F

There it is. I understand the actual questions. I don't get their description of the intensity of the lamp just.

Steve

7. Mar 29, 2016

### Staff: Mentor

So their description is a bit off, depending on what the DC resistance (DCR) of the inductor is. As the switch is closed, the bulb will have some initial brightness, and that brightness will dim to some point were the inductor current is constant (limited by its DCR). When the switch is opened, the inductive kickback will make the bulb briefly brighter, and then the bulb will go out. If the inductor has a low DCR, it will basically short out the bulb after a few time constants, and will generate a very bright flash when the switch is opened. If the inductor has a high DCR, then the bulb will be brighter than the low DCR case, but you will get less of a flash at the switch opening. Does that make sense?

8. Mar 29, 2016

### sophiecentaur

If the battery is 'ideal', having a coil in parallel will have no effect on the brightness of the bulb because the battery Volts will be unaffected. (there may be a lot of current flowing in the coil but that will not affect the volts across a perfect battery). A weak battery can, of course, be affected by connecting the low resistance coil across it - but unless the problem states more about the properties of the battery and the resistance the coil, you have to assume an ideal battery.
However, when you disconnect the switch, the magnetic field energy in the coil will cause an induced emf which will be in a direction to maintain the current that was originally flowing. This current will flow through the lamp (the only available path). There can be a very high initial emf, depending upon the DC resistance of the coil (which sets the initial current flowing through it) and the Inductance of the coil. The battery can no longer keep the bulb supply voltage to what it should be. This will cause a lot of current to flow through the lamp bulb and could cause it to blow.
The statement in the question, that the lamp increases in brightness to a steady glow is correct because that's what happens to all bulbs when you connect them - in a very short time. It has nothing to do with the inductor.

9. Mar 29, 2016

### Steve F

I understand what you are saying berkeman, I just wasn't convinced by their description of the lamp INCREASING to a steady glow.

When it says the lamp increases to a steady glow...do you think this means that it increases in so far as all lamps increase to a steady glow when switched on?

Steve

10. Mar 29, 2016

### Staff: Mentor

Beats me why they say that. Maybe your guess is correct.

11. Mar 29, 2016

### Staff: Mentor

It won't do that, and now that you have provided the exact quote, we can see that it doesn't claim what you believed it said. It says that at switch-on the lamp will brighten and shine as normal. An incandescent lamp is just not an "instant" glow light source, that's all it's saying.

Providing it's a good battery (i.e., has not much resistance), and the inductor is a practical device (i.e., has some resistance), then the lamp will brighten and glow just the same as it does when the inductor is not present. The battery will have no problems supplying current to both of the elements in parallel: rated current to the bulb, and a [presumably heavier] current to the inductor.

Then following switch-off, battery current ceases but inductor current continues to flow because of the property of an inductor, and the only closed path now available to this inductor current is via the lamp. (Notice that this current passes through the lamp in reverse direction to current that the battery supplied to the lamp, but reversal of current in a filament lamp is of no consequence.)

Assuming before switchoff the battery had been supplying a greater current to the inductor than to the lamp, then after switch-off the lamp will find itself briefly carrying that heavier current, so it glows more brightly, and may even burn out.