I just thought if the lamp was missing there couldn't be a pulse because no current could flow because it would be an open circuit

The choke have energy equal to 1/2 L I^2 charged in the magnetic field (I = the value of the current at the very moment of the starter contact opening). This energy must dissipate out the moment the starter opens, there is no other option

In a really open circuit, the voltage pulse would be of 0 width and unlimited voltage. But the circuit is never really open. If there is the starter, there is an arc inside the glow lamp. If there are mechanical contacts, there is an arc between them in air. If nothing else, the voltage will be high enough to start breaking through the isolation in the choke's coil or lantern wiring, and that will set the peak voltage

The lower the votlage limit is, the lower is the power dissipated, and the wider the pulse will be (untill all the energy stored in the magnetic field goes)

Ash knows what's up. He really knows how to math and def knows how to lighting. 

Heeeeey, that was Medved on the last page

Hi guys sorry to redirect an old topic but carrying on from my original question how long does the choke ballast push energy back into the supply at the moment of zero crossing I'm sorry to ask such odd things I'm just trying to understand thanks for your patience

Hi guys sorry to redirect an old topic but carrying on from my original question how long does the choke ballast push energy back into the supply at the moment of zero crossing I'm sorry to ask such odd things I'm just trying to understand thanks for your patience

The energy goes back, when the load current is opposite to the supply voltage.

Assume we have a series inductor ballast (the most common concept for virtually everything, at least in Europe)

So assume we just have a positive voltage halfcycle, just before it's end. That means the current is flowing into the load. During this time the power flows from the mains into the ballast and load.
At one moment the voltage reverses, but the current still continues to flow, the coil keeps it that way.
It takes some time for the current to go back to zero and eventually reverse the polarity as well.
So all the time after the voltage reversal, before the current reversal the power is actually transferred from the ballast coil to the load and back to the mains.

Similar with series capacitor ballast:
When the voltage goes up (first half of the halfwave), the capacitor get charged, so the power flows from the mains into the capacitor and load.
When the voltage goes down, the capacitor gets discharged back to the mains, so is the energy transferred from the capacitor, again to the load and back to the mains.

Of course the amount of energy going from the mains is always bigger than the energy going back.

So how long it takes depends on the phase shift:
Zero phase shift means no reverse power at all.
45deg phase shift (that means power factor about 0.7) means 45deg/(360deg * 50Hz) = 2.5ms.

Thank-you Medved that makes it clear One half cycle of mains is  about 20 millisecond (I think) so its not really long at all. If the power factories lower say 0.3 lagging it presumably will be 4 or 5 milliseconds long am I on the right track here?

The mains voltage is a sinusoidal waveform with the period of 20ms (50Hz mains), so one half cycle (= half of it) is 10ms.
In other words 10ms positive voltage, 10ms negative.


So when the phase lags so the power factor is 0.3, it means
PF = cos(LagAngle) = 0.3
so the LagAngle would be then 72.54 degrees.
The full period (20ms) has 360deg, so the time duration would be
Tbackpower = 20ms * LaagAngle/360deg = 4.03ms

If the lag would be 5ms, that means angle of 90deg and that means half of each half cycle the power goes back. That means zero power factor, so just a reactive power (the amount of energy drawn in the first half of the halfcycle is then fully returned back during the other half of the half cycle)

I see thanks again it lasts a lot longer than I thought to be honest I was thinking it would be micro seconds not milliseconds  its a weird concept one difficult to grasp especially when you can't understand maths like me!

My first lesson on inductive kick was about 6th grade when I discovered that a pair of 1.5v D batteries (3VDC) momentarily struck onto the primary wires of a 120vac power transformer would create an arc similar to what a normal choke ballast would do, if momentarily shorted. I never tried to measure it but it would flash a small fluorescent tube no problem and could also cause a painful shock if not careful!

I did this with a transformer and a 9 volt battery, the only difference is that I did it on the 6 volt secondary, so I got a helluva shock from the 120 side.