Can someone tell me how long does the voltage kick from the choke in a preheat ballast last for? Is it milliseconds or longer?

I'd say while the lamp is preheating and the starter is flashing. Normal starters have caps in them to prevent contact arcing and they go in the order of nanofarads. It could last from a half a seconds to three to five depending on how long the lamp preheats for.

Thanks I actually meant the how long the voltage pulse from the choke not the preheat time sorry if I wasn't clear

I think the pulse is as long as the lamp preheats. The starter with cap causes an inductive kick and heats electrodes and starts the lamp.

The starter does not produce the inductive kick per se. What happens is the starter makes a complete circuit running through the ballast and the electrodes. This preheats the electrodes as previously mentioned. When the starter opens, the current is suddenly interrupted everywhere in the circuit. Due to the nature of the ballast (inductive properties ?) the current cannot stop instantly in the coils and continues to "push out" as the successive phases bump into each other. The nature of the coils amplify this and it raises the voltage. This is what is the inductive kick and is produced by the ballast when the starter opens. So the starter is what causes the ballast to make the kick. I would imagine this kick only lasts several milliseconds. It is repeated as many times as the starter contacts open. Finally, when the time comes to turn the light off, a similar kick is produced, but since the power is out, the lamp does not light. All inductors have this phenomenon.

For HID fixtures with an electronic photocell and magnetic ballast, it can kill the photocell if the light is regularly switched off by other means when it is lit. An example of this would be a lantern wired into a switch in the house. When you go to bed, you turn it off and the inductive kick is sent in the PC. Repeat every night and after a while the PC will be toast.

This is how I see this phenomenon. If others have corrections to make in my explanation, feel free to do so.

Thanks yes that's what I was getting at the length of the choke voltage kick. I remember many years ago the school where my dad worked installed some new preheat fluorescents and when they were switched off the anciant tumbler switch would flash over  and once or twice it also blew the 5 amp fuse in the main box the switch was soon replaced. For those who might not know UK stopped using tumbler switches in the 1950's it was probably installed in the 30's

I've never heard of fluorescents behaving like that. What are tumbler switches.

A tumbler switch is a round Bakelite dome with a little hole in the !middle in which is mounted the toggle you use to operate it. I think its best if you google it you mite be able to find a picture. Early ones were rated at 5 amps later ones were 10 amps

I saw one on eBay I think a while back and it apparently glowed green.

Glowed green? Really I've never seen one like that

I think it was with a little neon bulb.

Yes probably was the one I'm on about never had anything like that they was just plane switches

I'll try to see ifu can find the listing later.

Can someone tell me how long does the voltage kick from the choke in a preheat ballast last for? Is it milliseconds or longer?

It depends, how high the voltage is allowed to go and on the exact current at the instant the starter opens.
To find it out exactly, you need to go through some math (and physics).
The ballast is an inductor, where the induced voltage is given by the faradays law of induction, which could be flipped to describe an inductor as a component would look like:
V = L * dI/dt.

When the inductor is connected to a voltage source (so the voltage is forced externally), the current will evolve in such way, the induced voltage equals the connected voltage source (well, at this point I neglect the wire resistance). So when you want to know the current, just integrate the equation, you will end up in an equation for the current:

I = 1/L * int(V * dt) + I(0)
The I(0) is the coil current at the "t=zero" moment, mathematically it is the integration constant. Practically it depend on the history before the evaluated event.

From that you may see, once the connected voltage becomes a nonzero constant (= DC voltage source), the current will rise over time to infinity, hence the "short circuit to DC" behavior).

Now in the case of the ignition event in a preheat ballast, at the moment the starter opens (let's assume that happens at t=0), the I(0) would be the instant current in the ballast at that moment (it could be anything between zero and peak current during the preheat, depend on when exactly the starter opens within the current AC cycle). Just when the starter opens, the current will find other way than the closed contacts (the immediate collapse can not occure, as it would mean an infinite voltage), normally it is either the lamp (some 100's V) or discharge in the starter (nearly a kV at this current). The pulse ends at the moment, when the current reaches zero. By evaluating the integral above, you end up:

Tpulse = I(0) * L / Vpk
Tpulse is the voltage pulse duration, Vpk is the voltage of that pulse; Assume constant voltage over the complete pulse.
 
Now what is the exact L and maximum I(0)?
Now as the current was formed by the mains (so an AC voltage source), it's peak value would correspond to:
I(0) = 1.414 * Vmainsrms / (2*Pi*Freq) / L

When replacing the I(0) for the equation for the current:

Tpulse = 1.414 * Vmainsrms / (2*Pi*Freq) / L * L / Vpk = 1.414 * Vmainsrms / Vpk / (2 * Pi * Freq)

And you see the ballast inductance had disappeared, so we don't care about it's value (ballast rating,...) nor changes with current (saturation,...), just the voltages and mains frequency.

So with 230V/50Hz mains and about 500V pulse (the cold electrode discharge across a not yet warmed fluorescent tube), you end up with about 2ms pulse.
With 120V/60Hz and 300V across the lamp (assume shorter lamp, used in 120V circuits), it would be about 1.5ms.
But when the lamp is missing and the pulse is limited by the starter (assume 1kV), you end up about 300us (for the 120V case).

Thank you medved I will not pretend to understand all the maths but I understand what your getting at you've been very helpful thanks again x Hannah