Trevor Smith wrote:

I know the current that the kit draws will be needed to get the right
capacitor for the right time but I haven't checked this yet, and I
wouldn't the formula to work it out any way so I was going to go for
the biggest. Any advice would be appreciated

I echo the comments from NT and Dave, but if the current is low enough
it might work. You need to know the current (I), the time for which the
power will be off (t) and what drop in voltage might be acceptable
(delta_V). Then the required capacitance is I * t / delta_V (with units
in amps, seconds and volts, the answer will be in farads).

So for example if I = 100 mA (a wild guess), t is say 3 seconds and
assume delta_V of one volt then you'd need 0.1 * 3 farads, i.e. 300,000
uF, which is rather a large capacitor - large enough, possibly, for the
initial charging current to blow a fuse, unless you take precautions.

--
Andy