[chbot] Choc Fish Challenge #5

[Cass Jones]

Perhaps then they've given you a 48v battery and two LED's soldered in
parallel in oposite polarities with a 2k resistor in series,
All behind a smoked glass lens so you can't tell which is lighting.
The LEDs will still light visibly with 4.4kOhms at 48 volts and will
work without burning out at 2kOhms.
Ranging from 24 mA to 10.9 mA current through the LEDs.
The same device scales up to 300 cables too if you use 2 48v batteries
in series, perhaps they've glued them together and soldered the
terminals together as well.

If you want to go poking holes in the problem.
=P

On Mon, Sep 5, 2011 at 10:47 AM, hamster <hamster at snap.net.nz> wrote:
> Humm, interesting....
>
> Telephone cable has > 20 ohms/kilometre, so if you separate into two
> bundles, jumper pairs together allowing you to form into one long chain
> that you can analyse you will have at least a 2400O Ohms in series with the
> bulb (i.e. the bulb won't light)
>
> You also need some way to disconnect the battery so you don't short it
> out...
>
> Mike
>
>
> On Mon, 05 Sep 2011 10:13:20 +1200, m.beckett at amuri.net wrote:
>> Upon reflection we have walked too far.
>>
>> We believe only 2km is required, and a third to remove the jumpers.
>>
>>
>> Mark
>>
>>
>
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On Mon, Sep 5, 2011 at 12:36 PM,  <m.beckett at amuri.net> wrote:
> Okay,
> My esteemed college has asked the question.
> Can you use the exchange in this solution...if so he believes its only
> 1km
>
>
> I don't think its fair, as it requires inside knowledge for that
> one....
> Mark
>
>
> On Mon, 5 Sep 2011 11:54:02 +1200, Synco Reynders wrote:
>> To clarify: Wire resistance, battery voltages, cross-talk, ... can be
>> ignored.
>>
>>
>
>
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