Cox glow battery

If a 1V2 voltage source gives 2A5 current, that means the resistance of the plug is about 0R5. To drop from 2V2 to 1V2 you will therefore require something along the lines of 0R47 or 0R33 resistance in series with the plug.

Please note - the resistance actually varies with the plug temperature. The resistance is lowest when the plug is cold (just like with any glowing element, like a lightbulb for example).

The old books always say that you should have a variable resistance - resistive wire, or just any long wire.

Thank you both for quick and helpful reply.
Think I have got my head round it.
I'm using the same cell as Mr Cox, i.e. 5ah 2.2v. and want to drop
voltage to 1.5v, same as old dry cells we used to use.
Worked it out as follows:
glow head resistance 1.2/2.5 = 0.48ohm
Total resistance 1.5/2.5 = 0.60ohm
Dropping resistance 0.60 - 0.48 = 0.12ohm
Now I like to have ammeter in the circuit as it tells me when element is flooded
burnt out or glowing nicely.
So would like to get a variable resistor that will go down from a few ohms to about
0.1 ohm so I can fit it in series with my existing ammeter which probably not accurate
but know it will be OK when it reads 2.5amps.
A bit of resistance wire shortened till I get 2.5 amps should do nicely.
Thanks again for getting me there. John

I'm not sure where you got this from:
Total resistance 1.5/2.5 = 0.60ohm
but it doesn't seem right.

What you need to do is to drop 2V2 - 1V2 = 1V at 2A5. This gives you roughly 0.4Ohm, unless I'm missing something.

Golly, I'm a simpleton but I use a Cox battery pack to start my engines. 2 Dura-cell D cell flashlight batteries connected in parallel fires Cox engines just fine. If you REALLY understand Ohm's law you wouldn't bother with resistor wires.

You guys are making a simple thing difficult!

SuperDave wrote:Golly, I'm a simpleton but I use a Cox battery pack to start my engines. 2 Dura-cell D cell flashlight batteries connected in parallel fires Cox engines just fine. If you REALLY understand Ohm'd law you wouldn't bother with resistor wires.

You guys are making a simple thing difficult!

I think I might have to upgrade to two D-cells in parallels. I'm using a single AA battery... Barely works.

Dude:

That's because your smaller batteries don't have the amperage (amps) of "D cells".

(Amperage is a measure of a battery's abilty to sustain a load)

SuperDave wrote:Golly, I'm a simpleton but I use a Cox battery pack to start my engines. 2 Dura-cell D cell flashlight batteries connected in parallel fires Cox engines just fine. If you REALLY understand Ohm'd law you wouldn't bother with resistor wires.

You guys are making a simple thing difficult!

Using the D-cells is the most easy way. However in the netherlands you pay about $8,- for two good Duracell batteries. A 2V 5AH battery cost twice that amount, and after a night of charing, i can use it for a long...long time.

Those D-cell batteries where always empty, when I was not able to buy new ones!

Baastian:

Here in the US Dura-cell batteries are not nearly expensive so we buy them in quatity just to have them on-hand for our flashlights during power outages which are frequent during the winter months.

Bought in quantity, D-cells are lest than a dollar each.

Sorry Dirk Gently,
What I was trying to do was show total resistance in circuit with a 1.5v
cell such that current was 2.5amp.
i.e. R = V / I, 1.5 divided by 2.5 = 0.6ohm
Less resistance of glow head, 0.48ohm = 0.12ohm
If I got that wrong I'm sorry, please excuse this old duffer.
Anyway I now have my cyclone cell rigged up with a bit of resistance wire
and it is glowing nicely.
Promise not to bug anybody else with any more of my dodgy sums.
john

With those prices, it is a different story;)

You are right to keep a flashlight in the home!

Actually we have several three-D cell flashlights in our home.

They would also make excellent "attitude adjustment tools" should I ever encounter a prowler.

SuperDave wrote:Actually we have several three-D cell flashlights in our home.

They would also make excellent "attitude adjustment tools" should I ever encounter a prowler.

Now that is worth a or two!

Since I have so many glow engines, I decided to invest in one of these. Adustable voltage, 5200 mah li-ion. It feeds the glow plug the juice it needs by determining the health of the plug (resistance, ect). really works nice.

https://www.youtube.com/watch?v=0tP73WuHSOA
A quick vid I shot a couple months ago.

proctor wrote:Sorry Dirk Gently,
What I was trying to do was show total resistance in circuit with a 1.5v
cell such that current was 2.5amp.
i.e. R = V / I, 1.5 divided by 2.5 = 0.6ohm
Less resistance of glow head, 0.48ohm = 0.12ohm
If I got that wrong I'm sorry, please excuse this old duffer.
Anyway I now have my cyclone cell rigged up with a bit of resistance wire
and it is glowing nicely.
Promise not to bug anybody else with any more of my dodgy sums.
john

I drew the circuit out and I got about the same values you did.

RknRusty wrote:

proctor wrote:Sorry Dirk Gently,
What I was trying to do was show total resistance in circuit with a 1.5v
cell such that current was 2.5amp.
i.e. R = V / I, 1.5 divided by 2.5 = 0.6ohm
Less resistance of glow head, 0.48ohm = 0.12ohm
If I got that wrong I'm sorry, please excuse this old duffer.
Anyway I now have my cyclone cell rigged up with a bit of resistance wire
and it is glowing nicely.
Promise not to bug anybody else with any more of my dodgy sums.
john

I drew the circuit out and I got about the same values you did.

Of course, if you use the same input values, you will get the same output. But this:
i.e. R = V / I, 1.5 divided by 2.5 = 0.6ohm
should be
i.e. R = V / I, 2.2 divided by 2.5 = 0.9ohm
because you are computing the resistance of the whole circuit, and then subtract the resistance of the plug.

Thanks Dirk, got it, I need to calculate resistance of cell I,m using.
Will sleep tonight! John

Actually, that wasn't my point at all.
The resistance of any half-decent battery is very small an can be ignored.
(Sorry, I suck at explaining).

You have 2.2V (the voltage of your battery)
you need 1.2V (nominal operating voltage of the plug)
You need to drop 1V.
So you form a voltage divider: the plug an a resistor (and the battery resistance which is neglible, so let's skip it)
The voltage dropped across the plug / the voltage dropped across the resistor = resistance of the plug / resistance of the resistor
1.2V / 1V = 0.48V / x
Where x is the resistor value.
this means x equals around 0.4Ohms.
Typical resistors have 0.33 or 0.47Ohms (at least in Europe, in the US it used to be different, not sure how it looks like nowadays).

http://en.wikipedia.org/wiki/Voltage_divider
Hope this helps,
Cheers

Thanks Dirk,
I really do understand now, you explanation was very clear.
Sorry to have taken up so much of your time,and so much space on here. John

Not to be too picky but I too think the battery being discussed is a great choice as rechargable and inexpensive

BUT a lead acid CELL will always be 2 VDC and not the 2.2 being discussed. ( I looked up the data sheets on 6 different sizes of the Cyclon SLA cells from Enersys Cyclon)

I believ the voltage drop from 36 inches of 14 guage wire to the glow plug clip will bring the 2VDC into a safe 1.6 to 1.7VDC that will not fry a healthy glow plug element

http://www.tnrbattery.com/brands/Enersys-Cyclon-.html

"D CELL"
2V 2.5AH, Sealed Lead Acid Battery $7.25

The Hawker Cyclon 0810-0004 battery is unique in design. Its cylindrical shape allows it to overcome the limitations most lead acid systems experience. The cylinder shape of the 0810-0004 battery provides top performance without sacrificing cost effectiveness, reliability and long life. The 0810-0004 battery offers a sealed design which offers many benefits.

Yes the nominal voltage for lead acid cells is 2v and they are
all advertised as such but a fully charged cell can be anything up
to 2.2v I measured my Cyclon 5ah cell at 2.14v and I can't be
sure it is fully charged.
Ideally these cells should be charged at constant voltage, at a
float charge voltage of 2.3V until current falls to C/20 but if
you wanted such a charger for a single cell you would probably
have to make it.
I charge mine at 1amp until my $20 Hobbyking charger times out
and if I do this regularly after flying sessions I hope this will
keep it topped up.
John