Depends on what solenoids you are using

 

Yep but most systems if you figure 20 per stage you will have be O-K , this figure is for average consumption, reactance, overcoming the nitrous pressure acting on the seal etc will cause a momentary demand above this in many cases.

PS the "20" is amps the voltage it uses will be whatever the battery has available minus the voltage lost in the wiring (thats why using adequate thickness wire is so important) the multiple of voltage times amperage will equal current in watts so 12v at 20 amps will be 240 watts.

 

Ohms Law I=V/R

I = Inductance(Amperage)
V = Volts
R=Resistance(Ohms)

Get a voltmeter, set it to measure ohms. Measure the resistance between the two wires on the solenoid.

Divide your battery voltage at the solenoid by the measured ohms. This will give you the amperage used for the measured solenoid.

Do this with each solenoid, add them together and that is the total amperage.

This takes the guessing completely out of the equation, pun intended.

Some people will tell you it takes more power to overcome higher bottle pressures. It is electrically impossible. The only thing the solenoid cares about is the resistance to ground.

 

Ohms Law I=V/R

Get a voltmeter, set it to measure ohms. Measure the resistance between the two wires on the solenoid.

Divide your battery voltage at the solenoid by the measured ohms. This will give you the amperage used for the measured solenoid.

Do this with each solenoid, add them together and that is the total amperage.

This takes the guessing completely out of the equation, pun intended.

Some people will tell you it takes more power to overcome higher bottle pressures. It is electrically impossible. The only thing the solenoid cares about is the resistance to ground.

I disagree with most of that statement. Resistance can change drastically with heat. Put your multi meter on a 100 W light bulb and see if the above holds true?

 

Its ohms law, take it or leave it that is how it works. Unless you know something that no one else knows about the properties of electric.

You can throw as many assumptions into it as you want, temp constant or temp varying. The deal is if resistance and or voltage changes, then amperage changes too. There is no way around it. If resistance goes down like in your light bulb example, the amperage will change too.

Here is a link to a site that has any way you want to solve for amperage/current, wattage/power, resistance and voltage.
http://www.the12volt.com/ohm/page2.asp

The Temperature Coefficient of Copper (near room temperature) is +0.393 percent per degree C. This means if the temperature increases 1°C the resistance will increase 0.393%.

Here is a table for temp coefficient based on materials
http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv.html#c1

 

Its ohms law, take it or leave it that is how it works. Unless you know something that no one else knows about the properties of electric.

Not trying to start a pissing match, just don't like it when people hand out incorrect information. Measuring the resistance of a coil tells you nothing about the current draw of that coil after the first few milliseconds.

I agree it is not voodoo, but it can seem like it when people use equations incorrectly.

As stated above, if you want to know for sure, put an ammeter on it. You cannot measure resistance of the coil when it is powered. Not only will you not get the correct reading, you may damage the multimeter.

 

Really I grew up on a farm and have a computer info systems degree. What I want to show people is that amperage draw is not some vodoo shit, its all based on simple math.

 

It isn't Vodoo, it just has way more variables that will come into play then just punching #'s into a calculator. Cause if calculators were always right then Desktop Dyno says my stock 302 with a cam swap will make power to 9600rpm. lol

 

Yea if the desktop dyno was accurate, there wouldn't be much use for flywheel dynos.

The math for electric is pretty simple especially compared to deriving a torque number from a system of parts. It was a couple of years ago I was working on the Toyota FC (Scelzi was driving it at the time). We had TRD and Swift engineering put together some formulas to 'predict' the cars ET based on how we tuned the car. It was pretty impressive to see all the inputs. We used it mainly on the Winston dragster. It was pretty damn accurate. It didn't work to well on the FC, mainly there wasn't an input into the formula to take into consideration the body was a POS lol .

 

i may be wrong but here is what i thought , the only thing that causes the heat in a wire , is when you demand the flow of electrons to be greater than expected causing friction and thus causing heat?

Dan

 

Explain it for everyone then if that is incorrect.

 

I'm on a slow connection in a hotel room tonite and lost connection while making the last post.

Not being a wise ass or starting a pissing match. I understand the meaning of using an ammeter for sure, not all have a very accurate meter to measure resistance at low levels. Like was mentioned earlier temp does play a role and I listed the number to use to account for it. I didn't mention the inductance of the coil that does cause an initial 'spike' in amperage draw. However this is milliseconds in time and then it goes completely into the simple Ohms law equation, with temp varying the resistance of course. This spike has nothing to do with the pressure against the poppet valve, its the electric charging the coil.

If I am missing something, fill myself in and everyone else too. By no means am I an engineer as I caught hell for earlier lol.

 

OK, so whats the temperature of the windings?

It would be much easier to calculate the resistance of the coil by measuring voltage and amperage than to measure the amp draw by measuring voltage, initial resistance, initial temperature, and working temperature. The OP wanted to know amp draw.

But yes, you are completely correct in everything you said. My disagreement was not with your math, it was in your application of it.