Need some electrical engineering assistance

[BuckeyeDennis]

A 1ohm resistor in series with a 12volt supply will draw 12amps of current or said another way will dissipate 12watts of power.

E=IR while P=IE Where E = volts dc, P= power in watts, I=current in amps and R=resistance in ohms.

Example: I=E/R = 12v/1ohm = 12 amps

What am I missing?

Yes, if the 1 Ohm resistor were the ONLY load, it would draw 12A. And 12A x 12V = 144W = SMOKE!.

But the electrolysis tank is also in series with the 1 Ohm resistor, and the plan is to tune it to draw less than 500mA, as that is the most that the wall warts can supply. So the worst case power dissipation in the resistor is 0.5A * 0.5A * 1.0 Ohm = 0.25W.

[JPG]

A 1ohm resistor in series with a 12volt supply will draw 12amps of current or said another way will dissipate 12watts of power.

E=IR while P=IE Where E = volts dc, P= power in watts, I=current in amps and R=resistance in ohms.

Example: I=E/R = 12v/1ohm = 12 amps

What am I missing?

The resistor is connected in series between the source and the load.

Supply+........Resistor.......Anode......Electrolyte.......Cathode.......Supply-

The voltage drop across the resistor will be determined by the source voltage and the ratio of the resistor and load.

VR = Vs x (RΩ / RΩ+LΩ)
V=Voltage
R=Resistor
L=Load
Ω=resistance
s=source

[JPG]

Yes, if the 1 Ohm resistor were the ONLY load, it would draw 12A. And 12A x 12V = 144W = SMOKE!.

But the electrolysis tank is also in series with the 1 Ohm resistor, and the plan is to tune it to draw less than 500mA, as that is the most that the wall warts can supply. So the worst case power dissipation in the resistor is 0.5A * 0.5A * 1.0 Ohm = 0.25W.

Put another way,

P = IE

E = IR

Soooo P = I²R



Also I = E/R

Soooo P = E²/R

Just make sure all those parameters are all relevant to the same component(s).

[heathicus]

I REALLY wish I understood electronics better!

I should have added to the original post, "Explain it to me like I'm 5."

But, I think I've gathered some important points. Multimeter in series with the electrolysis set, then adjust the distance between anode and cathode until mA reading is ~400 or so. That makes sense to me. "P = IE" just makes me hungry for pie.

I'm going to read all the above posts a dozen more times and hope more of it starts to click.

[frank81]

Have you considered making the power source one of the controls rather than choosing the most difficult variable? You could change your variable to the amount of water in the tank (dispersal), amount of sodium carbonate (concentration), or changing the catalyst?

Electrolysis can occur with any salt, doesn't have to be washing soda that just gives the quickest and cheapest result for removing rust from ferrous metals. And you can choose any two conductive materials for the electrodes. Just be careful you don't choose something that creates a toxic byproduct, like stainless steel. Also don't forget, the form of electrolysis you are planning gives off hydrogen gas bubbles so careful of ventilation or leaving the experiment running in a classroom overnight.

A quick example...ocean water + aluminum hull boat + steel bolts to hang the motor - any insulation from eachother = warranty claim..errr I mean electrolysis.

In the end you just need a science experiement he can write a report on to show he understands it...the teacher doesn't care about rust removal like we do!

[spiderclimber]

I may be off base here, but I thought you were looking for a way to adjust the voltage using one source. The best way I know to do this is by using a computer power supply. You have 3 v 5 v and 12 v coming out of this. You also have a few other odd voltage wires in there as well. It has a built in short circuit breaker that shuts it down if you have a short or an issue. Just a thought as you can get one pretty cheap if not free. It will take you about 10 minutes to hack it into a usable supply. You can't regulate the amps this way but the voltage you can.

your wires you need are yellow, orange and red for your voltage. the black will be the neutral for each. You will also need to trick the CPU into thinking it has a motherboard attached. Take the one green wire and pick any black wire in the harness and connect them. Should give you fairly consistent and even voltage to your project. As well, most supplies are rated to 10 to 15amps so you get better output than a wall wart.

Hope this helps. Probably won't be back on here in time to see a question before your weekend is over, so good luck. Youtube hacking a computer power supply if you need help doing it.

[JPG]

OK I hit you with a lot of ifo.

Here is the simple approach as I see it.

Keep everything except electrode separation the same.

Common power supply.

Common container(3 of them)

Common electrolyte mixture and volume of each

Identical test pieces and sacrificial parts.(3 each)

For current setting and monitoring, 'momentarily' insert a low value resistor in series with each test sample(either + or - as long as multimeter polarity is observed when connecting the meter across the resistor).

The current should be variable by changing the electrode spacing.

Remember (-) to test sample, (+) to sacrificial part.

I leave the power supply choice to you, but wall warts are likely too small.

Battery charger is likely safer than a car battery, but having both connected will produce the fastest results.

BTW Oxygen is also being released as well as hydrogen.

[heathicus]

I had considered a computer power supply. I actually have a couple stashed back just for the purpose of converting to a "lab" power supply. But, different amps are supplied with the different voltages and I felt that would invalidate the test. If differences were notable, then was it the voltage or the amperage that made the difference? So we thought it best to make sure one of the values stayed the same across the experiment. We had decided on voltage as the variable and amperage as a constant, but I'm still unsure about that decision. Time to change it is just about out, though.

I did present some of the other variables to him as possibilities for the experiment, such as those you mentioned, frank81, but this is what he wanted to test.

I'm going to ruminate on everything that's been said here.

Quick question, though. How do I tell which wire from the wall warts is positive and which is negative?

[JPG]

I had considered a computer power supply. I actually have a couple stashed back just for the purpose of converting to a "lab" power supply. But, different amps are supplied with the different voltages and I felt that would invalidate the test. If differences were notable, then was it the voltage or the amperage that made the difference? So we thought it best to make sure one of the values stayed the same across the experiment. We had decided on voltage as the variable and amperage as a constant, but I'm still unsure about that decision. Time to change it is just about out, though.

I did present some of the other variables to him as possibilities for the experiment, such as those you mentioned, frank81, but this is what he wanted to test.

I'm going to ruminate on everything that's been said here.

Quick question, though. How do I tell which wire from the wall warts is positive and which is negative?

Doing that will prove that the voltage will not affect the result as long as the current is the same!!! i.e. There will be only incidental differences in the results of all three samples.

ALL the work is performed by the current. More current = more results etc.

The hardest thing to do when determining what to 'do' is to make sure only one parameter is varied. You already got the list of things(parameters) to control/vary between samples.

Put yer multimeter on a high dc voltage range. If analog touch the leads to the output terminals/wires BRIEFLY. Observe needle deflection. You can figure it out from there. If digital, it will indicate polarity or an error.

Only do that after determining the wall wart does not tell you on a label.

The +12v portion of a desktop computer supply should provide more than enough current.

If not. the +5 volt portion will.

All else being equal, either the voltage or the spacing will affect the current. By using a common power supply (voltage) only the spacing will affect the current. Even if the voltage varies during the test, all three samples will be equally affected and any current changes will share a common ratio to each other.

I do think as the rust is depleted, the current on each sample will vary, but that will also be a constant for each sample, just not at the same time interval.

[dusty]

I had considered a computer power supply. I actually have a couple stashed back just for the purpose of converting to a "lab" power supply. But, different amps are supplied with the different voltages and I felt that would invalidate the test. If differences were notable, then was it the voltage or the amperage that made the difference? So we thought it best to make sure one of the values stayed the same across the experiment. We had decided on voltage as the variable and amperage as a constant, but I'm still unsure about that decision. Time to change it is just about out, though.

I did present some of the other variables to him as possibilities for the experiment, such as those you mentioned, frank81, but this is what he wanted to test.

I'm going to ruminate on everything that's been said here.

Quick question, though. How do I tell which wire from the wall warts is positive and which is negative?

Just be aware that if you change the voltage the current will change, if you change the solution the current changes, if you change the electrodes the current changes.