Calculating resistor rating for simulated load of X amps.

You do know the other variable - you want to draw 35 amps. So, 4/35, or a little more than a tenth of an ohm. Since you'll be pulling 35 amps x 4 volts, that will need to be a 140 watt resistor.

To make things a little easier, I would try for a 10 to 20 amp load. Since you're comparing packs, that should give you just about the same results.

I doubt the charging port is meant to handle the level of current you're talking about.

- David Herron, http://www.7gen.com/

In any case that will be a big resistor. lets see 15 amps at 3.6 volts. R=E/I 3.6/15=.24 ohms and 3.6*15=54 watts. http://www.digikey.com/scripts/dksearch/dksus.dll?Detail?Ref=169700&Row=780895&Site=US two of these in parallel would work out nicely. Kind of expensive at 7 bucks a piece, and becareful theyll get hot.
Joe

power resistors typically have lousy temperature coefficients, so when the resistor gets hot its value will change thus throwing off your measurement. as the battery voltage lowers the current will drop, another source of error. it is very easy to build a load that will be constant and not move around with temp and voltage. all you need is an op amp and a fet that can dissipate the power you need, and a sense resistor. Suppose you have a 10 milliohm sense resistor, that would have 350 mv across it at 35A. you use a fet to control the current through the sense resistor, then you feed the voltage across the resistor to the summing input of the op amp and 350 mv to the other input and you have a programmable load. you can cut down on the power dissipation of the FET(s) by using a series dropping resistor. 4 volts will push 35A through .114 ohms, so if you used a .1 ohm 150W resistor in series with the FET then the FET would only have to dissipate 20W so a small heat sink would be enough. The FET would keep the current at a constant 35A because the sense resistor has a better tempco and it will not heat up anyway.

-bob

Good idea building a constant current source, i didnt even consider the fact that the load wouldnt be constant. Would be a lot more expensive than just using a resistor. I usually just use a zener diode supplying a constant voltage to the base of a transistor to do this.
A bipolar transistor would serve better than a fet using this method. The minimum voltage you can apply to the CC load is the sense resistor voltage plus the gate to drain or base to emitter voltage. In most fets you either have 10 volts or 5 volts gate to drain plus a certain hundred or so millivolts in the resistor. With a bipolar you have 1.2~ volts plus the sense resistor. Most big bipolar transistors also have low gain so a seperate transistor hooked up in darlington will probably be needed. Your method using an op-amp would get rid of this problem but the op-amp would end up needed a power supply other than the batteries being drained themselves, a 9 volt battery would suffice and would power an op-amp for a long time. Just a thought.
Joe

Similar but different question. Apologies in advance for my limited knowledge.
I'd like to simulate a load from between 40 - 150 watts @ 110v. I need to trick my shop vac into thinking it's got a power tool plugged into it that will trigger it's auto start feature. I do this with a foot actuated switch which I will plug the resistor(s) into.
Many thanks,

-C

Use an ordinary incandescent light bulb!

For a larger dummy-load, you can use a toaster or other plain resistive appliance - i.e one without a motor.