stix2it

If a geared motor prop combo pulls 8.0 amps with a 9.6 volt pack, is there a formula for figuring out what the amps would be if you increase the pack voltage to say 11.1 volts?

bianca121

The formula is V=R*I, v=volts, R=resistance from the load in Ohms, I=current in A, so in your case, you have to find R with 8.0 Amps and 9.6volts,
9.6v=R*8.0A, so R=9.6V/8.0A , R=1.2ohms.....
with the 11.1 pack, 11.1v=1.2ohms*I, so to find I, 11.1v/1.2ohms=9.25A..

So it should be 9.25A

Somebody can correct me if i'm wrong......

exeter_acres

My Feedback: (2)

http://www.the12volt.com/ohm/ohmslaw.asp

Semi Retired Aviator

When I did electronics/radio school in the Air Force about 40 years ago, I found this a simple way to figure it out:

E = elctromotive force (EMF) or Volts, I = Amps, R = Resistance in Ohms

I (Amps) = E/R i.e, EMF (Volts) divided by Resistance (Ohms)

E (Volts) = I X R i.e., Current (Amps) multiplied by Resistance (Ohms)

R (Ohms) = E/I i.e., EMF (Volts) divided by Current (Amps)

I never did figure out why Amps weren't A instead of I though, and worth remembering, 5 amps is written 5 A, not 5 I !!

rcs r us

oh jusus!please im 12 years old speak english!

rcs r us

i know this isnt the right room for this but like i have a transall c-160 its white.it has landing gear,but its a park flyer.is that still good its also only 2 channel???[]

Semi Retired Aviator

rcs r us, Are the formulae still good?? Or what's still good??

The formulae will always be good, no matter what you fly, with or without gear, flaps or anything else; Volts are volts, ohms are ohms and amps are amps, always!

And another one, the most useful formula for Power is P (Power) = I squared X R, i.e., Power (Watts) = Amps squared multiplied by Resistance (ohms), but Power also can be arrived at by multiplying Volts by Amps.

If you use the values of 12 volts, 4 amps, and 3 ohms, you can play around with the numbers until you become proficient.

Oh, and this is English.

Lazyflier

We must have learned from the same books! I had one that gave it this way E
over
I R
That's in english also.
<G>
E= Volts
I = Amps
R- Resistance

Don't know why I was A either.

Matt Kirsch

My Feedback: (21)

As I recall it from college:

E stands for Electromotive Force, another term for Volts.
I may stand for... oh crap, I can't remember.

rcs r us, don't worry about it not understanding. 12 is 7th grade, and you're just getting into algebra.

Have you done proportions? You know 1/2 = X/4? Figuring out what the Amps would be with a higher voltage can be expressed as a proportion too:

8A/9.6V = XA/11.1V

and the answer is?

CGRetired

My Feedback: (1)

Whoa... hold on a sec. First of all, let's think of watts, or power consumed. Just like a light-bulb.

The question was 8 amps with 9.6 volts.

Watts (P) = Volts (V) times Amps (A) , or That comes out to 8 Amps X 9.6 Volts = 76.8 watts.

Work out the formula with algebra.

Now, suppose you want to operate at 8 amps but with 11.1 volt LiPo. Then, using that formula, P=VxA, 11.1 volts x 8 amps = 88.8 watts, so you can operate a motor that has about 15% more power at the same current of 8 amps but with an 11.1 volt battery pack (three cell LiPo for instance).

flyingace451

This is English, and the question didn't involve you. As you get into higher planes outside of 2 or 3 channel rtfs, you will need to learn about Amps and Volts. Please wait until you take science class before you let your temper flare here.

As for your 2ch plane, any plane that has only 2 channels is way on the low end of the spectrum here, meaning that is the most basic you can get. It might be a decent flyer, I wouldn't know because I've never flown it. Get yourself something a little better when you step up to 3 channel like a HobbyZone Aerobird Challenger or other plane similar.

saitofreak

The relationship between V and I for an electric motor is reactive (ie. not just resistive). Depending on the motor, you may end up with large errors if attempting to calculate just using ohms law. Not to mention that the load varies with prop size and altitude. I'd invest in a range of props, a good watt-meter and measure the system in use to determine the parameters.

styrenejunkie

My Feedback: (3)

I think that the "I" in Ohm's Law (expressed relationships between current, voltage and resistance) stands for "intensity"...at least that's what we teach the students here in our A&P school.