On Wed, Jun 03, 2009 at 02:39:29PM -0700, Don White wrote: > > Wouldn't the wattage be lower than that with the resistor in series > with the motor. The resistance of the circuit is the resistor plus > the motor.
Rough calculations: # E / I = R 12v / 1.5A = 8 ohms (motor alone) To drop a third of the voltage, I'd need a 4 ohm resistor; to drop half, I'd need an 8 ohm one. Assuming the smaller resistor first: # Series R for motor + resistor 8 + 4 = 12 # Series current 12v / 12 ohms = 1A # P = EI 1A * 4 ohms = 4W # Safety factor 4W * 2 = 8W For the 8 ohm resistor: 8 + 8 = 16 # Series current 12v / 16 ohms = 9A 12v * 0.75A = 9W # Safety factor 9W * 2 = 18W So, from 8 to 18W. -- * Ben Okopnik * Editor-in-Chief, Linux Gazette * http://LinuxGazette.NET * _______________________________________________ Liveaboard mailing list [email protected] To adjust your membership settings over the web http://www.liveaboardnow.org/mailman/listinfo/liveaboard To subscribe send an email to [email protected] To unsubscribe send an email to [email protected] The archives are at http://www.liveaboardnow.org/pipermail/liveaboard/ To search the archives http://www.mail-archive.com/[email protected] The Mailman Users Guide can be found here http://www.gnu.org/software/mailman/mailman-member/index.html
