# Minimum limit for current and maximum limit for voltage while calculating power

I have a brushless Motor then i am using in conjunction with a ESC, powered by a Lithium Polymer Battery(3S 1300mah). I just got a new Lithium Polymer battery (6S 15000mah) from a friend and am planning on using it to power my motor.

From the motor details shown by the manufacturer here:- http://www.emaxmodel.com/views.asp?hw_id=15 , it shows the motor being rated at 3S 17A MAX. If i were to calculate it's wattage, it would be (4.2 * 3) * 17 = 219.3W, assuming a voltage of 4.2V per LiPo Cell. However, as 17A is the max rated current, and that the max continuous current is not given, i would be using 15A as an estimate, which would give me a wattage of 189 Watts.

While the motor is shown to be using 3S LiPo batteries, would i be able to power it with a 6S LiPo Battery with a lower current draw?( I'm assuming that the motor will draw less current since more voltage is supplied to it).

Am I right in thinking that i should be able to power the motor with a variety of voltages, as long as the power(voltage * current) does not exceed the max rated wattage of the motor?

If so, given an extreme example, would i be able to power the motor at 100V @ 1.89A(not that it would be feasible or would i ever do it), or is there a max limit to the voltage and/or a min limit to what the current supplied must be for the motor to work?

All thoughts, opinions and answers would be greatly appreciated.

• Have you got access to a proper pdf data sheet? The detail on the web page is very poor. For a start 17 amps can only be taken for less than 1 minute and no other current value is offered. This makes estimating normal running maximum power difficult. I will add that increasing the voltage doesn't mean current going down but up. A motor has no idea what power it is taking. Nov 12 '14 at 13:45
• Hi andy, i don't have the pdf data sheet for the motor, the best info i have is here :- modelaccessories.co.uk/emax-brushless-motor-specifications.html under GT 2210/13 where the no load current and power is shown. Am i wrong in thinking that the motor will only draw as much power as is needed? If i increase the throttle, it should draw more power and vice versa if i reduce the throttle. If so, if the votlage supplied is increased, shouldnt the current draw drop? Nov 12 '14 at 13:51
• The way to run a motor off twice the voltage, taking half the current, is ... rewind it with twice as many turns. Nov 12 '14 at 14:53

Also note that the speed of the motor is dependent on voltage ($K_V$) and that is 1270 rpm per volt applied. The more applied voltage the faster it spins and, for a fixed mechanical load, the power will increase proportionally with RPM squared. So, for a given mechanical load, power quadruples for a doubling of voltage.
This cannot be avoided because the motor spins faster with more voltage applied. Output power is $2\pi n T$ where n is revs per second and T is torque to the load and for a fixed mechanical load, a doubling in n also means a doubling in T hence power output to the load quadruples with a doubling of voltage.