DC cirquit quiz. Impact of Voltage on current

[elias]

Hi everybody!
Lets say that we are providing
a) an electric motor (e.g. windlass or pump) that is rated 1000w with dc current and voltage 12.5 V. This means that a current of 80 Amps will flow through the connections. If we provide it with a voltage of 14V will it drop 71.42 Amps right? The other way around if we provide 12v then 83.33 Amps?

So when we are calculating the DC cirquit should we assume that the rated power (1000w in the example) of a motor is fixed and the cirquit will do what is necessary to provide the energy so the current will raise or drop depending on the voltage?

b) is this apply to simple resistance loads eg. an electric water heater element?

[Rampage]

I always use the nominal voltage (ie 12 volts) for circuit calculations. 
However, for the windlass, which is the biggest draw on the boat, I use 11.5 volts for calculating the fuse size so as to provide a margin to cope with depleted batteries after a few days on the hook.
Resistive loads use the same calculations but don’t have a “start up” surge in the way motors do.

[cegri]

...
b) is this apply to simple resistance loads eg. an electric water heater element?

"P=V*I" does apply to simple resistive loads. So if you increase the voltage, the current and the power will increase. (I=V/R)
(Unless there is a current limiting component installed. Very rare for water heaters.)

For motors it'a a bit more complicated. P=VI applies. However, the power also depends heavily on the torque on and speed. (For a pump, this corresponds to pressure and flow rate.) Under the same conditions, if you increase the voltage, the current and power will again increase, and the motor will settle at a new torque and speed equilibrium.

For your windlass; the power output will increase if you increase the supply from 12.0 volts to 13.0 volts. You can easily hear and feel the difference between running it with engine running versus batteries only.

[elias]

Hmmm enlighting...
After this comment i made some digging in an electronics forum and found this


" What may be causing your confusion is the mistaken belief that V is the same as V, that I is the same as I, that R is the same as R, and the P is the same as P. What I mean is that you need to understand what specific values of voltage, current, resistance, and power are being considered in any given equation. The V you use in one equation might relate to the voltage at the energy source, or the voltage dropped along the circuit wires, or the voltage at the load location. Just showing the letter V does not convey all the information that is needed.

The other thing you need to keep in mind is that for different types of equipment, the thing that might change, the thing that might respond to that change, and the thing that stays constant throughout will be different. For example, in a resistive load, such as a light bulb or a space heater, it is the resistance that always remains the same. (OK, so the resistance changes as the device heats up, but for the sake of the present discussion let's ignore that.) So if you were to double the voltage applied to the heater, you would double the current. On the other hand, for motors, it is generally (again not precisely, but it's good enough an approximation for this discussion) the value of power (P) that is constant. So if you were to double the voltage applied to the motor, you would get half the current flowing.
 "