a DC electric motor moving a car. will it experience the same torque with current provided from a 4V battery with a capacity of 1000 Ah as with current provided from a 400V battery with a capacity of 100 Ah? according to math, P=V x A it should be the same, but would 4V be enough to move that car, even though it is supposedly moving so much more current (same C rates)? i have read about "energy efficiency" where higher voltage batteries are more efficient deliverers of energy as opposed to higher capacity ones, but don't understand why.
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2 Answers
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Any given motor will develop torque proportional to current. However, within some reasonable power range motors can be built with considerable latitude in their voltage versus current tradeoff. For example, a 1 kW motor could be made for 100 V 10 A operation, or 200 V 5 A operation. Once the motor has been chosen though, you're stuck with that tradeoff.
This tradeoff doesn't go infinitely in both directions due to practical limitations. If you use too low of a voltage, then the current becomes so high that conductors need to be thick and expensive. Note the losses is wires goes with the square of the current, or put another way, for the same power loss you need copper proportional to the square of the current with DC. With AC skin effect gets in the way and makes things more complicated.
At the other end, high voltages become difficult to handle and keep safe. Above a few 100 Volts, electronic components get a lot more expensive and you have significant safety issues.
answered Jun 27, 2013 at 12:17
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\$\begingroup\$ so is it current (A=C/s) that moves the electric motor? what is the meaning of power here? P=V*I so a higher voltage would need less current for the same power output. BUT, if the battery provides a higher voltage (given the same resistors/wiring are used) it also provides higher currents so it seems to me it is the current that flows through the motor (which creates the magnetic field which actually turns the axle) that causes the car to move. the voltage provided by the battery only need to be high enough to ensure a high enough rate of current. \$\endgroup\$ Commented Jun 27, 2013 at 13:06
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\$\begingroup\$ @guywi: For any one motor, higher voltage means higher current (all else being equal), which means higher torque. My point was that motors can be made for the same power with some latitude in their voltage and current combinations. \$\endgroup\$ Commented Jun 27, 2013 at 13:14
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\$\begingroup\$ thank you for the explanation. i have a lot to learn. in your example, how can 5A provide the same torque as 10A? \$\endgroup\$ Commented Jun 27, 2013 at 13:22
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\$\begingroup\$ ok i guess if the radius of the motor is varied, 5A can deliver the same torque as 10A. I still don't understand what the voltage rating means from the point of view of the electric motor, since the only thing it seems to care about is electric current which establishes its torque. Isn't the voltage only important on the delivery side (for the battery, such that it can deliver the current needed for the motor)? In that case, the motor always wants the most current possible, but the delivery system (due to losses) always wants the smallest current possible. Am I right? \$\endgroup\$ Commented Jun 27, 2013 at 14:11
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\$\begingroup\$ @guywi: Coils can be made that are about the same size, but with less turns of thicker wire or more turns of thinner wire. The thicker wire will require more current but at a lower voltage than the thin wire, when both are creating the same size magnetic field. This same concept can be applied to the windings in a motor to target it for high current and low voltage operation, or low current and high voltage, with both being able to deliver the same output power. For the same motor, 5A will result in half the torque as 10A, but I'm talking about different motors with the same power. \$\endgroup\$ Commented Jun 27, 2013 at 16:11
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Transferring the same power at higher voltage means less current is flowing, P=V*I.
Given that the loss is due to (I^2)R due to resistance of the cables, internal resistance of the battery and resistances in the motor windings, you want to keep the current as low as possible, therefore you want to keep the voltage as high as possible.
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\$\begingroup\$ thank you. i am confused about this, even though i am familiar with the concept. if we keep the resistors fixed, based on I=V/R, a higher voltage will draw a higher intensity of current. so then how then can a higher voltage command less current in P=V*I? \$\endgroup\$ Commented Jun 27, 2013 at 13:01
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\$\begingroup\$ @guy: do the sums with
P=100V=2 andP=100V=20. WhatIdo you get fromP=V*Iin the two cases? Note that this requires two distinct motor designs of the same wattage. \$\endgroup\$ Commented Jun 27, 2013 at 13:22 -
\$\begingroup\$ yes i get that. based on P=V*I, 20V will only need 5A, whereas 2V will need 50A to achieve the same 100W of power. my question is, given the same wiring/resistance, what will stop the imaginary battery operating at 20V from delivering 10x current, so 500A instead of 5A? \$\endgroup\$ Commented Jun 27, 2013 at 13:32
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\$\begingroup\$ The 20V battery will only deliver the current into the load as given by I = V/R. \$\endgroup\$– MartinCommented Jun 27, 2013 at 14:26
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\$\begingroup\$ @Martin what does that mean? V=20V, R is the same as it was for the 2V battery. Doesn't the 20V battery give substantially higher current? \$\endgroup\$ Commented Jun 27, 2013 at 14:38