I have a question concerning motors and voltage drop. It is said that a motor will draw more as the voltage is decreased ( say from 120V to 115 volts and less to a point) The motor will try to maintain the design HP output. What happens when you have a very significant resistance such as a smaller and longer extension cord, How does this all come into play with total impedance of the motor with the addition of another significant resistance. I was told that the ampacity would increase so much that a breaker would flip, due to the fact the motor will draw more. On the other hand I am looking at the line loss on the way to the motor that would limit current as a whole. Other than physically trying this (which I may do)This all stems from a friend of mine telling me too small of a conductor will cause excess current in a line with a motor. Does anybody have any comments on this?
The voltage available at the motor will be less, so the current will be higher. That can be enough to trip the breaker.
Now onward to the confusion... When STARTING, the current draw at a lower voltage will be less (that's why we have reduced voltage starters). When the motor is RUNNING, the current will be more. At some point in that whole process; you'll likely exceed the rating of the breaker and whammo.
Start with a hypothetical 100% efficient motor. The electrical power input exactly equals the mechanical power output. Now reduce the input voltage. If the mechanical output were to remain constant, and the efficiency were to remain constant, the current drawn would need to increase.
_Real_ motors are not 100% efficient. Change the voltage and you change the efficiency. Real motors draw reactive power; change the voltage and you change the reactive current flow. Real motors will change speed when the supply voltage changes, possibly reducing the mechanical load. Real motors can _stall_, ending up right at that starting current situation described above.
Depending upon the specifics of the situation, a reduction in supply voltage can do anything from slightly reduce the current drawn by the motor, all the way up to causing the motor to stall and draw many times more current.
Think of a simple series circuit containing 3 resistances: R1 can represent the resistance of one lead wire (L1), R3 can represent the resistance of the other lead wire (L2 or Neutral), and R2 can represent the impedance of the motor. As you know, the total voltage drop accross the entire curcuit (equal to the source voltage) will be split proportionally accross the 3 resistances.
So if the resistance values of R1 and R3 are increased, say by providing smaller wire, then the voltage drop accross R1 and R3 will increase proportionally, and the voltage drop accross R2 (the motor) will decrease proportionally.
The result is the motor will see less voltage, and will draw more current to maintain the connected power requirement.
Hope that helps some. Radar
There are 10 types of people. Those who know binary, and those who don't.
Yes I can see where you all are coming from. I just was thinking that with a significant voltage drop along the lines to the load that there would not be a large draw as a whole due to the resistance of the entire circuit. I guess if the voltage drop is large enough, and if it was not for the oc device we would have major heating of the winding... and this is a motor and not a baseboard heater...........
Yes, motor loads react differently. The issue is, motor torque is reduced by the square of the voltage reduction. So if you end up with just a 20% voltage drop for whatever reason, you have 80% voltage and your motor output torque just dropped to .80 x .80 = 64% of rated torque! Since it isn't likely that the load changed, the lower torque allows the motor to slow down more, meaning increased slip. More slip means more current draw to try to regain speed. In the worst case, more current draw means more voltage drop, means more torque loss, means more slip, means more current, more voltage drop etc. etc. until the breaker or overload relay trips.