Originally Posted by rwidman
It is correct that a series wound motor develops the most torque and draws the most current at zero RPM
but only if the input voltage remains relatively constant. If there are corroded connections (resistance), this resistance is in series with the motor and the voltage is divided between the resistance of the connections and the motor. Introducing series resistance into the circuit will make the current in the circuit less, not more.
This has always been (for me) an interesting conundrum.
Simple circuit analysis suggests that an increase in circuit resistance will decrease current flow and of course this is true - ohms law is always true!
However when considering a series wound motor in a simple circuit, there is a sweet spot or perhaps better put a "sour spot" when the inclusion of some series resistance causes a short term increase in current flow to the point where the breaker operates.
As best as I can understand/explain it, this is what happens.
operation has full supply voltage at the motor terminals, the current flow is a function of the load resistance of the motor which is directly proportional to the RPM (higher RPM, higher load resistance, lower current).
When a full working load is applied to the winch
, the designed RPM is such that the resultant current flow just a little lower than the circuit breaker. This is lowest RPM the winch should normally be allowed to operate at.
Now add some smallish series resistance by way of say corroded terminals, the voltage available at the motor terminals is reduced and therefore current flow decreases thus reducing torque, this causes the RPM to drop significantly (as load on winch remains the same). This significant drop in RPM causes a significant drop in the load resistance and therefore total circuit resistance has momentarily decreased
(even with the addition of the "corrosion resistance"). The current flow increases to a point beyond that of the breaker capacity.
Of course, add too much "corrosion resistance" and Ohms law will dictate a small current flow regardless.
But maybe I'm wrong