# The effect of the resistance of coils on DC motor

I have some doubts on how the resistance of coil affects the efficiency of a DC motor.

I assume there are two DC motors operating under the same supplied voltage, and the resistance of the coil $$R_1$$ is higher than the other $$R_2$$.

The current through coil 1 $$I_1$$ then should be lower than $$I_2$$ as the voltage is the same.

Then the torque produced on coil 1 is less than on coil 2.

Then the rotational speed of coil 1 is less than on coil 2.

However, the lower rotational speed leads to a smaller back emf, then the current in coil 1 increases again.

Then I am unable to analyse the effect of resistance of coil on the efficiency of DC motor.

• "The current through coil 1 I1 then should be lower than I2 as the voltage is the same." - are both DC motors stalled? Oct 5, 2018 at 2:08
• @AlfredCentauri Assume they are working normally. Oct 5, 2018 at 2:14
• Then I don't follow your reasoning. The quoted statement follows if the only cosideration is the coil resistance which is true only in the static case. Why don't you put a DC motor equivalent circuit diagram in your question and, with that, flesh out your reasoning. Oct 5, 2018 at 2:35
• @AlfredCentauri I am wrong. But how can I determine whose current is higher under the same applied voltage when operating normally? Oct 5, 2018 at 2:37
• Related? [DC motors, back emf, mechanical power output of a DC motor]{physics.stackexchange.com/a/276657/104696} Oct 5, 2018 at 6:36

This is not correct "However, the lower rotational speed leads to a smaller back emf, then the current in coil 1 increases again." Each motor will reach equilibrium and stay at that point.

As you say M1 is less efficient but in practice the resistance produces heat as you load the motor, M1 will overheat first. You can evaluate efficiency by knowing that at any current I^2 x R will give you energy loss due to heat. At higher current (higher load) M1 will get much warmer.

• How can I determine which coil has a higher current if I use P(loss)=I^2R? Also, the coil resistance is different either. Oct 5, 2018 at 2:54
• I^2R is correct, thanks. Stall the motor and use a low low voltage and measure the current and voltage. R=V/I. Oct 5, 2018 at 12:43