# Is the resistance in a DC Motor constant if the voltage is?

Is resistance in a DC motor constant if both the load and voltage are constant as well?

• Is the motor a mechanically commutated motor, where the electromagnets (coils) spin, or is the electromagnet stationary, and the magnets spin? Jun 5, 2016 at 9:16
• I think you and your answerers mean different things by 'resistance of a motor'. Your answerer correctly is referring to the resistance of the motor, which is determined by the copper wire in it. I think you mean 'terminal volts divided by terminal amps', which has dimensions of resistance, but is not 'the resistance of the motor'. Amps will increase as the load increases. Some of the terminal voltage drives current through the 'resistance of the motor'. The rest of the terminal voltage opposes the voltage generated by its rotation. Jun 5, 2016 at 13:48

The resistance in a DC motor is determined by the motor coil winding length and wire gauge. This doesn't change other than a small increase in resistance as the motor temperature increases.

Is resistance in a DC motor constant if both the load and voltage are constant as well?

The resistance will not change (other than as noted above). For constant voltage current will increase with load.

Current and back-emf

Some basic DC motor concepts: (Try and grasp each one before going on to the next.)

• DC motors spin when a voltage is applied to the terminals.
• When unloaded the motor speed will be proportional to the voltage applied.
• If you disconnect the motor (from the supply) and spin it then voltage appears at the terminals. It is acting as a dynamo.
• The voltage that is generated in dynamo mode is proportional to speed. We call this "back electro-motive force" or "back-emf" for short. simulate this circuit – Schematic created using CircuitLab

Figure 1. Ideal motor and it's series resistance with power circuit.

Let's have a very simplified look at what happens when the motor is switched on.

• The motor is at rest. Back-emf will be zero.
• SW1 is pressed. A current will flow into the motor and we can determine this by Ohm's Law, $I = \frac {V_{BAT}}{R}$.
• Once the motor starts to turn it will generate a back-emf. The "back" refers to the fact that it opposes the driving voltage. This means we have to modify our current calculation to $I = \frac {V_{BAT}-V_{BEMF}}{R}$.
• The faster the motor spins the higher will be the back-emf and the current will reduce. This is why DC motors take highest current at start or if stalled.
• The motor will settle down at whatever speed allows enough current to flow to overcome losses in heat and friction.
• If you increase the load then the "losses" increase. (You're doing work, remember, and you don't get it for free.) This will cause the motor to slow down. This in turn will reduce the back-emf so more current will flow.
• Thanks but, why will the current increase with increase of load? Jun 5, 2016 at 10:33
• With increasing load, the motor speed will fall and hence the back emf will reduce. The current will increase since the effective emf in the loop (supply emf - back emf) increases.
– Chu
Jun 5, 2016 at 19:10
• @SamRolls: See update "Current and back-emf". Jun 5, 2016 at 19:52
• V = RI (ohm's law). if resistance will not change, how can current increase with load for constant voltage ? I'm probably missing something. Aug 29, 2018 at 19:01
• @tigrou: Read the last line of my answer and Chu's comment. We've addressed that already. Aug 29, 2018 at 19:46

If the motor is a brushed motor, a difference in resistance will be seen for the motor based on the position of the shaft. This thread, Measuring the Internal Resistance of a Brushed DC Motor for Use in Speed Control (IR Compensation), discusses ways to measure these resistances. A common way that I have experienced in lab to measure the motor resistance is to use a ohmmeter to measure the resistance while slowly turning the shaft and shorting out the terminals to remove the back emf.