I guess I get confused because things don't work like they would in an RL circuit. In that case the emf at first 'jumps' to match the source's voltage and decays over time reaching zero eventually.

When we have a motor though it's the opposite. Why is that? My guess is the induction of the emf is not 'directly' related to the current. It's caused by the change in ΔΦ which this time is caused by the rotor's movement and it's not created by the current. The current is definitely what causes the movement (Lorentz force) but I hope you get what I'm trying to say. Am I correct in my thinking? The motor accelerates and ΔΦ/Δt increases(I'm not sure but it has to so that the emf increases too) and at some point the latter becomes stable.

Does this mean that emf is still there opposing the source's voltage but now they are equal? I read somewhere that acceleration stops when the torque is the same as the load torque. I haven't exactly understood what the load is in the case of the motor. I also found that if you increase it, the rotor slows down even more.. but how? What is the load torque? I would really appreciate it if you could also clarify this part. Thank you.

  • \$\begingroup\$ In a RL circuit the inductor voltage is accessible and can measured. In a motor, the R and L are distributed through the winding and cannot be physically separated out. In both cases the terminal voltage will be equal to the supply voltage at all times. \$\endgroup\$
    – Chu
    Jan 25, 2017 at 7:23

2 Answers 2


Just like it's simplest to learn about a lossless inductor first, so let's start with more or less lossless motor. We'll take account of losses when we have to, but they are not essential for basic understanding.

A motor is also a generator. Spin it, and it generates volts on the armature. It doesn't matter whether it's spinning because it's a motor, or spinning because you're driving it as a generator, speed = armature_volts/k.

Pass a current through it and it generates a torque. Torque = armature_current.k

You can think of a motor as a mechanical transformer. Power in = power out. Volts x amps in = speed x torque out. When equating power, that pesky k has cancelled out. If you change the value of k, the torque constant, then the motor gets faster and delivers less torque, or vice versa, but the power balance is the same. If you run the same machine as a generator, then speed x torque in = volts x amps out.

What happens if you apply a voltage source to a motor at rest?

Two things happen, at different speeds, the first so quickly you may not notice, the second rather more slowly.

A motor armature has inductance, Larm, and resistance Rarm. At the moment of switch on, we apply V to the armature. The current starts to increase. It initially increases at such a rate that Larm x dI/dt generates a back EMF equal to the terminal voltage. The current flowing through Rarm generates a voltage IRarm which opposes V, so there is less voltage across the inductance to drive an increase of current, so the rate of current increase slows down. Eventually, the current has increased to settle at V/Rarm, with a time constant of Larm/Rarm, typically in a matter of mS.

With a 'good' motor with a low Rarm, this current will typically be very large. It's known as the 'starting' current, for obvious reasons. Small motors are rated to be started like this safely. Big motors cannot be started like this, and need some sort of soft start controller.

So far, the motor still hasn't moved, the mechanical inertia means it's not rotating yet, or has barely started. But now there is a big armature current flowing, which generates a torque, and the motor accelerates.

Once it is turning, at any speed, it generates a back EMF proportional to its speed. This back EMF reduces the effective terminal voltage available to drive current through Rarm. The armature current therefore falls (with a time constant of Larm/Rarm), and so generates less torque.

Eventually the motor reaches a balance, where it's at a speed where the generated back EMF balances off most of the input voltage, and the small difference in voltage that remains drives an armature current through Rarm, which generates enough torque in the motor to match the load torque, plus loss torques like friction and air resistance.

  • \$\begingroup\$ Very detailed and helpful answer. It really helped my understanding. Thank you sir! One more thing , that small current at the end (balanced condition) flowing through the armature creates a torque as you mentioned . Speed still increases therefore .. shouldn't emf do so as well since it is proportional to the speed ? Or is it that the speed remains the same because of the friction , the load etc ? \$\endgroup\$ Jan 25, 2017 at 7:55
  • \$\begingroup\$ The small torque left at the end is required to keep the motor turning against the load and friction torques. One assumes a motor has a load, is driving a fan or a food mixer or something. That requires a torque. Even if there's no external load, then there are friction, air resistance etc internal to the motor which provide a braking torque. After if, if you switched off the power, it would slow down, that shows there's a torque trying to stop the motor. \$\endgroup\$
    – Neil_UK
    Jan 25, 2017 at 8:02

There are two parts that go into equilibrium in a motor: There is electrical balance and there is mechanical balance.

Electrically the voltage connected to the motor equals the counter emf and the losses because of the resistance of the armature on a DC motor (or the voltage drop because of the AC impedance on an AC motor). Let's ignore for a bit any losses on the rotor, if there are. You can think of a motor as a 'transformer', it reflects in its 'primary' a voltage drop which is proportional to the mechanical torque it is delivering in its 'secondary'.

Mechanically the rotational force provided by the motor must overcome the mechanical load and the losses (friction, etc.). As in linear movement we have the mass, for rotational movement we have the moment which is the product of mass by rotational speed. As a mass tends to keep its linear movement status, a load tries to keep its rotational status. When the motor tries to rotate (faster or slower, namely, to change the rotational speed), the mass of the load (and that of the motor itself) is 'against' this change and equilibrium is found when the torque provided by the motor meets the load of anything connected to it (fan, lift, pulleys, etc.) and any losses like friction. Usually the mechanical load grows proportional to speed. The faster you want a load to rotate, the more force the motor will have to give to achieve a faster rotational speed.

  • \$\begingroup\$ Since the source voltage becomes equal to the emf and the losses the current stops flowing, right ? \$\endgroup\$ Jan 25, 2017 at 7:11
  • \$\begingroup\$ No, the source voltage is equal to the emf voltage PLUS the losses on the armature resistance. There is current all the time flowing on the armature while the motor rotates. Without current there would be zero power transfer and that is impossible (it would be like saying that the motor is creating rotating power from nothing, a thing that cannot be according phisics energy conservation). \$\endgroup\$ Jan 25, 2017 at 17:13

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