# How does the RL series circuit model the effect of load/backemf in an induction motor?

I know how induction motor works:
The rotor produces maximum backemf on no load and keeps the current to minimum.
On load, the backemf reduces and the rotor draws more current.

In power factor correction for an induction motor, a series RL circuit is presented with lagging current.
What are R,L here? How are they related to rotor?

My thinking:
If R is the winding resistance of the rotor coil, then this is a constant value.
If L is the inductance of the rotor coil, then this value is also constant as inductance depends only on geometry.

Then how can the current vary with load here?

If I reduce $$\R\$$, then current increases, but $$\I^2R\$$ decreases even though $$\I\$$ increases.
This means real power in the circuit is less with a smaller R.
So clearly reducing $$\R\$$ doesn't help power factor.

How is the mechanical work done by rotor modelled here? Clearly $$\R\$$ is resistive loss, not mechanical work.

• I have some thoughts and suggestions of reading material, but I'm not well-enough read on motors to attempt such a general question as this. (I wish you'd restricted this question to a specific type of motor.) But I'll +1 to encourage someone who is expert to try and address it. Have you looked up the concept of slip and how it increases the rate of flux change in a rotor in, say, synchronous motors? Might be a start. But I'm leaving this for my betters.
– jonk
Mar 6, 2022 at 19:30
• @jonk thank you. I have a conceptual understanding. When rpm is maximum, slip is minimum, backemf is maximum. When we increase load, rpm decreases, slip increases, backemf decreases. Mar 6, 2022 at 19:53
• backemf: emf produced in the rotor of a motor due to it spinning in the external magnetic field: $E\propto-\frac{d\phi}{dt}$ Mar 6, 2022 at 19:54
• If I have to be specific, I want to know how series RL circuit models the mechanical work. I think I know how electrical energy is converted to mechanical work when back emf is reduced. What I don't get is how this is modelled in the series RL circuit. R is winding resistance, not mechanical work right? And energy stored in inductor is given back to the supply and R. From the series RL circuit, I don't see energy converted to mechanical work... energy is either dissipated in R or stored in L. no mechanical work... Mar 6, 2022 at 20:00

## 1 Answer

Using just a RL circuit like this to model an induction motor is only accurate if the motor is running in steady state with no load changes. Things like stator and rotor resistance, inductance, slip, torque, ect... have all been simplified into the R and L values. You can get away with this because when a real single phase motor is running at steady state, and you simply measure the terminal voltage and the line current it would be indistinguishable from a simple RL circuit. From the picture you posted it looks like this simple model was used to not confuse the explanation on power factor correction.

It would take a more complicated model to capture the effects of changing back emf and rpm due to a changing load. Something closer to the following from Electrical 4U:

With this model the back emf is described by the transformer element that couples the Stator's equivalent circuit on the left and the rotor's equivalent circuit on the right. The 'S' parameter captures the effect of the motor's slip. Note that this model still does not capture the effects of a load change specifically, It simply calculates the mechanical power output with the R2/S resistive element. What the output power needed to reach a given RPM is not captured by the model.

So again this is a model that is only accurate if we can assume the load is unchanging, or at least the rpm and slip is a known parameter we simply enter into the model. To account for the mechanical system, the engineer would need to add the appropriate elements that describe the mechanical behavior. An explanation for how to do that is a large topic in its self.