Consider a very simple singe phase synchronous motor, such as the one shown in figure 1. This motor will not be self starting, but if an AC voltage is applied and the permanent magnet is given an initial spin, the magnet will continue to spin at the same frequency as the AC source.

Figure 1:

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If the mechanical load is increased, e.g the weight in the pulley system in figure 1 is increased, the current drawn by the motor will increase to provide the extra torque necessary. Source: bottom of page 428

What I do not understand is how the motor "knows" that it needs to increase the current drawn. There is no controller involved, the motor inherently does this. Somehow the information has been passed from the mechanical load to the electrical circuit even though nothing is connecting them. How does this work?

If I analyse the equivalent circuit:

The voltage across individual components is:

$$V_{R_1}(t) = i(t)*R_1$$ $$V_{L_1} = L_1*\frac{di(t)}{dt}$$ $$V_{L_2} = L_2*\frac{di(t)}{dt}$$

If connected to a sinusoidal voltage source



$$L_1 = L_2$$

The current can then be found by:

$$i(t) = V_{mag}*\frac{R (-e^{-\frac{R t}{2 L}})+2 L w \sin(t w)+R \cos(t w)}{4 L^2 w^2+R^2}$$


Using trig identities:

$$i(t) = V_{mag}*\frac{R (-e^{-\frac{R t}{2 L}})+ (\sqrt{(2Lw)^2+R^2})\sin(wt+\tan^{-1}(\frac{R}{2Lw}))}{4 L^2 w^2+R^2}$$

In steady state operation, the exponential goes to zero and the magnitude of the current is given by:

$$I_{mag}=\frac{V_{mag}}{sqrt(4 L^2 w^2+R^2)}$$

This is independent of the mechanical load. But this cant be true as the current is supposed to increase when the mechanical load is increased. What am I doing wrong?

  • \$\begingroup\$ Your initial assumption about the voltage across L1 and L2 is not correct: There will also be a voltage induced into those inductors by the changing magnetic field from the rotor. This voltage is in usually referred to as Back EMF. Mechanical load will change the back EMF and so change the current in those inductors. Kevin \$\endgroup\$ Commented Mar 22, 2015 at 23:19

2 Answers 2


From the practical viewpoint:

  • As the load increases, the motor continues to rotate at the same speed, but the phase of the rotating magnetic field lags the applied voltage, so the current increases in an attempt to reduce the phase angle.
  • Similarly, if the output shaft is driven from an external mechanical source the phase begins to lead, and the motor starts to generate power back into the supply.

I suspect that your circuit represents the motor under no-load conditions. If the motor was delivering say 100W of mechanical power to its output shaft, then there is no way your circuit can model this (unless R1 is a variable resistor).

I last did maths like this at university 40 years ago, so someone more academic will need to provide a more scholarly answer.


A perfectly unloaded sync motor will have its rotor coinciding rotor pole to stator pole with stator voltage. No power is taken in this theoretical motor because the back emf exactly cancels the applied stator voltage and no current flows. This is an ultra simplistic examination of course.

As load increases, the rotor is forced to have an angular lag and this means the back emf into the stator is not 100% of the stator voltage hence a current will flow. As load increases the angular lag increases and more current flows into the stator.

I'm not sure that anything else needs to be stated.


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