AC single phase induction motor - phase difference problem

Question

I want to build and AC single phase induction motor. There's the following schematic for its construction.

This motor can only work if there's a 90º phase difference between \$I_1\$ and \$I_2\$ as you can see written here, on the Construction section. I did some circuit analysis and I found the following relation:

$$\bar{I}_2=\frac{w L_1}{wL_2-\frac{1}{wC}}\bar{I}_1$$

So, according to this relation, there only can be a 0 or 180º phase difference between the two currents. Did I do something wrong?

Answer 1

If you know the resistance and inductance of the windings, you can calculate the impedance angles from Zm = Rm + jXm and Za = Ra + jXa - jXc. Main winding angle = arctan (Xm/Rm) and aux winding angle = arctan (Ra/(Xa-Xc)). These angles will only be true for the instant the motor is energized. Once the rotor begins to rotate, the rotor characteristics and load must be taken into account.

Answer 2

What I see from the equation you have given, you have only found out the magnitude equation and not the phase equation. From the phase equation you will see that there can be a wide range of angle difference between the two currents, depending on the values for the inductance, capacitance and frequency.

As mentioned by Brian Drummond earlier, there is no requirement for exact 90 degree phase shift, although it is more efficient and there is more starting torque. As long as there there is a phase difference in both time and phase of the 2 currents in the machine, there will be some starting torque. The phase shift in space means the actual difference in orientation of the windings in the machine, and is already present. The L and C are in the circuit to provide the time shift. Of course you can only see the time shift and not the phase shift in the equivalent circuit.

PS: Please correct me if you think I am wrong somewhere. Single phase IMs are harder to understand then 3 phase ones!