By Richard T. Smith (Auth.)

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The factor of 2 in the example is used to convert single-phase values to total quantities for the entire two-phase induction machine. 4 As a final example for this section, let us consider the elementary synchronous machine shown. We again assume inductance variations to be L aa — L + L cos20 L L hh = L — L cos20 L 2 0 2 0 — Mcos0 af — Msind hf — L sin20 L 2 ah The electrical torque is (1) Now, let us assume currents to be of the form i ~ ]f2I coscot a i — If a f (constant) (2) i = ]fl I smoot.

However, if the rotor is displaced slightly from this position, a torque is developed which tends to restore the rotor to position 0 = 0 . In other words, 0 = 0 is a stable equilibrium position of the rotor. To show this is true we calculate 2L / cos20, 2 2 d c and note that the small amount of torque produced during small displacement A0 is approximately given by A7 = L\6. dT/dO is negative for 0 close to zero or zero, so for a positive value of A0 we find AT is negative. This means the electric torque opposes the + displacement and tends to restore the rotor to 0 = 0 .

We also saw that the voltage equations of the windings may be written, without much difficulty, in terms of winding currents and machine parameters, for simple operating conditions. Before proceeding to a more general analysis of machines, we will consider several other elementary machines (induction and synchronous) operating in a well-behaved manner. 3 Consider the arrangement of four windings shown. , the rotating member is round and the stator surface is smooth and concentric with it. The stator (rotor) coils are 77/2 radians apart in their magnetic axes.