By Richard T. Smith (Auth.)

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**Example text**

Draw the usual equivalent circuit for a two winding transformer, showing all impedances referred to one winding. 5 (for the 2-winding case) in terms of the first equivalent circuit parameters, and turns ratio. 17. A transformer has three windings on the same core. Load impedance 3 + j4 is on winding 3; |/ | = 10; e, = 10/0°; /, = 3 0 / 3 0 ° ; #, = 100, and N =500. Find the phasor voltage and current of each winding. r 2 ? 18. For a certain electromagnet, the relation among current, plunger position, and flux linkages has been experimentally found to be where a, b, and d are positive constants.

Three identical single-phase transformers are connected to form a wye-delta three-phase bank. Accidentally, the polarity of one of the 52 Analysis of Electrical Machines secondary windings is reversed. Balanced three-phase voltages are applied to the wye side. Derive literal expressions for: (a) The voltage across the terminals of the delta if it is opened at one corner. (b) The current that will flow in the closed delta. (c) the current that will flow in the closed delta if a resistance R is inserted in series with it.

2 0 M = M c o s 0 + ... r and L —L r + L cos2d 0r 2r + ... Thus, dL/d6 = -2L sm26, 2 Msin0; and 2L sin20. 2r The torque is T — — L i sin26 2 e 2 — i i MsmO s r — L i sm20. 2 2r r 28 Analysis of Electrical Machines Assume i — I r and i = / c o s co/; then dc s w T — — L / s i n 2 0 — I I Msm8coscot 2 2r e d c m dc — L I cos 2 2 m co/sin20. If the various functions are expanded, we get T = - L / s i n 2 0 - I I M^ [sin(0 + at) + sin(0 - co/)] 2 2r e d c m dc sin20 + -j- (sin(20 + 2 co/) + sin(20 -2cof)} As before, we see that there is no average torque for positive steady speeds, except when 6 — cot + 8.