Page 437 - Handbook of Electrical Engineering

426 HANDBOOK OF ELECTRICAL ENGINEERING

The combined admittance of the rotor and magnetising impedances become,

R 2
− jnX 2
−j S n
Y m2 =
+
2
nX m R 2
2 2
+ n X 2
s n
= 0.01211 − j 1.8650 ohms
1
Z m2 = = 0.00348 + j 0.5362 ohms
Y m2
Add the stator impedance.
Z 1m2 = R 1 + jnX 1 + Z m2 = 0.00878 + j 0.7711 ohms

All the harmonics of the supply voltage have zero phase shift (except the triplens which are
√
anti-phase). The fifth harmonic supply voltage V 5n is 67.77/ 2 volts (rms). Supply this voltage to
the circuit. The supply current is,

V 5 67.77 + j 0.0
I 15 = = √
Z 1m2 2(0.00878 + j 0.7711)
= 0.7075 − j 62.1377 amps

The volt-drop across the stator impedance is,

V 1m5 = (0.0053 + j 0.2350)(0.7075 − j 62.1377)
= 14.606 − j 0.163 volts

The air-gap voltage V m5 becomes,

V m5 = V 15 − V 1m5
= 47.921 − 14.606 + j 0.163
= 33.315 + j 0.163 volts

The magnetising current I m5 is,

V m5 33.315 + j 0.163
I m5 = =
j n X m 0.0 + j 14.665
= 0.0111 − j 2.272 amps

Hence the rotor current I 25 becomes,

I 25 = I 15 − I m5
= 0.7075 − j 62.1377 − 0.0111 + j 2.272

= 0.6964 − j 59.866 amps