A universal motor (ac-operated) has a 2-pole armature with 960 conductors. At a certain load the motor speed is 5000 rpm and the armature current is 4.6 A; the armature terminal voltage and input are respectively 100 V and 300 W.
Compute the following, assuming an armature resistance of 3.5 \Omega .
(a) Effective armature reactance
(b) Maximum value of useful flux/pole
The operating conditions in terms of voltage and current of the armature circuit are shown in Fig. 10.49.
100 \times 4.6 \cos \phi=330 WOr \phi=49.3^{\circ}
(lagging because of reactive nature of the circuit).
(a) From the circuit the following can be written
\frac{100 \angle 49.3^{\circ}-E_{a} \angle 0^{\circ}}{3.5+j X_{a}}=4.6 \angle 0^{\circ}
E_{a} is in-phase with \bar{I}_{a}
Or 65.2+j 75.8-E_{a}=16.1+j 4.6 X_{a}
Equating real and imaginary parts,
E_{a}=65.2-16.1=49.1 VX_{a}=\frac{75.8}{4.6}=165 \Omega
(b) E_{a}=\frac{1}{\sqrt{2}} \cdot \frac{\Phi n Z}{60}\left(\frac{P}{A}\right)
\Phi=\frac{\sqrt{2} \times 49.1 \times 60}{5000 \times 960}=0.868 mWb
