A 4 pole, lap wound d.c motor has 540 conductors and 230 V supply. The armature resistance is 0.8Ω. Its speed found to be 1000 r.p.m. the flux per pole is 25 mWb. The Induced e.m.f.and Armature current will be
A 4 pole, lap wound d.c motor has 540 conductors and 230 V supply. The armature resistance is 0.8Ω. Its speed found to be 1000 r.p.m. the flux per pole is 25 mWb. The Induced e.m.f.and Armature current will be
Right Answer is:
225 V, 6.25 A
SOLUTION
Number of Poles P = 4
The number of parallel Path A = 4 (In Lap wound the number of brushes required by this winding is always equal to the number of poles.)
Total number of armature conductor Z = 540
Flux per Pole Φ = 20 mWb = 25 x 10-3 Wb,
Armature current Ia =?
Back EMF Eb =?
(i) Back EMF of DC motor
${E_b} = \dfrac{{P\Phi ZN}}{{60A}}$
= (PΦZN)/(60A) = (25 x 10 -3 x 4 x 1000 x 540)/(60 x 4) = 225 V
Eb = 225 V
(ii) Armature current
From the voltage equation, the back EMF of the DC motor is
Eb = V − IaRa
225 = 230 − 0.8Ia
Ia = 6.25 A
Hence the back EMF is 225 V and the Armature current is 6.25 A