# 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

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