A 250 V, DC shunt motor takes a line current of 20 A. The resistance of shunt field winding is 200Ω and resistance of the armature is 0.3 Ω Find the armature current and the back EMF.

A 250 V, DC shunt motor takes a line current of 20 A. The resistance of shunt field winding is 200Ω and resistance of the armature is 0.3 Ω Find the armature current and the back EMF.

Right Answer is:

18.75 A, 245 V

SOLUTION

Supply Voltage V = 250 V
Line Current IL = 20A
Shunt Resistance Rsh = 200Ω
Armature Resistance Ra = 0.3 Ω
Armature Current Ia =?
Back EMF Eb =?

The Diagram of the given question is shown in figure

(i) Armature current

The Line current of DC shunt motor is the sum of Armature current and Shunt field current

IL = Ia + Ish

And Shunt Field Resistance is given as

Ish = V/Rsh = 250/200 = 1.25 A

∴ Ia = IL − Ish

Ia = 20 − 1.25 = 18.75 A

(ii) Back EMF

The Back EMF of DC Motor is given by

Eb = V − IaRa

= 250 − 18.75 × 0.3
= 250 − 5.625

= 244.375 V ≅ 245 V

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