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For a given compound DC machine, Net field current obtained from characteristic is equal to 7.5 A, where shunt field current is equal to 5 A, armature current is 505 A, demagnetizing current equal to 0.95 A and shunt field winding of 1000 turns at rated speed of 1000 rpm. What will be the series field turns?

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For a given compound DC machine, Net field current obtained from characteristic is equal to 7.5 A, where shunt field current is equal to 5 A, armature current is 505 A, demagnetizing current equal to 0.95 A and shunt field winding of 1000 turns at rated speed of 1000 rpm. What will be the series field turns?

(a) 7

(b) 8

(c) 5

(d) 9

I got this question during an internship interview.

The above asked question is from Graphical Representation of External Characteristics in section DC Generator of DC Machines

1 Answer

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by (1.2m points)
Correct answer is (a) 7

Easiest explanation: From the excitation balance equation, If + [Nse/Nf] Ia – Ifd =If (net).

5 + 505(Nse /1000) – 0.95 = If (net).

Calculating for Nse by substituting If (net), we get Nse = 6.8 that is 7 turns.

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