A shunt generator has an induced e.m.f. of 224 V. When supplying a load, the terminal voltage falls to 204 V. The armature and shunt field resistances are 0.05 Ω and 20 Ω, respectively. The load current, neglecting the armature reaction, is

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ESE Electronics 2018: Official Paper

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Answer 1

Generator induced emf E_g = 224 V

Terminal voltage V = 204 V

Field resistance R_f = 20 Ω

Field current \({I_f} = \frac{{204}}{{20}} = 10.2\;A\)

Armature resistance, R_a = 0.05 Ω

Armature drop = 224 – 204 = 20 V

Armature current \({I_a} = \frac{{20}}{{0.05}} = 400\;A\)

Load current I_L = I_a – I_f = 400 – 10.2 = 389.8 A