Class 4 (4th Engineer) MET 📅 Dec 2023

Exam Question

A 4-pole lap wound DC shunt generator has an open e.m..f of 250V when the flux per pole is 0.08 Wb and the speed is 10 rev/sec. The speed of the generator is reduced to 10 per cent and the flux per pole is increased by 5% when the generator supplies a load of 100A. Determine the terminal voltage, if the armature resistance is 0.06 ohm and the new total field circuit reisitance is 200 ohm. (16)

Reference Answer

### Principle
The fundamental principle governing this problem is the Electromotive Force (E.M.F.) equation for a DC generator and its relationship to the terminal voltage under load. The generated E.M.F. (Eg) is directly proportional to the magnetic flux per pole (Φ) and the speed of rotation (N). Any change in these parameters will result in a change in the generated E.M.F. When the generator is loaded, the terminal voltage (Vt) available to the load is the generated E.M.F. minus the voltage drop across the armature resistance due to the armature current (Ia).
For a DC shunt generator, the armature current is the sum of the load current (IL) and the shunt field current (Ish).
### Given Data

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