The e.m.f. generated by a shunt wound D.C. generator is E. Now while pole flux remains constant, if the speed of the generator is doubled, the e.m.f. generated will be

$$rac{{ ext{E}}}{2}$$
2E
generator is E. Now while pole flux remains constant, if the speed of the generator is doubled, the e.m.f. generated will be A. $$rac{{ ext{E}}}{2}$$ B. 2E C. Slightly less than E
E

The correct answer is B. 2E.

The e.m.f. generated by a shunt wound D.C. generator is given by the equation:

$$E = K\phi N$$

where:

  • $E$ is the e.m.f. generated
  • $K$ is the generator constant
  • $\phi$ is the magnetic flux
  • $N$ is the speed of rotation

If the pole flux remains constant, and the speed of rotation is doubled, then the e.m.f. generated will be doubled. This is because the speed of rotation is in the numerator of the equation, so doubling the speed will double the e.m.f.

Option A is incorrect because it is half of the correct answer. Option C is incorrect because it is slightly less than the correct answer. Option D is incorrect because it is the same as the original e.m.f.

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