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A d.c. generator has an armature e.m.f. of 100 V when the useful flux per pole is 20 mWb and the speed is 800 r.p.m. calculate the generated e.m.f. (i) with the same flux and a speed of 1000 r.p.m. (ii) with a flux per pole of 24 mWb and a speed of 900 r.p.m.?
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A d.c. generator has an armature e.m.f. of 100 V when the useful flux ...
Answer:


Given Data:

Armature e.m.f = 100 V
Flux per pole = 20 mWb
Speed = 800 rpm

Calculation:

The generated e.m.f. in a DC generator is given by the formula:

Generated e.m.f = ΦNZ/60A

Where,
Φ = Flux per pole in Wb
N = Speed of the armature in rpm
Z = Total number of armature conductors
A = Number of parallel paths in the armature

(i) With the same flux and a speed of 1000 rpm:

Given,
Flux per pole = 20 mWb
Speed = 1000 rpm

The generated e.m.f. can be calculated as follows:
Generated e.m.f = ΦNZ/60A
= (20 x 1000 x 1)/60 x 2
= 166.67 V

Therefore, the generated e.m.f. with the same flux and a speed of 1000 rpm is 166.67 V.

(ii) With a flux per pole of 24 mWb and a speed of 900 rpm:

Given,
Flux per pole = 24 mWb
Speed = 900 rpm

The generated e.m.f. can be calculated as follows:
Generated e.m.f = ΦNZ/60A
= (24 x 900 x 1)/60 x 2
= 270 V

Therefore, the generated e.m.f. with a flux per pole of 24 mWb and a speed of 900 rpm is 270 V.

Explanation:

A DC generator works on the principle of Faraday's law of electromagnetic induction. According to this law, whenever there is a relative motion between a magnetic field and a conductor, an e.m.f. is induced in the conductor. In a DC generator, the armature rotates in a magnetic field, and the relative motion between the magnetic field and the armature conductors induces an e.m.f. in the conductors.


The generated e.m.f. depends on the flux per pole, the speed of the armature, the number of conductors, and the number of parallel paths in the armature. Therefore, by changing any of these parameters, we can change the generated e.m.f. of a DC generator.


In the given problem, we are given the armature e.m.f., flux per pole, and speed of the armature. Using the formula for generated e.m.f., we can calculate the generated e.m.f. for different values of flux per pole and speed of the armature.
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A d.c. generator has an armature e.m.f. of 100 V when the useful flux per pole is 20 mWb and the speed is 800 r.p.m. calculate the generated e.m.f. (i) with the same flux and a speed of 1000 r.p.m. (ii) with a flux per pole of 24 mWb and a speed of 900 r.p.m.?
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A d.c. generator has an armature e.m.f. of 100 V when the useful flux per pole is 20 mWb and the speed is 800 r.p.m. calculate the generated e.m.f. (i) with the same flux and a speed of 1000 r.p.m. (ii) with a flux per pole of 24 mWb and a speed of 900 r.p.m.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A d.c. generator has an armature e.m.f. of 100 V when the useful flux per pole is 20 mWb and the speed is 800 r.p.m. calculate the generated e.m.f. (i) with the same flux and a speed of 1000 r.p.m. (ii) with a flux per pole of 24 mWb and a speed of 900 r.p.m.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A d.c. generator has an armature e.m.f. of 100 V when the useful flux per pole is 20 mWb and the speed is 800 r.p.m. calculate the generated e.m.f. (i) with the same flux and a speed of 1000 r.p.m. (ii) with a flux per pole of 24 mWb and a speed of 900 r.p.m.?.
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