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A 4-pole DC generator runs at 650 rpm and generates an EMF of 200 V. The armature is wave wound and has 750 conductors. If the total flux from each pole is 0.015 wb, then what is the leakage coefficient ?
1. 4.2
2. 3.2
3. 2.2
4. 1.2

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Correct Answer - Option 4 : 1.2

Concept:

Emf induced in a DC generator is directly proportional to Flux per pole and the rotor speed i.e.

E ∝ ϕ ⋅ ω 

or. E = Ka ϕ ⋅ ω 

Where Ka = Proportionality constant \(=\frac{ZP}{2\pi A}\)

Where, Z = No. of conductors

P = No. of Poles

A = No. of Parallel paths

ϕ = Flux per pole

ω = Angular velocity in rad/sec

Converting it in rotation per minute,

\(\omega = \frac{2\pi N}{60}\)

By Putting the above values in EMF equation, we will get

\(E=\frac{ϕ ZN}{60}\times \frac{P}{A}\)

Calculation:

Given,

P = 4, N = 650 rpm

E = 200 V, Z = 750

A = 2, ϕ = 0.015 wb

\(E = \frac{\phi Z N}{60}\times \frac{P}{A}\)

\(E=\frac{0.015 \times 750 \times 650}{60} \times \frac{4}{2}\)

E = 243.75 V

Given, generation rating of convertor is 200.

But it is generating 243.75 V

Hence, leakage coefficient \(=\frac{243.75}{200}\)

= 1.21875

Hence, correct option will be option (4)

No. of parallel paths in wave wound machine = 2

No. of parallel paths in lap wound machine = No. of poles i.e. A = P

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