How Does a Single-Phase Generator Calculate EMF?

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A single-phase generator with 200 turns, a coil length of 14 cm, a width of 9 cm, and a magnetic field of 0.15T is analyzed for its electromotive force (emf) production at 3000 rpm. The calculated maximum emf is 37.8π, leading to an average emf of approximately 83.97V. However, an alternative method using the change in magnetic flux over a 90-degree turn yields a different emf of 75.6V. The discussion highlights discrepancies between methods for calculating emf, with one relying on instantaneous values and the other on average values. The conversation underscores the complexity and varying approaches to understanding emf in generators.
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A simple single-phase generator has coils of 200 turns. The coil is 14cm long and 9cm wide. The magnetic field in the generator is 0.15T. The generator coil is turned at a rate of 3000rpm.

Calculate the emf produced by this generator.


<br /> \begin{array}{c} \\<br /> \frac{{d\theta }}{{dt}} = 3000rpm = 100\pi {\rm{ rad/sec}} \\<br /> \varepsilon = - N\frac{d}{{dt}}(AB\cos \theta ) \\<br /> = \frac{{d\theta }}{{dt}}NAB\sin \theta \\<br /> = (100\pi )(200)(0.09)(0.14)(0.15)\sin (100\pi t) \\<br /> = 37.8\pi \sin (100\pi t) \\<br /> \end{array}<br />

Now it asks for 'the emf', which is a bit hard to give as a numerical answer, since the emf varies with time. I looked at the answer, and they give an average emf.

So i proceeded as follows:

<br /> \varepsilon _{{\rm{av}}} = \frac{{\varepsilon _{\max } }}{{\sqrt 2 }} = \frac{{37.8\pi }}{{\sqrt 2 }} = 83.97V<br />

Now the book does it differently. They find the change in flux over a 90 degree turn (and i think are assuming that the rate of change of flux is constant), and then find the change in time.

<br /> \begin{array}{c}<br /> \Phi _B = AB\cos \theta \\ <br /> \Phi _i = 0.00189\cos \frac{\pi }{2} = 0 \\ <br /> \Phi _f = 0.00189\cos 0 = 0.00189 \\ <br /> \Delta \Phi _B = 0.00189 \\ <br /> \Delta t = 0.005 \\ <br /> \varepsilon = - N\frac{{\Delta \Phi _B }}{{\Delta t}} = 75.6V \\ <br /> \end{array}<br />


Which way is the correct way?

Thanks in advance,
Dan.
 
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Ahh everything seems to be contradicting everything else.

My teacher gave us the formula:
<br /> \begin{array}{l}<br /> \varepsilon(t) = 2\pi fNBA\sin (2\pi ft) \\ <br /> \therefore \varepsilon _{\max } = 2\pi fNBA \\ <br /> \end{array}<br />

For a coil of N turns with cross sectional area A rotating through a magnetic field of flux density B, with a rotational frequency of f. Now i completely agree with this formula, and am able to derive it using some basic calculus and faradays law. However, another textbook uses another method, which gives a different answer.
 
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