How Does a Single-Phase Generator Calculate EMF?

[danago]

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.

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

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:

$$\varepsilon _{{\rm{av}}} = \frac{{\varepsilon _{\max } }}{{\sqrt 2 }} = \frac{{37.8\pi }}{{\sqrt 2 }} = 83.97V$$

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.

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


Which way is the correct way?

Thanks in advance,
Dan.

 

[danago]

Ahh everything seems to be contradicting everything else.

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

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.

 

[m2003]

Both methods are correct, but they are approaching the problem from slightly different perspectives. The first method calculates the instantaneous emf at any given point in time, taking into account the varying rate of change of flux. The second method calculates the average emf over a 90 degree turn, assuming a constant rate of change of flux. Both methods will give you a valid answer, but they may differ slightly due to the approximations made in the second method.

The first method is more accurate as it takes into account the changing rate of flux, but it may be more complicated to calculate. The second method is simpler, but it assumes a constant rate of change of flux, which may not be entirely accurate.

In general, it is always better to use the first method if possible, as it will give you a more accurate result. However, if the calculations become too complex, the second method can be used as an approximation. It is important to understand the assumptions and limitations of each method.