danago
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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.
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.