Question

Find the relation used to calculate emf (average) for dynamo in 3/4 period

Answer 1

@Michele_Laino

Answer 2

here we have to start from the equation which expresses the f.e.m. E(t) in function of time t

Answer 3

\[emf = -N \frac{ d \phi }{ dt }\]

Answer 4

I mean this function:
\[E\left( t \right) = {E_0}\sin \left( {\omega t + \phi } \right)\]

Answer 5

where:
the period T is such that:
\[T = \frac{{2\pi }}{\omega }\]

Answer 6

Now ?

Answer 7

by definition, the requested average value is given by the subsequent formula:
\[\Large \frac{1}{{\frac{{3T}}{4}}}\int_0^{\frac{{3T}}{4}} {{E_0}\sin \left( {\omega t + \phi } \right)\;dt} \]

Answer 8

Now we will get the integration, right ?

Answer 9

that's right!

Answer 10

for simplicity we can assume \[\Large \phi = 0\]

Answer 11

Well, nice trick.

Answer 12

\[1/3T/4 \int\limits_{0}^{3T/4} E _{0}\sin(\omega t) dt\]

That's what we have.

Answer 13

ok!

Answer 14

now we change variable, namley:
\[\omega t = \tau \]
where \tau is the new variable

Answer 15

so after a simple integration, we get:
\[\Large \frac{1}{{\frac{{3T}}{4}}}\int_0^{\frac{{3T}}{4}} {{E_0}\sin \left( {\omega t} \right)\;dt} = - \frac{{4{E_0}}}{{3\omega T}}\left. {\cos \left( {\omega t} \right)} \right|_0^{\frac{{3T}}{4}} = ...?\]

Answer 16

\[\frac{ -4 }{ 3 } E_{0} 1/2\pi \cos(wt) \]

Answer 17

no, you have to evaluate the function \cos(\omega*t), between 0 and 3*T/4

Answer 18

Hmm, I am confused.

Answer 19

\[\Large \left. {\cos \left( {\omega t} \right)} \right|_0^{\frac{{3T}}{4}} = \cos \left( {\omega \frac{{3T}}{4}} \right) - \cos \left( {\omega \cdot 0} \right) = ...?\]

Answer 20

keep in mind that:
\[\Large T = \frac{{2\pi }}{\omega }\]

Answer 21

cos(3 * 2* pi / 4 ) - cos 0 = cos(3pi/2) - cos(0) = -1

Answer 22

ok!

Answer 23

now you have to compute this multiplication:
\[\Large - \frac{{4{E_0}}}{{3\omega T}} \times \left( { - 1} \right) = ...?\]

Answer 24

Can I replace omega T with 2 pi ?

Answer 25

yes!

Answer 26

2/3pi * E_{0}

Answer 27

that's right!

Answer 28

so now we will assume that the number of turns of the dynamo is N so it will be
2/3pi NE what else ?

Answer 29

yes! correct!

Answer 30

what more ?

Answer 31

N are the turns on the rotor of the alternator.
we have finished!

Answer 32

so the final relation is \[\frac{ 2 }{ 3 \Pi}NE _{0}\] can we replace E_{0} with BAomega ? and reaplce omega with 2piv so it can be 2/3pi * 2pi * v * B *A * N = 4/3 NBAv ?

Answer 33

I got this:
\[\Large \frac{{2NBA\omega }}{{3\pi }}\]

Answer 34

And we can replace the omega with 2 pi (nu) , as nu = 1/T

Answer 35

ok!