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Aladdin
Aladdin
07:58
imgur.com/a/YdAf2qG
@JohnRennie when u are free, can u help with 73 and 75
John Rennie
John Rennie
08:09
@Aladdin hi
In a coil the voltage is $V = -d\Phi/dt$ where $\Phi$ is the total flux linked. Yes?
Aladdin
Aladdin
Yed
John Rennie
John Rennie
We are told the maximum flux density is 1T ($10^4$ gauss = 1T).
And the area is $0.0006 m^2$
So the flux is $1 \times 0.0006 = 0.0006$ call this $P_m$
Aladdin
Aladdin
Ok
John Rennie
John Rennie
So the flux is $P(t) = P_m \sin(\omega t)$ and therefore $dP/dt = P_m \omega \cos(\omega t)$
$F = 50$ Hz so $\omega = 100\pi$
Aladdin
Aladdin
Ok
John Rennie
John Rennie
08:21
That would make $P_m\omega = 0.0006 \times 100\pi = 0.188$.
So that's the peak voltage. Divide by sqrt(2) to get the RMS voltage.
Aladdin
Aladdin
0.1329
John Rennie
John Rennie
0.1332863756 so it rounds to 0.133
Aladdin
Aladdin
It says turns per volt
But we calculated rms voltage of flux in the core
John Rennie
John Rennie
I've made two mistakes that cancelled each other out :-)
The total flux is what I calculated times the number of turns, then we divide by the number of turns at the end to get the volts per turn.
Aladdin
Aladdin
Okay
John Rennie
John Rennie
08:26
So we get a factor of $N/N$ that ends up being equal to 1 so I got the right answer anyway :-)
Aladdin
Aladdin
The total flux came to be 0.0006
John Rennie
John Rennie
$0.0006 N$ where $N = 100$ because we are calculating the value for the 100 turn secondary.
Aladdin
Aladdin
Ah Okay
John Rennie
John Rennie
So that's going to give an RMS voltage hundred times bigger than I calculated i.e. 13.3 V.
Aladdin
Aladdin
If we were calculating for primary N=1000
John Rennie
John Rennie
08:31
That would make the RMS voltage in the primary 133V.
This all got a bit messy. Do you want to go through it again, step by step?
Aladdin
Aladdin
so u meant to say we do 0.0006N/N to get total flux
for primary it is 1000
for secondary it is 100
no we gets turns/volt by dividing
@JohnRennie Can we do that
John Rennie
John Rennie
If you have N coils with an area A and a field B through the coils then the flux linked by the coils is $\Phi = NAB$. That's the staring point.
Aladdin
Aladdin
Yes
John Rennie
John Rennie
In a transformer the field $B$ will be oscillating sinusoidally. We are told the maximum field $B_m$, and we are told it is 1T.
So $B = \sin(100\pi t)$
So $\Phi(t) = NA\sin(100\pi t)$
($\omega = 2\pi f = 100\pi$)
Aladdin
Aladdin
08:48
okay
John Rennie
John Rennie
This applies to both coils because the flux is the same through both coils.
The difference between the coils is the number of turns $N$.
Aladdin
Aladdin
ok
John Rennie
John Rennie
We are being asked about the secondary so we'll do the calculation for the secondary so $N = 100$. OK so far?
Aladdin
Aladdin
ok
John Rennie
John Rennie
Now we have the equation for $\Phi(t)$ and Faraday's law tells us $V(t) = -d\Phi/dt$. Yes?
Aladdin
Aladdin
08:51
yes
emf induced in the coils
John Rennie
John Rennie
So we just differentiate $\Phi(t) = NA\sin(100\pi t)$
$$ V(t) = -\frac{d\Phi}{dt} = - N A 100\pi \cos(100\pi t) $$
Aladdin
Aladdin
ah ok
i can follow the steps now
John Rennie
John Rennie
Cool :-)
Q75 is basically the same but calculating the flux from the voltage.
Aladdin
Aladdin
yeah i wll try both of them now

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