ASAP explanation on this emf graph (1 Viewer)

Run hard@thehsc · Jun 15, 2022

Hey guys, I am just wondering as to why the given emf graph is more of a positive cos graph. If you graph and then apply the emf equation shouldn't you get a negative cos graph? Thanks!!!

Lith_30 · Jun 15, 2022

I think for the question you can think of a regular cos graph for the flux, then you shift it by pi cause the normal of the armature starts perpendicular to the mangetic field which gives a -sin graph.

This means that the flux is decreasing at a decreasing rate intially as t increases. Hence $\frac{d\phi}{dt}$ is negative when the motor starts. This means that emf is initially positive and starts at a maximum instead of a minimum.

Alternatively you could derive the equation itself... (hopefully I didn't make any mistakes)

Starting off with the equation $\epsilon=-\frac{d\phi}{dt}$ assuming that the number of coils is one.

using chain rule $\frac{d\phi}{dt}=\frac{d\phi}{d\theta}\times\frac{d\theta}{dt}$

$\frac{d\theta}{dt}=\frac{2\pi}{T}$ since the armature of the motor rotates 2pi every period

$\frac{d\phi}{d\theta}=-BA\sin\theta$

Therefore $\epsilon=\frac{2\pi{}BA}{T}\sin\theta$

However the graph is in terms of time therefore we must make it in terms of time

$\\\frac{d\theta}{dt}=\frac{2\pi}{T}\\\\\int{}d\theta=\int\frac{2\pi}{T}dt\\\\\theta=\frac{2\pi{t}}{T}+c$

At t=0 the angle is pi as we take the angle from the normal when calculating magnetic flux

hence c=pi

therefore $\\\epsilon=\frac{2\pi{}BA}{T}\sin\left(\frac{2\pi{t}}{T}+\pi\right)$

$\\\epsilon=\frac{2\pi{}BA}{T}\cos\left(\frac{2\pi{t}}{T}\right)$

hence we get a positive cos graph shape not a negative.

ASAP explanation on this emf graph (1 Viewer)

Run hard@thehsc

Well-Known Member

Lith_30

o_o

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