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I was trying to prove this mathematically from the definition of EMF being the change in magnetic flux over time.

Starting with

$$ \Phi=BAcos(\theta)$$

Then differentiate w.r.t time (assuming B field doesn't change with time)

$$\frac{d\Phi}{dt}=-BA\omega sin(\theta)$$

Where \$\omega\$ is angular velocity. Thus,

$$emf=BA\omega sin(\theta)$$

Now if we consider the motion of a single armature loop \$\theta\$ will be the angle between the normal vector to the plane of the armature then \$\theta\$ is going to constantly change as the loop rotates (hence why \$\theta\$ was treated as a function of time when differentiating to give \$\omega\$).

So the EMF follows a sine wave relationship as the loop rotates, therefore it is always changing.

Where have I gone wrong?

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  • \$\begingroup\$ Isn't back EMF usually defined as an RMS value? When we've used back EMF as a surrogate for motor speed (so that a separate tachometer is not needed), that's what was processed. \$\endgroup\$
    – SteveSh
    Commented Jan 29, 2022 at 12:54
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    \$\begingroup\$ "Now if we consider the motion of a single armature loop ..." Have you forgotten the commutator and brushes? \$\endgroup\$
    – Transistor
    Commented Jan 29, 2022 at 13:16
  • \$\begingroup\$ \$\theta=\omega t\$ \$\endgroup\$
    – Chu
    Commented Jan 29, 2022 at 13:16
  • \$\begingroup\$ @SteveSh if the brushes are correctly aligned, it's a peak value (and approximately DC) \$\endgroup\$
    – user16324
    Commented Jan 29, 2022 at 13:18

1 Answer 1

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The commutator changes the rotor connections after a certain angle of rotation. The angle is determined by the number of commutator segments. That reduces the amount of voltage variation. The average EMF is proportional to rotational speed, but there is some variation that is dependent on rotor angle.

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