- #1

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**Quick concept question!!**

Ok,

How does the current relate to the speed.

So lets say everything is normal and if I reduced the speed by 1/3 what would happen to my current?

Thanks!

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- Thread starter Winner
- Start date

- #1

- 94

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Ok,

How does the current relate to the speed.

So lets say everything is normal and if I reduced the speed by 1/3 what would happen to my current?

Thanks!

- #2

vsage

- #3

- 94

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- #4

- 94

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Any ideas anyone?? Thanks.

- #5

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So i really doubt anyone would help,unless you 'd be more precise.

Daniel.

- #6

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- #7

- 13,131

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Speed of what??What does this formula [itex] I=\frac{V-E}{R} [/itex] mean...?

Daniel.

Daniel.

- #8

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- #9

SpaceTiger

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[tex]B=\mu in[/tex]

where B is the magnetic field, [tex]\mu[/tex] is the permeability of the object around which the wires are wrapped, i is the current through the solenoid, and n is the number of turns per unit length. This magnetic field will vary linearly with the current, so cutting the current by a third should cut the magnetic field by a third.

As for the velocity, I would imagine you can approximate the two solenoids as a magnetic dipole in an external field created by the permanent magnet. The potential energy of such a setup would be given by

[tex]U=-\mu_{magnet} \bullet B_{ext}[/tex]

where this time mu is the magnetic dipole moment of the electromagnet. The maximum kinetic energy that can be extracted from such a setup is just equal to the maximum magnitude of the potential, so you'd expect

[tex]v_{max} \propto \sqrt{U} \propto \sqrt{\mu_{magnet}} \propto \sqrt{i}[/tex]

This is only a very rough argument, however, and I doubt that things are so nicely linear in practice.

- #10

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- #11

SpaceTiger

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- #12

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As you can see, I'm stuck at c)

a was easy, since it's just started we can say E(emf) is 0. So 12.20=117-0/R and solve for R, R=9.59 ohms.

in b) we just go 2.30=117-E/9.59, and E=95V.

c)and finally I don't know... lol

- #13

SpaceTiger

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Winner said:c)and finally I don't know...

The back emf generated by a motor is proportional to its speed, so for the current of the underlying circuit, you'll want to instead assume that one-third speed corresponds to one-third back emf. Then plug into your equation:

[tex]i=\frac{V-E}{R}[/tex]

and you're done.

- #14

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huh, that easy eh. Well, great stuff, thanks!

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