Capacitor efficiency problem

  • Thread starter map7s
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  • #1
map7s
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Homework Statement



A 0.45 µF capacitor is charged by a 1.5 V battery. After being charged, the capacitor is connected to a small electric motor. Assuming 100% efficiency, to what height can the motor lift a 4.7 g mass?

Homework Equations



I really need some direction to start off this problem. I think that I probably need to use U=1/2QV=1/2CV^2=Q^2/2C and/or Q=CV but I have no idea how to go about doing this.
 

Answers and Replies

  • #2
antonantal
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The whole idea of this problem is that if the efficiency is 100% no energy will be lost. So all http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html" [Broken] will be used to lift that mass. Now what would be the energy necessary to lift that mass to a height h?
 
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  • #3
map7s
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So...would I have to use Q=CV to find out Q and then use U=1/2QV to find out U ?
 
  • #4
antonantal
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So...would I have to use Q=CV to find out Q and then use U=1/2QV to find out U ?

You could use any of the 3 formulas you mention, though since you know [tex]C[/tex] and [tex]V[/tex] the fastest would be to use [tex]U=\frac{CV^2}{2}[/tex]
 

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