Power dissipation

Discussion in 'General Electronics Chat' started by kvi037, May 17, 2013.

  1. kvi037

    Thread Starter New Member

    May 17, 2013
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    Hi,

    I'm struggling with this task for my preperation for a exam next week.
    Can sombody please help me?

    An AC voltage of amplitude 150 volts is impressed across a pure resistance of 100 ohms. Calculate the current and the power dissipation when the frequency of the voltage is 60 cycles/sec.
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    I already gave you some tips.

    Why are you posting the same question again?
     
  3. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Since the load is purely resistive the frequency doesn´t matter at all.
    Amplitude means peak voltage, so you need to calulate the average voltage, then the equation is P=V*V/R and I=V/R
    The current you calculated is completely wrong, VR/R is still V, not I. (unless VR means Vavg as in average voltage, then it would be true)
     
    Last edited: May 17, 2013
  4. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Also, this is completely wrong too. V0*sin (wt) is the instantenous voltage at a certain point in time t, so where did the t go in the second part? It should be Io*R sin(wt) = Io R sin (2*pi*60*t) so if you wanted to know the voltage at t=260ms this equation would be useful.
     
  5. MrChips

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    I think when he writes VR he is referring to V across resistor R, i.e. V with subscript R.

    Sloppy notation.
     
  6. kvi037

    Thread Starter New Member

    May 17, 2013
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    Thanks for reply. So you can just say. I = 1.5A, and that the power disspation is
    P = IVcos 1 =225 W

    Is that correct?
     
  7. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    No, the average current is 150*0.707/100=1.06A and the power dissipation is average voltage times average current = 112W. And no cosinus nor sinus in there.
     
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  8. kvi037

    Thread Starter New Member

    May 17, 2013
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    Aha, thanks a lot for the help! :)
     
  9. MrChips

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    Oct 2, 2009
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    Power dissipation = 112W is correct.

    But do not say average voltage times average current.

    The correct answer is rms voltage times rms current.

    Average voltage and rms voltage are not the same.
     
  10. kvi037

    Thread Starter New Member

    May 17, 2013
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    Thanks for all the replies:)
     
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