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Calculating Power Loss

Discussion in 'Electronics Homework Help' started by JayJoe, Jan 14, 2017.

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  1. JayJoe

    JayJoe

    12
    1
    Jul 31, 2015
    Hi,

    While I am watching this video of GreatScott! **At 3:21 ,
    I took up my calculator and calculate the power by using P=(V^2)/R, and the answer is 72W.

    - This happens due to the voltage across the Peltier may not be accurately 12V as there's Req=Rpeltier+Rwire, is my statement correct?
    - 72W and 32W is totally a huge difference, why?
    - So, when to use P=(I^2)R and P=(V^2)/R to calculate power loss?

    Thanks in advanced.
    Jay
     

    Attached Files:

  2. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    You have a load of 2Ω.
    With 4A the power is I*I*R =32W and the voltage is 8V.
    With 12V the power is V*V/R = 144/2 = 72W and the current is 6A.

    Which voltage do you have?
     
  3. JayJoe

    JayJoe

    12
    1
    Jul 31, 2015
    Hi duke37,

    From the video, we can see that GreatScott! is supplying 12V to the peltier and the total current flow is 4A. So, basically 12V and 4A are given and constant.

    So, P=VI=48W is the power supply, while Ploss=(I^2)R=32W. My question is why P=(V^2)/R can't be used to calculate the Ploss?

    Jay.
     
  4. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    I do not have video most of the time.

    You need to get a consistent set of data.
    12V at 4A means the resistance is 3Ω, not 2Ω as previously assumed.
    P = 12*12/3 = 48W
    P = 4*4*3 = 48W
    everything is consistent with 12V, 4A, 3Ω and 48W
     
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