Calculating Power Loss

[JayJoe]

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

 

[duke37]

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?

 

[JayJoe]

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.

 

[duke37]

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