AL phy max power 概念問題

The crucial point lies on the current I.
Since power P dissipated on the external resistance R is given by,
P = I^2.(R)
Increasing R will lower the current , but decreasing R will increase the current.(Because I = E/(R + r), where E is the emf of the battery, and r its internal resistance).

Therefore, in order to have maximum power P, I and R should be compromised such that the product (I^2.R) is maximum. This can be shown to occur when R = r

Be aware that the power delivered by the battery when R = r, and when R >> r are not the same. Because power P depends on the square of current, because r is fixed, hence the current when R = r is is larger than the current when R >> r. Thus the power delivered by the battery in the former situation (i.e. R = r) is expected to be larger than that in the latter (R >> r).
You cannot use "efficieny" to compare the power dissipated in the resistance R under different conditions (i.e. power delivered by battery when R=r and R>>r are different).

你將Max Power Output 和 Efficiency兩個概念混亂了..
這是兩種不同的意思,

Max Porwer Output 是指when R = r時
resistance有最大Power(最大功率..即放得最多能量..)
V = I(R+r)
Output = I^2 R = V^2 R / (R+r)^2

而Efficiency是指效能...
即所用的電能有多少%落係output,
Efficiency = P output / P input


打個比如...
當你R = r 時,Power output可以達到最高...(每秒用到非常多的能量)
但同時,你亦有好多能量浪費左..(有50%用左係internal resistance)

相反...你R>>r...你會得到非常高的efficiency...(當90.9%)
雖然你大部份能量沒有浪費...但你Power Output非常低..(即每秒只有好少能量便用出來)


when r = 50 , V = 100
when R = 50
Power Output = V^2 R /(R+r)^2 = (100^2)(50)/(50+50)^2 = 50W
Efficiency = 50/50+50 = 50%

when R = 500
Power Output = V^2 R /(R+r)^2 = (100^2)(500)/(500+50)^2 = 16.5W
Efficiency = 500/500+50 = 90.9%