Optimization Problem in Relation to Battery, Resistor and Internal Resistance?
P = E^2*R / (R+r)^2 (P equals E square multiplied with R then divided with R plus r in brackets squared)
If E and r are fixed but R varies, what is the maximum value of the power?
I'm kind of stuck with this question and the only thing I can think about now is to differentiate it once. Thanks for the help!
Answer:
HINT: The max power dissipated by R will be when r=R
[REPOST] Okay, another hint, the curve the value of R versus the power it will dissipate is a downward-opening parabola with its vertex at r=R.
For example, let's first assume both r and R equal 1 ohm and the battery is 1 volt. Then current will be
1V/(r+R)= 1/2 = 0.5 amps, and the power dissipated by R will be, according to P=I^2R, 0.25W
Now let r still equal 1, but let R =0.9 ohms. Then
PsubR = I^2R = (V/(r+R))^2*R =~ 0.2493W, a slightly-lower value than at R=1. You will get the exact same result if you select R=1.1 ohms.
If this is a calculus question, differentiating
PsubR= (V/(r+R))^2*R will give a zero at r=R
Well I wonder if you couldn't take the limit of this equation? then evaluate it over R=0 and R= inf. Since E and r are fixed constants you will end up with P=1/R. A engineering professor's favorite, P will be maxed out when R approaches zero, but be careful here because it will also be maxed out when it approaches zero from the negative side.
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