A battery develops a maxim power Pmax=9.0W and the intensity I=3.0A. Find the electromotive voltage and it's internal resistance.

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The power is RI^2=R[E/(R+r)^2]=f(R)

The extrem condition is to annule the first derivative:

f'(R)=[E/(R+r)^2]-R*2[E/(R+r)][E/(R+r)^2]=

=[E^2/(R+r)^3](r-R)=0, so R=r

The adjustment theorem says that the extreme nature could be found from the sign of the second derivative:

f"(R)=2[E^2/(R+r)^4](R-2r). For the r root of the first derivative is negative, so the extrem is maxim.

The result could be found right away, without derivatives:

P=E^2/(r/[(sqrt R)+sqrtR]

At denominator we have the sum of 2 terms, which product is constant, so the sum is minim when the terms are equal

r/(sqrt R)=sqrtR, meaning R=r.

The denominator is minim, so the ratio, power, will be maxim.

E=2Pm:Im=6V, where Im=E/2r

Pm=rIm^2=(1/2)EIm

r=Pm/Im^2=1 Ohm

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