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If a resistor of R ohms is connected across a battery of E colts with internal resistance r ohms, then the power (in watts) in the external resistor is \[P = \frac{E^2R}{(R+r)^2}\]

If E and r are fixed but R varies, what is the maximum value of the power?

Verified Answer :

syed514

First, let's calculate it's value: P(r) = (E² * r) / (r + r )² = E² / (4r) Now to check whether this is a maximum, we can compare it with other values of the function, in points between R = r and the other zeroes of P'(R), which here is R = -r. If P reaches a maximum in R=r, that would mean that (1) we can evaluate P in any point R>r and its value should be less than P(r), you can do this for, e.g. R=2r, which yields P(2r) = (2E²)/(9r), which is obviously smaller than P(r).
Also it means that (2) we can evaluate P in any point R so that -r < R < r, and this value should be smaller than P(r). For example, if we take R to be 0, P(R)=0, which is also smaller than P(r). Thus we have proven that P reaches a maximum in r, with corresponding value E² /(4r).

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