I\'m having trouble with the following problem. I don\'t have problems with taki

Question

I'm having trouble with the following problem. I don't have problems with taking the derivative but the steps afterward don't make sense. Why does P'(R) = 0 -> R = r? What does that mean? Why do we plug r into the derivative?

43. If a resistor of R ohms is connected across a battery of E volts with internal resistance r ohms, then the power (in watts) in the external resistor is If E and r are fixed but R varies, what is the maximum value of the power? The expression for P?(R) shows that P?(R) > 0 for R

Explanation / Answer

The power in external resistor depends on its value 'R'. So, P(R) is function where the variable is R. Now to find the value of R for which the power in external resistor is maximum or in other words where the function P(R) has a global maximum, we simply put P'(R) = 0.

Now on solving we get R = r is a point of extremum, i.e., it can be either a global maxima point (point where P(R) has a maximum) or a global minima point (point where P(R) has a minimum). Now, to verify whether R = r it is a maxima or minima, we check the signs of P'(R) around the point R = r.

We see that P'(R)>0 for R<r and P'(R)<0 for R>r. Hence R = r is a point of maxima of the function P(R). OR in other words P(R) attains its maximum value at the point R=r, i.e., power in external resistor is maximum when its value is equal to r, i.e. R=r.

And this maximum value can be found by plugging R = r in the expression of P(R). Hence the result.

Hope it helps.

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