Chapter 4.6, Problem 31E

To determine

To calculate: The maximum value of the power if a resister 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 $P=\frac{E^{2}R}{{\left(R+r\right)}^2}$ if $E$ and $r$ are fixed but $R$ varies.

Answer to Problem 31E

The maximum value of the power if a resister of $R$ ohms is connected across a battery of $E$ volts with internal resistance $r$ ohms is $\frac{E^2}{4r}$

Explanation of Solution

Given information:

A resister 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 $P=\frac{E^{2}R}{{\left(R+r\right)}^2}$ if $E$ and $r$ are fixed but $R$ varies.

Formula used:

Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

A resister 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 $P=\frac{E^{2}R}{{\left(R+r\right)}^2}$ if $E$ and $r$ are fixed but $R$ varies.

The external resistor is $P=\frac{E^{2}R}{{\left(R+r\right)}^2}\mathrm{......}\left(1\right)$

Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if $f^{\prime }(a)=0$ and
1. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $\begin{array}{l}\frac{d}{dR}P\left(R\right)=\frac{{\left(R+r\right)}^2\frac{d}{dR}{E}^{2}R-E^{2}R\frac{d}{dR}{\left(R+r\right)}^2}{{\left\{{\left(R+r\right)}^2\right\}}^2}\\ =\frac{{\left(R+r\right)}^2{E}^2-2E^{2}R\left(R+r\right)}{{\left(R+r\right)}^4}\\ =\frac{E^2\left(R+r-2R\right)}{{\left(R+r\right)}^3}\\ =\frac{E^2\left(r-R\right)}{{\left(R+r\right)}^3}\mathrm{......}\left(2\right)\end{array}$

Solve for $\frac{d}{dR}P\left(R\right)=0$ , and simplified

  $\begin{array}{l}\frac{E^2\left(r-R\right)}{{\left(R+r\right)}^3}=0\\ \Rightarrow r-R=0\\ \Rightarrow r=R\end{array}$

Differentiate equation $\left(2\right)$ with respect to $R$

  $\frac{d^2}{d{R}^2}P\left(R\right)=\frac{d}{dR}\frac{E^2\left(r-R\right)}{{\left(R+r\right)}^3}$

Recall that, Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

  $\begin{array}{l}\frac{d^2}{d{R}^2}P\left(R\right)=\frac{{\left(R+r\right)}^3\frac{d}{dR}{E}^2\left(r-R\right)-E^2\left(r-R\right)\frac{d}{dR}{\left(R+r\right)}^3}{{\left\{{\left(R+r\right)}^3\right\}}^2}\\ =\frac{{\left(R+r\right)}^3\times -E^2-3E^2\left(r-R\right){\left(R+r\right)}^2}{{\left\{{\left(R+r\right)}^3\right\}}^2}\\ =\frac{-E^2\left(R+r+3r-3R\right)}{{\left\{{\left(R+r\right)}^3\right\}}^2}\\ =\frac{-E^2\left(4r-2R\right)}{{\left(R+r\right)}^3}\end{array}$

Substitute $R=r$ and solve,

  $\begin{array}{l}=\frac{-E^2\left(4r-2r\right)}{{\left(r+r\right)}^3}\\ =\frac{-2E^{2}r}{8r^3}\\ =\frac{-E^2}{4r^2}\end{array}$

Therefore,

  $\frac{d^2}{d{R}^2}P\left(R\right)=\frac{-E^2}{4r^2}<0$

For $R=r$ the power is maximum

Substitute $R=r$ in equation $\left(1\right)$ and simplified

  $\begin{array}{l}P=\frac{E^{2}R}{{\left(R+r\right)}^2}\\ =\frac{E^{2}r}{{\left(r+r\right)}^2}\\ =\frac{E^{2}r}{4r^2}\\ =\frac{E^2}{4r}\end{array}$

Conclusion:

Thus the maximum value of the power if a resister of $R$ ohms is connected across a battery of $E$ volts with internal resistance $r$ ohms is $\frac{E^2}{4r}$