Question

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

\(P = \frac{{{E^2}R}}{{{{\left( {R + r} \right)}^2}}}\)

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

Short answer

The obtained value of maximum power is: \(P = \frac{{{E^2}}}{{4r}}\).

Step-by-step solution

Optimization of a Function

TheOptimization of a Function\(f\left( x \right)\)for any real value of\(c\)can be given as:

\(\begin{aligned}{l}f\left( x \right)\left| {_{x = c} \to \max \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\\{\rm{and}}\\f\left( x \right)\left| {_{x = c} \to \min \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\end{aligned}\)

Optimising the given function of power.

The given function for Power is:

\(P = \frac{{{E^2}R}}{{{{\left( {R + r} \right)}^2}}}\)

Since,\(R\)is variable. Then, for maximization, we have:

\(\begin{aligned}{c}P'\left( R \right) = \frac{d}{{dR}}\left( {\frac{{{E^2}R}}{{{{\left( {R + r} \right)}^2}}}} \right) = 0\\{E^2}\left( {\frac{{{{\left( {R + r} \right)}^2} - 2R\left( {R + r} \right)}}{{{{\left( {R + r} \right)}^4}}}} \right) = 0\\{E^2}\left( {\frac{{\left( {r - R} \right)}}{{{{\left( {R + r} \right)}^3}}}} \right) = 0\\r = R\end{aligned}\)

Now, we have:

\(\begin{aligned}{l}{\rm{for }}R > r \to P'\left( R \right) < 0\\{\rm{for }}R < r \to P'\left( R \right) > 0\end{aligned}\)

Therefore, the function will have absolute maximumvalue at \(r = R\). For this, we have the maximum power as:

\(\begin{aligned}{c}P\left| {_{\max }} \right. = \frac{{{E^2}r}}{{{{\left( {r + r} \right)}^2}}}\\ = \frac{{{E^2}}}{{4r}}\end{aligned}\)

Hence, the maximum power obtained is \(P = \frac{{{E^2}}}{{4r}}\).