What is the maximum power output that can be obtained from a cell of emf $E$ and internal resistance $r$?
A. $\dfrac{{2{E^2}}}{r}$
B. $\dfrac{{{E^2}}}{{2r}}$
C. $\dfrac{{{E^2}}}{{4r}}$
D. None of these

Hint:Calculate the current in a circuit with a battery having external resistance and external resistance and then calculate the maximum power by using the maxima and differentiation property of function.

Complete step-by-step solution:
According to the question it is given that the a cell has emf of value $E$ and the internal resistance value is $r$.
Assume a battery with emf $E$ and internal resistance $r$ is connected with an external resistance.
Let the External resistance is $R$ and the value of current is $i$.
The value of current in the circuit is calculated as,
$i = \dfrac{E}{{r + R}}$
Write the formula of power across the external resistance $R$.
$P = {i^2}R$
Substitute $\dfrac{E}{{r + R}}$ for $i$ in the above formula.
$P = {\left( {\dfrac{E}{{r + R}}} \right)^2}R$ …..(1)
If we have to find the maximum power output then it will be differentiated with respect to $R$ and then make it equal to zero and substitute the value of $R$ to get the maximum value of power output.
Differentiate the equation (1) with respect to $R$.
\[\dfrac{{dP}}{{dR}} = \dfrac{d}{{dR}}\left[ {{{\left( {\dfrac{E}{{r + R}}} \right)}^2}R} \right]\] …..(2)
Use the property of differentiation as shown below.
$d\dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{g\left( x \right)df\left( x \right) - f\left( x \right)dg\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}$
Use the property of differentiation in the equation (2).
\[
   \Rightarrow \dfrac{{dP}}{{dR}} = \dfrac{d}{{dR}}\left[ {{{\left( {\dfrac{E}{{r + R}}} \right)}^2}R} \right] \\
   \Rightarrow \dfrac{{dP}}{{dR}} = {E^2}\left[ {\dfrac{{{{\left( {r + R} \right)}^2} - 2\left( {r + R} \right)R}}{{{{\left( {r + R} \right)}^4}}}} \right] \\
 \]
For the condition of maxima put \[\dfrac{{dP}}{{dR}}\] equals to zero.
\[
   \Rightarrow {E^2}\left[ {\dfrac{{{{\left( {r + R} \right)}^2} - 2\left( {r + R} \right)R}}{{{{\left( {r + R} \right)}^4}}}} \right] = 0 \\
   \Rightarrow {\left( {r + R} \right)^2} - 2\left( {r + R} \right)R = 0 \\
 \]
Simplify the above expression to get the relation.
\[
   \Rightarrow {\left( {r + R} \right)^2} = 2\left( {r + R} \right)R \\
   \Rightarrow \left( {{r^2} + {R^2} + 2rR} \right) = 2rR + 2{R^2} \\
   \Rightarrow {R^2} = {r^2} \\
   \Rightarrow R = r \\
 \]
So, the power will be maximum when the value of external resistance is equal to the value of internal resistance.
Substitute $r$ for $R$ in the equation (2) to get the maximum power output.
\[
  P = {\left( {\dfrac{E}{{r + r}}} \right)^2}r \\
  P = \dfrac{{{E^2}}}{{4{r^2}}} \times r \\
  P = \dfrac{{{E^2}}}{{4r}} \\
 \]
So, the maximum power output that can be obtained from a cell of emf $E$ and internal resistance $r$ is \[\dfrac{{{E^2}}}{{4r}}\].
Hence, the correct option is C.

Note:-
In the case of finding the maxima of a function always differentiate with variable and get the value of variable. Always solve the equation in the known variable and eliminate the unknown variables.