A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when:
A. \[R = 1000r\]
B. \[R = 0.001r\]
C. \[R = 2r\]
D. \[R = r\]

Hint: Use the formula for the current in the cell in terms of the emf, internal resistance and the external resistance. Also, use the formula of the power delivered by cell to the external resistance and the condition for which the power delivered by the cell to the external resistance is maximum.
Formula used:
The current \[i\] in the circuit is given by
\[i = \dfrac{E}{{R + r}}\] …… (1)
Here, \[E\] is the emf of the cell, \[R\] is the external resistance of the circuit and \[r\] is the internal resistance of the circuit.
The power \[P\] delivered by the cell to the external resistance is given by
\[P = {i^2}R\] …… (2)
Here, \[i\] is the current and \[R\] is the external resistance.
If a function is \[y = \dfrac{u}{v}\] then the derivative \[\dfrac{{dy}}{{dx}}\] is given by
\[\dfrac{{dy}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\] …… (3)

Complete step by step answer:
The current in the given cell is
\[i = \dfrac{E}{{R + r}}\]
Calculate the power \[P\] delivered by the cell to the external resistance \[R\].
Substitute \[\dfrac{E}{{R + r}}\] for \[i\] in equation (2).
\[P = {\left( {\dfrac{E}{{R + r}}} \right)^2}R\]
\[ \Rightarrow P = \dfrac{{{E^2}R}}{{{{\left( {R + r} \right)}^2}}}\]
The power in the circuit is maximum when the derivative of the power \[P\] with respect to the external resistance \[R\] is zero.
\[\dfrac{{dP}}{{dR}} = 0\]
Substitute \[\dfrac{{{E^2}R}}{{{{\left( {R + r} \right)}^2}}}\] for \[P\] in the above equation and take derivative of the expression of power \[P\] with respect to the external resistance \[R\].
\[\dfrac{{d\left[ {\dfrac{{{E^2}R}}{{{{\left( {R + r} \right)}^2}}}} \right]}}{{dR}} = 0\]
\[ \Rightarrow {E^2}\left[ {\dfrac{{{{\left( {R + r} \right)}^2}\dfrac{{dR}}{{dR}} - R\dfrac{{d{{\left( {R + r} \right)}^2}}}{{dR}}}}{{{{\left( {{{\left( {R + r} \right)}^2}} \right)}^2}}}} \right] = 0\]
\[ \Rightarrow \dfrac{{{{\left( {R + r} \right)}^2}\left( 1 \right) - R\left( {2\left( {R + r} \right)\left( {1 + \dfrac{{dr}}{{dR}}} \right)} \right)}}{{{{\left( {R + r} \right)}^4}}} = 0\]
\[ \Rightarrow \dfrac{{{{\left( {R + r} \right)}^2} - R\left( {2\left( {R + r} \right)\left( {1 + 0} \right)} \right)}}{{{{\left( {R + r} \right)}^4}}} = 0\]
\[ \Rightarrow {\left( {R + r} \right)^2} - 2R\left( {R + r} \right) = 0\]
\[ \Rightarrow \left( {R + r} \right) = 2R\]
\[ \Rightarrow R = r\]

From the above equation, it can be concluded that
Therefore, the power \[P\] delivered by the cell to the external resistance is maximum when the internal resistance and the external resistance are equal.

So, the correct answer is “Option D”.

Note:
Since the value of the internal resistance \[r\] for the given cell is constant, the derivative of the internal resistance \[r\] with respect to the external resistance \[R\] is zero (\[\dfrac{{dr}}{{dR}} = 0\]).