A cell having EMF E and internal resistance r is connected to a load R. Find the value of R such that power dissipated by the cell is maximum.

Hint:In order to calculate the value of R such that power dissipated by the cell is maximum,we need to use the following formula for current as well as power dissipated as mentioned below.

Complete Solution
Current Is calculated by,
\[I=\dfrac{E}{R+r}\]
Power dissipated by cell is,
\[~P={{I}^{2}}R\]
\[=\dfrac{{{E}^{2}}R}{{{(R+r)}^{2}}}\]
For maximum power by cell,
\[\dfrac{dP}{dR}=0\]
\[\Rightarrow {{E}^{2}}\dfrac{{{(R+r)}^{2}}-2R(R+r)}{{{(R+r)}^{4}}}=0\]
\[\Rightarrow R=r\]

Thus, that is R is internal resistance.

Additional Information:An “electric power supply” is also an Electric cell. Cells generate electricity and also derive chemical reactions. One or more electrochemical cells are batteries.EMF is Electromotive Force, which is measured in coulombs of charge. It is pressure developed or an electric intensity from an electrical energy or a source. It is a device which converts any form of energy into electrical energy which is then measured with coulombs of charge.When there is current present in the device or the electrical circuit and there’s a voltage drop in source voltage or source battery is internal resistance. It is caused due to electrolytic material in batteries or other voltage sources.

Note:While solving this question, we should know that the EMF is Electromotive Force, which is measured in coulombs of charge and other basic concepts of it. Also, about the internal resistance of it. It is caused due to electrolytic material in batteries or other voltage sources. One should be careful about the various formulae used here.