Question

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

Answer 1

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,
$i={\displaystyle \frac{E}{r+R}}$
Write the formula of power across the external resistance $R$.
$P=i^{2}R$
Substitute ${\displaystyle \frac{E}{r+R}}$ for $i$ in the above formula.
$P={\left({\displaystyle \frac{E}{r+R}}\right)}^{2}R$ …..(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 $R$.
$${\displaystyle \frac{dP}{dR}}={\displaystyle \frac{d}{dR}}\left[{\left({\displaystyle \frac{E}{r+R}}\right)}^{2}R\right]$$ …..(2)
Use the property of differentiation as shown below.
$d{\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}={\displaystyle \frac{g\left(x\right)df\left(x\right)-f\left(x\right)dg\left(x\right)}{{\left(g\left(x\right)\right)}^2}}$
Use the property of differentiation in the equation (2).
$$\begin{gathered} \Rightarrow {\displaystyle \frac{dP}{dR}}={\displaystyle \frac{d}{dR}}\left[{\left({\displaystyle \frac{E}{r+R}}\right)}^{2}R\right] \\ \Rightarrow {\displaystyle \frac{dP}{dR}}=E^2\left[{\displaystyle \frac{{\left(r+R\right)}^2-2\left(r+R\right)R}{{\left(r+R\right)}^4}}\right] \end{gathered}$$
For the condition of maxima put $${\displaystyle \frac{dP}{dR}}$$ equals to zero.
$$\begin{gathered} \Rightarrow E^2\left[{\displaystyle \frac{{\left(r+R\right)}^2-2\left(r+R\right)R}{{\left(r+R\right)}^4}}\right]=0 \\ \Rightarrow {\left(r+R\right)}^2-2\left(r+R\right)R=0 \end{gathered}$$
Simplify the above expression to get the relation.
$$\begin{gathered} \Rightarrow {\left(r+R\right)}^2=2\left(r+R\right)R \\ \Rightarrow \left(r^2+R^2+2rR\right)=2rR+2R^2 \\ \Rightarrow R^2=r^2 \\ \Rightarrow R=r \end{gathered}$$
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.
$$\begin{gathered} P={\left({\displaystyle \frac{E}{r+r}}\right)}^{2}r \\ P={\displaystyle \frac{E^2}{4r^2}}\times r \\ P={\displaystyle \frac{E^2}{4r}} \end{gathered}$$
So, the maximum power output that can be obtained from a cell of emf $E$ and internal resistance $r$ is $${\displaystyle \frac{E^2}{4r}}$$.
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.