Solveeit Logo
Login

Question

Question: What is the maximum power output that can be obtained from a cell of emf \(E\) and internal resistan...

What is the maximum power output that can be obtained from a cell of emf EE and internal resistance rr?
A. 2E2r\dfrac{{2{E^2}}}{r}
B. E22r\dfrac{{{E^2}}}{{2r}}
C. E24r\dfrac{{{E^2}}}{{4r}}
D. None of these

Explanation

Solution

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 EE and the internal resistance value is rr.
Assume a battery with emf EE and internal resistance rr is connected with an external resistance.
Let the External resistance is RR and the value of current is ii.
The value of current in the circuit is calculated as,
i=Er+Ri = \dfrac{E}{{r + R}}
Write the formula of power across the external resistance RR.
P=i2RP = {i^2}R
Substitute Er+R\dfrac{E}{{r + R}} for ii in the above formula.
P=(Er+R)2RP = {\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 RR and then make it equal to zero and substitute the value of RR to get the maximum value of power output.
Differentiate the equation (1) with respect to RR.
dPdR=ddR[(Er+R)2R]\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.
df(x)g(x)=g(x)df(x)−f(x)dg(x)(g(x))2d\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).

⇒dPdR=ddR[(Er+R)2R] ⇒dPdR=E2[(r+R)2−2(r+R)R(r+R)4]  \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 dPdR\dfrac{{dP}}{{dR}} equals to zero.

⇒E2[(r+R)2−2(r+R)R(r+R)4]=0 ⇒(r+R)2−2(r+R)R=0  \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.

⇒(r+R)2=2(r+R)R ⇒(r2+R2+2rR)=2rR+2R2 ⇒R2=r2 ⇒R=r  \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 rr for RR in the equation (2) to get the maximum power output.

P=(Er+r)2r P=E24r2×r P=E24r  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 EE and internal resistance rr is E24r\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.