Question

A cell of constant emf first connected to a resistance ${R_1}$ and then connected to a resistance ${R_2}$. If power delivered in both cases is same, then the internal resistance of the cell is :
A. $\sqrt {{R_1}{R_2}} $
B. $\sqrt {\dfrac{{{R_1}}}{{{R_2}}}} $
C. $\dfrac{{{R_1} - {R_2}}}{2}$
D. $\dfrac{{{R_1} + {R_2}}}{2}$

Answer 1

Hint: To solve this problem, it is important to know the concept of terminal voltage. The relationship between the emf and terminal voltage is given by –
$V = E - Ir$
where V = terminal voltage, E = emf, I = current and r = internal resistance.
This terminal voltage should be applied to calculate the power in both the above cases with different resistances and equated to calculate the internal resistance.

Complete step-by-step answer:
EMF is known as the electromotive force, which represents the potential difference between the terminals of the cell.
It is the maximum potential difference that the cell is designated to offer. However, due to several factors inside the cell, there is some voltage loss inside the cell. This voltage loss is due to the resistance offered by the components inside the cell. To overcome this resistance, some potential difference is lost and the potential difference that is actually obtained from the cell is lesser than the marked emf.
This voltage is called the terminal voltage.
If V is the terminal voltage and E is the emf of the cell, their relationship is given by –
$V = E - Ir$
where I = current and r = internal resistance.
In the first case, when we connect the cell to resistance ${R_1}$, the power dissipated is given by –
$P = \dfrac{{{V^2}}}{{{R_1}}}$
Substituting the expression for terminal voltage V, we get –
$\Rightarrow$$P = \dfrac{{{{\left( {E - Ir} \right)}^2}}}{{{R_1}}}$
In the second case, when we connect the cell to resistance ${R_2}$, the power dissipated is given by –
$\Rightarrow$$P = \dfrac{{{V^2}}}{{{R_2}}}$
Substituting the expression for terminal voltage V, we get –
$\Rightarrow$$P = \dfrac{{{{\left( {E - Ir} \right)}^2}}}{{{R_2}}}$
Given that the power dissipated is the same, we get –
$\Rightarrow$$\dfrac{{{{\left( {E - Ir} \right)}^2}}}{{{R_2}}} = \dfrac{{{{\left( {E - Ir} \right)}^2}}}{{{R_1}}}$
However, the current through the resistors are not the same. The formula for current is given by –
$I = \dfrac{E}{{R + r}}$
Applying this equation of current in the above expression, we get –
$\Rightarrow$$\dfrac{{{{\left( {E - \dfrac{E}{{{R_2} + r}}r} \right)}^2}}}{{{R_2}}} = \dfrac{{{{\left( {E - \dfrac{E}{{{R_1} + r}}r} \right)}^2}}}{{{R_1}}}$
Rearranging,
${R_1}{\left( {E - \dfrac{E}{{{R_2} + r}}r} \right)^2} = {R_2}{\left( {E - \dfrac{E}{{{R_1} + r}}r} \right)^2}$
$ \Rightarrow {R_1}{\left( {\dfrac{{E{R_2} + Er - Er}}{{{R_2} + r}}} \right)^2} = {R_2}{\left( {\dfrac{{E{R_1} + Er - Er}}{{{R_1} + r}}} \right)^2}$
$ \Rightarrow {R_1}{\left( {\dfrac{{E{R_2}}}{{{R_2} + r}}} \right)^2} = {R_2}{\left( {\dfrac{{E{R_1}}}{{{R_1} + r}}} \right)^2}$
$ \Rightarrow \dfrac{{{R_1}{E^2}R_2^2}}{{R_2^2 + {r^2} + 2{R_2}r}} = \dfrac{{{R_2}{E^2}R_1^2}}{{R_1^2 + {r^2} + 2{R_1}r}}$
$ \Rightarrow \dfrac{{{R_2}}}{{R_2^2 + {r^2} + 2{R_2}r}} = \dfrac{{{R_1}}}{{R_1^2 + {r^2} + 2{R_1}r}}$
$ \Rightarrow {R_2}R_1^2 + {R_2}{r^2} + 2{R_1}{R_2}r = {R_1}R_2^2 + {R_1}{r^2} + 2{R_1}{R_2}r$
$ \Rightarrow {R_2}{r^2} - {R_1}{r^2} = {R_1}R_2^2 - {R_2}R_1^2$
$ \Rightarrow {r^2}\left( {{R_2} - {R_1}} \right) = {R_1}{R_2}\left( {{R_2} - {R_1}} \right)$
Thus,
${r^2} = {R_1}{R_2}$
$r = \sqrt {{R_1}{R_2}} $

Hence, the correct option is A.

Note: The formula for power mentioned in the equation is derived from the Joule’s law of heating, which gives the relation,
Electric power, $P = {I^2}R$
By Ohm’s law, $V = IR \to I = \dfrac{V}{R}$
Substituting,
$P = \dfrac{{{V^2}}}{{{R^2}}}R$
$\therefore P = \dfrac{{{V^2}}}{R}$