Question

A cell sends a current through a resistance \[R\] for time \[t\]. Now the same cell sends current through another resistance \[r\] for the same time. If same amount of heat is developed in both the resistance, then the internal resistance of cell is
(A) \[\dfrac{{R + r}}{2}\]
(B) \[\dfrac{{R - r}}{2}\]
(C) \[\dfrac{{Rr}}{2}\]
(D) \[\sqrt {Rr} \]

Answer 1

Hint: The electrical energy consumed by the resistance is equal to the heat generated. Since, the time through them is constant, then the power through the resistances are the same.
Formula used: In this solution we will be using the following formulae;
\[I = \dfrac{E}{{R + {r_{in}}}}\] where \[E\] is the emf generated by the cell, \[I\] is the current flowing through the circuit and \[{r_{in}}\] is the internal resistance of the cell, \[R\] is the external resistance.
\[P = {I^2}R\] where \[P\] is the power consumed by a resistor with resistance \[R\].

Complete Step-by-Step Solution:
When a resistance \[R\] is connected to the cell, the current through it is \[{I_R}\], then can be given as \[{I_R} = \dfrac{E}{{R + {r_{in}}}}\]where \[E\] is the emf generated by the cell, and \[{r_{in}}\] is the internal resistance of the cell.
Similarly, when the second resistance is connected, the current flowing in the circuit will be \[{I_r} = \dfrac{E}{{r + {r_{in}}}}\]The heat developed is equal to the electric energy consumed, hence electric power since the times are equal.
Electric power is given by
\[P = {I^2}R\] where \[P\] is the power consumed by a resistor with resistance \[R\], \[I\] is the current flowing in the resistor.
Hence,
\[{\left( {\dfrac{E}{{R + {r_{in}}}}} \right)^2}R = {\left( {\dfrac{E}{{r + {r_{in}}}}} \right)^2}r\]
\[ \Rightarrow \dfrac{{{E^2}R}}{{{{\left( {R + {r_{in}}} \right)}^2}}} = \dfrac{{{E^2}r}}{{{{\left( {r + {r_{in}}} \right)}^2}}}\]
Cancelling \[{E^2}\], square-rooting both sides, and inverting, we have
\[\dfrac{{R + {r_{in}}}}{{\sqrt R }} = \dfrac{{r + {r_{in}}}}{{\sqrt r }}\]
Cross-multiplying, we have
\[R\sqrt r + {r_{in}}\sqrt r = r\sqrt R + {r_{in}}\sqrt R \]
Collecting all like terms and making \[{r_{in}}\] subject, we get
\[{r_{in}} = \dfrac{{R\sqrt r - r\sqrt R }}{{\sqrt R - \sqrt r }}\]
Hence, by multiplying numerator and denominator by the conjugate of the denominator we have,
\[{r_{in}} = \dfrac{{R\sqrt r - r\sqrt R }}{{\sqrt R - \sqrt r }} \times \dfrac{{\sqrt R + \sqrt r }}{{\sqrt R + \sqrt r }}\]
Multiplying through and simplifying, we have
\[{r_{in}} = \sqrt {Rr} \dfrac{{\left( {R - r} \right)}}{{R - r}}\]
\[ \Rightarrow {r_{in}} = \sqrt {Rr} \]

Hence, the correct option is D

Note: For clarity, the current the circuit when a resistance is connected to the cell given by \[I = \dfrac{E}{{R + {r_{in}}}}\] can be proven from Kirchhoff’s voltage law, as in
\[E - I{r_{in}} - IR = 0\] where \[I{r_{in}}\] drop across the internal resistance (recall that it is modelled as though it is connected in series to the emf) , and \[IR\] is the voltage drop across the external resistance.