Question

A cell of emf E is connected to a resistance ${{R}_{1}}$ for time t and the amount of heat generated in it is H. If the resistance ${{R}_{1}}$ is replaced by another resistance ${{R}_{2}}$ and is connected to the cell for the same time t, the amount of heat generated in ${{R}_{2}}$ is 4H. The internal resistance of the cell is:
A. $\dfrac{2{{R}_{1}}+{{R}_{2}}}{2}$
B. $\sqrt{{{R}_{1}}{{R}_{2}}}\times \dfrac{2\sqrt{{{R}_{2}}}-\sqrt{{{R}_{1}}}}{\sqrt{{{R}_{2}}}-2\sqrt{{{R}_{1}}}}$
C. $\sqrt{{{R}_{1}}{{R}_{2}}}\times \dfrac{\sqrt{{{R}_{2}}}-2\sqrt{{{R}_{1}}}}{2\sqrt{{{R}_{2}}}-\sqrt{{{R}_{1}}}}$
D. $\sqrt{{{R}_{1}}{{R}_{2}}}\times \dfrac{\sqrt{{{R}_{2}}}-\sqrt{{{R}_{1}}}}{\sqrt{{{R}_{2}}}+\sqrt{{{R}_{1}}}}$

Answer 1

Hint: We have two cases in a circuit by connecting two different resistances in series with a cell of emf E and internal resistance r. You could recall the expression for heat produced for the two cases then you could rewrite the expression in terms of internal resistance and hence find the answer.
 Formula used:
Heat produced,
${{I}^{2}}R=H$
Current,
$I=\dfrac{E}{R+r}$

Complete step by step answer:
In the question, we are said that on connecting a cell of emf E to a resistance ${{R}_{1}}$ for a duration of t time produces heat H and on connecting another resistance ${{R}_{2}}$ for the same duration produces a heat of 4H. We are supposed to find the internal resistance of the cell using this information.
We know that heat generated in the first case could be given by,
${{I}_{1}}^{2}{{R}_{1}}=H$ …………………………………………… (1)
Now, for the second case it would be,
${{I}_{2}}^{2}{{R}_{2}}=4H$ ……………………………………………. (2)
But we know that, for a circuit with a cell of emf E and internal resistance r, the current would be given by,
$I=\dfrac{E}{R+r}$
(1) and (2) now becomes,
$\dfrac{{{E}^{2}}}{{{\left( {{R}_{1}}+r \right)}^{2}}}{{R}_{1}}=H$
$\dfrac{{{E}^{2}}}{{{\left( {{R}_{2}}+r \right)}^{2}}}{{R}_{2}}=4H$
Now, we have,
$\dfrac{{{R}_{2}}}{{{\left( {{R}_{2}}+r \right)}^{2}}}=4\dfrac{{{R}_{1}}}{{{\left( {{R}_{1}}+r \right)}^{2}}}$
$\Rightarrow \sqrt{{{R}_{2}}}\left( {{R}_{1}}+r \right)=2\sqrt{{{R}_{1}}}\left( {{R}_{2}}+r \right)$
$\therefore r=\dfrac{\sqrt{{{R}_{1}}{{R}_{2}}}\left[ \sqrt{{{R}_{1}}}-2\sqrt{{{R}_{2}}} \right]}{\left[ 2\sqrt{{{R}_{1}}}-\sqrt{{{R}_{2}}} \right]}$ …………………………………….. (3)
Therefore, we found the internal resistance of the cell to be,
$r=\dfrac{\sqrt{{{R}_{1}}{{R}_{2}}}\left[ 2\sqrt{{{R}_{2}}}-\sqrt{{{R}_{1}}} \right]}{\left[ \sqrt{{{R}_{2}}}-2\sqrt{{{R}_{1}}} \right]}$

So, the correct answer is “Option B”.

Note: We have simply multiplied the numerator and denominator by -1 so as to arrive at the final result. This is done so as to make the answer just as given in the options. So, there would be many cases like this where your final result wouldn’t look exactly as given in the options and you will have to make these minor changes.