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  • A battery of emf and internal resistance <img alt="\mathrm{r}" src="https://entrancecorner.oncodecogs.c

A battery of emf \varepsilon and internal resistance \mathrm{r} is connected to a resistor of resistance\mathrm{ R_1 \: and \: Q J}of heat is produced in a certain time \mathrm{t}. When the same battery is connected to another resistor of resistance \mathrm{R_2}, the same quantity of heat is produced in the same time \mathrm{ t}. Then, the value of \mathrm{ r} is:
 

Option: 1

\frac{\mathrm{R}_2^2}{\mathrm{R}_1}
 


Option: 2

\frac{1}{2}\left(\mathrm{R}_1+\mathrm{R}_2\right)
 


Option: 3

\sqrt{\mathrm{R}_1 \mathrm{R}_2}
 


Option: 4

\mathrm{\frac{R_1^2}{R_2}}


Answers (1)

best_answer

In the first case,

Current in the circuit, \mathrm{I}_1=\frac{\varepsilon}{\mathrm{R}_1+\mathrm{r}}
Amount of heat produced in resistor \mathrm{\mathrm{R}_1 \: in \: time \: \mathrm{t}} is
\mathrm{Q}_1=\mathrm{I}_1^2 \mathrm{R}_1 \mathrm{t}=\frac{\varepsilon^2}{\left(\mathrm{R}_1+\mathrm{r}\right)^2} \mathrm{R}_1 \mathrm{t}             (i)

In the second case,

Current in \mathrm{R}_1 the circuit, \mathrm{I}_2=\frac{\varepsilon}{\mathrm{r}+\mathrm{R}_2}
Amount of heat produced in resistor \mathrm{R_2} in the same time \mathrm{t}  is
\mathrm{Q}_2=\mathrm{I}_2^2 \mathrm{R}_2 \mathrm{t}=\frac{\varepsilon^2}{\left(\mathrm{r}+\mathrm{R}_2\right)^2} \mathrm{R}_2 \mathrm{t}                (ii)
As per question, \mathrm{Q}_1=\mathrm{Q}_2=\mathrm{Q}
From (i) and (ii), we get
\begin{aligned} & \frac{\varepsilon^2 \mathrm{R}_1 \mathrm{t}}{\left(\mathrm{r}+\mathrm{R}_1\right)^2}=\frac{\varepsilon^2 \mathrm{R}_2 \mathrm{t}}{\left(\mathrm{r}+\mathrm{R}_2\right)^2} \\ \end{aligned}

\begin{aligned} \frac{\mathrm{R}_1}{\left(\mathrm{r}+\mathrm{R}_1\right)^2}=\frac{\mathrm{R}_2}{\left(\mathrm{r}+\mathrm{R}_2\right)^2} \end{aligned}
On solving, we get

\mathrm{r=\sqrt{R_1 R_2}}

Hence option 3 is correct.

Posted by

sudhir.kumar

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