• # question_answer An accumulator of emf $\varepsilon$ and internal resistance r is first connected to an external resistance ${{R}_{1}}$ and then to an external resistance ${{R}_{2}}$ for the same time. For what value of r, the heats dissipated in ${{R}_{1}}$ and ${{R}_{2}}$ will De same?

$\sqrt{{{R}_{1}}{{R}_{2}}}$ When the accumulator is connected to ${{R}_{1}},$ heat dissipated is given by ${{H}_{1}}=I_{1}^{2}{{R}_{1}}t={{\left( \frac{\varepsilon }{{{R}_{1}}+r} \right)}^{2}}{{R}_{1}}t$ When it is connected to ${{R}_{2}},$ heat dissipated is given by                         ${{H}_{2}}=I_{2}^{2}{{R}_{2}}t={{\left( \frac{\varepsilon }{{{R}_{2}}+r} \right)}^{2}}{{R}_{2}}t$             Given,   ${{H}_{1}}={{H}_{2}}$ i.e. ${{\left( \frac{\varepsilon }{{{R}_{1}}+r} \right)}^{2}}{{R}_{1}}t={{\left( \frac{\varepsilon }{{{R}_{2}}+r} \right)}^{2}}{{R}_{2}}$                         $\frac{{{R}_{1}}}{{{({{R}_{1}}+r)}^{2}}}=\frac{{{R}_{2}}}{{{({{R}_{2}}+r)}^{2}}}$                         ${{R}_{1}}{{({{R}_{2}}+r)}^{2}}={{R}_{2}}{{({{R}_{1}}+r)}^{2}}$                         ${{R}_{1}}(R_{2}^{2}+{{r}^{2}}+2{{R}_{2}}r={{R}_{2}}(R_{1}^{2}+{{r}^{2}}+2{{R}_{1}}r)$ ${{R}_{1}}R_{2}^{2}+{{R}_{1}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r={{R}_{2}}R_{1}^{2}+{{R}_{2}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r$                         ${{R}_{1}}R_{2}^{2}-{{R}_{2}}R_{1}^{2}+{{R}_{1}}{{r}^{2}}-{{R}_{2}}{{r}^{2}}=0$                         ${{R}_{1}}{{R}_{2}}({{R}_{2}}-{{R}_{1}})={{r}^{2}}({{R}_{2}}-{{R}_{1}})$                                     $r=\sqrt{{{R}_{1}}{{R}_{2}}}$