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?

Answer:

\[\sqrt{{{R}_{1}}{{R}_{2}}}\] 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. \[{{\left( \frac{\varepsilon }{{{R}_{1}}+r} \right)}^{2}}{{R}_{1}}t={{\left( \frac{\varepsilon }{{{R}_{2}}+r} \right)}^{2}}{{R}_{2}}\]                         \[\frac{{{R}_{1}}}{{{({{R}_{1}}+r)}^{2}}}=\frac{{{R}_{2}}}{{{({{R}_{2}}+r)}^{2}}}\]                         \[{{R}_{1}}{{({{R}_{2}}+r)}^{2}}={{R}_{2}}{{({{R}_{1}}+r)}^{2}}\]                         \[{{R}_{1}}(R_{2}^{2}+{{r}^{2}}+2{{R}_{2}}r={{R}_{2}}(R_{1}^{2}+{{r}^{2}}+2{{R}_{1}}r)\] \[{{R}_{1}}R_{2}^{2}+{{R}_{1}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r={{R}_{2}}R_{1}^{2}+{{R}_{2}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r\]                         \[{{R}_{1}}R_{2}^{2}-{{R}_{2}}R_{1}^{2}+{{R}_{1}}{{r}^{2}}-{{R}_{2}}{{r}^{2}}=0\]                         \[{{R}_{1}}{{R}_{2}}({{R}_{2}}-{{R}_{1}})={{r}^{2}}({{R}_{2}}-{{R}_{1}})\]                                     \[r=\sqrt{{{R}_{1}}{{R}_{2}}}\]