Question

A cell of constant emf first connected to a resistance ${{R}_{1}}$ and then connected to a resistance ${{R}_{2}}$ . If power delivered in both cases is same, then the internal resistance of the cell is

• A. $\sqrt{{{R}_{1}}{{R}_{2}}}$
• B. $\sqrt{\frac{{{R}_{1}}}{{{R}_{2}}}}$
• C. $\frac{{{R}_{1}}-{{R}_{2}}}{2}$
• D. $\frac{{{R}_{1}}+{{R}_{2}}}{2}$

Answer 1

Answer: (A)

$\sqrt{{{R}_{1}}{{R}_{2}}}$

Answer 2

Current given by cell $I=\frac{E}{R+r}$ Power delivered in second case ${{P}_{2}}={{I}^{2}}{{R}_{2}}$ $={{\left( \frac{E}{{{R}_{2}}+r} \right)}^{2}}{{R}_{2}}$ Power delivered is same in the both cases ${{\left( \frac{E}{{{R}_{2}}+r} \right)}^{2}}{{R}_{1}}={{\left( \frac{E}{{{R}_{2}}+r} \right)}^{2}}$ ${{R}_{1}}(R_{2}^{2}+{{r}^{2}}+2{{R}_{2}}r)={{R}_{2}}(R_{1}^{2}{{r}^{2}}+{{r}^{2}}2{{R}_{1}}r)$ ${{R}_{1}}R_{2}^{2}+{{R}_{1}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r={{R}_{2}}R_{1}^{2}{{r}^{2}}+{{R}_{2}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r$ ${{R}_{1}}R_{2}^{2}-{{R}_{2}}R_{1}^{2}={{R}_{2}}{{R}^{2}}-{{R}_{1}}{{r}^{2}}$ ${{R}_{1}}{{R}_{2}}({{R}_{2}}-{{R}_{1}})={{R}_{2}}{{r}^{2}}-{{R}_{1}}{{r}^{2}}$ $r=\sqrt{{{R}_{1}}{{R}_{2}}}$