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)

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}}} $