A) \[\sqrt{{{R}_{1}}{{R}_{2}}}\]
B) \[\sqrt{\frac{{{R}_{1}}}{{{R}_{2}}}}\]
C) \[\frac{{{R}_{1}}-{{R}_{2}}}{2}\]
D) \[\frac{{{R}_{1}}+{{R}_{2}}}{2}\]
Correct Answer: A
Solution :
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 \[{{\left( \frac{E}{{{R}_{2}}+r} \right)}^{2}}{{R}_{1}}={{\left( \frac{E}{{{R}_{2}}+r} \right)}^{2}}\] \[{{R}_{1}}(R_{2}^{2}+{{r}^{2}}+2{{R}_{2}}r)={{R}_{2}}(R_{1}^{2}{{r}^{2}}+{{r}^{2}}2{{R}_{1}}r)\] \[{{R}_{1}}R_{2}^{2}+{{R}_{1}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r={{R}_{2}}R_{1}^{2}{{r}^{2}}+{{R}_{2}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r\] \[{{R}_{1}}R_{2}^{2}-{{R}_{2}}R_{1}^{2}={{R}_{2}}{{R}^{2}}-{{R}_{1}}{{r}^{2}}\] \[{{R}_{1}}{{R}_{2}}({{R}_{2}}-{{R}_{1}})={{R}_{2}}{{r}^{2}}-{{R}_{1}}{{r}^{2}}\] \[r=\sqrt{{{R}_{1}}{{R}_{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
