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 first case \[{{P}_{1}}={{I}^{2}}{{R}_{1}}\] \[={{\left( \frac{E}{{{R}_{1}}+r} \right)}^{2}}{{R}_{1}}\] 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 the cases \[{{\left( \frac{E}{{{R}_{1}}+r} \right)}^{2}}{{R}_{1}}={{\left( \frac{E}{{{R}_{2}}+r} \right)}^{2}}{{R}_{2}}\] \[\frac{{{R}_{1}}}{{{({{R}_{1}}+r)}^{2}}}=\frac{{{R}_{2}}}{{{({{R}_{2}}+r)}^{2}}}\] \[{{R}_{1}}(R_{2}^{2}+{{r}^{2}}+2{{R}_{2}}r)={{R}_{2}}(R_{1}^{2}+{{r}^{2}}+2{{R}_{1}}r)\] \[{{R}_{1}}R_{2}^{2}+{{R}_{1}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r={{R}_{2}}R_{1}^{2}+{{R}_{2}}{{r}^{2}}+2{{R}_{1}}{{R}_{2}}r\]\[{{R}_{1}}R_{2}^{2}-{{R}_{2}}R_{1}^{2}={{R}_{2}}{{r}^{2}}-{{R}_{1}}{{r}^{2}}\] \[{{R}_{1}}{{R}_{2}}({{R}_{2}}-{{R}_{1}})={{r}^{2}}({{R}_{2}}-{{R}_{1}})\] \[r=\sqrt{{{R}_{1}}{{R}_{2}}}\]
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