A) \[R=r\]
B) \[R<r\]
C) \[R>r\]
D) \[R=1/r\]
Correct Answer: A
Solution :
Current in the circuit \[I=\frac{E}{R+r}\] Power delivered to the resistance R is : \[P={{I}^{2}}\,R=\frac{{{E}^{2}}R}{{{\left( R+r \right)}^{2}}}\] It is maximum when \[\frac{d\,P}{d\,R}=0\] \[\frac{d\,P}{d\,R}={{E}^{2}}\left[ \frac{{{\left( r+R \right)}^{2}}-2R\left( r+R \right)}{{{\left( r+R \right)}^{4}}} \right]=0\] or \[{{\left( r+R \right)}^{2}}=2R\left( r+R \right)\]or \[R=r\]
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