A) \[r<R\]
B) \[r>R\]
C) \[r=1/R\]
D) \[r=R\]
Correct Answer: D
Solution :
We know that the current in the circuit\[I=\frac{E}{R+r}\]and power delivered to the resistance R is \[P={{I}^{2}}R=\frac{{{E}^{2}}R}{{{(R+r)}^{2}}}\] It is maximum when\[\frac{dP}{dR}=0\] \[\frac{dP}{dR}={{E}^{2}}\left[ \frac{{{(r+R)}^{2}}-2R(r+R)}{{{(r+R)}^{4}}} \right]=0\] or \[{{(r+R)}^{2}}=2R(r+R)\]or \[R=r\]
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