A) \[r=R\]
B) \[R=\frac{r}{2}\]
C) \[R=2\,r\]
D) \[R=0\]
Correct Answer: B
Solution :
Let I be the current through the circuit, then effective resistance \[R=R+\frac{r}{2}\] and \[I=\frac{E}{R+\frac{r}{2}}\] From Joules law \[P={{I}^{2}}R\] \[\Rightarrow \] \[P={{\left( \frac{E}{R+\frac{r}{2}} \right)}^{2}}R\] \[\Rightarrow \] \[P=\frac{4{{E}^{2}}R}{{{(r+2R)}^{2}}}\] For maximum power \[\frac{dP}{dR}=0\] \[\therefore \] \[\frac{dP}{dR}=0=\frac{4r{{E}^{2}}-8{{E}^{2}}R}{{{(r+2R)}^{3}}}\]. \[\Rightarrow \] \[R=\frac{r}{2}\]You need to login to perform this action.
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