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