A) 4/9
B) 8/9
C) 2
D) 18
Correct Answer: C
Solution :
The equivalent network is Clearly, the network of resistances is a balanced Wheatstone bridge. So \[{{R}_{AB}}\] is given by \[\frac{1}{{{R}_{AB}}}=\frac{1}{3R}+\frac{1}{6R}=\frac{2+1}{6R}=\frac{1}{2R}\] Þ \[{{R}_{AB}}=2R\] For maximum power transfer \[2R=4\Omega \]Þ \[R=\frac{4}{2}=2\Omega \]You need to login to perform this action.
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