A) \[2\Omega \]
B) \[3\,\Omega \]
C) \[4\,\Omega \]
D) \[5\,\Omega \]
Correct Answer: A
Solution :
[a] We have, \[r=\frac{2r}{r+2}+1,\] which gives , \[{{r}^{2}}-r-2=0\] or \[\left( r-2 \right)\left( r+\text{ }1 \right)=0\] \[\therefore \,\,r=2\,\Omega \] Here the rest of the infinite ladder has been replaced by r.You need to login to perform this action.
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