A) 40\[\Omega \], 90\[\Omega \]
B) 30\[\Omega \], 7.5\[\Omega \]
C) 20\[\Omega \], 6\[\Omega \]
D) 10\[\Omega \], 3\[\Omega \]
Correct Answer: B
Solution :
Let \[l\] be the distance of balancing point from left gap, then \[\frac{X}{Y}=\frac{l}{100-l}=\frac{80}{20}=4\] or \[X=4Y\] ?(i) Again in parallel, the net resistance is \[X=\frac{10X}{10+X}\] So, \[\frac{X}{Y}=\frac{50}{100-50}=1\] or \[\frac{10X}{10+X}=Y\] or \[10X=10Y+XY\] pr \[40Y=10Y+4{{Y}^{2}}\] [from Eq. (i)] or \[Y=7.5\Omega \] Putting in Eq. (i), we get \[X=30\Omega \]You need to login to perform this action.
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