A) 400, 90
B) 300, 7.50
C) 200, 60
D) 100, 30
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 \[10X=10Y+XY\] or \[40Y=10y+4{{Y}^{2}}\] [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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