A) \[50.0\,\Omega \]
B) \[10\,\Omega \]
C) \[36.7\,\Omega \]
D) \[26.7\,\Omega \]
Correct Answer: D
Solution :
[d] For ADE\[\frac{1}{R'}=\frac{1}{2x}+\frac{1}{10}\] or \[R'=\frac{20x}{10+2x}\] \[{{R}_{BC}}=\frac{20x}{10+2x}+20-x+20-x\] ?(i) or \[\frac{20x}{10+2x}+40=2x\] Solving we get \[x=10\Omega \] Putting the value of \[x=10\Omega \] in equation (i) We get \[{{R}_{BC}}=\frac{20\times 10}{10+2\times 10}+20-10+20-10\] \[=\frac{80}{3}=26.7\Omega \]You need to login to perform this action.
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