A) \[7\,\Omega \]
B) \[\frac{54}{15}\Omega \]
C) \[\frac{10}{3}\Omega \]
D) \[17\,\Omega \]
Correct Answer: C
Solution :
Key Idea: The given Wheatstone bridge is a balanced one. In the given circuit, the ratio of resistances in the arms of the bridge is \[\frac{P}{Q}=\frac{2}{3}\] ?(1) \[\frac{R}{S}=\frac{4}{6}=\frac{2}{3}\] ?(2) Since, \[\frac{P}{Q}=\frac{R}{S}=\frac{2}{3},\]hence the bridge is a balanced one. Hence,\[7\,\Omega \] resistance is avoided. The given circuit can be as follows: The \[2\,\Omega \] and \[3\,\Omega \] resistance are connected in Series, hence circuit reduces to. The \[5\,\Omega \] and \[10\,\Omega \] Resistances are in parallel \[\frac{1}{R}=\frac{1}{5}+\frac{1}{10}=\frac{3}{10}\] \[\Rightarrow \] \[R'=\frac{10}{3}\Omega \]You need to login to perform this action.
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