A) \[\frac{10}{3}\Omega \]
B) \[\frac{20}{3}\Omega \]
C) \[\frac{10}{5}\Omega \]
D) \[5\,\Omega \]
Correct Answer: A
Solution :
Key Idea: Given circuit is a balanced Wheat- stone bridge. Taking the ratio of resistance's in the opposite arms, we have \[\frac{P}{Q}=\frac{2}{3}\] \[\frac{R}{S}=\frac{4}{6}=\frac{2}{3}\] Since, \[\frac{P}{Q}=\frac{R}{S}=\frac{2}{3}\], the circuit is balanced. The given circuit now reduces is as show \[R'=2+3=5\,\Omega \] \[R''=4+6=10\,\Omega \] The resistances R' and R" are now in parallel \[\therefore \] \[\frac{1}{R}=\frac{1}{R'}+\frac{1}{R''}\] \[=\frac{1}{5}+\frac{1}{10}\] \[=\frac{2+1}{10}=\frac{3}{10}\] \[\Rightarrow \] \[R=\frac{10}{3}\Omega \]You need to login to perform this action.
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