A) \[10\,\Omega \]
B) \[60\,\Omega \]
C) \[5\,\Omega \]
D) \[40\,\Omega \]
Correct Answer: A
Solution :
Since given circuit is a balanced Wheatstone bridge, In the given circuit, the ratio of resistances in the opposite arms is same. \[\frac{P}{Q}=\frac{10}{10}=\frac{1}{1}\] \[\frac{R}{S}=\frac{10}{10}=\frac{1}{1}\] Thus, bridge is balanced. The given circuit now can be drawn as Here, \[10\Omega \]and \[10\Omega \] resistors are connected in series, Again, the two 200 resistors are combined in parallel, therefore equivalent resistance is given by: \[\frac{1}{R}=\frac{1}{20}+\frac{1}{20}=\frac{40}{20\times 20}=\frac{1}{10}\] \[\Rightarrow \] \[R=10\Omega \]You need to login to perform this action.
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