A) \[10\,\Omega \]
B) \[20\,\Omega \]
C) \[30\,\Omega \]
D) \[40\,\Omega \]
Correct Answer: A
Solution :
In the given circuit, the ratio of resistances in the opposite arms is same \[\frac{P}{Q}=\frac{10}{10}=\frac{1}{1}\] \[\frac{R}{S}=\frac{10}{10}=\frac{1}{1}\] Hence bridge is balanced. The given circuit now reduces to Here, \[10\Omega \] and \[10\Omega \] resistors are in series in both arms, therefore, reduced circuit is shown Now, the two \[20\Omega \] resistors are connected in parallel, hence equivalent resistance is \[\frac{1}{R}=\frac{1}{20}+\frac{1}{20}=\frac{4}{20\times 20}=\frac{1}{10}\]You need to login to perform this action.
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