question_answer

The equivalent resistance between A and B is

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                 \[\frac{1}{R}=\frac{1}{20}+\frac{1}{20}=\frac{4}{20\times 20}=\frac{1}{10}\]