question_answer

The equivalent resistance between the points P and Q in the network shown in the figure is given by:

A)  2.5\[\Omega \]           

B)  7.5\[\Omega \]

C)  10 \[\Omega \]            

D)  12.5 \[\Omega \]

Correct Answer: B

Solution :

Key Idea: The given circuit is a balanced Wheatstone bridge. The equivalent circuit is shown below : In the given circuit, the ratio of resistances in the opposite arms is same                 \[\frac{P}{Q}=\frac{10}{10}=\frac{1}{1}\]                 \[\frac{R}{S}=\frac{5}{5}=\frac{1}{1}\] Hence, bridge is balanced. The given circuit now reduces to Here, \[10\,\Omega \] and \[5\,\Omega \] resistors are in series, therefore Now, the two \[15\,\,\Omega \] resistors are connected in parallel, hence equivalent resistance is \[R=\frac{15\times 15}{15+15}=\frac{225}{30}=7.5\,\,\Omega \]