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. We can show the network as below From the circuit,\[\frac{10}{5}=\frac{10}{5},\]i.e.,\[2=2\] So, it is balanced Wheatstone's bridge. Therefore, resistance of its middle arm, will remain unaffected. The net resistance in upper arms \[{{R}_{U}}=10+5=15\,\Omega \]           (series) The net resistance in lower arms \[{{R}_{L}}=10+5=15\,\Omega \]            (series) Hence, equivalent resistance of the network \[R=\frac{{{R}_{U}}\times {{R}_{L}}}{{{R}_{U}}+{{R}_{L}}}\]                     (parallel)                 \[=\frac{15\times 15}{15+15}\]                 \[=\frac{15\times 15}{30}\]         \[=7.5\,\Omega \]