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 :

 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 behaves as open circuited (inactive). So 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)  \[=\frac{15\times 15}{15+15}=7.5\,\,\Omega \]