Question

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 $

Answer 1

Answer: (B)

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 $