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)
$=\frac{15\times 15}{15+15}$
$=\frac{15\times 15}{30}$
$=7.5\Omega$