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)

$7.5 \, \Omega$

Answer 2

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
$R=\frac{15\times 15}{15+15}$
$=\frac{225}{30}=7.5\,\Omega$