Question

The equivalent resistance of the arrangement of resistances shown in the given figure between points $P$ and $Q$ is

• A. $6\Omega$
• B. $8\Omega$
• C. $24\Omega$
• D. $16\Omega$

Answer 1

Answer: (B)

In the given circuit, the three resistors of $3\Omega ,16\Omega$ and $16\Omega$ are connected in parallel, hence equivalent resistance is
$\Rightarrow R^{\prime }=4\Omega$
This $4\Omega$ resistor is connected in series with $20\Omega$ resistor.
$R^{\prime \prime }=20\Omega +4\Omega =24\Omega$
Similarly in the lower arm, the $9\Omega$ and $18\Omega$ resistance's are connected in parallel, hence equivalent resistance is
$\frac{1}{R^{\prime \prime \prime }}=\frac{1}{9}+\frac{1}{18}=\frac{18+9}{18\times 9}$
$\Rightarrow R^{\prime \prime \prime }=6\Omega$
The two $6\Omega$ resistor's are connected in series, hence equivalent resistance is
$R^{\prime \prime \prime }=6\Omega +6\Omega =12\Omega$
Resultant equivalent resistance between $P$ and $Q$ is
$\frac{1}{R}=\frac{1}{24}+\frac{1}{12}$
$\Rightarrow R=8\Omega$