Explanation:
A 
$\mathrm{R}_{1}, \mathrm{R}_{2}, \mathrm{R}_{3}$ and $\mathrm{R}_{4}$ resistance are arranged according to Wheatstone bridge.
Hence, There is no current in $R_{5}$.
So, $R_{1}$ and $R_{3}$ in series-
$\mathrm{R}^{\prime} =2 \Omega+2 \Omega$
$=4 \Omega$
$\mathrm{R}_{2}$ and $\mathrm{R}_{4}$ in series-
$\mathrm{R}^{\prime \prime}=4 \Omega$
$\mathrm{R}^{\prime}$ and $\mathrm{R}^{\prime \prime}$ in parallel
$\frac{1}{\mathrm{R}^{\mathrm{\prime \prime}}} =\frac{1}{4}+\frac{1}{4}$
$\frac{1}{\mathrm{R}^{\mathrm{\prime}}} =\frac{1}{2}$
$\mathrm{R}^{\mathrm{\prime}} =2 \Omega$
$\mathrm{R}^{\prime \prime}$ and $\mathrm{R}_{5}$ is in parallel,
So $\quad \frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}^{\prime \prime \prime}}+\frac{1}{\mathrm{R}_{5}}$
$= \frac{1}{2}+\frac{1}{2}$
$\frac{1}{\mathrm{R}_{\text {eq }}} =1$
$\mathrm{R}_{\text {eq }} =1 \Omega$