The equivalent resistance of the following circuit is nearly

The given circuit is a Wheatstone bridge.

Let us check the balance condition of the bridge:
$\dfrac{R_{AB}}{R_{AD}} = \dfrac{5}{10} = \dfrac{1}{2}$

$\dfrac{R_{BC}}{R_{DC}} = \dfrac{15}{30} = \dfrac{1}{2}$

Since $\dfrac{R_{AB}}{R_{AD}} = \dfrac{R_{BC}}{R_{DC}}$, the Wheatstone bridge is balanced.

Therefore, no current flows through the $4\ \Omega$ resistor in the branch $BD$, and it can be removed from the circuit.

The circuit now consists of two parallel branches:
Upper branch resistance, $R_{upper} = 5\ \Omega + 15\ \Omega = 20\ \Omega$
Lower branch resistance, $R_{lower} = 10\ \Omega + 30\ \Omega = 40\ \Omega$

The equivalent resistance $R_{eq}$ of the circuit is given by the parallel combination of $R_{upper}$ and $R_{lower}$:
$\dfrac{1}{R_{eq}} = \dfrac{1}{R_{upper}} + \dfrac{1}{R_{lower}}$

$\dfrac{1}{R_{eq}} = \dfrac{1}{20} + \dfrac{1}{40} = \dfrac{2 + 1}{40} = \dfrac{3}{40}\ \Omega^{-1}$

$R_{eq} = \dfrac{40}{3}\ \Omega \approx 13.3\ \Omega$

Answer: $13.3\ \Omega$