Every resistance shown in the figure has a value of 20 . Determine...

The given circuit can be analyzed as a Wheatstone bridge.

Let the junction between the top two resistors be $C$ and the junction between the bottom two resistors be $D$. The diode is connected between $C$ and $D$.

The resistance of each resistor is $20 \Omega$.
The ratio of resistances in the left arms is $\dfrac{20}{20} = 1$, and the ratio in the right arms is also $\dfrac{20}{20} = 1$.

Since the ratios are equal, the Wheatstone bridge is balanced. Therefore, the potential at point $C$ is equal to the potential at point $D$ ($V_C = V_D$).

Because there is no potential difference across the diode ($V_C - V_D = 0$), no current flows through the middle branch. This condition remains true regardless of whether point $A$ or point $B$ is at a higher potential.

Since the diode branch carries no current, it can be removed from the circuit for the purpose of calculating the equivalent resistance. The circuit then simplifies to two parallel branches:
Resistance of the upper branch $= 20 \Omega + 20 \Omega = 40 \Omega$
Resistance of the lower branch $= 20 \Omega + 20 \Omega = 40 \Omega$

The equivalent resistance $R_{eq}$ between $A$ and $B$ is the parallel combination of these two branches:
$R_{eq} = \dfrac{40 \times 40}{40 + 40} = 20 \Omega$

This value is independent of the polarity of the applied voltage.