What is the equivalent resistance between the points $A$ and $B$ of the network in $\Omega$?

(B) $1$. Analyze the circuit diagram. Notice that the resistors $1.8 \, \Omega$ and $2.2 \, \Omega$ are connected in parallel to a short circuit,effectively bypassing them.
$2$. The resistors $3 \, \Omega$ and $1 \, \Omega$ are in series,giving an equivalent resistance of $3 \, \Omega + 1 \, \Omega = 4 \, \Omega$.
$3$. This $4 \, \Omega$ equivalent resistance is in parallel with the $2 \, \Omega$ resistor. The equivalent resistance is $\frac{4 \times 2}{4 + 2} = \frac{8}{6} = \frac{4}{3} \, \Omega$.
$4$. However,looking at the simplified diagram provided,the circuit reduces to a series combination of $2 \, \Omega$,$1 \, \Omega$,and $5 \, \Omega$ resistors.
$5$. Therefore,the total equivalent resistance $R_{AB} = 2 \, \Omega + 1 \, \Omega + 5 \, \Omega = 8 \, \Omega$.