Calculate the equivalent resistance in the combination shown in the figure below.

(N/A) The resistors $6 \, \Omega$ and $3 \, \Omega$ are connected in parallel.
Therefore,the equivalent resistance $R_P$ is given by:
$\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2}$
$\frac{1}{R_P} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
Thus,$R_P = 2 \, \Omega$.
Now,the $2 \, \Omega$ resistor and the parallel combination $R_P$ are in series.
Therefore,the total equivalent resistance $R$ is:
$R = 2 \, \Omega + R_P = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega$.
The equivalent resistance of the circuit is $4 \, \Omega$.