What is the net current flowing in the given circuit?

We know that the equivalent resistance of resistances connected in a series combination is given by 
${\mathrm{R}}_{\mathrm{S}}={\mathrm{R}}_1+{\mathrm{R}}_2$
From the circuit diagram, the resistors $3\mathit{ \Omega }$ and $2\mathit{ \Omega }$ are in series arrangement and therefore, their equivalent resistance is,
${\mathrm{R}}_{\mathrm{S}}=3+2$
$\Rightarrow {\mathrm{R}}_{\mathrm{S}}=5\mathrm{ \Omega }$
We also know that the equivalent resistance of resistances connected in a parallel combination is given by
$\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}$
The resistor $5\mathit{ \Omega }$ and R_s are in a parallel arrangement and therefore their equivalent resistance is given as
$\Rightarrow \frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1}{5}+\frac{1}{5}$
$\Rightarrow \frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{2}{5}$
${\Rightarrow \mathrm{R}}_{\mathrm{P}}=2.5\mathrm{ \Omega }$
The equivalent resistance of the circuit is $2.5\mathit{ \Omega }$.
Now, since the potential difference applied in the circuit is 6V therefore, 

we can use Ohm’s law, $\mathrm{V}=\mathrm{RI}$ where (V) is the potential difference across the wire, (R) is the resistance of the wire, and (I) is the current through it to find the net current in the circuit as follows:
$\Rightarrow \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$
$\Rightarrow \mathrm{I}=\frac{6}{2.5} \mathrm{A}$
$\Rightarrow \mathrm{I}=2.4 \mathrm{A}$
Hence, the net current in the circuit is 2.4 A.