The current $I$ shown in the figure is:

A) $1.33\,A$
B) $Zero$
C) $2.00\,A$
D) $1.00\,A$

Hint: To find the current flowing through the three branches in the parallel circuit, use the formula of the current given below and substitute the values of the resistance in the formula, The obtained result provides the answer for the current in the circuit.

Useful formula:
(1) The current in the parallel circuit is given by
$I = \dfrac{{{R_1} \times {R_2} \times {R_3}}}{{{R_1} + {R_2} + {R_3}}}$
Where $I$ is the current flowing through the circuit, ${R_1}$ is the resistance developed in the first branch, ${R_2}$ is the resistance developed in the second branch and ${R_3}$ is the resistance developed in the third branch.

Complete step by step solution:
It is given that the
The resistance of the circuit, ${R_1} = 2\,\Omega $
The potential difference of the first branch, ${V_1} = 2\,V$
The resistance of the second branch, ${R_2} = 2\,\Omega $
The potential difference of the first branch, ${V_1} = 2\,V$
The resistance of the second branch, ${R_1} = 2\,\Omega $
The potential difference of the first branch, ${V_1} = 2\,V$
Since it is the parallel circuit, the voltage of the potential developed across each branch is same. The formula of the current is taken.
$I = \dfrac{{{R_1} \times {R_2} \times {R_3}}}{{{R_1} + {R_2} + {R_3}}}$
Substituting the value of the resistance from three branches in the above equation.
$I = \dfrac{{2 \times 2 \times 2}}{{2 + 2 + 2}}$
By performing the simple arithmetic operation,
$I = \dfrac{8}{6}$
Dividing the numerator and the denominator in the above step.
$I = 1.33\,A$
Hence the value of the current obtained from the parallel circuit is $1.33\,A$ .

Thus the option (A) is correct.

Note: In the parallel circuit, the voltage which develops across each branch is equal so why does the formula only contain resistance. The current flowing through it is the sum of the current flowing through the individual component of the circuit.