Pls i want clear steps in English

We define three clockwise mesh currents:

• $I_1$: Current in the top-left mesh.
• $I_2$: Current in the bottom-left mesh.
• $I_3$: Current in the right mesh.

For Mesh 1 (top-left loop):
Starting from the bottom-left corner of Mesh 1 and moving clockwise:

$$-6 \text{ V} + (I_1)(2 \Omega) + (I_1 - I_2)(3 \Omega) = 0$$ $$2I_1 + 3I_1 - 3I_2 = 6$$ $$5I_1 - 3I_2 = 6 \quad (\text{Equation 1})$$

For Mesh 2 (bottom-left loop):
Starting from the bottom-left corner of Mesh 2 and moving clockwise:

$$-6 \text{ V} + (I_2 - I_1)(3 \Omega) + (I_2 - I_3)(4 \Omega) = 0$$ $$3I_2 - 3I_1 + 4I_2 - 4I_3 = 6$$ $$-3I_1 + 7I_2 - 4I_3 = 6 \quad (\text{Equation 2})$$

For Mesh 3 (right loop):
Starting from the top-right corner of Mesh 3 and moving clockwise:

$$(I_3 - I_1)(4 \Omega) + (I_3 - I_2)(4 \Omega) = 0$$ $$4I_3 - 4I_1 + 4I_3 - 4I_2 = 0$$ $$-4I_1 - 4I_2 + 8I_3 = 0$$

Dividing by 4:

$$-I_1 - I_2 + 2I_3 = 0 \quad (\text{Equation 3})$$

From Equation 3, we can express $2I_3$ in terms of $I_1$ and $I_2$:

$$2I_3 = I_1 + I_2$$

So, $4I_3 = 2(I_1 + I_2)$.

Substitute this into Equation 2:

$$-3I_1 + 7I_2 - 2(I_1 + I_2) = 6$$ $$-3I_1 + 7I_2 - 2I_1 - 2I_2 = 6$$ $$-5I_1 + 5I_2 = 6 \quad (\text{Equation 4})$$

Now we have a system of two equations with two variables ($I_1$ and $I_2$):

1. $5I_1 - 3I_2 = 6$
2. $-5I_1 + 5I_2 = 6$

Add Equation 1 and Equation 4:

$$(5I_1 - 3I_2) + (-5I_1 + 5I_2) = 6 + 6$$ $$2I_2 = 12$$ $$I_2 = \frac{12}{2} = 6 \text{ A}$$

Substitute the value of $I_2$ into Equation 1:

$$5I_1 - 3(6 \text{ A}) = 6$$ $$5I_1 - 18 = 6$$ $$5I_1 = 6 + 18$$ $$5I_1 = 24$$ $$I_1 = \frac{24}{5} = 4.8 \text{ A}$$

The $3 \Omega$ resistor is shared by Mesh 1 and Mesh 2. The current through it is the difference between $I_1$ and $I_2$. If we consider the current flowing downwards (in the direction of $I_1$), it is $I_1 - I_2$.

$$I_{3\Omega} = I_1 - I_2$$ $$I_{3\Omega} = 4.8 \text{ A} - 6 \text{ A}$$ $$I_{3\Omega} = -1.2 \text{ A}$$

The negative sign indicates that the actual current flows upwards through the $3 \Omega$ resistor, opposite to the assumed direction of $I_1$ through that branch. The "size" of the current refers to its magnitude.

The size of the current through the $3 \Omega$ resistor is $\boxed{\text{1.2 A}}$.