### Current flowing through the 1 ohm resistor using Thevenin Theorem:

In order to find the current flowing through the $1$ $\text{ohm}$ resistor we will use the Thevenin's theorem, we need to find the Thevenin equivalent circuit of the given circuit [6].

First, we need to find the equivalent resistance of the circuit. The $4$ $\text{ohm}$ resistor is in series with the $1$ $\text{ohm}$ and $9$ $\text{ohm}$ resistors in parallel, so we can simplify this part of the circuit to a single equivalent resistance, $\text{Req}$, using the formula:

$$\frac{1}{\text{Req}}=\frac{1}{1}+\frac{1}{9}=\frac{10}{9}$$ $$\Rightarrow \text{Req}=0.9 \ \text{ohm.}$$

Now, the equivalent resistance of the entire circuit is given by:

$$\text{Req}=4+0.9=4.9 \text{ ohm}.$$

Next, we need to find the Thevenin voltage. To do this, we can remove the $1$ $\text{ohm}$ resistor and calculate the voltage across its terminals. This voltage is the Thevenin voltage.

The voltage, $Volt$, across the $9$ $\text{ohm}$ resistor is:

$$Volt = 10 * \left(\frac{9}{1+9}\right) = 9 V.$$

The voltage across the $1$ $\text{ohm}$ resistor is:

$$Volt = 10 - 9 = 1 V.$$

Therefore, the Thevenin voltage is $1 V$.

Now, we can draw the Thevenin equivalent circuit, which consists of a voltage source of $1 V$ in series with a resistor of $4.9$ $\text{ohm}$.

To find the current flowing through the 1 ohm resistor, we can use Ohm's law:

$$I = \frac{V}{R} = \frac{1}{4.9 + 1} = 0.1818 A.$$

Therefore, the current flowing through the $1$ $\text{ohm}$ resistor is $0.1818 A$.

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