To calculate the voltage across the $2 \, \Omega$ resistor in the circuit with a supply voltage of $10 \, V$, let's go step by step:

1. Identify resistances and voltage

The circuit consists of three resistors:

• $R_1 = 2 \, \Omega$
• $R_2 = 1 \, \Omega$
• $R_3 = 2 \, \Omega$

The resistors $R_1$ and $R_2$ are in series, and the combination is in parallel with $R_3$. The total supply voltage is $10 \, V$.

2. Calculate the equivalent resistance of the series combination

The series combination of $R_1$ and $R_2$ gives: $R_{\text{series}} = R_1 + R_2 = 2 \, \Omega + 1 \, \Omega = 3 \, \Omega$

3. Find the total resistance of the circuit

The equivalent resistance of the circuit can be found by combining $R_{\text{series}}$ with $R_3$ (which are in parallel). The formula for parallel resistances is: $\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_3}$ Substituting values: $\frac{1}{R_{\text{total}}} = \frac{1}{3 \, \Omega} + \frac{1}{2 \, \Omega} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}$ Thus, the total resistance is: $R_{\text{total}} = \frac{6}{5} = 1.2 \, \Omega$

4. Calculate the total current in the circuit

Using Ohm's law, the total current $I$ is: $I = \frac{V}{R_{\text{total}}} = \frac{10 \, V}{1.2 \, \Omega} = 8.33 \, A$

5. Find the current through $R_3$ (the $2 \, \Omega$ resistor)

Since $R_{\text{series}}$ and $R_3$ are in parallel, the voltage across $R_3$ is the same as the voltage across $R_{\text{series}}$. Let’s first calculate the voltage across $R_{\text{series}}$, which will be the same as across $R_3$.

Using the current divider rule: [ V_{R_3} = I \times R_{\text{parallel}} = 8.33 , A \times \left( \frac{R_{\text{parallel}}}{R_{\text{parallel}} + R_3} \right) =8. thus