Here's the solution to the problem: i) the reading on the ammeter Step 1: Determine the voltage across the $5 \, \Omega$ resistor. In a series circuit, the total voltage is the sum of the voltages across individual components. The total voltage ($V_1$) is $30 \, \text{V}$. The voltage across $R_x$ ($V_2$) is $10 \, \text{V}$. Let $V_{5\Omega}$ be the voltage across the $5 \, \Omega$ resistor. $$V_1 = V_{5\Omega} + V_2$$ $$30 \, \text{V} = V_{5\Omega} + 10 \, \text{V}$$ $$V_{5\Omega} = 30 \, \text{V} - 10 \, \text{V}$$ $$V_{5\Omega} = 20 \, \text{V}$$ Step 2: Calculate the current using Ohm's Law. The ammeter measures the total current ($I$) flowing through the series circuit. Using Ohm's Law ($V=IR$) for the $5 \, \Omega$ resistor: $$I = \frac{V_{5\Omega}}{R}$$ $$I = \frac{20 \, \text{V}}{5 \, \Omega}$$ $$I = 4 \, \text{A}$$ The reading on the ammeter is $\boxed{\text{4 A}}$. ii) the value of the resistor $R_x$ Step 1: Calculate $R_x$ using Ohm's Law. We know the voltage across $R_x$ ($V_2 = 10 \, \text{V}$) and the current flowing through it ($I = 4 \, \text{A}$) from part (i). $$R_x = \frac{V_2}{I}$$ $$R_x = \frac{10 \, \text{V}}{4 \, \text{A}}$$ $$R_x = 2.5 \, \Omega$$ The value of the resistor $R_x$ is $\boxed{\text{2.5 \Omega}}$.