Here are the calculations for the given circuit: Problem 1: Resistors of $2 \Omega$, $4 \Omega$, and $6 \Omega$ are connected in series, and a voltage of $24 \text{ V}$ is applied across them. i) Calculate the total resistance. Step 1: For resistors connected in series, the total resistance $R_{total}$ is the sum of the individual resistances. $$R_{total} = R_1 + R_2 + R_3$$ $$R_{total} = 2 \Omega + 4 \Omega + 6 \Omega$$ $$R_{total} = 12 \Omega$$ $$\boxed{\text{Total resistance } R_{total} = 12 \Omega}$$ ii) Calculate the current flowing in the circuit. Step 1: Use Ohm's Law, $V = IR$, where $V$ is the total voltage and $R$ is the total resistance. Rearrange for current $I$: $$I = \frac{V_{total}}{R_{total}}$$ Step 2: Substitute the values. $$I = \frac{24 \text{ V}}{12 \Omega}$$ $$I = 2 \text{ A}$$ $$\boxed{\text{Current flowing in the circuit } I = 2 \text{ A}}$$ iii) Calculate the potential difference across each resistor. Step 1: The current is the same through all resistors in a series circuit ($I = 2 \text{ A}$). Use Ohm's Law ($V = IR$) for each resistor. For the $2 \Omega$ resistor ($R_1$): $$V_1 = I \times R_1$$ $$V_1 = (2 \text{ A}) \times (2 \Omega)$$ $$V_1 = 4 \text{ V}$$ $$\boxed{\text{Potential difference across } 2 \Omega \text{ resistor } V_1 = 4 \text{ V}}$$ For the $4 \Omega$ resistor ($R_2$): $$V_2 = I \times R_2$$ $$V_2 = (2 \text{ A}) \times (4 \Omega)$$ $$V_2 = 8 \text{ V}$$ $$\boxed{\text{Potential difference across } 4 \Omega \text{ resistor } V_2 = 8 \text{ V}}$$ For the $6 \Omega$ resistor ($R_3$): $$V_3 = I \times R_3$$ $$V_3 = (2 \text{ A}) \times (6 \Omega)$$ $$V_3 = 12 \text{ V}$$ $$\boxed{\text{Potential difference across } 6 \Omega \text{ resistor } V_3 = 12 \text{ V}}$$