1. Stating the problem: Find voltages $V_1$, $V_3$, $V_{ab}$, and the source current $I_s$ in the given circuit with resistors $R_1=5\ \Omega$, $R_2=3\ \Omega$, $R_3=6\ \Omega$, $R_4=2\ \Omega$, and voltage sources $E_1=6\ \text{V}$, $E_2=18\ \text{V}$. 2. Analyze the branches: The top branch has $R_3$ and $R_4$ in series; the bottom branch has $R_1$ and $R_2$ in series. Both branches are connected between the voltage sources $E_1$ and $E_2$. 3. Calculate total resistance of each branch: Top branch resistance: $$R_{top} = R_3 + R_4 = 6 + 2 = 8\ \Omega$$ Bottom branch resistance: $$R_{bottom} = R_1 + R_2 = 5 + 3 = 8\ \Omega$$ 4. Total current from $E_2$ will split into two branches with equal resistance. Calculate voltage difference across the entire parallel resistor combination: $$V_{ab} = E_2 - E_1 = 18 - 6 = 12\ \text{V}$$ 5. Calculate current through each branch (since resistors are equal, current divides equally): $$I_{top} = \frac{V_{ab}}{R_{top}} = \frac{12}{8} = 1.5\ \text{A}$$ $$I_{bottom} = \frac{V_{ab}}{R_{bottom}} = \frac{12}{8} = 1.5\ \text{A}$$ 6. Calculate voltages across individual resistors: For $V_3$ across $R_3$ (top branch): $$V_3 = I_{top} \times R_3 = 1.5 \times 6 = 9\ \text{V}$$ For $V_1$ across $R_1$ (bottom branch): $$V_1 = I_{bottom} \times R_1 = 1.5 \times 5 = 7.5\ \text{V}$$ 7. Confirm $V_{ab}$ by adding voltages across resistors in either branch: Top branch total voltage: $V_3 + V_4 = 9 + (1.5 \times 2) = 9 + 3 = 12\ \text{V}$ Bottom branch total voltage: $V_1 + V_2 = 7.5 + (1.5 \times 3) = 7.5 + 4.5 = 12\ \text{V}$ Matches $V_{ab}$. 8. Calculate source current $I_s$, which is total current coming from $E_2$: Since branches are in parallel and currents sum: $$I_s = I_{top} + I_{bottom} = 1.5 + 1.5 = 3\ \text{A}$$ Final answers: \[V_1 = 7.5\ \text{V}, \quad V_3 = 9\ \text{V}, \quad V_{ab} = 12\ \text{V}, \quad I_s = 3\ \text{A}\]