1. **State the problem**: Find voltages $V_1$, $V_3$, $V_{ab}$ and source current $I_s$ for the circuit with resistors $R_1=5\ \Omega$, $R_2=3\ \Omega$, $R_3=6\ \Omega$, $R_4=2\ \Omega$ and voltage sources $E_1=6$ V, $E_2=18$ V arranged as described. 2. **Analyze the circuit setup**: There are two loops sharing nodes $a$ and $b$. The bottom loop has $R_1$ and $R_2$ in series with voltage source $E_1$. The top loop has $R_3$ and $R_4$ in series with $E_2$. The voltage $V_{ab}$ is the voltage difference from node $a$ to $b$. 3. **Calculate equivalent resistances**: - Bottom branch: $R_{bottom}=R_1 + R_2 = 5 + 3 = 8\ \Omega$ - Top branch: $R_{top}=R_3 + R_4 = 6 + 2 = 8\ \Omega$ 4. **Calculate current in each branch**: For bottom branch, voltage source $E_1=6$ V drives current: $$I_{bottom} = \frac{E_1}{R_{bottom}} = \frac{6}{8} = 0.75\ \mathrm{A}$$ For top branch, voltage source $E_2=18$ V drives current: $$I_{top} = \frac{E_2}{R_{top}} = \frac{18}{8} = 2.25\ \mathrm{A}$$ 5. **Find voltages $V_1$ and $V_3$ across resistors $R_1$ and $R_3$ respectively** using Ohm's Law $V = IR$: $$V_1 = I_{bottom} \times R_1 = 0.75 \times 5 = 3.75\ \mathrm{V}$$ $$V_3 = I_{top} \times R_3 = 2.25 \times 6 = 13.5\ \mathrm{V}$$ 6. **Calculate voltage $V_{ab}$ (from node $a$ to $b$):** Node $a$ connects the junctions next to $R_1$ and $R_3$, node $b$ connects junctions next to $R_2$ and $R_4$. Bottom branch voltage drop from $a$ to $b$ is across $R_2$: $$V_{R_2} = I_{bottom} \times R_2 = 0.75 \times 3 = 2.25\ \mathrm{V}$$ Top branch voltage drop from $a$ to $b$ is across $R_4$: $$V_{R_4} = I_{top} \times R_4 = 2.25 \times 2 = 4.5\ \mathrm{V}$$ Since $V_{ab}$ is the node voltage difference, it equals the difference between the voltages drops across $R_4$ and $R_2$ (top minus bottom): $$V_{ab} = V_{R_4} - V_{R_2} = 4.5 - 2.25 = 2.25\ \mathrm{V}$$ 7. **Find the source current $I_s$ from $E_2$:** Current $I_s$ is the current flowing out of $E_2$, which is the top branch current: $$I_s = I_{top} = 2.25\ \mathrm{A}$$ **Final answers:** $$V_1 = 3.75\ \mathrm{V},\quad V_3 = 13.5\ \mathrm{V},\quad V_{ab} = 2.25\ \mathrm{V},\quad I_s = 2.25\ \mathrm{A}$$