1. **Problem 2.13: Use KCL to find the branch currents $I_1$ to $I_4$**. Given currents in the circuit: - Top branch current $I_2 = 2$ A (left to right) - Resistor branch with 7 A current - Vertical branch left: $I_1$ (unknown) - Vertical branch right: $I_3$ and $I_4$ with currents 3 A and 4 A downward 2. **Write KCL equations at essential nodes:** Assuming a node where currents meet, KCL states the sum of currents entering equals sum leaving. Node A (top-left): $$I_1 + 2 = 3 + I_3$$ Node B (top-right): $$I_4 + I_3 = 4 + 7$$ 3. **Express unknowns using the equations:** From Node B: $$I_4 + I_3 = 11$$ From Node A: $$I_1 + 2 = 3 + I_3 \,\Rightarrow \, I_1 = 1 + I_3$$ 4. **Due to currents and labels, generally: rewrite considering realistic constraints or more circuit data. But based on provided info, assume $I_3$ is the unknown; solve for $I_3$ by separate constraints if given. Lacking additional data, we provide expressions: $$I_1 = 1 + I_3$$ $$I_4 = 11 - I_3$$ 5. **Summary for 2.13:** With given partial data and labels, the currents relate by $$I_1 = 1 + I_3$$ $$I_4 = 11 - I_3$$ Rest known or defined by problem context. --- 6. **Problem 2.14: Use KVL to find branch voltages $V_1$ to $V_4$**. Given voltages at corners: - Top-left corner +3 V source - Bottom-left corner +4 V source - Top-right corner +2 V source in the middle horizontal - Bottom-right corner +5 V source 7. **Apply KVL in the rectangular loop:** Clockwise around the loop: $$ +3 + V_1 + 2 + V_2 - 5 - V_4 - 4 - V_3 = 0 $$ Rearranged: $$ V_1 + V_2 - V_3 - V_4 = 5 - 3 - 2 + 4 = 4 $$ 8. **Use another loop to write individual voltages if possible:** For example, left loop vertical voltage drop: $$ 3 + V_1 - 4 - V_3 = 0 \Rightarrow V_1 - V_3 = 1 $$ For right loop vertical voltage drop: $$ 2 + V_2 - 5 - V_4 = 0 \Rightarrow V_2 - V_4 = 3 $$ 9. **Use these two equations and combined equation:** - $V_1 - V_3 = 1$ - $V_2 - V_4 = 3$ - $V_1 + V_2 - V_3 - V_4 = 4$ From the third equation and the first two: Replace $V_3 = V_1 - 1$ and $V_4 = V_2 - 3$ into the third: $$ V_1 + V_2 - (V_1 - 1) - (V_2 - 3) = 4 $$ $$ V_1 + V_2 - V_1 + 1 - V_2 + 3 = 4 $$ $$ 4 = 4 $$ Which is consistent but dependent, showing infinite solutions related by above linear relations. 10. **Express voltages compactly:** Choose $V_1 = a$, $V_2 = b$ arbitrary, $$ V_3 = a - 1 $$ $$ V_4 = b - 3 $$ where $a,b$ are voltages based on other circuit details or measurement. --- **Final answers:** **For 2.13:** $$ I_1 = 1 + I_3,\quad I_4 = 11 - I_3 $$ where $I_3$ depends on more detail. **For 2.14:** $$ V_1 = a, \quad V_2 = b, \quad V_3 = a - 1, \quad V_4 = b - 3 $$ for arbitrary $a,b$ due to linear dependency in equations.