1. **State the problem:** We have a circuit with nodes a, b, and c (ground). Resistors are between nodes: $R_{ab}=2\ \Omega$, $R_{ac}=5\ \Omega$, $R_{bc}=10\ \Omega$. Current sources: 7 A entering node a, 5 A leaving node b. We use nodal analysis to find branch currents. 2. **Assign node voltages:** Let $V_a$ and $V_b$ be voltages at nodes a and b relative to ground node c ($V_c=0$). 3. **Write nodal equations using Kirchhoff's Current Law (KCL):** At node a: $$7 = \frac{V_a - V_b}{2} + \frac{V_a - 0}{5}$$ At node b: $$-5 = \frac{V_b - V_a}{2} + \frac{V_b - 0}{10}$$ 4. **Simplify equations:** Node a: $$7 = \frac{V_a - V_b}{2} + \frac{V_a}{5} = \frac{5(V_a - V_b) + 2V_a}{10} = \frac{7V_a - 5V_b}{10}$$ Multiply both sides by 10: $$70 = 7V_a - 5V_b$$ Node b: $$-5 = \frac{V_b - V_a}{2} + \frac{V_b}{10} = \frac{5(V_b - V_a) + V_b}{10} = \frac{6V_b - 5V_a}{10}$$ Multiply both sides by 10: $$-50 = 6V_b - 5V_a$$ 5. **Rewrite system:** $$7V_a - 5V_b = 70$$ $$-5V_a + 6V_b = -50$$ 6. **Solve system:** Multiply second equation by 7: $$-35V_a + 42V_b = -350$$ Multiply first equation by 5: $$35V_a - 25V_b = 350$$ Add: $$(35V_a - 25V_b) + (-35V_a + 42V_b) = 350 - 350$$ $$17V_b = 0 \implies V_b = 0$$ Substitute $V_b=0$ into first equation: $$7V_a = 70 \implies V_a = 10$$ 7. **Calculate branch currents:** - Current from a to b: $$I_{ab} = \frac{V_a - V_b}{R_{ab}} = \frac{10 - 0}{2} = 5\ \text{A}$$ - Current from a to c: $$I_{ac} = \frac{V_a - 0}{R_{ac}} = \frac{10}{5} = 2\ \text{A}$$ - Current from b to c: $$I_{bc} = \frac{V_b - 0}{R_{bc}} = \frac{0}{10} = 0\ \text{A}$$ 8. **Check currents at nodes:** At node a, incoming current 7 A equals outgoing currents $I_{ab} + I_{ac} = 5 + 2 = 7$ A, consistent. At node b, outgoing current 5 A equals incoming $I_{ab} + I_{bc} = 5 + 0 = 5$ A, consistent. **Final answer:** $$I_{ab} = 5\ \text{A},\quad I_{ac} = 2\ \text{A},\quad I_{bc} = 0\ \text{A}$$