1. **State the problem:** We have a circuit with nodes a, b, c, and d, resistors between nodes, and current sources. We need to find currents $I_{ab}$ (between a and b), $I_{bd}$ (between b and d), and $I_{bc}$ (between b and c) using nodal analysis. 2. **Label node voltages:** Let the voltages at nodes a, b, c, and d be $V_a$, $V_b$, $V_c$, and $V_d$. Assume $V_d=0$ as reference (ground). 3. **Given resistances:** - $R_{ab} = 0.5$ ohm - $R_{bc} = 0.5$ ohm - Vertical resistors: $R_{ad} = 0.25$ ohm, $R_{bd} = 0.5$ ohm, $R_{cd} = 0.25$ ohm 4. **Current sources:** - 6A into node a (upward) - 4A into node c (upward) 5. **Write nodal equations using Kirchhoff's Current Law (KCL):** At node a: $$\frac{V_a - V_b}{0.5} + \frac{V_a - V_d}{0.25} = 6$$ At node b: $$\frac{V_b - V_a}{0.5} + \frac{V_b - V_c}{0.5} + \frac{V_b - V_d}{0.5} = 0$$ At node c: $$\frac{V_c - V_b}{0.5} + \frac{V_c - V_d}{0.25} = 4$$ Recall $V_d=0$. 6. **Simplify equations:** Node a: $$2(V_a - V_b) + 4V_a = 6 \implies 6V_a - 2V_b = 6$$ Node b: $$2(V_b - V_a) + 2(V_b - V_c) + 2V_b = 0 \implies -2V_a + 6V_b - 2V_c = 0$$ Node c: $$2(V_c - V_b) + 4V_c = 4 \implies -2V_b + 6V_c = 4$$ 7. **Write system of equations:** $$\begin{cases} 6V_a - 2V_b = 6 \\ -2V_a + 6V_b - 2V_c = 0 \\ -2V_b + 6V_c = 4 \end{cases}$$ 8. **Solve for $V_a$, $V_b$, $V_c$:** From first equation: $$6V_a = 6 + 2V_b \implies V_a = 1 + \frac{1}{3}V_b$$ Substitute $V_a$ into second: $$-2(1 + \frac{1}{3}V_b) + 6V_b - 2V_c = 0 \implies -2 - \frac{2}{3}V_b + 6V_b - 2V_c = 0$$ Simplify: $$-2 + \frac{16}{3}V_b - 2V_c = 0 \implies \frac{16}{3}V_b - 2V_c = 2$$ Multiply by 3: $$16V_b - 6V_c = 6$$ From third equation: $$-2V_b + 6V_c = 4$$ Multiply third by 3: $$-6V_b + 18V_c = 12$$ 9. **Solve the two equations:** $$\begin{cases} 16V_b - 6V_c = 6 \\ -6V_b + 18V_c = 12 \end{cases}$$ Multiply first by 3: $$48V_b - 18V_c = 18$$ Add to second: $$(48V_b - 18V_c) + (-6V_b + 18V_c) = 18 + 12 \implies 42V_b = 30 \implies V_b = \frac{30}{42} = \frac{5}{7} \approx 0.7143$$ Substitute $V_b$ into third equation: $$-2(0.7143) + 6V_c = 4 \implies -1.4286 + 6V_c = 4 \implies 6V_c = 5.4286 \implies V_c = 0.9048$$ Substitute $V_b$ into $V_a$: $$V_a = 1 + \frac{1}{3}(0.7143) = 1 + 0.2381 = 1.2381$$ 10. **Calculate currents:** $$I_{ab} = \frac{V_a - V_b}{0.5} = \frac{1.2381 - 0.7143}{0.5} = \frac{0.5238}{0.5} = 1.0476\,A$$ $$I_{bc} = \frac{V_b - V_c}{0.5} = \frac{0.7143 - 0.9048}{0.5} = \frac{-0.1905}{0.5} = -0.3810\,A$$ (negative means current flows from c to b) $$I_{bd} = \frac{V_b - V_d}{0.5} = \frac{0.7143 - 0}{0.5} = 1.4286\,A$$ **Final answers:** - $I_{ab} = 1.05$ A (from a to b) - $I_{bc} = -0.38$ A (from c to b) - $I_{bd} = 1.43$ A (from b to d)