1. **State the problem:** We have a circuit with nodes a, b, and c (ground). Given conductances and current sources, we need to find node voltages $V_1$ (at node a) and $V_2$ (at node b) using nodal analysis, and then find the branch currents. 2. **Identify given data:** - Conductance between a and c: $G_{ac} = 0.5S$ - Conductance between b and c: $G_{bc} = 1S$ - Conductance between a and b: $G_{ab} = 0.25S$ - Current source at node a: $I_1 = 3A$ (injected into node a) - Current source at node b: $I_2 = 2A$ (injected into node b) - Node c is ground: $V_c = 0V$ 3. **Set node voltages:** Let $V_1$ be voltage at node a, $V_2$ at node b, and $V_c=0$. 4. **Write nodal equations using KCL:** At node a: $$I_1 = G_{ac}(V_1 - V_c) + G_{ab}(V_1 - V_2)$$ Substitute values: $$3 = 0.5(V_1 - 0) + 0.25(V_1 - V_2) = 0.5V_1 + 0.25V_1 - 0.25V_2 = 0.75V_1 - 0.25V_2$$ At node b: $$I_2 = G_{bc}(V_2 - V_c) + G_{ab}(V_2 - V_1)$$ Substitute values: $$2 = 1(V_2 - 0) + 0.25(V_2 - V_1) = 1V_2 + 0.25V_2 - 0.25V_1 = 1.25V_2 - 0.25V_1$$ 5. **Rewrite system of equations:** $$3 = 0.75V_1 - 0.25V_2$$ $$2 = -0.25V_1 + 1.25V_2$$ 6. **Solve the system:** Multiply second equation by 3 to align coefficients: $$6 = -0.75V_1 + 3.75V_2$$ Add to first equation: $$3 + 6 = (0.75V_1 - 0.25V_2) + (-0.75V_1 + 3.75V_2)$$ $$9 = 3.5V_2$$ $$V_2 = \frac{9}{3.5} = 2.5714V$$ Substitute $V_2$ back into first equation: $$3 = 0.75V_1 - 0.25(2.5714)$$ $$3 = 0.75V_1 - 0.6429$$ $$0.75V_1 = 3 + 0.6429 = 3.6429$$ $$V_1 = \frac{3.6429}{0.75} = 4.8572V$$ 7. **Calculate branch currents:** - Current through $G_{ac}$: $$I_{ac} = G_{ac}(V_1 - V_c) = 0.5 \times 4.8572 = 2.4286A$$ - Current through $G_{bc}$: $$I_{bc} = G_{bc}(V_2 - V_c) = 1 \times 2.5714 = 2.5714A$$ - Current through $G_{ab}$: $$I_{ab} = G_{ab}(V_1 - V_2) = 0.25 \times (4.8572 - 2.5714) = 0.25 \times 2.2858 = 0.5714A$$ 8. **Verify currents at nodes:** At node a, sum currents leaving: $$I_{ac} + I_{ab} = 2.4286 + 0.5714 = 3A$$ matches $I_1$. At node b, sum currents leaving: $$I_{bc} - I_{ab} = 2.5714 - 0.5714 = 2A$$ matches $I_2$. **Final answers:** $$V_1 = 4.8572V, \quad V_2 = 2.5714V$$ Branch currents: $$I_{ac} = 2.4286A, \quad I_{bc} = 2.5714A, \quad I_{ab} = 0.5714A$$