1. **State the problem:** We need to find the power dissipated in the 9-Ω resistor using the nodal voltage method for the given circuit. 2. **Identify nodes and assign voltages:** Let the bottom line be the reference node (ground, 0 V). Assign node voltages: - Node 1 at the junction after the 3-Ω resistor and before the parallel branches, voltage $V_1$. - Node 2 at the junction between the 4-Ω and 9-Ω resistors, voltage $V_2$. 3. **Write KCL equations at nodes:** At Node 1: Current from 27 V source through 3 Ω resistor: $\frac{27 - V_1}{3}$ Current from Node 1 to Node 2 through 4 Ω resistor: $\frac{V_1 - V_2}{4}$ Current from Node 1 to ground through 5 Ω resistor: $\frac{V_1 - 0}{5} = \frac{V_1}{5}$ Current from Node 1 to ground through 6 A current source downward (current source injects 6 A downward, so 6 A leaves Node 1): $6$ KCL at Node 1 (sum currents leaving node = 0): $$\frac{27 - V_1}{3} - \frac{V_1 - V_2}{4} - \frac{V_1}{5} - 6 = 0$$ At Node 2: Current from Node 2 to Node 1 through 4 Ω resistor: $\frac{V_2 - V_1}{4}$ Current from Node 2 to ground through 9 Ω resistor: $\frac{V_2 - 0}{9} = \frac{V_2}{9}$ KCL at Node 2: $$\frac{V_2 - V_1}{4} + \frac{V_2}{9} = 0$$ 4. **Simplify equations:** Node 1: $$\frac{27 - V_1}{3} - \frac{V_1 - V_2}{4} - \frac{V_1}{5} - 6 = 0$$ Multiply through by 60 (LCM of 3,4,5): $$20(27 - V_1) - 15(V_1 - V_2) - 12 V_1 - 360 = 0$$ $$540 - 20 V_1 - 15 V_1 + 15 V_2 - 12 V_1 - 360 = 0$$ $$540 - 47 V_1 + 15 V_2 - 360 = 0$$ $$180 - 47 V_1 + 15 V_2 = 0$$ Node 2: $$\frac{V_2 - V_1}{4} + \frac{V_2}{9} = 0$$ Multiply through by 36 (LCM of 4 and 9): $$9(V_2 - V_1) + 4 V_2 = 0$$ $$9 V_2 - 9 V_1 + 4 V_2 = 0$$ $$13 V_2 - 9 V_1 = 0$$ 5. **Solve the system:** From Node 2 equation: $$13 V_2 = 9 V_1 \implies V_2 = \frac{9}{13} V_1$$ Substitute into Node 1 equation: $$180 - 47 V_1 + 15 \times \frac{9}{13} V_1 = 0$$ $$180 - 47 V_1 + \frac{135}{13} V_1 = 0$$ Multiply through by 13: $$2340 - 611 V_1 + 135 V_1 = 0$$ $$2340 - 476 V_1 = 0$$ $$476 V_1 = 2340 \implies V_1 = \frac{2340}{476} = 4.916$$ Then, $$V_2 = \frac{9}{13} \times 4.916 = 3.404$$ 6. **Calculate power dissipated in 9-Ω resistor:** Voltage across 9 Ω resistor is $V_2 - 0 = V_2 = 3.404$ V. Power dissipated: $$P = \frac{V^2}{R} = \frac{3.404^2}{9} = \frac{11.58}{9} = 1.287 \text{ watts}$$ **Final answer:** The power dissipated in the 9-Ω resistor is approximately **1.29 watts**.