1. **State the problem:** We need to find the current through the 15 \(\Omega\) resistor in the given circuit using the nodal method. 2. **Identify nodes and assign voltages:** Let the bottom left node be the reference node (0 V). Assign node voltages \(V_1\) at the top left corner, \(V_2\) at the junction between the 20 \(\Omega\) and 15 \(\Omega\) resistors, and \(V_3\) at the junction between the 15 \(\Omega\) and 10 \(\Omega\) resistors. 3. **Write KCL equations at nodes:** - At node \(V_1\): $$\frac{V_1 - 400}{80} + \frac{V_1 - V_2}{20} + \frac{V_1 - 0}{15} = 0$$ - At node \(V_2\): $$\frac{V_2 - V_1}{20} + \frac{V_2 - V_3}{15} + \frac{V_2 - 0}{90} = 0$$ - At node \(V_3\): $$\frac{V_3 - V_2}{15} + \frac{V_3 - 0}{10} + \frac{V_3 - 200}{200} = 0$$ 4. **Simplify each equation:** - Node \(V_1\): $$\frac{V_1 - 400}{80} + \frac{V_1 - V_2}{20} + \frac{V_1}{15} = 0$$ Multiply through by 240 (LCM of 80, 20, 15): $$3(V_1 - 400) + 12(V_1 - V_2) + 16V_1 = 0$$ $$3V_1 - 1200 + 12V_1 - 12V_2 + 16V_1 = 0$$ $$31V_1 - 12V_2 = 1200$$ - Node \(V_2\): $$\frac{V_2 - V_1}{20} + \frac{V_2 - V_3}{15} + \frac{V_2}{90} = 0$$ Multiply through by 180 (LCM of 20, 15, 90): $$9(V_2 - V_1) + 12(V_2 - V_3) + 2V_2 = 0$$ $$9V_2 - 9V_1 + 12V_2 - 12V_3 + 2V_2 = 0$$ $$23V_2 - 9V_1 - 12V_3 = 0$$ - Node \(V_3\): $$\frac{V_3 - V_2}{15} + \frac{V_3}{10} + \frac{V_3 - 200}{200} = 0$$ Multiply through by 600 (LCM of 15, 10, 200): $$40(V_3 - V_2) + 60V_3 + 3(V_3 - 200) = 0$$ $$40V_3 - 40V_2 + 60V_3 + 3V_3 - 600 = 0$$ $$103V_3 - 40V_2 = 600$$ 5. **Write system of equations:** $$\begin{cases} 31V_1 - 12V_2 = 1200 \\ -9V_1 + 23V_2 - 12V_3 = 0 \\ -40V_2 + 103V_3 = 600 \end{cases}$$ 6. **Solve the system:** From first equation: $$V_1 = \frac{1200 + 12V_2}{31}$$ Substitute into second: $$-9\left(\frac{1200 + 12V_2}{31}\right) + 23V_2 - 12V_3 = 0$$ Multiply both sides by 31: $$-9(1200 + 12V_2) + 23 \times 31 V_2 - 12 \times 31 V_3 = 0$$ $$-10800 - 108V_2 + 713V_2 - 372V_3 = 0$$ $$605V_2 - 372V_3 = 10800$$ From third equation: $$-40V_2 + 103V_3 = 600$$ Multiply third equation by 372 and second by 103 to eliminate \(V_3\): Second equation \(\times 103\): $$605 \times 103 V_2 - 372 \times 103 V_3 = 10800 \times 103$$ $$62315 V_2 - 38316 V_3 = 1,112,400$$ Third equation \(\times 372\): $$-40 \times 372 V_2 + 103 \times 372 V_3 = 600 \times 372$$ $$-14880 V_2 + 38316 V_3 = 223,200$$ Add equations: $$(62315 - 14880) V_2 = 1,112,400 + 223,200$$ $$47435 V_2 = 1,335,600$$ $$V_2 = \frac{1,335,600}{47435} \approx 28.15\,V$$ Substitute \(V_2\) into third equation: $$-40(28.15) + 103 V_3 = 600$$ $$-1126 + 103 V_3 = 600$$ $$103 V_3 = 1726$$ $$V_3 = \frac{1726}{103} \approx 16.75\,V$$ Substitute \(V_2\) into first equation for \(V_1\): $$V_1 = \frac{1200 + 12(28.15)}{31} = \frac{1200 + 337.8}{31} = \frac{1537.8}{31} \approx 49.61\,V$$ 7. **Find current through 15 \(\Omega\) resistor:** The 15 \(\Omega\) resistor is between nodes \(V_2\) and \(V_3\). Current \(I = \frac{V_2 - V_3}{15} = \frac{28.15 - 16.75}{15} = \frac{11.4}{15} = 0.76\,A\). **Final answer:** The current through the 15 \(\Omega\) resistor is approximately \(0.76\) amperes.