1. **Problem Statement:** Write the nodal equilibrium equations for the circuit in Fig. 2.92 with nodes K1, K2, K3 having voltages $V_1$, $V_2$, and $V_3$ respectively, and find these voltages. All resistors are 1 Ω, and the voltage source is 10 V. 2. **Assign Node Voltages:** Let the reference node (ground) be at the negative terminal of the 10 V source. 3. **Write KCL Equations at Each Node:** Using Ohm's law, current through a resistor is $I=\frac{V_{node}-V_{adjacent}}{R}$. - At node $K_1$ ($V_1$): $$\frac{V_1 - 10}{1} + \frac{V_1 - V_2}{1} + \frac{V_1 - 0}{1} = 0$$ Simplify: $$ (V_1 - 10) + (V_1 - V_2) + V_1 = 0 $$ $$ 3V_1 - V_2 = 10 $$ - At node $K_2$ ($V_2$): $$\frac{V_2 - V_1}{1} + \frac{V_2 - V_3}{1} + \frac{V_2 - 0}{1} = 0$$ Simplify: $$ (V_2 - V_1) + (V_2 - V_3) + V_2 = 0 $$ $$ -V_1 + 3V_2 - V_3 = 0 $$ - At node $K_3$ ($V_3$): $$\frac{V_3 - V_2}{1} + \frac{V_3 - 0}{1} = 0$$ Simplify: $$ (V_3 - V_2) + V_3 = 0 $$ $$ -V_2 + 2V_3 = 0 $$ 4. **Write the system of equations:** $$\begin{cases} 3V_1 - V_2 = 10 \\ -V_1 + 3V_2 - V_3 = 0 \\ - V_2 + 2V_3 = 0 \end{cases}$$ 5. **Solve the system:** From the third equation: $$ 2V_3 = V_2 \implies V_3 = \frac{V_2}{2} $$ Substitute $V_3$ into the second equation: $$ -V_1 + 3V_2 - \frac{V_2}{2} = 0 $$ $$ -V_1 + \frac{5}{2} V_2 = 0 \implies V_1 = \frac{5}{2} V_2 $$ Substitute $V_1$ into the first equation: $$ 3 \times \frac{5}{2} V_2 - V_2 = 10 $$ $$ \frac{15}{2} V_2 - V_2 = 10 $$ $$ \left( \frac{15}{2} - 1 \right) V_2 = 10 $$ $$ \frac{13}{2} V_2 = 10 \implies V_2 = \frac{20}{13} $$ Calculate $V_1$: $$ V_1 = \frac{5}{2} \times \frac{20}{13} = \frac{100}{26} = \frac{50}{13} $$ Calculate $V_3$: $$ V_3 = \frac{V_2}{2} = \frac{20}{13} \times \frac{1}{2} = \frac{10}{13} $$ **Final voltages:** $$ V_1 = \frac{50}{13} \approx 3.85, \quad V_2 = \frac{20}{13} \approx 1.54, \quad V_3 = \frac{10}{13} \approx 0.77 $$