1. **State the problem:** We have a circuit with two voltage sources (25 V and 50 V) and resistors arranged with nodes A and B. We need to find the node voltages $V_1$ (at node A) and $V_2$ (at node B) and the branch currents using nodal analysis. 2. **Assign node voltages:** Let $V_1$ be the voltage at node A and $V_2$ be the voltage at node B with respect to ground (bottom reference). 3. **Write KCL equations at nodes:** At node A: $$\frac{V_1 - 25}{5} + \frac{V_1}{2} + \frac{V_1 - V_2}{10} = 0$$ Explanation: Current through 5-ohm resistor from node A to 25 V source is $(V_1 - 25)/5$, through 2-ohm resistor to ground is $V_1/2$, and through 10-ohm resistor to node B is $(V_1 - V_2)/10$. At node B: $$\frac{V_2 - V_1}{10} + \frac{V_2}{4} + \frac{V_2 - (-50)}{2} = 0$$ Explanation: Current through 10-ohm resistor from B to A is $(V_2 - V_1)/10$, through 4-ohm resistor to ground is $V_2/4$, and through 2-ohm resistor to 50 V source (bottom positive terminal means node voltage is -50 V) is $(V_2 + 50)/2$. 4. **Simplify equations:** Node A: $$\frac{V_1 - 25}{5} + \frac{V_1}{2} + \frac{V_1 - V_2}{10} = 0$$ Multiply all terms by 10: $$2(V_1 - 25) + 5V_1 + (V_1 - V_2) = 0$$ $$2V_1 - 50 + 5V_1 + V_1 - V_2 = 0$$ $$8V_1 - V_2 = 50$$ Node B: $$\frac{V_2 - V_1}{10} + \frac{V_2}{4} + \frac{V_2 + 50}{2} = 0$$ Multiply all terms by 20: $$2(V_2 - V_1) + 5V_2 + 10(V_2 + 50) = 0$$ $$2V_2 - 2V_1 + 5V_2 + 10V_2 + 500 = 0$$ $$17V_2 - 2V_1 = -500$$ 5. **Solve the system:** From first equation: $$8V_1 - V_2 = 50 \Rightarrow V_2 = 8V_1 - 50$$ Substitute into second: $$17(8V_1 - 50) - 2V_1 = -500$$ $$136V_1 - 850 - 2V_1 = -500$$ $$134V_1 = 350$$ $$V_1 = \frac{350}{134} \approx 2.61\,V$$ Then: $$V_2 = 8(2.61) - 50 = 20.88 - 50 = -29.12\,V$$ 6. **Calculate branch currents:** - Current through 5-ohm resistor: $$I_{5\Omega} = \frac{V_1 - 25}{5} = \frac{2.61 - 25}{5} = \frac{-22.39}{5} = -4.48\,A$$ (flows from node A to 25 V source) - Current through 2-ohm resistor at node A: $$I_{2\Omega A} = \frac{V_1}{2} = \frac{2.61}{2} = 1.305\,A$$ (to ground) - Current through 10-ohm resistor: $$I_{10\Omega} = \frac{V_1 - V_2}{10} = \frac{2.61 - (-29.12)}{10} = \frac{31.73}{10} = 3.173\,A$$ (from A to B) - Current through 4-ohm resistor at node B: $$I_{4\Omega} = \frac{V_2}{4} = \frac{-29.12}{4} = -7.28\,A$$ (negative means direction opposite to assumed) - Current through 2-ohm resistor at node B: $$I_{2\Omega B} = \frac{V_2 - (-50)}{2} = \frac{-29.12 + 50}{2} = \frac{20.88}{2} = 10.44\,A$$ (from B to 50 V source) **Final answers:** $$V_1 \approx 2.61\,V, \quad V_2 \approx -29.12\,V$$ Branch currents as above with directions noted.