1. **Problem Statement:** Determine the node voltages at points A, B, and C using node D as the reference node in the given circuit. 2. **Given:** - Current source of 3 A entering node A. - Resistors: 2Ω (A-B), 7Ω (B-C), 2Ω (B-D), 3Ω (B-C), 1Ω (C-D). - Voltage source of 3 V between nodes D and B. - Node D is reference (0 V). 3. **Step 1: Assign node voltages:** Let $V_D=0$ V (reference), $V_A$, $V_B$, and $V_C$ be the voltages at nodes A, B, and C respectively. 4. **Step 2: Use Kirchhoff's Current Law (KCL) at each node (except reference):** - At node A: Current entering node A is 3 A (from current source). Current leaving through 2Ω resistor to node B is $\frac{V_A - V_B}{2}$. KCL at A: $3 = \frac{V_A - V_B}{2}$ - At node B: Currents leaving node B: To A: $\frac{V_B - V_A}{2}$ To C via 7Ω: $\frac{V_B - V_C}{7}$ To C via 3Ω: $\frac{V_B - V_C}{3}$ To D via 2Ω: $\frac{V_B - V_D}{2} = \frac{V_B - 0}{2} = \frac{V_B}{2}$ Also, voltage source of 3 V between D and B means $V_B = V_D + 3 = 0 + 3 = 3$ V. - At node C: Currents leaving node C: To B via 7Ω: $\frac{V_C - V_B}{7}$ To B via 3Ω: $\frac{V_C - V_B}{3}$ To D via 1Ω: $\frac{V_C - V_D}{1} = V_C$ KCL at C: Sum of currents leaving = 0 5. **Step 3: Use voltage source relation:** $V_B = 3$ V 6. **Step 4: Solve for $V_A$ from node A equation:** $3 = \frac{V_A - 3}{2} \implies 6 = V_A - 3 \implies V_A = 9$ V 7. **Step 5: Write KCL at node C:** $$\frac{V_C - 3}{7} + \frac{V_C - 3}{3} + V_C = 0$$ Multiply both sides by 21 (LCM of 7 and 3): $$3(V_C - 3) + 7(V_C - 3) + 21 V_C = 0$$ $$3V_C - 9 + 7V_C - 21 + 21 V_C = 0$$ $$31 V_C - 30 = 0$$ $$31 V_C = 30$$ $$V_C = \frac{30}{31} \approx 0.9677 \text{ V}$$ **Final node voltages:** $$V_A = 9 \text{ V}, \quad V_B = 3 \text{ V}, \quad V_C \approx 0.968 \text{ V}, \quad V_D = 0 \text{ V (reference)}$$