Let's **solve the nodal voltages \( V_A \), \( V_B \), \( V_C \), and \( V_D \)** using **nodal analysis** step by step. --- ## **Step 1: Label Nodes & Assign Voltages** - \( V_A \): at the top-left node (above 6V) - \( V_B \): next, to the right (above .4A current source) - \( V_C \): next, to the right (junction of 4Ω, 8Ω, 4V) - \( V_D \): far right (junction of 8Ω, 7Ω, above 4V source) - Bottom node is **ground (V)**. --- ## **Step 2: Write Nodal Equations** ### **Node \( V_A \):** Sum currents leaving node: - Through 3Ω to \( V_B \): \( \frac{V_A - V_B}{3} \) - Through 7Ω to ground: \( \frac{V_A - }{7} \) - Through voltage source (6V): \( V_A = 6V \) (since the - terminal is at ground) So, **\( V_A = 6V \)**. --- ### **Node \( V_B \):** Sum currents leaving node: - Through 3Ω from \( V_A \): \( \frac{V_B - V_A}{3} \) - Through 4Ω to \( V_C \): \( \frac{V_B - V_C}{4} \) - Through current source (down to ground): **Current source tells us the sum of currents leaving must be .4A downwards.** So: \[ \frac{V_B - V_A}{3} + \frac{V_B - V_C}{4} = -.4 \] (The negative because .4A is entering node, so leaving is -.4A) --- ### **Node \( V_C \):** Sum currents leaving: - Through 4Ω from \( V_B \): \( \frac{V_C - V_B}{4} \) - Through 8Ω to \( V_D \): \( \frac{V_C - V_D}{8} \) - Through 4Ω to ground: \( \frac{V_C - }{4} \) \[ \frac{V_C - V_B}{4} + \frac{V_C - V_D}{8} + \frac{V_C}{4} = \] --- ### **Node \( V_D \):** Sum currents leaving: - Through 8Ω from \( V_C \): \( \frac{V_D - V_C}{8} \) - Through 7Ω to ground: \( \frac{V_D - }{7} \) - Through 4V source: \( V_D = 4V \) (since bottom is at ground) So, **\( V_D = 4V \)**. --- ## **Step 3: Substitute Known Values** Recall: - \( V_A = 6V \) - \( V_D = 4V \) Update equations: ### **Node \( V_B \):** \[ \frac{V_B - 6}{3} + \frac{V_B - V_C}{4} = -.4 \] ### **Node \( V_C \):** \[ \frac{V_C - V_B}{4} + \frac{V_C - 4}{8} + \frac{V_C}{4} = \] --- ## **Step 4: Simplify and Solve** ### **Equation 1 (Node \( V_B \)):** \[ \frac{V_B - 6}{3} + \frac{V_B - V_C}{4} = -.4 \] Multiply both sides by 12: \[ 4(V_B - 6) + 3(V_B - V_C) = -4.8 \] \[ 4V_B - 24 + 3V_B - 3V_C = -4.8 \] \[ 7V_B - 3V_C = 19.2 \] --- ### **Equation 2 (Node \( V_C \)):** \[ \frac{V_C - V_B}{4} + \frac{V_C - 4}{8} + \frac{V_C}{4} = \] Multiply both sides by 8: \[ 2(V_C - V_B) + (V_C - 4) + 2V_C = \] \[ 2V_C - 2V_B + V_C - 4 + 2V_C = \] \[ (2V_C + V_C + 2V_C) - 2V_B - 4 = \] \[ 5V_C - 2V_B - 4 = \] \[ 5V_C - 2V_B = 4 \] --- ## **Step 5: Solve Simultaneous Equations** 1. \( 7V_B - 3V_C = 19.2 \) 2. \( 5V_C - 2V_B = 4 \) Let's solve for \( V_B \) and \( V_C \): From (2): \[ 5V_C - 2V_B = 4 \implies 2V_B = 5V_C - 4 \implies V_B = \frac{5V_C - 4}{2} \] Substitute into (1): \[ 7\left(\frac{5V_C - 4}{2}\right) - 3V_C = 19.2 \] \[ \frac{35V_C - 28}{2} - 3V_C = 19.2 \] \[ \frac{35V_C - 28 - 6V_C}{2} = 19.2 \] \[ \frac{29V_C - 28}{2} = 19.2 \] \[ 29V_C - 28 = 38.4 \] \[ 29V_C = 66.4 \] \[ V_C = \frac{66.4}{29} \approx 2.29~\text{V} \] Now, plug back to find \( V_B \): \[ V_B = \frac{5 \times 2.29 - 4}{2} = \frac{11.45 - 4}{2} = \frac{7.45}{2} = 3.725~\text{V} \] --- ## **Step 6: ** Final Answers \[ \boxed{ \begin{align*} V_A &= 6~\text{V} \\ V_B &\approx 3.73~\text{V} \\ V_C &\approx 2.29~\text{V} \\ V_D &= 4~\text{V} \end{align*} } \] --- ### **Summary Table** | Node | Voltage (V) | |------|-------------| | \( V_A \) | 6 | | \( V_B \) | 3.73 | | \( V_C \) | 2.29 | | \( V_D \) | 4 | --- **Let me know if you need step-by-step substitution details or have questions about any step!**