Let's solve the node voltages step by step using **nodal analysis**. ## **Step 1: Identify Nodes and Assign Voltages** Let: - Bottom node (ground) = 0 V - \( v_1 \): voltage at the node between \( R_1, R_3, R_4 \) - \( v_2 \): voltage at the node between \( R_5, R_4, R_6 \) - \( v_3 \): voltage at the node between \( R_2, R_6 \) and positive terminal of current source Given: - \( R_1 = 15\,\Omega \) - \( R_2 = 5\,\Omega \) - \( R_3 = 20\,\Omega \) - \( R_4 = 10\,\Omega \) - \( R_5 = 8\,\Omega \) - \( R_6 = 4\,\Omega \) - \( I_s = 5\,A \) --- ## **Step 2: Write KCL Equations** ### **Node \( v_1 \):** \[ \frac{v_1}{R_3} + \frac{v_1 - v_2}{R_4} + \frac{v_1}{R_1} = 0 \] But notice \( v_1 \) also connects to \( v_3 \) via current source, so let's write for all nodes. ### **Node \( v_1 \):** \[ \frac{v_1}{R_1} + \frac{v_1}{R_3} + \frac{v_1 - v_2}{R_4} = 0 \] ### **Node \( v_2 \):** \[ \frac{v_2}{R_5} + \frac{v_2 - v_1}{R_4} + \frac{v_2 - v_3}{R_6} = 0 \] ### **Node \( v_3 \):** At \( v_3 \), the current leaving through \( R_2 \) and \( R_6 \) must equal \( I_s \) entering: \[ \frac{v_3}{R_2} + \frac{v_3 - v_2}{R_6} = I_s \] --- ## **Step 3: Substitute Values** ### **Equation 1: (Node \( v_1 \))** \[ \frac{v_1}{15} + \frac{v_1}{20} + \frac{v_1 - v_2}{10} = 0 \] ### **Equation 2: (Node \( v_2 \))** \[ \frac{v_2}{8} + \frac{v_2 - v_1}{10} + \frac{v_2 - v_3}{4} = 0 \] ### **Equation 3: (Node \( v_3 \))** \[ \frac{v_3}{5} + \frac{v_3 - v_2}{4} = 5 \] --- ## **Step 4: Simplify the Equations** ### **Equation 1:** \[ \frac{v_1}{15} + \frac{v_1}{20} + \frac{v_1}{10} - \frac{v_2}{10} = 0 \] Find LCM: 60 \[ \frac{4v_1}{60} + \frac{3v_1}{60} + \frac{6v_1}{60} - \frac{6v_2}{60} = 0 \] \[ \frac{(4 + 3 + 6)v_1 - 6v_2}{60} = 0 \] \[ \frac{13v_1 - 6v_2}{60} = 0 \] \[ 13v_1 = 6v_2 \] \[ v_1 = \frac{6}{13}v_2 \] --- ### **Equation 2:** \[ \frac{v_2}{8} + \frac{v_2 - v_1}{10} + \frac{v_2 - v_3}{4} = 0 \] Multiply by 40 to clear denominators: \[ 5v_2 + 4(v_2 - v_1) + 10(v_2 - v_3) = 0 \] \[ 5v_2 + 4v_2 - 4v_1 + 10v_2 - 10v_3 = 0 \] \[ (5 + 4 + 10)v_2 - 4v_1 - 10v_3 = 0 \] \[ 19v_2 - 4v_1 - 10v_3 = 0 \] --- ### **Equation 3:** \[ \frac{v_3}{5} + \frac{v_3 - v_2}{4} = 5 \] Multiply by 20: \[ 4v_3 + 5(v_3 - v_2) = 100 \] \[ 4v_3 + 5v_3 - 5v_2 = 100 \] \[ 9v_3 - 5v_2 = 100 \] --- ## **Step 5: Substitute \( v_1 \) into Other Equations** Recall: \( v_1 = \frac{6}{13}v_2 \) ### **Equation 2 (Substitute \( v_1 \)):** \[ 19v_2 - 4v_1 - 10v_3 = 0 \] \[ 19v_2 - 4\left(\frac{6}{13}v_2\right) - 10v_3 = 0 \] \[ 19v_2 - \frac{24}{13}v_2 - 10v_3 = 0 \] \[ \left(19 - \frac{24}{13}\right)v_2 - 10v_3 = 0 \] \[ \left(\frac{247}{13} - \frac{24}{13}\right)v_2 - 10v_3 = 0 \] \[ \frac{223}{13}v_2 - 10v_3 = 0 \] \[ \frac{223}{13}v_2 = 10v_3 \] \[ v_3 = \frac{223}{130}v_2 \] --- ### **Equation 3 (Substitute \( v_3 \)):** \[ 9v_3 - 5v_2 = 100 \] \[ 9\left(\frac{223}{130}v_2\right) - 5v_2 = 100 \] \[ \frac{2007}{130}v_2 - 5v_2 = 100 \] \[ \frac{2007 - 650}{130}v_2 = 100 \] \[ \frac{1357}{130}v_2 = 100 \] \[ v_2 = 100 \times \frac{130}{1357} \] \[ v_2 = \frac{13000}{1357} \] \[ v_2 \approx 9.58\ \text{V} \] --- ### **Now find \( v_1 \) and \( v_3 \):** \[ v_1 = \frac{6}{13}v_2 = \frac{6}{13} \times 9.58 \approx 4.42\ \text{V} \] \[ v_3 = \frac{223}{130}v_2 = \frac{223}{130} \times 9.58 \approx 16.44\ \text{V} \] --- ## **Final Answers** \[ \boxed{ \begin{align*} v_1 &\approx 4.42\ \text{V} \\ v_2 &\approx 9.58\ \text{V} \\ v_3 &\approx 16.44\ \text{V} \end{align*} } \] --- If you need all steps or a check, let me know!