Let's solve the circuit step-by-step as requested: --- ## Step 1: **Redraw the Circuit with Node Labels** Let's denote: - \( V_1 \): Voltage at the node between \( 1\Omega \), \( 4\Omega \), and \( 2\Omega \) resistors (left). - \( V_2 \): Voltage at the node between \( 2\Omega \), \( 2\Omega \), and \( 2\Omega \) resistors (middle). - \( V_3 \): Voltage at the node between \( 2\Omega \), \( 2\Omega \), and \( 2A \) source (right). --- ### **Node Assignment** - The leftmost node (ground reference) is at 0V. - The rightmost node (bottom, after \( 2A \) current source) is taken as reference (0V). --- ## Step 2: **Source Transformation** The 40V voltage source in series with \( 1\Omega \) can be transformed into a current source: \[ I = \frac{40V}{1\Omega} = 40A \] So, this becomes a \( 40A \) current source in parallel with \( 1\Omega \). However, for nodal analysis, we can keep it as is for simplicity. --- ## Step 3: **Nodal Analysis** Let's assign the nodes as shown in the image: - Node 1: \( V_1 \) - Node 2: \( V_2 \) - Node 3: \( V_3 \) (rightmost node before the current source) ### **Write KCL at Each Node** #### **Node 1 (\( V_1 \)):** Connected to: - \( 1\Omega \) to ground (left) - \( 4\Omega \) to \( V_2 \) - \( 2\Omega \) to top (which connects to \( V_2 \)) Current leaving node: \[ \frac{V_1}{1} + \frac{V_1 - V_2}{4} + \frac{V_1 - V_2}{2} = 0 \] Simplify: \[ \frac{V_1}{1} + \frac{V_1 - V_2}{4} + \frac{V_1 - V_2}{2} = 0 \] \[ V_1 + \frac{V_1 - V_2}{4} + \frac{V_1 - V_2}{2} = 0 \] \[ V_1 + \frac{V_1 - V_2 + 2(V_1 - V_2)}{4} = 0 \] \[ V_1 + \frac{3(V_1 - V_2)}{4} = 0 \] \[ 4V_1 + 3(V_1 - V_2) = 0 \] \[ 4V_1 + 3V_1 - 3V_2 = 0 \] \[ 7V_1 - 3V_2 = 0 \implies 7V_1 = 3V_2 \implies V_1 = \frac{3}{7}V_2 \] #### **Node 2 (\( V_2 \)):** Connected to: - \( 4\Omega \) to \( V_1 \) - \( 2\Omega \) to \( V_1 \) (top) - \( 2\Omega \) to \( V_3 \) - \( 2\Omega \) to top right node - \( 2A \) current source (leaving node) Write KCL: \[ \frac{V_2 - V_1}{4} + \frac{V_2 - V_1}{2} + \frac{V_2 - V_3}{2} + 2 = 0 \] Simplify: \[ \frac{V_2 - V_1}{4} + \frac{V_2 - V_1}{2} + \frac{V_2 - V_3}{2} = -2 \] \[ \frac{V_2 - V_1 + 2(V_2 - V_1) + 2(V_2 - V_3)}{4} = -2 \] \[ \frac{3(V_2 - V_1) + 2(V_2 - V_3)}{4} = -2 \] \[ 3(V_2 - V_1) + 2(V_2 - V_3) = -8 \] \[ 3V_2 - 3V_1 + 2V_2 - 2V_3 = -8 \] \[ (3V_2 + 2V_2) - 3V_1 - 2V_3 = -8 \] \[ 5V_2 - 3V_1 - 2V_3 = -8 \] #### **Node 3 (\( V_3 \)):** Connected to: - \( 2\Omega \) to \( V_2 \) - \( 2\Omega \) to ground KCL: \[ \frac{V_3 - V_2}{2} + \frac{V_3}{2} = 0 \] \[ \frac{V_3 - V_2 + V_3}{2} = 0 \] \[ V_3 - V_2 + V_3 = 0 \] \[ 2V_3 - V_2 = 0 \] \[ 2V_3 = V_2 \implies V_3 = \frac{V_2}{2} \] --- ### **Now substitute relationships:** Recall: - \( V_1 = \frac{3}{7}V_2 \) - \( V_3 = \frac{V_2}{2} \) Plug into equation for Node 2: \[ 5V_2 - 3V_1 - 2V_3 = -8 \] \[ 5V_2 - 3\left(\frac{3}{7}V_2\right) - 2\left(\frac{V_2}{2}\right) = -8 \] \[ 5V_2 - \frac{9}{7}V_2 - V_2 = -8 \] \[ 5V_2 - V_2 - \frac{9}{7}V_2 = -8 \] \[ 4V_2 - \frac{9}{7}V_2 = -8 \] \[ \left(4 - \frac{9}{7}\right)V_2 = -8 \] \[ \left(\frac{28}{7} - \frac{9}{7}\right)V_2 = -8 \] \[ \frac{19}{7}V_2 = -8 \] \[ V_2 = -8 \cdot \frac{7}{19} \] \[ V_2 = -\frac{56}{19} \approx -2.95V \] --- ### **Now calculate \( V_1 \) and \( V_3 \):** \[ V_1 = \frac{3}{7}V_2 = \frac{3}{7} \times -2.95 \approx -1.27\text{ V} \] \[ V_3 = \frac{V_2}{2} = \frac{-2.95}{2} = -1.475\text{ V} \] --- ## **Final Answers for (a):** \[ \boxed{ V_1 \approx -1.27\text{ V} \V_2 \approx -2.95\text{ V} \\ V_3 \approx -1.48\text{ V} } \] --- ## Step 4: **Find Power at the \( 4\Omega \) Resistor (b):** The current through the \( 4\Omega \) resistor: \[ I_{4\Omega} = \frac{V_1 - V_2}{4} \] Plug in the values: \[ I_{4\Omega} = \frac{-1.27 - (-2.95)}{4} = \frac{1.68}{4} = 0.42\text{ A} \] Power dissipated: \[ P = I^2 R = (0.42)^2 \times 4 \approx 0.1764 \times 4 = 0.71\text{ W} \] --- ## **Final Answer for (b):** \[ \boxed{ \text{Power at the } 4\Omega \text{ resistor} = 0.71\text{ W} } \] --- ### **Summary Table** | | Value | |---|------------| | \(V_1\) | -1.27 V | | \(V_2\) | -2.95 V | | \(V_3\) | -1.48 V | | \(P_{4\Omega}\) | 0.71 W | --- **If you need clarification or more details at any step, let me know!**