Let's solve the circuit **step-by-step**: ### **Given Circuit** - **Voltage source:** 40V (left) - **Current source:** 2A (right) - **Resistors:** As shown - **Nodes:** \( V_1 \) (bottom left), \( V_2 \) (middle right), \( V_3 \) (bottom right) --- ## **Step 1: Label all nodes and assign ground** Let the bottommost node be ground (0V). --- ## **Step 2: Nodal Analysis** Let: - Node at left of \(1\,\Omega\): \( V_1 \) - Node at right of \(4\,\Omega\): \( V_2 \) - Node at left of \(2\,\Omega\) (bottom): \( V_3 \) --- ## **Step 3: Write Nodal Equations** ### **Node 1 (\( V_1 \)):** \[ \frac{V_1 - 40}{1} + \frac{V_1 - V_2}{4} + \frac{V_1 - V_3}{2} = 0 \] ### **Node 2 (\( V_2 \)):** \[ \frac{V_2 - V_1}{4} + \frac{V_2}{2} + \frac{V_2 - V_3}{2} = 0 \] ### **Node 3 (\( V_3 \)), with 2A current source:** Current flows into the node: \[ 2 = \frac{V_3 - V_1}{2} + \frac{V_3 - V_2}{2} + \frac{V_3}{2} \] --- ## **Step 4: Simplify Each Equation** ### **Equation 1:** \[ (V_1 - 40) + \frac{V_1 - V_2}{4} + \frac{V_1 - V_3}{2} = 0 \] Multiply both sides by 4: \[ 4(V_1 - 40) + (V_1 - V_2) + 2(V_1 - V_3) = 0 \] \[ 4V_1 - 160 + V_1 - V_2 + 2V_1 - 2V_3 = 0 \] \[ (4V_1 + V_1 + 2V_1) - V_2 - 2V_3 = 160 \] \[ 7V_1 - V_2 - 2V_3 = 160 \] --- ### **Equation 2:** \[ \frac{V_2 - V_1}{4} + \frac{V_2}{2} + \frac{V_2 - V_3}{2} = 0 \] Multiply both sides by 4: \[ (V_2 - V_1) + 2V_2 + 2(V_2 - V_3) = 0 \] \[ V_2 - V_1 + 2V_2 + 2V_2 - 2V_3 = 0 \] \[ (V_2 + 2V_2 + 2V_2) - V_1 - 2V_3 = 0 \] \[ 5V_2 - V_1 - 2V_3 = 0 \] \[ - V_1 + 5V_2 - 2V_3 = 0 \] --- ### **Equation 3:** \[ 2 = \frac{V_3 - V_1}{2} + \frac{V_3 - V_2}{2} + \frac{V_3}{2} \] Multiply both sides by 2: \[ 4 = (V_3 - V_1) + (V_3 - V_2) + V_3 \] \[ 4 = V_3 - V_1 + V_3 - V_2 + V_3 \] \[ 4 = 3V_3 - V_1 - V_2 \] \[ 3V_3 - V_1 - V_2 = 4 \] --- ## **Step 5: Solve the System** We have: 1. \( 7V_1 - V_2 - 2V_3 = 160 \) 2. \( -V_1 + 5V_2 - 2V_3 = 0 \) 3. \( -V_1 - V_2 + 3V_3 = 4 \) --- Let's solve step by step. ### **From Equation 2:** \[ -V_1 + 5V_2 - 2V_3 = 0 \implies V_1 = 5V_2 - 2V_3 \] ### **Substitute \( V_1 \) into Equations 1 & 3:** #### **Equation 1:** \[ 7V_1 - V_2 - 2V_3 = 160 \] \[ 7(5V_2 - 2V_3) - V_2 - 2V_3 = 160 \] \[ 35V_2 - 14V_3 - V_2 - 2V_3 = 160 \] \[ (35V_2 - V_2) + (-14V_3 - 2V_3) = 160 \] \[ 34V_2 - 16V_3 = 160 \] \[ 17V_2 - 8V_3 = 80 \quad (4) \] --- #### **Equation 3:** \[ - V_1 - V_2 + 3V_3 = 4 \] \[ -(5V_2 - 2V_3) - V_2 + 3V_3 = 4 \] \[ -5V_2 + 2V_3 - V_2 + 3V_3 = 4 \] \[ -6V_2 + 5V_3 = 4 \quad (5) \] --- Now solve equations (4) and (5): - (4) \( 17V_2 - 8V_3 = 80 \) - (5) \( -6V_2 + 5V_3 = 4 \) Multiply (5) by 8: \[ -48V_2 + 40V_3 = 32 \] Multiply (4) by 5: \[ 85V_2 - 40V_3 = 400 \] Now add: \[ (85V_2 -40V_) + (-48V_2 + 40V_3) = 400 + 32 \] \[ V_2 - 48V_2) + (-40V_3 + 40V_3) = 432 \] \[ 37V_ = 432 \] \[ V_2 = \frac{432}{37} \approx 11.68\,V \] Now plug \( V_2 \) into (4): \[ 17V_2 - 8V_3 = 80 \] \[ 17 \times 11.68 - 8V_3 = 80 \] \[ 198.56 - 8V_3 = 80 \] \[ 198.56 - 80 = 8V_3 \] \[ 118.56 = 8V_3 \] \[ V_3 = \frac{118.56}{8} \approx 14.82\,V \] Now plug \( V_2 \) and \( V_3 \) into the earlier expression for \( V_1 \): \[ V_1 = 5V_2 - 2V_3 \] \[ = 5 \times 11.68 - 2 \times 14.82 \] \[ = 58.40 - 29.64 = 28.76\,V \] --- ## **Step 6: Final Answers** \[ \boxed{ \begin{align*} V_1 &\approx 28.76\,V \\ V_2 &\approx 11.68\,V \\ V_3 &\approx 14.82\,V \\ \end{align*} } \] --- ## **Step 7: Power in the \(4\,\Omega\) Resistor** The current through the \(4\,\Omega\) resistor: \[ I_{4\Omega} = \frac{V_1 - V_2}{4} \] \[ = \frac{28.76 - 11.68}{4} = \frac{17.08}{4} \approx 4.27\,A \] Power: \[ P = I^2 R = (4.27)^2 \times 4 \approx 18.23 \times 4 \approx 72.92\,W \] --- ## **Final Answers (Boxed):** \[ \boxed{ \begin{align*} \text{a) } & V_1 \approx 28.76\,V,\quad V_2 \approx 11.68\,V,\quad V_3 \approx 14.82\,V \\ \text{b) } & \text{Power at the } 4\Omega \text{ resistor} \approx 72.92\,W \end{align*} } \] --- **If you need the source transformation steps, let me know!**