Let's solve the problem step by step. ## a. **Node-Voltage Method** Let’s label the node voltages as shown in the diagram: - \( v_1 \): at the top left node (between 3 Ω, 1 Ω, and 40 Ω resistor) - \( v_2 \): at the top middle node (between 2 Ω, 40 Ω, and 4 Ω resistor) - \( v_3 \): at the bottom middle node (between 40 Ω, 2 Ω, and 1 Ω resistor) Let the bottom node (connected to the negative terminal of the 40V source and the 28A source) be our reference (0 V). ### 1. **Node 1: \( v_1 \)** At node 1, apply KCL (sum of outgoing currents = 0): \[ \frac{v_1 - 40}{3} + \frac{v_1 - v_2}{40} + \frac{v_1 - v_3}{1} = 0 \] ### 2. **Node 2: \( v_2 \)** At node 2, apply KCL: \[ \frac{v_2 - v_1}{40} + \frac{v_2 - v_3}{2} + \frac{v_2}{4} = 0 \] ### 3. **Node 3: \( v_3 \)** At node 3, apply KCL (don't forget the current source): \[ \frac{v_3 - v_1}{1} + \frac{v_3 - v_2}{2} + \frac{v_3}{2} = -28 \] (The current source is going **out** of the node, so it is negative.) --- Now, let's simplify and solve the system. --- #### **Equation 1:** \[ \frac{v_1 - 40}{3} + \frac{v_1 - v_2}{40} + (v_1 - v_3) = 0 \] Multiply both sides by 120 (LCM of denominators): \[ 40(v_1 - 40) + 3(v_1 - v_2) + 120(v_1 - v_3) = 0 \] \[ 40v_1 - 1600 + 3v_1 - 3v_2 + 120v_1 - 120v_3 = 0 \] \[ (40 + 3 + 120)v_1 - 3v_2 - 120v_3 = 1600 \] \[ 163v_1 - 3v_2 - 120v_3 = 1600 \] --- #### **Equation 2:** \[ \frac{v_2 - v_1}{40} + \frac{v_2 - v_3}{2} + \frac{v_2}{4} = 0 \] Multiply both sides by 40: \[ (v_2 - v_1) + 20(v_2 - v_3) + 10v_2 = 0 \] \[ v_2 - v_1 + 20v_2 - 20v_3 + 10v_2 = 0 \] \[ (-v_1) + (v_2 + 20v_2 + 10v_2) - 20v_3 = 0 \] \[ - v_1 + 31v_2 - 20v_3 = 0 \] --- #### **Equation 3:** \[ \frac{v_3 - v_1}{1} + \frac{v_3 - v_2}{2} + \frac{v_3}{2} = -28 \] Multiply both sides by 2: \[ 2(v_3 - v_1) + (v_3 - v_2) + v_3 = -56 \] \[ 2v_3 - 2v_1 + v_3 - v_2 + v_3 = -56 \] \[ (2v_3 + v_3 + v_3) - 2v_1 - v_2 = -56 \] \[ 4v_3 - 2v_1 - v_2 = -56 \] --- Now, we have a system of three equations: 1. \(163v_1 - 3v_2 - 120v_3 = 1600\) 2. \(- v_1 + 31v_2 - 20v_3 = 0\) 3. \(-2v_1 - v_2 + 4v_3 = -56\) --- ## **Solving the System** Let’s use equations 2 and 3 to express \(v_1\) and \(v_2\) in terms of \(v_3\): From (2): \(-v_1 + 31v_2 - 20v_3 = 0\) \[ v_1 = 31v_2 - 20v_3 \] From (3): \(-2v_1 - v_2 + 4v_3 = -56\) Plug in \(v_1\) from above: \[ -2(31v_2 - 20v_3) - v_2 + 4v_3 = -56 \] \[ -62v_2 + 40v_3 - v_2 + 4v_3 = -56 \] \[ -63v_2 + 44v_3 = -56 \] \[ 63v_2 = 44v_3 + 56 \] \[ v_2 = \frac{44v_3 + 56}{63} \] Now plug \(v_2\) into \(v_1\): \[ v_1 = 31v_2 - 20v_3 = 31 \left(\frac{44v_3 + 56}{63}\right) - 20v_3 \] \[ v_1 = \frac{31 \cdot 44v_3 + 31 \cdot 56}{63} - 20v_3 \] \[ v_1 = \frac{1364v_3 + 1736}{63} - 20v_3 \] \[ v_1 = \frac{1364v_3 + 1736 - 1260v_3}{63} \] \[ v_1 = \frac{104v_3 + 1736}{63} \] Now, substitute \(v_1\) and \(v_2\) in terms of \(v_3\) into equation 1: \[ 163v_1 - 3v_2 - 120v_3 = 1600 \] \[ 163 \left(\frac{104v_3 + 1736}{63}\right) - 3\left(\frac{44v_3 + 56}{63}\right) - 120v_3 = 1600 \] \[ \frac{163 \times 104v_3 + 163 \times 1736 - 3 \times 44v_3 - 3 \times 56}{63} - 120v_3 = 1600 \] \[ \frac{16952v_3 + 283768 - 132v_3 - 168}{63} - 120v_3 = 1600 \] \[ \frac{16952v_3 - 132v_3 + 283768 - 168}{63} - 120v_3 = 1600 \] \[ \frac{16820v_3 + 283600}{63} - 120v_3 = 1600 \] Multiply both sides by 63: \[ 16820v_3 + 283600 - 120 \times 63 v_3 = 1600 \times 63 \] \[ 16820v_3 + 283600 - 7560v_3 = 100800 \] \[ (16820 - 7560)v_3 + 283600 = 100800 \] \[ 9260v_3 = 100800 - 283600 \] \[ 9260v_3 = -182800 \] \[ v_3 = -\frac{182800}{9260} = -19.74\, \text{V} \] --- Now, back to \(v_2\): \[ v_2 = \frac{44v_3 + 56}{63} = \frac{44 \times -19.74 + 56}{63} \] \[ = \frac{-868.56 + 56}{63} = \frac{-812.56}{63} = -12.90\, \text{V} \] Now, \(v_1\): \[ v_1 = \frac{104v_3 + 1736}{63} = \frac{104 \times -19.74 + 1736}{63} \] \[ = \frac{-2053.0 + 1736}{63} = \frac{-317.0}{63} = -5.03\, \text{V} \] --- ## **Final Answers:** \[ \boxed{ v_1 = -5.03\, \text{V}, \quad v_2 = -12.90\, \text{V}, \quad v_3 = -19.74\, \text{V} } \] --- ## b. **Power delivered by the 40V source** The current delivered by the 40V source is the current through the 3Ω resistor (from 40V to \(v_1\)): \[ I = \frac{40 - v_1}{3} = \frac{40 - (-5.03)}{3} = \frac{45.03}{3} = 15.01\, \text{A} \] Power delivered: \[ P = VI = 40 \times 15.01 = 600.4\, \text{W} \] --- ### **Boxed Answers** \[ \boxed{ \text{a) } v_1 = -5.03\, \text{V},\ v_2 = -12.90\, \text{V},\ v_3 = -19.74\, \text{V} } \] \[ \boxed{ \text{b) Power delivered by 40V source:}\ 600.4\, \text{W} } \]