Let's solve the circuit step by step using the **node-voltage method**. --- ## **Step 1: Assign Node Voltages** Let: - \( v_1 \): Voltage at the top left node (above 40V source) - \( v_2 \): Voltage at the top right node (after 2Ω resistor) - \( v_3 \): Voltage at the bottom middle node (between 1Ω, 40Ω, 2Ω resistors) Set the **bottom node** (bottom of 40V source and 1Ω resistor) as the **reference (ground)**, \( v = 0 \). --- ## **Step 2: Write Node Equations** ### **Node 1 (\( v_1 \)):** Connected to: 3Ω, 1Ω, 40Ω, and 2Ω resistors \[ \frac{v_1 - 40}{3} + \frac{v_1}{1} + \frac{v_1 - v_3}{40} + \frac{v_1 - v_2}{2} = 0 \] ### **Node 2 (\( v_2 \)):** Connected to: 2Ω, 4Ω, 2Ω resistors, and the 28A current source \[ \frac{v_2 - v_1}{2} + \frac{v_2}{4} + \frac{v_2 - v_3}{2} = 28 \] ### **Node 3 (\( v_3 \)):** Connected to: 40Ω, 2Ω, 2Ω resistors \[ \frac{v_3 - v_1}{40} + \frac{v_3 - v_2}{2} + \frac{v_3}{2} = 0 \] --- ## **Step 3: Simplify Equations** ### **Node 1:** \[ \frac{v_1 - 40}{3} + v_1 + \frac{v_1 - v_3}{40} + \frac{v_1 - v_2}{2} = 0 \] Least common denominator is 120: \[ 40(v_1 - 40) + 120v_1 + 3(v_1 - v_3) + 60(v_1 - v_2) = 0 \] \[ 40v_1 - 1600 + 120v_1 + 3v_1 - 3v_3 + 60v_1 - 60v_2 = 0 \] \[ (40 + 120 + 3 + 60)v_1 - 60v_2 - 3v_3 = 1600 \] \[ 223v_1 - 60v_2 - 3v_3 = 1600 \] --- ### **Node 2:** \[ \frac{v_2 - v_1}{2} + \frac{v_2}{4} + \frac{v_2 - v_3}{2} = 28 \] \[ \frac{v_2 - v_1 + v_2/2 + v_2 - v_3}{1} = 28 \] Multiply by 4 to clear denominators: \[ 2(v_2 - v_1) + v_2 + 2(v_2 - v_3) = 112 \] \[ 2v_2 - 2v_1 + v_2 + 2v_2 - 2v_3 = 112 \] \[ (2 + 1 + 2)v_2 - 2v_1 - 2v_3 = 112 \] \[ 5v_2 - 2v_1 - 2v_3 = 112 \] --- ### **Node 3:** \[ \frac{v_3 - v_1}{40} + \frac{v_3 - v_2}{2} + \frac{v_3}{2} = 0 \] Multiply by 40 to clear denominators: \[ v_3 - v_1 + 20(v_3 - v_2) + 20v_3 = 0 \] \[ v_3 - v_1 + 20v_3 - 20v_2 + 20v_3 = 0 \] \[ (1 + 20 + 20)v_3 - v_1 - 20v_2 = 0 \] \[ 41v_3 - v_1 - 20v_2 = 0 \] --- ## **Step 4: Write Matrix Form** \[ \begin{cases} 223v_1 - 60v_2 - 3v_3 = 1600 \\ -2v_1 + 5v_2 - 2v_3 = 112 \\ - v_1 - 20v_2 + 41v_3 = 0 \end{cases} \] --- ## **Step 5: Solve the System** Let's solve using substitution/elimination or matrix. Let's rewrite: \[ \begin{bmatrix} 223 & -60 & -3 \\ -2 & 5 & -2 \\ -1 & -20 & 41 \\ \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 1600 \\ 112 \\ 0 \end{bmatrix} \] Solving this system (using calculator, Cramer's rule or any method): --- ### **Using Python/Calculator:** Let's solve: **First, from equation 3:** \[ - v_1 - 20v_2 + 41v_3 = 0 \implies v_1 = -20v_2 + 41v_3 \] Plug into the other equations. **Equation 2:** \[ -2v_1 + 5v_2 - 2v_3 = 112 \] Substitute \( v_1 \): \[ -2(-20v_2 + 41v_3) + 5v_2 - 2v_3 = 112 \\ 40v_2 - 82v_3 + 5v_2 - 2v_3 = 112 \\ (40 + 5)v_2 + (-82 - 2)v_3 = 112 \\ 45v_2 - 84v_3 = 112 \] **Equation 1:** \[ 223v_1 - 60v_2 - 3v_3 = 1600 \] Substitute \( v_1 \): \[ 223(-20v_2 + 41v_3) - 60v_2 - 3v_3 = 1600 \\ -4460v_2 + 9143v_3 - 60v_2 - 3v_3 = 1600 \\ (-4460 - 60)v_2 + (9143 - 3)v_3 = 1600 \\ -4520v_2 + 9140v_3 = 1600 \] So our reduced system is: \[ -4520v_2 + 9140v_3 = 1600 \\ 45v_2 - 84v_3 = 112 \] Solve equation 2 for \( v_2 \): \[ 45v_2 - 84v_3 = 112 \\ 45v_2 = 112 + 84v_3 \\ v_2 = \frac{112 + 84v_3}{45} \] Plug into equation 1: \[ -4520 \left(\frac{112 + 84v_3}{45}\right) + 9140v_3 = 1600 \\ -4520 \times \frac{112}{45} - 4520 \times \frac{84v_3}{45} + 9140v_3 = 1600 \\ -11248.89 - 8432v_3/45 + 9140v_3 = 1600 \\ -11248.89 - 187.378v_3 + 9140v_3 = 1600 \\ -11248.89 + (9140 - 187.378)v_3 = 1600 \\ -11248.89 + 8952.62v_3 = 1600 \\ 8952.62v_3 = 1600 + 11248.89 \\ 8952.62v_3 = 12848.89 \\ v_3 = \frac{12848.89}{8952.62} \approx 1.435 \] Now, \( v_2 = \frac{112 + 84 \times 1.435}{45} = \frac{112 + 120.54}{45} = \frac{232.54}{45} \approx 5.17 \) Now, \( v_1 = -20v_2 + 41v_3 = -20 \times 5.17 + 41 \times 1.435 = -103.4 + 58.835 = -44.565 \) --- ## **Final Answers (rounded to 2 decimals):** \[ \boxed{v_1 = -44.57\, \text{V}} \\ \boxed{v_2 = 5.17\, \text{V}} \\ \boxed{v_3 = 1.44\, \text{V}} \] --- ## **Step 6: Power Delivered by 40V Source** Current through source is \(\frac{v_1 - 40}{3} + \frac{v_1}{1}\): \[ I = \frac{v_1 - 40}{3} + \frac{v_1}{1} \] Plug in \( v_1 = -44.57 \): \[ I = \frac{-44.57 - 40}{3} + (-44.57) = \frac{-84.57}{3} - 44.57 = -28.19 - 44.57 = -72.76\, \text{A} \] Power delivered: \[ P = V \cdot I = 40 \times (-72.76) = -2910.4\, \text{W} \] **Negative sign means power is absorbed by the circuit (delivered by the source).** --- ### **Final Answers:** **a)** \( v_1 = -44.57\,\text{V},\ v_2 = 5.17\,\text{V},\ v_3 = 1.44\,\text{V} \) **b)** **Power delivered by 40V source:** \(\boxed{2910.4\,\text{W}}\)

Certainly! Here's a detailed step-by-step solution with clear explanations for each part. --- # **Given Circuit Data** - Voltage source: \( 40\,V \) - Resistors: - \( R_1 = 3\,\Omega \) - \( R_2 = 1\,\Omega \) - \( R_3 = 40\,\Omega \) - \( R_4 = 4\,\Omega \) - \( R_5 = 2\,\Omega \) - \( R_6 = 2\,\Omega \) - Current source: \( 28\,A \) --- # **Circuit Description** The circuit consists of these elements connected as per the figure (referenced from the problem). Our goal: ### Part (a): Find **node voltages** \( v_y \), \( v_5 \), and \( v_3 \). --- # **Part (a): Use Node-Voltage Method** --- ## **Step 1: Assign Node Voltages** - Choose the **bottom node** as ground (0 V). - Define: - \( v_1 \): Voltage at the node connected to the 40V source (top left). - \( v_2 \): Voltage at the node connected after the 2Ω resistor (top right). - \( v_3 \): Voltage at the node between the resistors \( R_3 \) (40Ω), \( R_5 \) (2Ω), and \( R_6 \) (2Ω). From the diagram and the problem statement, the key nodes are: - Node \( v_1 \): connected to the 40V source and other resistors. - Node \( v_2 \): connected to the 28A current source and resistors. - Node \( v_3 \): the node at the junction of the resistors. --- ## **Step 2: Write Node Equations** ### **At Node \( v_1 \):** Connections: - To 40V source (via a resistor) - To \( v_3 \) through \( R_3 \) (40Ω) - To ground via \( R_2 \) (1Ω) - To \( v_2 \) through \( R_4 \) (4Ω) Assuming the 40V source is connected to \( v_1 \) (which is typical), the voltage at that node is: \[ \frac{v_1 - 40}{R_{source}} \quad \text{(but since it's a voltage source, } v_1 = 40\,V \text{)}. \] **Note:** The problem states the voltage source is 40V; if connected directly, then: \[ v_1 = 40\,V \] However, since the problem asks to find node voltages, we treat the source as a voltage source fixing \( v_1 \). ### **Answer for Part (a):** - \( v_1 = 40\,V \) (fixed by the source). --- ### **At Node \( v_2 \):** Connections: - To \( v_1 \) via \( R_4 = 4\,\Omega \) - To ground via \( R_5 = 2\,\Omega \) - To \( v_3 \) via \( R_6 = 2\,\Omega \) - Plus the current source of \( 28\,A \) injecting current into this node. Applying KCL at \( v_2 \): \[ \frac{v_2 - v_1}{4} + \frac{v_2 - v_3}{2} + \frac{v_2 - 0}{2} = 28 \] Plug in \( v_1 = 40\,V \): \[ \frac{v_2 - 40}{4} + \frac{v_2 - v_3}{2} + \frac{v_2}{2} = 28 \] Multiply through by 4 to clear denominators: \[ (v_2 - 40) + 2(v_2 - v_3) + 2v_2 = 112 \] Expand: \[ v_2 - 40 + 2v_2 - 2v_3 + 2v_2 = 112 \] Combine like terms: \[ (1v_2 + 2v_2 + 2v_2) - 2v_3 = 112 + 40 \] \[ 5v_2 - 2v_3 = 152 \] (Equation 1) --- ### **At Node \( v_3 \):** Connections: - To \( v_1 \) via \( R_3 = 40\,\Omega \) - To \( v_2 \) via \( R_6 = 2\,\Omega \) - To ground through \( R_5 = 2\,\Omega \) Applying KCL: \[ \frac{v_3 - v_1}{40} + \frac{v_3 - v_2}{2} + \frac{v_3 - 0}{2} = 0 \] Substitute \( v_1 = 40\,V \): \[ \frac{v_3 - 40}{40} + \frac{v_3 - v_2}{2} + \frac{v_3}{2} = 0 \] Multiply through by 40: \[ v_3 - 40 + 20(v_3 - v_2) + 20v_3 = 0 \] Expand: \[ v_3 - 40 + 20v_3 - 20v_2 + 20v_3 = 0 \] Combine: \[ (1v_3 + 20v_3 + 20v_3) - 20v_2 = 40 \] \[ 41v_3 - 20v_2 = 40 \] (Equation 2) --- ## **Step 3: Solve the Equations** From Equation 2: \[ 41v_3 = 20v_2 + 40 \] \[ v_3 = \frac{20v_2 + 40}{41} \] Plug into Equation 1: \[ 5v_2 - 2 \times \left( \frac{20v_2 + 40}{41} \right) = 152 \] Multiply both sides by 41 to clear denominator: \[ 41 \times 5v_2 - 2(20v_2 + 40) = 152 \times 41 \] Calculate: \[ 205v_2 - 40v_2 - 80 = 6232 \] Simplify: \[ (205v_2 - 40v_2) = 6232 + 80 \] \[ 165v_2 = 6312 \] \[ v_2 = \frac{6312}{165} \approx 38.24\,V \] Now, find \( v_3 \): \[ v_3 = \frac{20 \times 38.24 + 40}{41} = \frac{764.8 + 40}{41} = \frac{804.8}{41} \approx 19.63\,V \] --- ## **Final Node Voltages:** | Node | Voltage | |---------|--------------| | \( v_1 \) | **40 V** (fixed by source) | | \( v_2 \) | **38.24 V** | | \( v_3 \) | **19.63 V** | --- # **Part (a) Final Answer:** \[ \boxed{ \begin{aligned} v_y &= v_2 = \mathbf{38.24\,V} \\ v_5 &= v_3 = \mathbf{19.63\,V} \\ v_3 &= \text{(already found as } 19.63\,V) \\ \end{aligned} } \] *(Note: The problem labels may differ, but in general, these are the node voltages.)* --- # **Part (b): Power Delivered by the 40V Voltage Source** --- ## **Step 4: Determine the Current from the 40V Source** Since the source is connected to node \( v_1 \) (voltage \( 40\,V \)), and the resistor connecting it to the node (say \( R_1 \)) has value 3Ω: The current flowing **from** the source into the circuit: \[ I_{source} = \frac{v_1 - v_{next}}{R} \] But the key is: **the current leaving the source**: \[ I_{source} = \frac{40\,V - v_{node}\text{ connected to source}} \] Given that \( v_1 = 40\,V \), and considering the resistor between the source and node \( v_1 \), the current through that resistor: \[ I_{R1} = \frac{40 - v_{node}}{R} \] However, since the source is ideal and fixed at 40V, and the node voltage is also 40V, the **current** is determined by the network. Alternatively, because the source is a **voltage source**, the **power** delivered is: \[ P = V \times I \] Where \( I \) is the current **delivered by the source** into the circuit. --- ## **Step 5: Calculate the Current from the Source** From the node voltages, the **current leaving the source** (assuming the positive direction out of the source): - The source supplies current into the node \( v_1 \). - The current through the resistor \( R_1 = 3Ω \) from source to node: \[ I_{R1} = \frac{40 - v_1}{3} = \frac{40 - 40}{3} = 0\,A \] - The current through the resistor \( R_2 = 1Ω \), connected from \( v_1 \) to ground: \[ I_{R2} = \frac{v_1 - 0}{1} = 40\,A \] - Since the node voltage is fixed at 40V, the entire current must flow through \( R_2 \): \[ I_{source} = I_{R2} = 40\,A \] --- ## **Step 6: Calculate Power** Power delivered by the voltage source: \[ P = V \times I = 40\,V \times 40\,A = 1600\,W \] **Positive sign** indicates power delivered to the circuit. --- # **Final Answer:** ### **Part (a):** \[ \boxed{ \begin{aligned} v_y &= v_2 \approx 38.24\,V \\ v_5 &= v_3 \approx 19.63\,V \\ v_3 &= 19.63\,V \end{aligned} } \] ### **Part (b):** \[ \boxed{ \text{Power delivered by 40V source} = \mathbf{1600\,W} } \] --- # **Summary:** - Node voltages are approximately \( v_1 = 40\,V \), \( v_2 \approx 38.24\,V \), \( v_3 \approx 19.63\,V \). - The 40V source supplies approximately 1600W power into the circuit. --- Let me know if you'd like me to clarify any specific part further!

Certainly! Let's carefully analyze the circuit step-by-step to find the node voltages \( v_1 \), \( v_2 \), and \( v_3 \) using the node-voltage method. --- ## **Step 1: Understanding the Circuit** Based on the figure and description, the circuit components are: - A **40 V voltage source** connected to node \( v_1 \) - Resistors between nodes: - \( R_1 = 3\,\Omega \) (between \( 40\,V \) source and \( v_1 \)) - \( R_2 = 1\,\Omega \) - \( R_3 = 40\,\Omega \) - \( R_4 = 4\,\Omega \) - \( R_5 = 2\,\Omega \) - \( R_6 = 2\,\Omega \) - A **28 A current source** injecting current into the node \( v_2 \). Assuming the bottom node is ground (0 V). --- ## **Step 2: Assign Node Voltages** - The node connected directly to the 40 V source is \( v_1 \). Since it's connected directly to the voltage source, **we know**: \[ v_1 = 40\,V \] - Other nodes: - \( v_2 \): the node connected after some resistors, connected to the current source and other resistors. - \( v_3 \): the node at the junction of resistors \( R_3, R_5, R_6 \). --- ## **Step 3: Write Node Equations** ### **Node \( v_2 \):** - Connected to \( v_1 \) via \( R_4 = 4\,\Omega \) - Connected to ground via \( R_2 = 1\,\Omega \) - Connected to \( v_3 \) via \( R_6 = 2\,\Omega \) - Has a **current source** \( 28\,A \) injecting into it. Applying **KCL** at \( v_2 \): \[ \frac{v_2 - v_1}{4} + \frac{v_2 - 0}{1} + \frac{v_2 - v_3}{2} = 28 \] Plug in \( v_1 = 40\,V \): \[ \frac{v_2 - 40}{4} + v_2 + \frac{v_2 - v_3}{2} = 28 \] Multiply through by 4 to clear denominators: \[ (v_2 - 40) + 4 v_2 + 2 (v_2 - v_3) = 112 \] Expand: \[ v_2 - 40 + 4 v_2 + 2 v_2 - 2 v_3 = 112 \] Combine like terms: \[ (1 v_2 + 4 v_2 + 2 v_2) - 2 v_3 = 112 + 40 \] \[ 7 v_2 - 2 v_3 = 152 \] --- ### **Node \( v_3 \):** - Connected to \( v_1 \) via \( R_3 = 40\,\Omega \) - Connected to \( v_2 \) via \( R_6 = 2\,\Omega \) - Connected to ground via \( R_5 = 2\,\Omega \) Apply **KCL** at \( v_3 \): \[ \frac{v_3 - v_1}{40} + \frac{v_3 - v_2}{2} + \frac{v_3 - 0}{2} = 0 \] Substitute \( v_1 = 40\,V \): \[ \frac{v_3 - 40}{40} + \frac{v_3 - v_2}{2} + \frac{v_3}{2} = 0 \] Multiply through by 40: \[ v_3 - 40 + 20 (v_3 - v_2) + 20 v_3 = 0 \] Expand: \[ v_3 - 40 + 20 v_3 - 20 v_2 + 20 v_3 = 0 \] Combine \( v_3 \) terms: \[ (1 v_3 + 20 v_3 + 20 v_3) - 20 v_2 = 40 \] \[ 41 v_3 - 20 v_2 = 40 \] --- ## **Step 4: Solve the System of Equations** We have: \[ \begin{cases} 7 v_2 - 2 v_3 = 152 \quad \text{(Equation 1)} \\ 41 v_3 - 20 v_2 = 40 \quad \text{(Equation 2)} \end{cases} \] Let's express \( v_3 \) from Equation 1: \[ 7 v_2 - 2 v_3 = 152 \implies 2 v_3 = 7 v_2 - 152 \implies v_3 = \frac{7 v_2 - 152}{2} \] Substitute into Equation 2: \[ 41 \left( \frac{7 v_2 - 152}{2} \right) - 20 v_2 = 40 \] Multiply through by 2: \[ 41 (7 v_2 - 152) - 40 v_2 = 80 \] Expand: \[ 287 v_2 - 41 \times 152 - 40 v_2 = 80 \] Calculate \( 41 \times 152 \): \[ 41 \times 152 = 6232 \] So: \[ 287 v_2 - 6232 - 40 v_2 = 80 \] Combine like terms: \[ (287 v_2 - 40 v_2) = 80 + 6232 \] \[ 247 v_2 = 6312 \] Solve for \( v_2 \): \[ v_2 = \frac{6312}{247} \approx 25.56\,V \] Now, find \( v_3 \): \[ v_3 = \frac{7 \times 25.56 - 152}{2} = \frac{178.92 - 152}{2} = \frac{26.92}{2} = 13.46\,V \] --- ## **Step 5: Find \( v_1 \)** Given the source voltage: \[ v_1 = 40\,V \] --- ## **Final Node Voltages** | Node | Voltage (V) | |---------|--------------| | \( v_1 \) | **40.00** (fixed) | | \( v_2 \) | **25.56** | | \( v_3 \) | **13.46** | --- # **Summary** **Using node-voltage method, the node voltages are:** \[ \boxed{ \begin{aligned} v_1 &= 40\,V \quad \text{(by source)} \\ v_2 &\approx 25.56\,V \\ v_3 &\approx 13.46\,V \end{aligned} } \] --- If you need further clarification on any step, feel free to ask!