Let's solve **step by step** using the **node-voltage method**. --- ## **Step 1: Label and Assign Node Voltages** Let’s assign: - \( v_1 \): Top left node - \( v_2 \): Top right node - \( v_3 \): Bottom middle node (reference node, \( v_3 = 0 \)) - The bottom of the voltage source and all bottom wires are ground. --- ## **Step 2: Write Node Equations** ### **Node 1** (\( v_1 \)): Connected to: - 40V source (left) - 3Ω (top to \( v_2 \)) - 1Ω (bottom to ground) - 4Ω (middle to \( v_3 \)) #### **KCL at \( v_1 \):** \[ \frac{v_1 - 40}{1} + \frac{v_1 - v_2}{3} + \frac{v_1 - v_3}{4} = 0 \] But \( v_3 = 0 \): \[ \frac{v_1 - 40}{1} + \frac{v_1 - v_2}{3} + \frac{v_1}{4} = 0 \] --- ### **Node 2** (\( v_2 \)): Connected to: - 3Ω (left to \( v_1 \)) - 2Ω (top to ground) - 4Ω (bottom to \( v_3 \)) - 28A current source (to ground) #### **KCL at \( v_2 \):** \[ \frac{v_2 - v_1}{3} + \frac{v_2}{2} + \frac{v_2 - v_3}{4} = 28 \] (28A is *leaving* the node) With \( v_3 = 0 \): \[ \frac{v_2 - v_1}{3} + \frac{v_2}{2} + \frac{v_2}{4} = 28 \] --- ### **Node 3** (\( v_3 \)): Let’s check if this node is necessary. \( v_3 \) is connected to: - 4Ω (to \( v_1 \)) - 4Ω (to \( v_2 \)) - 2Ω (to ground) But since \( v_3 \) is ground (\( 0V \)), all connections resolve to ground potential. --- ## **Step 3: Simplify Equations** ### **Equation 1:** \[ \frac{v_1 - 40}{1} + \frac{v_1 - v_2}{3} + \frac{v_1}{4} = 0 \] Find common denominator (12): \[ 12(v_1 - 40) + 4(v_1 - v_2) + 3v_1 = 0 \] \[ 12v_1 - 480 + 4v_1 - 4v_2 + 3v_1 = 0 \] \[ (12v_1 + 4v_1 + 3v_1) - 480 - 4v_2 = 0 \] \[ 19v_1 - 4v_2 = 480 \] \[ 19v_1 - 4v_2 = 480 \] --- ### **Equation 2:** \[ \frac{v_2 - v_1}{3} + \frac{v_2}{2} + \frac{v_2}{4} = 28 \] Find common denominator (12): \[ 4(v_2 - v_1) + 6v_2 + 3v_2 = 28 \times 12 \] \[ 4v_2 - 4v_1 + 6v_2 + 3v_2 = 336 \] \[ (4v_2 + 6v_2 + 3v_2) - 4v_1 = 336 \] \[ 13v_2 - 4v_1 = 336 \] --- ## **Step 4: Solve the System of Equations** We have: 1. \( 19v_1 - 4v_2 = 480 \) 2. \( -4v_1 + 13v_2 = 336 \) Let's solve for \( v_1 \) and \( v_2 \): Multiply equation 1 by 13: \[ 13(19v_1 - 4v_2) = 13 \times 480 \] \[ 247v_1 - 52v_2 = 6240 \] Multiply equation 2 by 4: \[ 4(-4v_1 + 13v_2) = 4 \times 336 \] \[ -16v_1 + 52v_2 = 1344 \] Now, add the two equations: \[ 247v_1 - 52v_2 \\ + (-16v_1 + 52v_2) = 6240 + 1344 \] \[ (247 - 16)v_1 = 7584 \] \[ 231v_1 = 7584 \] \[ v_1 = \frac{7584}{231} = 32.83 \text{ V} \] Now substitute \( v_1 \) into equation 1: \[ 19v_1 - 4v_2 = 480 \\ 19 \times 32.83 - 4v_2 = 480 \\ 623.77 - 4v_2 = 480 \\ -4v_2 = 480 - 623.77 \\ -4v_2 = -143.77 \\ v_2 = \frac{-143.77}{-4} = 35.94 \text{ V} \] \( v_3 = 0 \) V (reference node). --- ## **Step 5: Power Delivered by the 40V Source** Current through the voltage source is: \[ I = \frac{40 - v_1}{1} = \frac{40 - 32.83}{1} = 7.17 \text{ A} \] So, power delivered: \[ P = V \times I = 40 \times 7.17 = 287 \text{ W} \] --- ## **Final Answers** ### **a. Node Voltages** \[ \boxed{ v_1 = 32.8~\text{V}, \quad v_2 = 35.9~\text{V}, \quad v_3 = 0~\text{V} } \] ### **b. Power Delivered** \[ \boxed{ P = 287~\text{W} } \] --- **If you need full calculation steps or want to check the answer, let me know!**

Certainly! Let's go through the detailed step-by-step solution for the circuit using the node-voltage method, including all intermediate calculations and explanations. --- ## **Given Circuit Overview:** - Voltage source: 40 V - Resistors: 1Ω, 2Ω, 3Ω, 4Ω - Current source: 28 A - Nodes: We'll label nodes as \( v_1 \) (top left), \( v_2 \) (top right), and ground (bottom node, \( v_3 = 0 \) V). --- ## **Step 1: Assign Node Voltages** - \( v_1 \): Voltage at the top left node (relative to ground) - \( v_2 \): Voltage at the top right node (relative to ground) - \( v_3 = 0 \) V (ground node) --- ## **Step 2: Write Node Equations (KCL)** ### **Node \( v_1 \):** Connections: - To the 40 V source (via a 1Ω resistor) - To \( v_2 \) through 3Ω resistor - To ground through 1Ω resistor - To ground through 4Ω resistor (via \( v_3 \)) - The 40 V source sets \( v_1 \) relative to ground. Since the 40 V source is connected directly to the node, **it's more straightforward to treat the voltage source as a supernode** encompassing the source and the node \( v_1 \). --- ### **Supernode Approach:** - The 40 V source connects between a known voltage (40 V) and node \( v_1 \). - Because the source is between a fixed voltage (40 V) and \( v_1 \), the voltage at \( v_1 \) is fixed **if the source is ideal and connected directly**. **But in the given circuit diagram:** - The 40 V source appears to be connected between the node \( v_1 \) and the 40 V reference (ground). - The node \( v_1 \) is therefore at 40 V **minus any voltage drop across the resistor**? - Or is the source connected between node \( v_1 \) and a 40 V terminal? **Assuming** the 40 V source is connected between the node \( v_1 \) and ground (or a reference point), then: \[ v_1 = 40 \text{ V} \] **However**, the initial diagram suggests the 40 V source is connected to node \( v_1 \) and ground, so: \[ v_1 = 40 \text{ V} \] --- ## **Important Correction:** **Given the circuit diagram,** the 40 V source is directly connected to node \( v_1 \), so: \[ v_1 = 40 \text{ V} \] This simplifies our problem significantly because \( v_1 \) is known, and we only need to find \( v_2 \). --- ## **Step 3: Simplify circuit with known \( v_1 \)** - \( v_1 = 40 \text{ V} \) Now, focus on node \( v_2 \). --- ## **Step 4: Write KCL at node \( v_2 \):** Connections: - To \( v_1 \) via 3Ω resistor - To ground via 2Ω resistor - To \( v_3 \) (ground) via 4Ω resistor - Current source of 28 A (assumed leaving the node) **Note:** The current source of 28 A is connected **from node \( v_2 \) to ground**, so: \[ \text{Current leaving } v_2 = 28\,A \] **Applying KCL at \( v_2 \):** \[ \frac{v_2 - v_1}{3} + \frac{v_2 - 0}{2} + \frac{v_2 - 0}{4} = 28 \] Plug in \( v_1 = 40V \): \[ \frac{v_2 - 40}{3} + \frac{v_2}{2} + \frac{v_2}{4} = 28 \] --- ## **Step 5: Solve for \( v_2 \)** **Combine the fractions:** - First term: \( \frac{v_2 - 40}{3} \) - Second term: \( \frac{v_2}{2} \) - Third term: \( \frac{v_2}{4} \) Find common denominator: 12 \[ \frac{4(v_2 - 40)}{12} + \frac{6 v_2}{12} + \frac{3 v_2}{12} = 28 \] Simplify numerator: \[ 4v_2 - 160 + 6v_2 + 3v_2 = 28 \times 12 \] \[ (4v_2 + 6v_2 + 3v_2) - 160 = 336 \] \[ 13v_2 - 160 = 336 \] Solve for \( v_2 \): \[ 13v_2 = 336 + 160 = 496 \] \[ v_2 = \frac{496}{13} \approx 38.15 \text{ V} \] --- ## **Step 6: Summary of node voltages** - \( v_1 = 40\,V \) - \( v_2 \approx 38.15\,V \) - \( v_3 = 0\,V \) (ground) --- ## **Step 7: Calculate the current from the 40 V source** Since the 40 V source is connected to node \( v_1 \), and the resistor between it and the node is 1Ω, the current flowing **from the source into the circuit**: \[ I_{source} = \frac{40\,V - v_1}{1\,Ω} = \frac{40 - 40}{1} = 0\,A \] **But this suggests no current flows through this resistor**, which makes sense because the node voltage matches the source voltage. **However,** the current supplied by the source is determined by the resistors connected to node \( v_1 \). Let's check the actual circuit: - The 40 V source maintains \( v_1 = 40\,V \). - The current from the source through the 1Ω resistor is zero because \( v_1 \) is at 40 V, matching the source voltage. --- ## **Step 8: Power delivered by the 40 V source** Since **no current flows from the source** (as \( v_1 = 40\,V \)), the power delivered is: \[ P = V \times I \] But \( I = 0\,A \), so: \[ \boxed{ \text{Power} = 0\,W } \] **This indicates** the source is not delivering any power because the node voltage is at the source voltage. --- ## **Final Results:** | Quantity | Value | |------------|---------| | \( v_1 \) | **40 V** | | \( v_2 \) | **38.15 V** | | Power delivered by the 40 V source | **0 W** | --- ## **Summary:** - **Node \( v_1 \):** 40 V (fixed by the voltage source) - **Node \( v_2 \):** approximately 38.15 V - **Power delivered by the source:** zero watts, because the voltage at the source matches the node voltage, resulting in zero current flow. --- **If the circuit or assumptions differ, or if you want to verify other quantities (like currents through resistors or power in other elements), please specify!**