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Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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**Problem 1:** ### Circuit Diagram: The circuit consists of three voltage nodes (V1, V2, V3) connected with resistors and voltage sources. The resistors have values of 400Ω, 200Ω, and 500Ω, and there is a current source of 100 mA. Node 0 is grounded. ### Node-Voltage Method: 1. **Identify Nodes:** - V1, V2, and V3 are the voltage nodes. We calculate currents entering and exiting each node. 2. **Calculate Current Equations for Each Node:** - **Node V1:** - V1 = 10V (connected to the voltage source, no currents calculated). - **Node V2:** Use Ohm’s law (I = V/R) and KCL: \[ \frac{V2 - V1}{400} + \frac{V2 - V3}{500} + \frac{V2 - 10}{200} = 0 \] 3. **Node 3 (V3):** - Equation: \[ \frac{V3 - V2}{500} + \frac{V3 - 0}{100} + (-0.1) = 0 \] - The (-0.1) term is the branch current for the 100 mA source. 3. **Solve Linear Equations:** - Known V1 = 10V. Two unknowns (V2 and V3) and two equations. 4. **Calculate Required Parameter:** - Current across the 200Ω resistor: \[ \frac{V2 - 10}{200} \] ### Mesh Analysis: Using KVL to calculate currents in closed loops: - **Circuit with Currents i1, i2, i3:** - Mesh currents are assumed in clockwise directions. This approach utilizes algebraic techniques to determine unknown voltages and currents using the principles of nodal analysis and Kirchhoff's voltage law (KVL).
Transcribed Image Text:**Problem 1:** ### Circuit Diagram: The circuit consists of three voltage nodes (V1, V2, V3) connected with resistors and voltage sources. The resistors have values of 400Ω, 200Ω, and 500Ω, and there is a current source of 100 mA. Node 0 is grounded. ### Node-Voltage Method: 1. **Identify Nodes:** - V1, V2, and V3 are the voltage nodes. We calculate currents entering and exiting each node. 2. **Calculate Current Equations for Each Node:** - **Node V1:** - V1 = 10V (connected to the voltage source, no currents calculated). - **Node V2:** Use Ohm’s law (I = V/R) and KCL: \[ \frac{V2 - V1}{400} + \frac{V2 - V3}{500} + \frac{V2 - 10}{200} = 0 \] 3. **Node 3 (V3):** - Equation: \[ \frac{V3 - V2}{500} + \frac{V3 - 0}{100} + (-0.1) = 0 \] - The (-0.1) term is the branch current for the 100 mA source. 3. **Solve Linear Equations:** - Known V1 = 10V. Two unknowns (V2 and V3) and two equations. 4. **Calculate Required Parameter:** - Current across the 200Ω resistor: \[ \frac{V2 - 10}{200} \] ### Mesh Analysis: Using KVL to calculate currents in closed loops: - **Circuit with Currents i1, i2, i3:** - Mesh currents are assumed in clockwise directions. This approach utilizes algebraic techniques to determine unknown voltages and currents using the principles of nodal analysis and Kirchhoff's voltage law (KVL).
## Analysis of Electrical Circuit with Kirchhoff's Voltage Law (KVL) In this example, we consider a circuit with three closed loops where currents are flowing: \(i_1, i_2,\) and \(i_3\). The assumed direction for all currents is clockwise, which simplifies the formulation of the equations. By applying Kirchhoff's Voltage Law (KVL), we acknowledge that the sum of voltages in a closed loop is always zero. ### Circuit Diagram Explanation The circuit consists of: - A voltage source of 10V on the left. - Resistors with values of 400Ω, 200Ω, 500Ω, and 100Ω. - A current source providing 100mA to the loop on the right. ### KVL Equations #### First Loop with Current \(i_1\) \[ -10 + 400i_1 + 200(i_1 - i_2) + 10 = 0 \] - Here, \(-10\) corresponds to the left voltage source, indicating current flow from negative to positive. - The \(+10\) at the end is from the right voltage source, showing flow from positive to negative. #### Second Loop with Current \(i_2\) \[ -10 + 200(i_2 - i_1) + 500i_2 + 100(i_2 - i_3) = 0 \] #### Third Loop with Current \(i_3\) \[ i_3 = -0.1\, \text{A} \] - This shows that the current from the 100mA source flows opposite to the assumed \(i_3\). ### Solution Strategy There are two unknowns, \(i_1\) and \(i_2\), and two equations that can be solved linearly. - The current across the 200Ω resistor is \(i_1 - i_2\) when flowing downward. If specified otherwise, it is \((i_2 - i_1)\). This instructional piece aids in understanding how to apply KVL to determine currents in a circuit. It provides foundational skills in circuit analysis, crucial for electronics and electrical engineering studies.
Transcribed Image Text:## Analysis of Electrical Circuit with Kirchhoff's Voltage Law (KVL) In this example, we consider a circuit with three closed loops where currents are flowing: \(i_1, i_2,\) and \(i_3\). The assumed direction for all currents is clockwise, which simplifies the formulation of the equations. By applying Kirchhoff's Voltage Law (KVL), we acknowledge that the sum of voltages in a closed loop is always zero. ### Circuit Diagram Explanation The circuit consists of: - A voltage source of 10V on the left. - Resistors with values of 400Ω, 200Ω, 500Ω, and 100Ω. - A current source providing 100mA to the loop on the right. ### KVL Equations #### First Loop with Current \(i_1\) \[ -10 + 400i_1 + 200(i_1 - i_2) + 10 = 0 \] - Here, \(-10\) corresponds to the left voltage source, indicating current flow from negative to positive. - The \(+10\) at the end is from the right voltage source, showing flow from positive to negative. #### Second Loop with Current \(i_2\) \[ -10 + 200(i_2 - i_1) + 500i_2 + 100(i_2 - i_3) = 0 \] #### Third Loop with Current \(i_3\) \[ i_3 = -0.1\, \text{A} \] - This shows that the current from the 100mA source flows opposite to the assumed \(i_3\). ### Solution Strategy There are two unknowns, \(i_1\) and \(i_2\), and two equations that can be solved linearly. - The current across the 200Ω resistor is \(i_1 - i_2\) when flowing downward. If specified otherwise, it is \((i_2 - i_1)\). This instructional piece aids in understanding how to apply KVL to determine currents in a circuit. It provides foundational skills in circuit analysis, crucial for electronics and electrical engineering studies.
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