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Calculate the power factor of the er source. What is the average power supplied by the 102 4020° Vrms j42

Calculate the power factor of the er source. What is the average power supplied by the 102 4020° Vrms j42

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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### Calculate the Power Factor and Average Power Supplied by the Source **Problem Statement:** Calculate the power factor of the entire circuit below as seen by the source. What is the average power supplied by the source? ![Circuit Diagram] - **Source Voltage:** \(40 \angle 0^\circ V_{rms}\) - **Components in Series:** - Resistor \(R_1 = 10\Omega\) - Resistor \(R_2 = 8\Omega\) - **Components in Parallel:** - Inductor with reactance \(j4\Omega\) - Capacitor with reactance \(-j6\Omega\) ### Solution Steps 1. **Calculate the Total Impedance (Z):** - Find the impedance of the parallel combination of \(j4\Omega\) and \(-j6\Omega\). \[ \frac{1}{Z_{\text{parallel}}} = \frac{1}{j4} + \frac{1}{-j6} \] \[ \frac{1}{Z_{\text{parallel}}} = \frac{-j6 + j4}{j(-j6 \cdot j4)} = \frac{-j2}{-24} = \frac{j2}{24} = j \frac{1}{12} \] So, \[ Z_{\text{parallel}} = -j12\Omega \] 2. **Total Series Impedance:** \[ Z_{\text{total}} = 10\Omega + 8\Omega + (-j12\Omega) = 18 - j12 \Omega \] 3. **Calculate the Magnitude of the Total Impedance:** \[ |Z_{\text{total}}| = \sqrt{18^2 + (-12)^2} = \sqrt{324 + 144} = \sqrt{468} = 21.63\Omega \] 4. **Power Factor Calculation:** \[ \text{Power Factor} = \cos(\theta) = \frac{R_{\text{total}}}{|Z_{\text{total}}|} = \frac{18}{21.63} = 0.832 \] 5. **Calculate the Average Power Supplied by the Source:**
Transcribed Image Text:### Calculate the Power Factor and Average Power Supplied by the Source **Problem Statement:** Calculate the power factor of the entire circuit below as seen by the source. What is the average power supplied by the source? ![Circuit Diagram] - **Source Voltage:** \(40 \angle 0^\circ V_{rms}\) - **Components in Series:** - Resistor \(R_1 = 10\Omega\) - Resistor \(R_2 = 8\Omega\) - **Components in Parallel:** - Inductor with reactance \(j4\Omega\) - Capacitor with reactance \(-j6\Omega\) ### Solution Steps 1. **Calculate the Total Impedance (Z):** - Find the impedance of the parallel combination of \(j4\Omega\) and \(-j6\Omega\). \[ \frac{1}{Z_{\text{parallel}}} = \frac{1}{j4} + \frac{1}{-j6} \] \[ \frac{1}{Z_{\text{parallel}}} = \frac{-j6 + j4}{j(-j6 \cdot j4)} = \frac{-j2}{-24} = \frac{j2}{24} = j \frac{1}{12} \] So, \[ Z_{\text{parallel}} = -j12\Omega \] 2. **Total Series Impedance:** \[ Z_{\text{total}} = 10\Omega + 8\Omega + (-j12\Omega) = 18 - j12 \Omega \] 3. **Calculate the Magnitude of the Total Impedance:** \[ |Z_{\text{total}}| = \sqrt{18^2 + (-12)^2} = \sqrt{324 + 144} = \sqrt{468} = 21.63\Omega \] 4. **Power Factor Calculation:** \[ \text{Power Factor} = \cos(\theta) = \frac{R_{\text{total}}}{|Z_{\text{total}}|} = \frac{18}{21.63} = 0.832 \] 5. **Calculate the Average Power Supplied by the Source:**
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