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Compute the equivalent impedance of the circuit below when the frequency is 20kHz: 10mH 602 202 30μF 402 ee 10mH

Compute the equivalent impedance of the circuit below when the frequency is 20kHz: 10mH 602 202 30μF 402 ee 10mH

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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**Title: Calculation of Equivalent Impedance at 20kHz** **Objective:** Compute the equivalent impedance of the given circuit when the frequency is 20 kHz. **Diagram Explanation:** The circuit consists of the following elements: 1. A 60Ω resistor connected in series at the left side of the circuit. 2. A combination of components arranged in a more complex structure: - Two inductors (10mH each) connected in parallel. - A 40Ω resistor connected in parallel with the inductors. 3. This parallel combination is then followed by a 20Ω resistor and a 30µF capacitor connected in series with each other and the combination. **Detailed Steps and Calculations:** Given elements and values: - Resistor: \( R_1 = 60Ω \) - Inductors: \( L_1 = 10mH \), \( L_2 = 10mH \) - Resistor in parallel: \( R_2 = 40Ω \) - Serial elements: \( R_3 = 20Ω \) - Capacitor: \( C = 30µF \) - Frequency: \( f = 20kHz \) The angular frequency \( ω = 2πf \): \[ ω = 2π \times 20000 ≈ 1.2566 \times 10^5 \ rad/s \] Inductive reactance: \[ X_L = ωL = 1.2566 \times 10^5 \times 10 \times 10^{-3} = 1.2566 \times 10^3 Ω = 1256.6Ω \] Capacitive reactance: \[ X_C = \frac{1}{ωC} = \frac{1}{1.2566 \times 10^5 \times 30 \times 10^{-6}} ≈ 26.5258Ω \] Parallel Impedance Calculation (for inductors and resistor \( R_2 \)): - Inductor impedance \( Z_{L1} = 1256.6Ω \) - Inductor impedance \( Z_{L2} = 1256.6Ω \) - Resistor \( R_2 = 40Ω \) Parallel combination: \[ \frac{1}{Z_{parallel}} = \frac{1}{R_{2}} + \frac{1}{Z_{L
Transcribed Image Text:**Title: Calculation of Equivalent Impedance at 20kHz** **Objective:** Compute the equivalent impedance of the given circuit when the frequency is 20 kHz. **Diagram Explanation:** The circuit consists of the following elements: 1. A 60Ω resistor connected in series at the left side of the circuit. 2. A combination of components arranged in a more complex structure: - Two inductors (10mH each) connected in parallel. - A 40Ω resistor connected in parallel with the inductors. 3. This parallel combination is then followed by a 20Ω resistor and a 30µF capacitor connected in series with each other and the combination. **Detailed Steps and Calculations:** Given elements and values: - Resistor: \( R_1 = 60Ω \) - Inductors: \( L_1 = 10mH \), \( L_2 = 10mH \) - Resistor in parallel: \( R_2 = 40Ω \) - Serial elements: \( R_3 = 20Ω \) - Capacitor: \( C = 30µF \) - Frequency: \( f = 20kHz \) The angular frequency \( ω = 2πf \): \[ ω = 2π \times 20000 ≈ 1.2566 \times 10^5 \ rad/s \] Inductive reactance: \[ X_L = ωL = 1.2566 \times 10^5 \times 10 \times 10^{-3} = 1.2566 \times 10^3 Ω = 1256.6Ω \] Capacitive reactance: \[ X_C = \frac{1}{ωC} = \frac{1}{1.2566 \times 10^5 \times 30 \times 10^{-6}} ≈ 26.5258Ω \] Parallel Impedance Calculation (for inductors and resistor \( R_2 \)): - Inductor impedance \( Z_{L1} = 1256.6Ω \) - Inductor impedance \( Z_{L2} = 1256.6Ω \) - Resistor \( R_2 = 40Ω \) Parallel combination: \[ \frac{1}{Z_{parallel}} = \frac{1}{R_{2}} + \frac{1}{Z_{L
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