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Determine the input impedance and an equivalent circuit model for the following network at 50 kHz. Zin 100 Ω 100 ΜΗ 10 nF

Determine the input impedance and an equivalent circuit model for the following network at 50 kHz. Zin 100 Ω 100 ΜΗ 10 nF

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
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**Determine the Input Impedance and Equivalent Circuit Model at 50 kHz** ### Problem Statement: Determine the input impedance and an equivalent circuit model for the following network at 50 kHz. ### Circuit Description: The given circuit consists of three components connected in series: - A resistor of 100 ohms (Ω) - An inductor of 100 microhenries (μH) - A capacitor of 10 nanofarads (nF) These components are connected from the top to the bottom, and the input impedance (\( Z_{in} \)) is measured across the entire series combination. ### Explanation: To determine the input impedance (\( Z_{in} \)) at a specific frequency (50 kHz), we need to consider the impedance contributions of each component: 1. **Resistor (R):** - The impedance of a resistor at any frequency is purely real and given by the resistance value \( R = 100 \Omega \). 2. **Inductor (L):** - The impedance of an inductor is purely imaginary and given by \( j \omega L \), where \( \omega \) is the angular frequency and \( L \) is the inductance. - Here, \( \omega = 2 \pi f \) and \( f = 50 \text{kHz} = 50,000 \text{Hz} \). - Therefore, \( \omega = 2 \pi \times 50,000 \approx 314,159 \text{rad/s} \). - The impedance of the inductor is \( j \times 314,159 \text{rad/s} \times 100 \mu H \). - Converting \( 100 \mu H \) to Henry (H): \( 100 \mu H = 100 \times 10^{-6} H \). - So, the inductor impedance is \( j 31.416 \Omega \). 3. **Capacitor (C):** - The impedance of a capacitor is purely imaginary and given by \( \frac{1}{j \omega C} \), where \( C \) is the capacitance. - The capacitance \( C = 10 nF = 10 \times 10^{-9} \text{F} \). - The impedance is \( \frac{1}{j \times 314,159 \text{
Transcribed Image Text:**Determine the Input Impedance and Equivalent Circuit Model at 50 kHz** ### Problem Statement: Determine the input impedance and an equivalent circuit model for the following network at 50 kHz. ### Circuit Description: The given circuit consists of three components connected in series: - A resistor of 100 ohms (Ω) - An inductor of 100 microhenries (μH) - A capacitor of 10 nanofarads (nF) These components are connected from the top to the bottom, and the input impedance (\( Z_{in} \)) is measured across the entire series combination. ### Explanation: To determine the input impedance (\( Z_{in} \)) at a specific frequency (50 kHz), we need to consider the impedance contributions of each component: 1. **Resistor (R):** - The impedance of a resistor at any frequency is purely real and given by the resistance value \( R = 100 \Omega \). 2. **Inductor (L):** - The impedance of an inductor is purely imaginary and given by \( j \omega L \), where \( \omega \) is the angular frequency and \( L \) is the inductance. - Here, \( \omega = 2 \pi f \) and \( f = 50 \text{kHz} = 50,000 \text{Hz} \). - Therefore, \( \omega = 2 \pi \times 50,000 \approx 314,159 \text{rad/s} \). - The impedance of the inductor is \( j \times 314,159 \text{rad/s} \times 100 \mu H \). - Converting \( 100 \mu H \) to Henry (H): \( 100 \mu H = 100 \times 10^{-6} H \). - So, the inductor impedance is \( j 31.416 \Omega \). 3. **Capacitor (C):** - The impedance of a capacitor is purely imaginary and given by \( \frac{1}{j \omega C} \), where \( C \) is the capacitance. - The capacitance \( C = 10 nF = 10 \times 10^{-9} \text{F} \). - The impedance is \( \frac{1}{j \times 314,159 \text{
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