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5 Find the steady-state expression for v, in the circuit = = 200 cos 5000t mA. of Fig. P9.26 if i g Figure P9.26 Y ig + 240 Ω Σ 0 Ω . - 2.5 μF 80 ΩΣ 48 mH

5 Find the steady-state expression for v, in the circuit = = 200 cos 5000t mA. of Fig. P9.26 if i g Figure P9.26 Y ig + 240 Ω Σ 0 Ω . - 2.5 μF 80 ΩΣ 48 mH

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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Circuits 1 HW 6 Q9
### Educational Resource: Steady-State Analysis in AC Circuits **Problem 9.26:** Find the steady-state expression for \( v_o \) in the circuit of Fig. P9.26 if \( i_g = 200 \cos 5000t \) mA. **Figure P9.26 Description:** The diagram illustrates an electrical circuit consisting of the following components: 1. **Current Source (\(i_g\)):** Provides a current of \( i_g = 200 \cos 5000t \) mA. 2. **Resistor (240 \(\Omega\)):** Positioned immediately after the current source. 3. **Capacitor (2.5 \(\mu\)F):** Connected in parallel, below the 240 \(\Omega\) resistor, with voltage across it labeled as \( v_o \), with the positive terminal at the top and negative at the bottom. 4. **Resistor (80 \(\Omega\)):** Positioned to the right of the capacitor, in series. 5. **Inductor (48 mH):** Connected in series with both the 80 \(\Omega\) resistor and the capacitor. **Objective:** Determine the steady-state voltage expression \( v_o \) across the capacitor given the input current source \( i_g \). **Approach:** To find \( v_o \), use techniques such as: - **Phasor analysis** to convert the time-domain equations into the frequency domain. - **Impedance calculations** for each circuit element: - Resistor: \( Z_R = R \) - Capacitor: \( Z_C = \frac{1}{j\omega C} \) - Inductor: \( Z_L = j\omega L \) - **Use of Ohm’s Law** and **Kirchhoff’s Voltage Law** to analyze phasor circuits and solve for the voltage across the capacitor. This problem involves understanding AC circuit analysis, focusing on impedances and phasor relationships in a simple RLC circuit.
Transcribed Image Text:### Educational Resource: Steady-State Analysis in AC Circuits **Problem 9.26:** Find the steady-state expression for \( v_o \) in the circuit of Fig. P9.26 if \( i_g = 200 \cos 5000t \) mA. **Figure P9.26 Description:** The diagram illustrates an electrical circuit consisting of the following components: 1. **Current Source (\(i_g\)):** Provides a current of \( i_g = 200 \cos 5000t \) mA. 2. **Resistor (240 \(\Omega\)):** Positioned immediately after the current source. 3. **Capacitor (2.5 \(\mu\)F):** Connected in parallel, below the 240 \(\Omega\) resistor, with voltage across it labeled as \( v_o \), with the positive terminal at the top and negative at the bottom. 4. **Resistor (80 \(\Omega\)):** Positioned to the right of the capacitor, in series. 5. **Inductor (48 mH):** Connected in series with both the 80 \(\Omega\) resistor and the capacitor. **Objective:** Determine the steady-state voltage expression \( v_o \) across the capacitor given the input current source \( i_g \). **Approach:** To find \( v_o \), use techniques such as: - **Phasor analysis** to convert the time-domain equations into the frequency domain. - **Impedance calculations** for each circuit element: - Resistor: \( Z_R = R \) - Capacitor: \( Z_C = \frac{1}{j\omega C} \) - Inductor: \( Z_L = j\omega L \) - **Use of Ohm’s Law** and **Kirchhoff’s Voltage Law** to analyze phasor circuits and solve for the voltage across the capacitor. This problem involves understanding AC circuit analysis, focusing on impedances and phasor relationships in a simple RLC circuit.
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