Vs L C R The switch is closed at t = 0. a) Derive a differential equation in the voltage across the cap for t≥ 0, with the source V. b) Find the resistance R such that when the switch is closed t the voltage across the capacitor oscillates at a frequency w 250 TT.
Vs L C R The switch is closed at t = 0. a) Derive a differential equation in the voltage across the cap for t≥ 0, with the source V. b) Find the resistance R such that when the switch is closed t the voltage across the capacitor oscillates at a frequency w 250 TT.
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![## RC Circuit Analysis
### Circuit Description
This section describes an RC (Resistor-Capacitor) circuit under study, illustrated in the accompanying diagram. The circuit components are as follows:
- **Capacitance (C):** 1 µF (microfarad)
- **Inductance (L):** 1 H (henry)
- **Source Voltage (\(V_s\)):** 10 V (volts)
- **Switch**: Closed at \( t = 0 \)
### Diagram Explanation
The diagram depicts a simple series circuit comprising the following components:
- A DC voltage source (\(V_s\))
- An inductor (L)
- A capacitor (C)
- A resistor (R)
The circuit is shown with the switch in the closed position, initiating the study at time \( t = 0 \).
### Problem Statements
**(a)** Derive a differential equation for the voltage across the capacitor (\(V_C\)) for \( t \geq 0 \), with the source voltage \(V_s\).
**(b)** Determine the resistance (R) such that the voltage across the capacitor oscillates at a frequency \( \omega = 250 \pi \) rad/s.
### Analysis Approach
**(a) Differential Equation Derivation**
To derive the differential equation for the voltage across the capacitor \(V_C\):
1. Apply Kirchhoff's Voltage Law (KVL) around the loop.
2. Utilize the relationships for the inductor and capacitor:
- Voltage across the inductor: \(V_L = L \frac{dI}{dt}\)
- Voltage across the capacitor: \(V_C = \frac{1}{C} \int I dt \)
3. Combine these relationships and the voltage source to form the differential equation.
**(b) Finding Resistance for Desired Frequency**
Given the desired angular frequency \( \omega = 250 \pi \), use the characteristic equation of the resulting second-order differential equation from part (a) to determine the resistance \(R\):
\[ \omega_0 = \sqrt{\frac{1}{LC}} \]
\[ \beta = \frac{R}{2L} \]
\[ \omega = \sqrt{\omega_0^2 - \beta^2} \]
By solving for \(R\), ensure that the specific frequency condition is met.
### Summary
This analysis requires applying fundamental electrical engineering principles to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9addf70b-7b61-4f39-8ae4-a2b3648accb6%2Fce89abda-ed1b-4cca-bd0e-f034df6fb1cf%2Fgh4ynwt_processed.png&w=3840&q=75)
Transcribed Image Text:## RC Circuit Analysis
### Circuit Description
This section describes an RC (Resistor-Capacitor) circuit under study, illustrated in the accompanying diagram. The circuit components are as follows:
- **Capacitance (C):** 1 µF (microfarad)
- **Inductance (L):** 1 H (henry)
- **Source Voltage (\(V_s\)):** 10 V (volts)
- **Switch**: Closed at \( t = 0 \)
### Diagram Explanation
The diagram depicts a simple series circuit comprising the following components:
- A DC voltage source (\(V_s\))
- An inductor (L)
- A capacitor (C)
- A resistor (R)
The circuit is shown with the switch in the closed position, initiating the study at time \( t = 0 \).
### Problem Statements
**(a)** Derive a differential equation for the voltage across the capacitor (\(V_C\)) for \( t \geq 0 \), with the source voltage \(V_s\).
**(b)** Determine the resistance (R) such that the voltage across the capacitor oscillates at a frequency \( \omega = 250 \pi \) rad/s.
### Analysis Approach
**(a) Differential Equation Derivation**
To derive the differential equation for the voltage across the capacitor \(V_C\):
1. Apply Kirchhoff's Voltage Law (KVL) around the loop.
2. Utilize the relationships for the inductor and capacitor:
- Voltage across the inductor: \(V_L = L \frac{dI}{dt}\)
- Voltage across the capacitor: \(V_C = \frac{1}{C} \int I dt \)
3. Combine these relationships and the voltage source to form the differential equation.
**(b) Finding Resistance for Desired Frequency**
Given the desired angular frequency \( \omega = 250 \pi \), use the characteristic equation of the resulting second-order differential equation from part (a) to determine the resistance \(R\):
\[ \omega_0 = \sqrt{\frac{1}{LC}} \]
\[ \beta = \frac{R}{2L} \]
\[ \omega = \sqrt{\omega_0^2 - \beta^2} \]
By solving for \(R\), ensure that the specific frequency condition is met.
### Summary
This analysis requires applying fundamental electrical engineering principles to
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