Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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![### Circuit Analysis Problem
#### Circuit Diagram Description
The given circuit diagram consists of the following components:
1. A voltage source \( V_1 \)
2. A switch that has been closed for a long time and is opened at \( t = 0 \)
3. Resistor \( R_1 \)
4. Inductor \( L \)
5. Current source \( I_1 \)
6. Resistor \( R_2 \)
This setup is commonly used to analyze transient responses in circuits, especially involving inductors, where the switch plays a crucial role in changing the circuit configuration at a specific time.
#### Given Component Values
- \( I_1 = 4 \, \text{A} \) (Current source value)
- \( V_1 = 8 \, \text{V} \) (Voltage source value)
- \( L = 2 \, \text{H} \) (Inductance)
- \( R_1 = 1 \, \Omega \) (Resistance of \( R_1 \))
- \( R_2 = 4 \, \Omega \) (Resistance of \( R_2 \))
#### Task
Determine the time constant \( \tau \) after the switch is opened. Round your answer to the nearest single digit decimal.
##### Calculation:
The time constant \( \tau \) for an LR circuit is given by:
\[ \tau = \frac{L}{R_{\text{total}}} \]
Where:
- \( R_{\text{total}} \) is the equivalent resistance seen by the inductor after the switch is opened. In this setup, the total resistance \( R_{\text{total}} = R_2 = 4 \, \Omega \).
- Substitute the values:
\[ \tau = \frac{2 \, \text{H}}{4 \, \Omega} = 0.5 \, \text{seconds} \]
#### Conclusion
- **Time Constant \( \tau \):** 0.5 seconds
This time constant indicates how quickly the current through the inductor decays after the switch is opened.](https://content.bartleby.com/qna-images/question/79b292ff-e294-4445-84a4-2da9891a32be/b7c4ce7f-080e-4834-9604-2f96846a0e81/ed0wwfl_processed.jpeg)
Transcribed Image Text:### Circuit Analysis Problem
#### Circuit Diagram Description
The given circuit diagram consists of the following components:
1. A voltage source \( V_1 \)
2. A switch that has been closed for a long time and is opened at \( t = 0 \)
3. Resistor \( R_1 \)
4. Inductor \( L \)
5. Current source \( I_1 \)
6. Resistor \( R_2 \)
This setup is commonly used to analyze transient responses in circuits, especially involving inductors, where the switch plays a crucial role in changing the circuit configuration at a specific time.
#### Given Component Values
- \( I_1 = 4 \, \text{A} \) (Current source value)
- \( V_1 = 8 \, \text{V} \) (Voltage source value)
- \( L = 2 \, \text{H} \) (Inductance)
- \( R_1 = 1 \, \Omega \) (Resistance of \( R_1 \))
- \( R_2 = 4 \, \Omega \) (Resistance of \( R_2 \))
#### Task
Determine the time constant \( \tau \) after the switch is opened. Round your answer to the nearest single digit decimal.
##### Calculation:
The time constant \( \tau \) for an LR circuit is given by:
\[ \tau = \frac{L}{R_{\text{total}}} \]
Where:
- \( R_{\text{total}} \) is the equivalent resistance seen by the inductor after the switch is opened. In this setup, the total resistance \( R_{\text{total}} = R_2 = 4 \, \Omega \).
- Substitute the values:
\[ \tau = \frac{2 \, \text{H}}{4 \, \Omega} = 0.5 \, \text{seconds} \]
#### Conclusion
- **Time Constant \( \tau \):** 0.5 seconds
This time constant indicates how quickly the current through the inductor decays after the switch is opened.
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