IL(t) R1 R2 the above circuit, the switch has been closed for a very long time and then at t=0 the switch is opened. Given the following component values for the circuit: = 4 A = 8 V =27 H 2= 40 Determine the time constant T after the switch happens. Round your answer to the nearest single digit decimal. Find t in secs (due not enter the units).

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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.
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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