The switch in Fig. 03 remained closed for an extended time. (a) Give iL for t < 0. (b) Just after opening the switch, find iL(0+). (c) Determine iL(∞). after opening the switch, find iL(0+). (c) Determine iL(∞). (d) Obtain the expression of iL(t) relative to t > 0.

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## Understanding Inductor Behavior in RLC Circuits

The diagram illustrates a switching RL circuit. Here, we'll analyze the behavior of the inductor current \( i_L \) over time, given different switching conditions.

### Problem Statement

The switch in Fig. 03 remained closed for an extended time. 
1. **For \( t < 0 \)**: Determine \( i_L \).
2. **Just after opening the switch**: Find \( i_L(0^+) \).
3. **For \( t \to \infty \)**: Determine \( i_L(\infty) \).
4. **Expression for \( i_L(t) \) relative to \( t > 0 \)**: Obtain the expression.

### Circuit Description

- **Resistors**: 20Ω and 5Ω
- **Inductor**: 0.5 H
- **Voltage Sources**: Two 100V sources
- **Switch Action**: Opens at \( t = 0 \)

### Analysis and Solution

#### (a) \( i_L \) for \( t < 0 \)

When the switch is closed for an extended time, the inductor acts like a short circuit in steady state.
- The series combination of resistors forms a voltage divider.
- The voltage across the inductor is zero in steady state (short circuit).
- The current \( i_L \) can be calculated using the resistors in series.

#### (b) \( i_L(0^+) \)

Just after the switch opens:
- The inductor resists any sudden change in current.
- Therefore, \( i_L(0^+) = i_L(0^-) \).

#### (c) \( i_L(\infty) \)

As \( t \to \infty \):
- The inductor will again reach a steady state where it acts like a short circuit.
- The current will be determined by the remaining part of the circuit.

#### (d) Expression of \( i_L(t) \) for \( t > 0 \)

To find \( i_L(t) \) for \( t > 0 \):
- Use the natural response of the RL circuit considering the resistor and the inductor.
- Write the differential equation governing \( i_L(t) \).
- Solve it with initial conditions to find the time-varying current.

### Diagram Explanation

The figure shows:
- A switch that opens at \(
Transcribed Image Text:## Understanding Inductor Behavior in RLC Circuits The diagram illustrates a switching RL circuit. Here, we'll analyze the behavior of the inductor current \( i_L \) over time, given different switching conditions. ### Problem Statement The switch in Fig. 03 remained closed for an extended time. 1. **For \( t < 0 \)**: Determine \( i_L \). 2. **Just after opening the switch**: Find \( i_L(0^+) \). 3. **For \( t \to \infty \)**: Determine \( i_L(\infty) \). 4. **Expression for \( i_L(t) \) relative to \( t > 0 \)**: Obtain the expression. ### Circuit Description - **Resistors**: 20Ω and 5Ω - **Inductor**: 0.5 H - **Voltage Sources**: Two 100V sources - **Switch Action**: Opens at \( t = 0 \) ### Analysis and Solution #### (a) \( i_L \) for \( t < 0 \) When the switch is closed for an extended time, the inductor acts like a short circuit in steady state. - The series combination of resistors forms a voltage divider. - The voltage across the inductor is zero in steady state (short circuit). - The current \( i_L \) can be calculated using the resistors in series. #### (b) \( i_L(0^+) \) Just after the switch opens: - The inductor resists any sudden change in current. - Therefore, \( i_L(0^+) = i_L(0^-) \). #### (c) \( i_L(\infty) \) As \( t \to \infty \): - The inductor will again reach a steady state where it acts like a short circuit. - The current will be determined by the remaining part of the circuit. #### (d) Expression of \( i_L(t) \) for \( t > 0 \) To find \( i_L(t) \) for \( t > 0 \): - Use the natural response of the RL circuit considering the resistor and the inductor. - Write the differential equation governing \( i_L(t) \). - Solve it with initial conditions to find the time-varying current. ### Diagram Explanation The figure shows: - A switch that opens at \(
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