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Why for t = infinity, is the circuit drawn without the capacitor or inductor? Also, why is the current at t = infinity 0, and why is the voltage at infinity 12v?

At \( t = \infty \), the circuit reaches its steady-state condition, where certain simplifications can be made due to the behavior of the inductor and the capacitor:

1. **Behavior of the Inductor at \( t = \infty \):**
   - At steady state (when \( t \to \infty \)), an inductor behaves like a short circuit. This is because the inductor opposes changes in current, and after a long time, the current becomes constant (i.e., \( di/dt = 0 \)), which means there's no voltage drop across the inductor.
   - Therefore, in the steady-state analysis, we can replace the inductor with a simple wire (short circuit).

2. **Behavior of the Capacitor at \( t = \infty \):**
   - At steady state, a capacitor behaves like an open circuit. This is because once the capacitor is fully charged, no more current flows through it (i.e., \( i_C = 0 \)), and the voltage across the capacitor remains constant.
   - Consequently, for steady-state analysis, the capacitor can be removed from the circuit, as it no longer affects current flow.

3. **Current \( i(\infty) = 0 \):**
   - Since the capacitor is open and the inductor is short-circuited, no current flows through the circuit at \( t = \infty \). Hence, the current \( i(\infty) = 0 \).

4. **Voltage \( v(\infty) = 12 \text{ V} \):**
   - The voltage across the capacitor at \( t = \infty \) is simply the supply voltage. Since the current is zero and the capacitor holds its charge, the voltage across the capacitor equals the source voltage of 12 V.

In summary, at \( t = \infty \):
- The inductor acts as a short circuit.
- The capacitor acts as an open circuit.
- The current is zero because there is no closed path for current to flow.
- The voltage across the capacitor is 12 V because it matches the supply voltage.

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Would you like more detailed explanations on any of these points? Here are five related questions you might find useful:

1. What is the behavior of inductors and capacitors at \( t = 0^+ \) and \( t = \infty \)?
2. How do you derive the time constants for circuits with both inductors and capacitors?
3. What is the effect of the resistor values on the transient response of this circuit?
4. Why does the capacitor hold the supply voltage at \( t = \infty \) instead of discharging?
5. How does the current and voltage evolve over time from \( t = 0^+ \) to \( t = \infty \)?

**Tip:** In transient circuit analysis, always remember that inductors resist changes in current and capacitors resist changes in voltage, which leads to their characteristic behaviors over time.

Explore the steady-state behavior of inductors and capacitors in an RL-Circuit at t = infinity. Learn why the inductor acts as a short circuit, the capacitor as an open circuit, and why the current becomes zero and voltage equals 12V. This problem covers key concepts in electrical circuit analysis and is ideal for undergraduate electrical engineering students.

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