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Let's analyze and solve each part of the question for the given RL circuit.

### Given Information:
- **Supply Voltage, \( E = 50 \, V \)**
- **Resistance, \( R = 20 \, \Omega \)**
- **Inductance, \( L = 20 \, \text{mH} = 0.02 \, H \)**

### Part (a): Time Constant of the Circuit
The time constant \( \tau \) of an RL circuit is given by:
\[
\tau = \frac{L}{R}
\]

Substitute the values:
\[
\tau = \frac{0.02}{20} = 0.001 \, \text{s} = 1 \, \text{ms}
\]

So, the **time constant** of the circuit is **1 ms**.

---

### Part (b): Time to Reach Steady State
In an RL circuit, the current reaches approximately 99% of its steady-state value after \( t = 5\tau \). Therefore, the time to reach steady-state is:
\[
t = 5 \tau = 5 \times 0.001 = 0.005 \, \text{s} = 5 \, \text{ms}
\]

So, the **time for the current to reach its steady value is 5 ms**.

---

### Part (c): Charging Equations for \( i_L \) and \( v_L \) in Terms of \( t \)
When the switch is closed at \( t = 0 \), the current \( i_L(t) \) and the voltage across the inductor \( v_L(t) \) follow exponential behaviors due to the RL circuit dynamics.

1. **Current through the inductor, \( i_L(t) \):**
   The expression for current in an RL circuit after the switch is closed is:
   \[
   i_L(t) = \frac{E}{R} \left( 1 - e^{-\frac{t}{\tau}} \right)
   \]

   Substitute \( E = 50 \, V \), \( R = 20 \, \Omega \), and \( \tau = 0.001 \, \text{s} \):
   \[
   i_L(t) = \frac{50}{20} \left( 1 - e^{-\frac{t}{0.001}} \right) = 2.5 \left( 1 - e^{-1000t} \right) \, \text{A}
   \]

2. **Voltage across the inductor, \( v_L(t) \):**
   The voltage across the inductor is given by:
   \[
   v_L(t) = L \frac{di_L(t)}{dt}
   \]

   Taking the derivative of \( i_L(t) \) with respect to \( t \):
   \[
   \frac{di_L(t)}{dt} = 2.5 \times 1000 \times e^{-1000t} = 2500 e^{-1000t}
   \]

   Therefore,
   \[
   v_L(t) = 0.02 \times 2500 e^{-1000t} = 50 e^{-1000t} \, \text{V}
   \]

---

### Summary of Answers:
1. **Time constant, \( \tau \)**: \( 1 \, \text{ms} \)
2. **Time to reach steady state**: \( 5 \, \text{ms} \)
3. **Current \( i_L(t) \)**: \( i_L(t) = 2.5 \left( 1 - e^{-1000t} \right) \, \text{A} \)
4. **Voltage \( v_L(t) \)**: \( v_L(t) = 50 e^{-1000t} \, \text{V} \)

Would you like further clarification on any part of this solution?

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### Follow-up Questions:
1. How does the time constant affect the rate of current change in the circuit?
2. What happens to the inductor voltage \( v_L(t) \) as time approaches infinity?
3. How would the time constant change if the resistance was halved?
4. What role does the inductor play in opposing changes in current initially?
5. How would the equations change if the inductor value was increased?

---

### Tip:
In an RL circuit, the inductor initially opposes the sudden rise in current, resulting in a gradual increase. The rate at which current reaches its maximum is governed by the time constant \( \tau = \frac{L}{R} \).

The switch of the following circuit in Fig.Q2 is closed at t = 0 s.
(a) What is the time constant of the circuit?
(b) What is the time for the current to reach its steady value?
(c) Express the charging equations of i_L and v_L in terms of time t.

This solution explains how to find the time constant, time to reach steady-state current, and charging equations for current and voltage in an RL circuit with a 50V source, a 20 Ohm resistor, and a 20 mH inductor. Detailed step-by-step calculations cover exponential growth and decay in circuit analysis.

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