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The problem asks to calculate the current in the circuit after the switch is closed, considering the resistor and inductor present in the circuit. 

This is a classic problem involving an RL (Resistor-Inductor) circuit, where the circuit is energized by a DC voltage source \(V = 20 \, \text{V}\), and the following values are provided:
- Resistance \(R = 100 \, \Omega\)
- Inductance \(L = 5 \, \text{H}\)

### Steps to Solve:

1. **Determine the Time Constant \(\tau\):**
   - The time constant \(\tau\) for an RL circuit is given by:
     \[
     \tau = \frac{L}{R}
     \]
     Substituting the given values:
     \[
     \tau = \frac{5 \, \text{H}}{100 \, \Omega} = 0.05 \, \text{s}
     \]

2. **Calculate the Current as a Function of Time:**
   - The current \(I(t)\) in an RL circuit after the switch is closed follows the equation:
     \[
     I(t) = \frac{V}{R} \left(1 - e^{-\frac{t}{\tau}}\right)
     \]
     where \(V\) is the supply voltage, \(R\) is the resistance, and \(\tau\) is the time constant.
   - The steady-state current (when \(t \to \infty\)) is:
     \[
     I(\infty) = \frac{V}{R} = \frac{20 \, \text{V}}{100 \, \Omega} = 0.2 \, \text{A}
     \]

3. **Voltage Across the Resistor:**
   - The voltage across the resistor \(V_R(t)\) can be found using Ohm's law:
     \[
     V_R(t) = I(t) \times R = 100 \times 0.2 \times \left(1 - e^{-\frac{t}{0.05}}\right) \, \text{V}
     \]
   - At \(t = 0\), the current \(I(0) = 0\) A, so the voltage across the resistor is \(0 \, \text{V}\).
   - As \(t\) increases, the voltage across the resistor increases exponentially until it reaches \(V = 20 \, \text{V}\).

4. **Voltage Across the Inductor:**
   - The voltage across the inductor \(V_L(t)\) is given by:
     \[
     V_L(t) = L \frac{dI(t)}{dt} = V e^{-\frac{t}{\tau}} = 20 e^{-\frac{t}{0.05}} \, \text{V}
     \]
   - At \(t = 0\), the voltage across the inductor is \(20 \, \text{V}\), and it decreases to \(0 \, \text{V}\) as time goes to infinity.

### Conclusion:
- At \(t = 0\), the current \(I(0) = 0\) A.
- As time increases, the current approaches \(0.2 \, \text{A}\) in the steady state.
- The voltage across the resistor increases from \(0 \, \text{V}\) to \(20 \, \text{V}\).
- The voltage across the inductor decreases from \(20 \, \text{V}\) to \(0 \, \text{V}\).

Would you like more details or have any other questions?

### Relative Questions:
1. What happens to the current in an RL circuit immediately after the switch is closed?
2. How does the inductor behave in an RL circuit when the switch is first closed?
3. What is the significance of the time constant \(\tau\) in an RL circuit?
4. How would the current response change if the resistance \(R\) was increased?
5. How can we determine the energy stored in the inductor after a long time?

### Tip:
In RL circuits, the inductor opposes changes in current when the switch is first closed, resulting in an exponential increase in current over time. Understanding this behavior is crucial when analyzing transient responses in electrical circuits.

Learn how to calculate the current in an RL circuit after the switch is closed. This problem involves analyzing an RL circuit with specific values for resistance and inductance, providing a detailed solution including the time constant calculation and current response over time.

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