لمس R = 400 ε = 12V 2μF 1mH1uF: a) In the circuit, both capacitors are initially uncharged. The itch S is closed at t = 0. Calculate iR, iL at t = 0. b) When the current in the circuit reached steady state, calculate the energy stored in the 2uF capacitor and in the inductor. c) After the current reached steady state, S is opened and time is reset to t = 0. Calculate the maximum charge the 2μF capacitor can get.

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### Circuit Analysis Problem

#### Circuit Diagram

The given circuit consists of:
- A resistor \( R = 40 \, \Omega \)
- A voltage source \( \epsilon = 12 \, V \)
- Two capacitors with capacitances \( 2 \, \mu F \) and \( 1 \, \mu F \)
- An inductor \( 1 \, mH \)
- A switch \( S \)

The capacitors are initially uncharged, and the switch \( S \) is closed at \( t = 0 \).

#### Problem Statement

a) In the circuit, both capacitors are initially uncharged. The switch \( S \) is closed at \( t = 0 \). Calculate \( i_R \), \( i_L \) at \( t = 0 \).

b) When the current in the circuit reaches steady state, calculate the energy stored in the \( 2 \, \mu F \) capacitor and in the inductor.

c) After the current reached steady state, \( S \) is opened and time is reset to \( t = 0 \). Calculate the maximum charge the \( 2 \mu F \) capacitor can get.

#### Explanation of Diagram:

The provided circuit diagram shows a series-parallel arrangement where:
- The resistor \( R \) and voltage source \( \epsilon \) are in series with the switch \( S \).
- The \( 2 \, \mu F \) capacitor is in parallel with the series combination of the \( 1 \, mH \) inductor and the \(1 \, \mu F \) capacitor.

For the calculations:

1. **At \( t = 0 \):**
   - \( i_R \): Current through the resistor immediately after the switch is closed.
   - \( i_L \): Current through the inductor immediately after the switch is closed.
   
2. **At Steady State:**
   - Calculate the energy stored in the \( 2 \, \mu F \) capacitor.
   - Calculate the energy stored in the inductor.
   
3. **Reset to \( t = 0 \) after Steady State:**
   - Determine the maximum charge that can be stored in the \( 2 \, \mu F \) capacitor when switch \( S \) is reopened and the circuit time is reset to \( t = 0 \).

This diagram and the given
Transcribed Image Text:### Circuit Analysis Problem #### Circuit Diagram The given circuit consists of: - A resistor \( R = 40 \, \Omega \) - A voltage source \( \epsilon = 12 \, V \) - Two capacitors with capacitances \( 2 \, \mu F \) and \( 1 \, \mu F \) - An inductor \( 1 \, mH \) - A switch \( S \) The capacitors are initially uncharged, and the switch \( S \) is closed at \( t = 0 \). #### Problem Statement a) In the circuit, both capacitors are initially uncharged. The switch \( S \) is closed at \( t = 0 \). Calculate \( i_R \), \( i_L \) at \( t = 0 \). b) When the current in the circuit reaches steady state, calculate the energy stored in the \( 2 \, \mu F \) capacitor and in the inductor. c) After the current reached steady state, \( S \) is opened and time is reset to \( t = 0 \). Calculate the maximum charge the \( 2 \mu F \) capacitor can get. #### Explanation of Diagram: The provided circuit diagram shows a series-parallel arrangement where: - The resistor \( R \) and voltage source \( \epsilon \) are in series with the switch \( S \). - The \( 2 \, \mu F \) capacitor is in parallel with the series combination of the \( 1 \, mH \) inductor and the \(1 \, \mu F \) capacitor. For the calculations: 1. **At \( t = 0 \):** - \( i_R \): Current through the resistor immediately after the switch is closed. - \( i_L \): Current through the inductor immediately after the switch is closed. 2. **At Steady State:** - Calculate the energy stored in the \( 2 \, \mu F \) capacitor. - Calculate the energy stored in the inductor. 3. **Reset to \( t = 0 \) after Steady State:** - Determine the maximum charge that can be stored in the \( 2 \, \mu F \) capacitor when switch \( S \) is reopened and the circuit time is reset to \( t = 0 \). This diagram and the given
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