E7.6 Consider the network in Fig. E7.6. If the switch opens at t = 0), find the output voltage v (1) fort > 0. Figure E7.6 12 V ww 20 2025 2H t=0 4 V 2023 vo(t) ANSWER: v (1) =6- 10 3 e² V.

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### Example Problem on Electrical Network Analysis

#### Problem Statement (E7.6):
Consider the network in Fig. E7.6. If the switch opens at \( t = 0 \), find the output voltage \( v_o(t) \) for \( t \geq 0 \).

#### Solution:
**Given that the switch opens at \( t = 0 \), the output voltage \( v_o(t) \) is determined as follows:**

\[ v_o(t) = 6 - \frac{10}{3}e^{-3t} \text{ V} \]

#### Circuit Description:
The figure (Figure E7.6) depicts an electrical circuit containing the following components:
- A 12V voltage source.
- A 4V voltage source.
- Three resistors of value 2Ω each.
- An inductor with an inductance of 2H.
- A switch that opens at \( t = 0 \).

#### Detailed Analysis:
1. **Initial Setup:**
    - At \( t < 0 \), with the switch closed, the circuit reaches a steady-state condition.
    - At \( t = 0 \), the switch opens, changing the circuit dynamics.
  
2. **Solving for \( v_o(t) \):**
    - Utilizing Kirchhoff's laws and the natural response for RL circuits, the transient response occurs due to the inductor and resistors.

3. **Mathematical Formulation:**
    - The voltage across the inductor is described by an exponential decay \( e^{-3t} \), where the constant derived is influenced by the resistor and inductor values.

### Graph / Diagram Explanation:
- **Circuit Diagram:**
  - **Top Branch:**
    - Begins with a 12V source and a 2Ω resistor in series.
    - Encountering the switch (which opens at \( t=0 \)).
    - Followed by a 2H inductor.
  - **Middle Branch:**
    - In parallel with the top branch.
    - Contains a 4V source and 2Ω resistor in series.
  - **Bottom Branch:**
    - Contains only a 2Ω resistor and connects back to the negative terminals of both the 12V and 4V sources.
  - **Output Voltage \( v_o(t) \):**
    - Measured across the rightmost 2Ω resistor in the circuit's
Transcribed Image Text:### Example Problem on Electrical Network Analysis #### Problem Statement (E7.6): Consider the network in Fig. E7.6. If the switch opens at \( t = 0 \), find the output voltage \( v_o(t) \) for \( t \geq 0 \). #### Solution: **Given that the switch opens at \( t = 0 \), the output voltage \( v_o(t) \) is determined as follows:** \[ v_o(t) = 6 - \frac{10}{3}e^{-3t} \text{ V} \] #### Circuit Description: The figure (Figure E7.6) depicts an electrical circuit containing the following components: - A 12V voltage source. - A 4V voltage source. - Three resistors of value 2Ω each. - An inductor with an inductance of 2H. - A switch that opens at \( t = 0 \). #### Detailed Analysis: 1. **Initial Setup:** - At \( t < 0 \), with the switch closed, the circuit reaches a steady-state condition. - At \( t = 0 \), the switch opens, changing the circuit dynamics. 2. **Solving for \( v_o(t) \):** - Utilizing Kirchhoff's laws and the natural response for RL circuits, the transient response occurs due to the inductor and resistors. 3. **Mathematical Formulation:** - The voltage across the inductor is described by an exponential decay \( e^{-3t} \), where the constant derived is influenced by the resistor and inductor values. ### Graph / Diagram Explanation: - **Circuit Diagram:** - **Top Branch:** - Begins with a 12V source and a 2Ω resistor in series. - Encountering the switch (which opens at \( t=0 \)). - Followed by a 2H inductor. - **Middle Branch:** - In parallel with the top branch. - Contains a 4V source and 2Ω resistor in series. - **Bottom Branch:** - Contains only a 2Ω resistor and connects back to the negative terminals of both the 12V and 4V sources. - **Output Voltage \( v_o(t) \):** - Measured across the rightmost 2Ω resistor in the circuit's
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