The switch has been in position a for a long time. At t = 0, the switch is moved to position b. 400ΚΩ 2092 a 90V € 609 40V 0.5μF a. What is the initial value of vc? What is the final value of vc? b. c. What is the time constant of the circuit when the switch is in position b? d. What is the expression for vc(t) when t≥ 0? e. What is the expression for i(t) when t ≥ 0? f. How long after the switch is in position b does the capacitor voltage pass through OV?

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### Electrical Circuit Problem

#### Problem Statement:
The switch has been in position $a$ for a long time. At $t = 0$, the switch is moved to position $b$.

#### Circuit Diagram:
![Circuit Diagram](path_to_circuit_image)

The circuit consists of the following components:
1. A 90V voltage source connected to:
    - A resistor of 400kΩ in series with the circuit when the switch is in position $b$.
2. A parallel configuration of:
    - A capacitor of 0.5μF
    - A 60Ω resistor and a 20Ω resistor in series with a 40V voltage source when the switch is in position $b$.

**Questions:**
a. What is the initial value of \( v_C \)?
b. What is the final value of \( v_C \)?
c. What is the time constant of the circuit when the switch is in position $b$?
d. What is the expression for \( v_C(t) \) when \( t \geq 0 \)?
e. What is the expression for \( i(t) \) when \( t \geq 0 \)?
f. How long after the switch is in position $b$ does the capacitor voltage pass through 0V?
g. Plot \( v_C(t) \) and \( i(t) \) versus \( t \).

#### Detailed Explanation:

a. **Initial Value of \( v_C \):**
   - When the switch is in position $a$ for a long time, the capacitor will be fully charged to the supply voltage.
   - Since the capacitor is charged through a series resistance with a 90V source, \( v_C(0^-) = 90V \).

b. **Final Value of \( v_C \):**
   - When the switch is moved to position $b$, the capacitor will discharge through the resistors and reach the voltage source of 40V. Thus the final steady-state voltage across the capacitor \( v_C(\infty) = 40V \).

c. **Time Constant of the Circuit:**
   - The time constant \( \tau \) is determined by the resistance and capacitance in the discharge path.
   - Equivalent resistance \( R_{eq} = 60Ω + 20Ω = 80Ω \).
   - Capacitance \( C = 0.5μF \).
Transcribed Image Text:### Electrical Circuit Problem #### Problem Statement: The switch has been in position $a$ for a long time. At $t = 0$, the switch is moved to position $b$. #### Circuit Diagram: ![Circuit Diagram](path_to_circuit_image) The circuit consists of the following components: 1. A 90V voltage source connected to: - A resistor of 400kΩ in series with the circuit when the switch is in position $b$. 2. A parallel configuration of: - A capacitor of 0.5μF - A 60Ω resistor and a 20Ω resistor in series with a 40V voltage source when the switch is in position $b$. **Questions:** a. What is the initial value of \( v_C \)? b. What is the final value of \( v_C \)? c. What is the time constant of the circuit when the switch is in position $b$? d. What is the expression for \( v_C(t) \) when \( t \geq 0 \)? e. What is the expression for \( i(t) \) when \( t \geq 0 \)? f. How long after the switch is in position $b$ does the capacitor voltage pass through 0V? g. Plot \( v_C(t) \) and \( i(t) \) versus \( t \). #### Detailed Explanation: a. **Initial Value of \( v_C \):** - When the switch is in position $a$ for a long time, the capacitor will be fully charged to the supply voltage. - Since the capacitor is charged through a series resistance with a 90V source, \( v_C(0^-) = 90V \). b. **Final Value of \( v_C \):** - When the switch is moved to position $b$, the capacitor will discharge through the resistors and reach the voltage source of 40V. Thus the final steady-state voltage across the capacitor \( v_C(\infty) = 40V \). c. **Time Constant of the Circuit:** - The time constant \( \tau \) is determined by the resistance and capacitance in the discharge path. - Equivalent resistance \( R_{eq} = 60Ω + 20Ω = 80Ω \). - Capacitance \( C = 0.5μF \).
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