Use the phasor concept to find the steady-state expression for i, (t) in the circuit below. (10 points) vs (t) = 25 cos( 4000 t - 90°) + 5.02 www. io (t) 2.5mH- 10Ω www- 2002 www 12.5μF

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**Title: Phasor Analysis in AC Circuits**

**Objective:**
Use the phasor concept to find the steady-state expression for \( i_o(t) \) in the given circuit.

**Circuit Description:**
The circuit consists of the following components in a series-parallel configuration:

1. **Voltage Source:**
   - \( v_s(t) = 25 \cos(4000t - 90^\circ) \)

2. **Resistors:**
   - \( 5\Omega \)
   - \( 10\Omega \)
   - \( 20\Omega \)

3. **Inductor:**
   - \( 2.5 \, \text{mH} \)

4. **Capacitor:**
   - \( 12.5 \, \mu F \)

**Configuration:**
- The voltage source is connected to a 5\(\Omega\) resistor in series with an inductor of 2.5 mH.
- In parallel with the 2.5 mH inductor, there is a series combination of a 10\(\Omega\) resistor, a 20\(\Omega\) resistor, and a capacitor of 12.5 \(\mu F\).

**Analysis Task:**
- Derive the steady-state expression for the output current \( i_o(t) \).

**Approach:**
Utilize phasor analysis by converting all circuit elements to their phasor equivalents:
- The resistors remain unchanged in the phasor domain.
- The inductor is represented by its impedance \( Z_L = j\omega L \).
- The capacitor is represented by its impedance \( Z_C = \frac{1}{j\omega C} \).
- Calculate the total impedance and use Ohm’s Law to find the phasor current.
- Convert the phasor current back to the time domain for the expression of \( i_o(t) \).

**Note:** This approach will provide insights into AC circuit behavior, focusing on amplitude and phase relationships.
Transcribed Image Text:**Title: Phasor Analysis in AC Circuits** **Objective:** Use the phasor concept to find the steady-state expression for \( i_o(t) \) in the given circuit. **Circuit Description:** The circuit consists of the following components in a series-parallel configuration: 1. **Voltage Source:** - \( v_s(t) = 25 \cos(4000t - 90^\circ) \) 2. **Resistors:** - \( 5\Omega \) - \( 10\Omega \) - \( 20\Omega \) 3. **Inductor:** - \( 2.5 \, \text{mH} \) 4. **Capacitor:** - \( 12.5 \, \mu F \) **Configuration:** - The voltage source is connected to a 5\(\Omega\) resistor in series with an inductor of 2.5 mH. - In parallel with the 2.5 mH inductor, there is a series combination of a 10\(\Omega\) resistor, a 20\(\Omega\) resistor, and a capacitor of 12.5 \(\mu F\). **Analysis Task:** - Derive the steady-state expression for the output current \( i_o(t) \). **Approach:** Utilize phasor analysis by converting all circuit elements to their phasor equivalents: - The resistors remain unchanged in the phasor domain. - The inductor is represented by its impedance \( Z_L = j\omega L \). - The capacitor is represented by its impedance \( Z_C = \frac{1}{j\omega C} \). - Calculate the total impedance and use Ohm’s Law to find the phasor current. - Convert the phasor current back to the time domain for the expression of \( i_o(t) \). **Note:** This approach will provide insights into AC circuit behavior, focusing on amplitude and phase relationships.
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