9.29 Find the impedance Zab in the circuit seen in Fig. P9.29. Express Zab in both polar and rectangu- lar form. will of Figure P9.29 TARDIDE a1 10 a. Zab be 101106080 10 Ω www -j40 Ω He wodą lupis om be <50 5Ω – –j10 Ω 2002 10 Ω j30 Ω j20 Ω

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Circuits 1 HW 6 Q10
## Problem 9.29

**Objective:** Find the impedance \( Z_{ab} \) in the circuit seen in Fig. P9.29. Express \( Z_{ab} \) in both polar and rectangular form.

### Figure P9.29: Circuit Diagram

- **Components:**
  - A series of components between points \( a \) and \( b \).
  - **Resistor:** 10 Ω
  - **Inductor (represented by impedance):** -j40 Ω

- **Parallel Branches:**
  - **Branch 1:**
    - **Resistor:** 5 Ω
    - **Capacitor (represented by impedance):** -j10 Ω
  - **Branch 2:**
    - **Resistor:** 10 Ω
    - **Inductor (represented by impedance):** j30 Ω
  - **Branch 3:**
    - **Resistor:** 20 Ω
    - **Inductor (represented by impedance):** j20 Ω

### Explanation of the Circuit:
- From point \( a \) to point \( b \), the series connection includes a 10 Ω resistor and an inductor with an impedance of -j40 Ω.
- There are three parallel branches:
  - The first branch consists of a resistor (5 Ω) in series with a capacitor (-j10 Ω).
  - The second branch consists of a resistor (10 Ω) in series with an inductor (j30 Ω).
  - The third branch consists of a resistor (20 Ω) in series with an inductor (j20 Ω).
  
The goal is to calculate the total impedance \( Z_{ab} \) between points \( a \) and \( b \), taking into account both series and parallel connections, and to express it in both rectangular and polar forms.
Transcribed Image Text:## Problem 9.29 **Objective:** Find the impedance \( Z_{ab} \) in the circuit seen in Fig. P9.29. Express \( Z_{ab} \) in both polar and rectangular form. ### Figure P9.29: Circuit Diagram - **Components:** - A series of components between points \( a \) and \( b \). - **Resistor:** 10 Ω - **Inductor (represented by impedance):** -j40 Ω - **Parallel Branches:** - **Branch 1:** - **Resistor:** 5 Ω - **Capacitor (represented by impedance):** -j10 Ω - **Branch 2:** - **Resistor:** 10 Ω - **Inductor (represented by impedance):** j30 Ω - **Branch 3:** - **Resistor:** 20 Ω - **Inductor (represented by impedance):** j20 Ω ### Explanation of the Circuit: - From point \( a \) to point \( b \), the series connection includes a 10 Ω resistor and an inductor with an impedance of -j40 Ω. - There are three parallel branches: - The first branch consists of a resistor (5 Ω) in series with a capacitor (-j10 Ω). - The second branch consists of a resistor (10 Ω) in series with an inductor (j30 Ω). - The third branch consists of a resistor (20 Ω) in series with an inductor (j20 Ω). The goal is to calculate the total impedance \( Z_{ab} \) between points \( a \) and \( b \), taking into account both series and parallel connections, and to express it in both rectangular and polar forms.
**Problem 9.27: Analysis of a Sinusoidal Steady-State Circuit**

**Task:** 
The circuit in Figure P9.27 is operating in the sinusoidal steady state. Find the steady-state expression for \( v_o(t) \) if \( v_g = 64 \cos 8000t \, \text{V} \).

**Circuit Description (Figure P9.27):**

- **Voltage Source (\( v_g \))**: Provides an input voltage of \( 64 \cos 8000t \, \text{V} \).
  
- **Capacitor**: Connected in series with the source, with a capacitance of 31.25 nF (nanoFarads).

- **Resistor**: Connected in parallel with the inductor and output voltage, having a resistance of 2 kΩ (kiloOhms).

- **Inductor**: Connected in parallel with the resistor and across the output voltage, with an inductance of 500 mH (milliHenrys).

The task involves finding the steady-state output voltage \( v_o(t) \) across the parallel combination of the resistor and the inductor.
Transcribed Image Text:**Problem 9.27: Analysis of a Sinusoidal Steady-State Circuit** **Task:** The circuit in Figure P9.27 is operating in the sinusoidal steady state. Find the steady-state expression for \( v_o(t) \) if \( v_g = 64 \cos 8000t \, \text{V} \). **Circuit Description (Figure P9.27):** - **Voltage Source (\( v_g \))**: Provides an input voltage of \( 64 \cos 8000t \, \text{V} \). - **Capacitor**: Connected in series with the source, with a capacitance of 31.25 nF (nanoFarads). - **Resistor**: Connected in parallel with the inductor and output voltage, having a resistance of 2 kΩ (kiloOhms). - **Inductor**: Connected in parallel with the resistor and across the output voltage, with an inductance of 500 mH (milliHenrys). The task involves finding the steady-state output voltage \( v_o(t) \) across the parallel combination of the resistor and the inductor.
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