Determine the power factor of the entire circuit below as seen by the source. Calculate the average power delivered by the source. -j22 3020° Vrms

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**Problem Statement:**

Determine the power factor of the entire circuit below as seen by the source. Calculate the average power delivered by the source.

**Circuit Diagram:** 

The circuit consists of the following components:

1. An AC voltage source of \(30 \angle 0^\circ \, V_{rms}\)
2. A series resistor of \(6 \Omega\)
3. A parallel combination of:
   - Capacitive reactance of \(-j2 \Omega\)
   - A resistor of \(4 \Omega\)

**Detailed Description of the Circuit:**

- The voltage source is represented by a sinusoidal symbol.
- The series resistor has a resistance of \(6 \Omega\).
- The parallel combination follows after the \(6 \Omega\) resistor and consists of:
  - A capacitor with a capacitive reactance of \(-j2 \Omega\) (indicating a reactive component with \( \text{Capacitive reactance} = \frac{1}{\omega C} \))
  - A resistor with a resistance of \(4 \Omega\)

**Task:**

1. **Determine the Power Factor:**
   - The power factor of an electrical circuit is a measure of how effectively the current is being converted into useful work power and is defined as the ratio of real power to apparent power.

2. **Calculate the Average Power Delivered by the Source:**
   - The average power (also known as real power) delivered by the source is calculated using the formula:
     \[
     P = V_{rms} \cdot I_{rms} \cdot \cos(\phi)
     \]
     where \(\phi\) is the phase angle between the voltage and the current.

**Note:**

To solve this problem, take the following steps:
- Determine the equivalent impedance of the circuit as seen by the source.
- Calculate the current flowing through the circuit using Ohm's Law.
- Compute the power factor.
- Calculate the average power delivered by the source.

**Explanation of Diagram:**

The circuit diagram shows an AC voltage source connected in series with a 6-ohm resistor. This series connection is then in parallel with a 4-ohm resistor and a capacitive reactance of -j2 ohms. The components are connected as follows:

1. The voltage source symbol is at the leftmost part, with the magnitude and phase angle denoted as \(30 \angle
Transcribed Image Text:**Problem Statement:** Determine the power factor of the entire circuit below as seen by the source. Calculate the average power delivered by the source. **Circuit Diagram:** The circuit consists of the following components: 1. An AC voltage source of \(30 \angle 0^\circ \, V_{rms}\) 2. A series resistor of \(6 \Omega\) 3. A parallel combination of: - Capacitive reactance of \(-j2 \Omega\) - A resistor of \(4 \Omega\) **Detailed Description of the Circuit:** - The voltage source is represented by a sinusoidal symbol. - The series resistor has a resistance of \(6 \Omega\). - The parallel combination follows after the \(6 \Omega\) resistor and consists of: - A capacitor with a capacitive reactance of \(-j2 \Omega\) (indicating a reactive component with \( \text{Capacitive reactance} = \frac{1}{\omega C} \)) - A resistor with a resistance of \(4 \Omega\) **Task:** 1. **Determine the Power Factor:** - The power factor of an electrical circuit is a measure of how effectively the current is being converted into useful work power and is defined as the ratio of real power to apparent power. 2. **Calculate the Average Power Delivered by the Source:** - The average power (also known as real power) delivered by the source is calculated using the formula: \[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \] where \(\phi\) is the phase angle between the voltage and the current. **Note:** To solve this problem, take the following steps: - Determine the equivalent impedance of the circuit as seen by the source. - Calculate the current flowing through the circuit using Ohm's Law. - Compute the power factor. - Calculate the average power delivered by the source. **Explanation of Diagram:** The circuit diagram shows an AC voltage source connected in series with a 6-ohm resistor. This series connection is then in parallel with a 4-ohm resistor and a capacitive reactance of -j2 ohms. The components are connected as follows: 1. The voltage source symbol is at the leftmost part, with the magnitude and phase angle denoted as \(30 \angle
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