2. Find magnitude of each branch, current, and the total current. What is the phase angle between the applied voltage and the total current? 46) Vs 540°V f=50kHz 10.047μF 0.011MF 18012 L3 15052

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Find magnitude of each branch, current, and the total current. What is the phase angle between the applied voltage and the total current
The image contains a circuit problem and a schematic diagram.

### Problem:
2. Find the magnitude of each branch current, and the total current. What is the phase angle between the applied voltage and the total current?

### Circuit Details:
- **Voltage Source:** \[ V_s = 5 \angle 0^\circ \text{V} \]
- **Frequency:** \( f = 50 \text{kHz} \)

The circuit is a parallel RLC circuit with the following components:
- **Capacitor 1:** \( 0.047 \mu\text{F} \)
- **Capacitor 2:** \( 0.011 \mu\text{F} \)
- **Resistor 1:** \( 180 \Omega \)
- **Resistor 2:** \( 150 \Omega \)

### Diagram Explanation:
The circuit diagram shows an AC source connected to four parallel branches:
1. First branch contains a \( 0.047 \mu\text{F} \) capacitor.
2. Second branch contains a \( 0.011 \mu\text{F} \) capacitor.
3. Third branch contains a resistor labeled \( 180 \Omega \).
4. Fourth branch contains a resistor labeled \( 150 \Omega \).

This parallel arrangement suggests analysis using complex impedance for phasor analysis of AC circuits, to calculate the currents through each branch and determine the phase angle between the voltage and total current.
Transcribed Image Text:The image contains a circuit problem and a schematic diagram. ### Problem: 2. Find the magnitude of each branch current, and the total current. What is the phase angle between the applied voltage and the total current? ### Circuit Details: - **Voltage Source:** \[ V_s = 5 \angle 0^\circ \text{V} \] - **Frequency:** \( f = 50 \text{kHz} \) The circuit is a parallel RLC circuit with the following components: - **Capacitor 1:** \( 0.047 \mu\text{F} \) - **Capacitor 2:** \( 0.011 \mu\text{F} \) - **Resistor 1:** \( 180 \Omega \) - **Resistor 2:** \( 150 \Omega \) ### Diagram Explanation: The circuit diagram shows an AC source connected to four parallel branches: 1. First branch contains a \( 0.047 \mu\text{F} \) capacitor. 2. Second branch contains a \( 0.011 \mu\text{F} \) capacitor. 3. Third branch contains a resistor labeled \( 180 \Omega \). 4. Fourth branch contains a resistor labeled \( 150 \Omega \). This parallel arrangement suggests analysis using complex impedance for phasor analysis of AC circuits, to calculate the currents through each branch and determine the phase angle between the voltage and total current.
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