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Phasor 60 Hz powergui oltage Source 480 V Line Current u² 90.76 A 0.18+j0.24 Ohms 1483 PLoss, W 4+j3 Ohms •Melle Load Voltage

Phasor 60 Hz powergui oltage Source 480 V Line Current u² 90.76 A 0.18+j0.24 Ohms 1483 PLoss, W 4+j3 Ohms •Melle Load Voltage

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Can you please draw me the phasor diagram for this circuit. I think it should look similar to the second pic but not sure. Thank you I will like

### Circuit Analysis Diagram This schematic represents an electrical circuit with power calculations using a phasor model at 60 Hz. The layout includes several components and measurements, which are detailed as follows: 1. **Voltage Source:** - **Specification:** 480 V - This is the primary AC voltage source supplying the circuit. 2. **Line Impedance:** - **Value:** \(0.18 + j0.24 \, \Omega\) - This represents a series combination of resistance and inductive reactance. 3. **Load:** - **Impedance:** \(4 + j3 \, \Omega\) - This is the load connected to the circuit, comprising both resistive and reactive components. 4. **Line Current Measurement:** - The current flowing through the line is detected and fed into further computations involving power loss. 5. **Calculations for Power Loss (P\(_{\text{Loss}}\)):** - The power loss is computed using the formula with components: - **Line Current (A):** 90.76 (observation from measurement block). - **Squared Current:** This value is squared as \( I^2 \). - **Resistance Component:** Multiplied by 0.18. - **Final Power Loss:** Resulting in 1483 Watts, shown in the computation block. 6. **Load Voltage:** - **Magnitude:** 453.8 V - **Phase Angle:** \(-0.9101\) radians - This is measured across the load, indicating both amplitude and phase shift. 7. **Software Interface:** - **Phasor Block (powergui):** Used to facilitate phasor calculations, specifically for a 60 Hz system. This diagram illustrates the fundamental principles of AC circuit behavior, including phasor relationships, complex impedance, and power loss calculations. This setup is typically used in teaching environments to demonstrate how complex impedances affect current flow and power dissipation in AC circuits, utilizing software for simulation and analysis.
Transcribed Image Text:### Circuit Analysis Diagram This schematic represents an electrical circuit with power calculations using a phasor model at 60 Hz. The layout includes several components and measurements, which are detailed as follows: 1. **Voltage Source:** - **Specification:** 480 V - This is the primary AC voltage source supplying the circuit. 2. **Line Impedance:** - **Value:** \(0.18 + j0.24 \, \Omega\) - This represents a series combination of resistance and inductive reactance. 3. **Load:** - **Impedance:** \(4 + j3 \, \Omega\) - This is the load connected to the circuit, comprising both resistive and reactive components. 4. **Line Current Measurement:** - The current flowing through the line is detected and fed into further computations involving power loss. 5. **Calculations for Power Loss (P\(_{\text{Loss}}\)):** - The power loss is computed using the formula with components: - **Line Current (A):** 90.76 (observation from measurement block). - **Squared Current:** This value is squared as \( I^2 \). - **Resistance Component:** Multiplied by 0.18. - **Final Power Loss:** Resulting in 1483 Watts, shown in the computation block. 6. **Load Voltage:** - **Magnitude:** 453.8 V - **Phase Angle:** \(-0.9101\) radians - This is measured across the load, indicating both amplitude and phase shift. 7. **Software Interface:** - **Phasor Block (powergui):** Used to facilitate phasor calculations, specifically for a 60 Hz system. This diagram illustrates the fundamental principles of AC circuit behavior, including phasor relationships, complex impedance, and power loss calculations. This setup is typically used in teaching environments to demonstrate how complex impedances affect current flow and power dissipation in AC circuits, utilizing software for simulation and analysis.
### Transformer Phasor Diagram The image presents a phasor diagram which is used to analyze transformer voltages and currents in electrical engineering. The diagram includes several vectors representing different electrical quantities: - **Vectors and Angles**: - \( I_s \) is the current phasor, depicted at an angle \( \theta \) from the base axis. - \( V_s \) is the secondary voltage phasor, horizontal on the graph. - \( V_p/a \) represents the primary voltage phasor normalized by the turns ratio (\( a \)). - **Components of Voltage Rise**: - \( R_{eq} I_s \cos \theta \) is the real component of the internal voltage drop due to equivalent resistance. - \( X_{eq} I_s \sin \theta \) is the reactive component of the internal voltage drop due to equivalent reactance. - \( jX_{eq} I_s \) (orthogonal to \( V_s \)) represents the imaginary part of the reactance. - **Equation**: The equation displayed is: \[ \frac{V_p}{a} \approx V_s + R_{eq} I_s \cos \theta + X_{eq} I_s \sin \theta \] This equation represents the approximate voltage transformation relationship factoring in impedance. - **Triangles and Projections**: - The diagram uses right-angled triangles to visually demonstrate how the impedance components add vectorially to the secondary voltage \( V_s \). - \( R_{eq} I_s \) and \( X_{eq} I_s \) are resolved into their horizontal and vertical components, respectively. This visualization is key in understanding the dynamics of voltage transformation in transformers, especially in analyzing the effects of internal impedance.
Transcribed Image Text:### Transformer Phasor Diagram The image presents a phasor diagram which is used to analyze transformer voltages and currents in electrical engineering. The diagram includes several vectors representing different electrical quantities: - **Vectors and Angles**: - \( I_s \) is the current phasor, depicted at an angle \( \theta \) from the base axis. - \( V_s \) is the secondary voltage phasor, horizontal on the graph. - \( V_p/a \) represents the primary voltage phasor normalized by the turns ratio (\( a \)). - **Components of Voltage Rise**: - \( R_{eq} I_s \cos \theta \) is the real component of the internal voltage drop due to equivalent resistance. - \( X_{eq} I_s \sin \theta \) is the reactive component of the internal voltage drop due to equivalent reactance. - \( jX_{eq} I_s \) (orthogonal to \( V_s \)) represents the imaginary part of the reactance. - **Equation**: The equation displayed is: \[ \frac{V_p}{a} \approx V_s + R_{eq} I_s \cos \theta + X_{eq} I_s \sin \theta \] This equation represents the approximate voltage transformation relationship factoring in impedance. - **Triangles and Projections**: - The diagram uses right-angled triangles to visually demonstrate how the impedance components add vectorially to the secondary voltage \( V_s \). - \( R_{eq} I_s \) and \( X_{eq} I_s \) are resolved into their horizontal and vertical components, respectively. This visualization is key in understanding the dynamics of voltage transformation in transformers, especially in analyzing the effects of internal impedance.
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