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What is the phase sequence of each of the following sets of voltages? a)$$\begin{array}{l} v_{\mathrm{a}}=120 \cos \left(\omega t+54^{\circ}\right) \mathrm{V} \\ v_{\mathrm{b}}=120 \cos \left(\omega t-66^{\circ}\right) \mathrm{V} \\ v_{\mathrm{c}}=120 \cos \left(\omega t+174^{\circ}\right) \mathrm{V} \end{array}$$ b)$$\begin{aligned} &v_{\mathrm{a}}=3240 \cos \left(\omega t-26^{\circ}\right) \mathrm{V}\\\ &v_{b}=3240 \cos \left(\omega t+94^{\circ}\right) \mathrm{V}\\\ &v_{\mathrm{c}}=3240 \cos \left(\omega t-146^{\circ}\right) \mathrm{V} \end{aligned}$$

Short Answer

Expert verified
a) The phase sequence is \( v_b \), \( v_a \), \( v_c \). b) The phase sequence is \( v_c \), \( v_a \), \( v_b \).

Step by step solution

01

Extract Angles from Voltages (Set a)

Identify the phase angles from each voltage equation in set (a): - For \( v_{a} \), the angle is \( +54^{\circ} \).- For \( v_{b} \), the angle is \( -66^{\circ} \).- For \( v_{c} \), the angle is \( +174^{\circ} \).
02

Compare Angles (Set a)

Compare the angles: \( -66^{\circ} \), \( +54^{\circ} \), and \( +174^{\circ} \).- The smallest angle is \( -66^{\circ} \).- Next is \( +54^{\circ} \).- Largest is \( +174^{\circ} \).Thus, the phase sequence is \( v_b \), \( v_a \), \( v_c \).
03

Extract Angles from Voltages (Set b)

Identify the phase angles from each voltage equation in set (b): - For \( v_{a} \), the angle is \( -26^{\circ} \).- For \( v_{b} \), the angle is \( +94^{\circ} \).- For \( v_{c} \), the angle is \( -146^{\circ} \).
04

Compare Angles (Set b)

Compare the angles: \( -146^{\circ} \), \( -26^{\circ} \), and \( +94^{\circ} \).- The smallest angle is \( -146^{\circ} \).- Next is \( -26^{\circ} \).- Largest is \( +94^{\circ} \).Thus, the phase sequence is \( v_c \), \( v_a \), \( v_b \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alternating Current (AC) Circuits
Alternating Current (AC) circuits are fundamental in modern electrical systems. Unlike direct current (DC), where the flow of electric charge is unidirectional, AC circuits utilize electricity that changes its direction periodically. This is more efficient for transferring electricity over large distances. Here's why:

  • **Efficiency in Power Transmission:** AC circuits can be easily transformed to higher or lower voltages using transformers. This ability minimizes energy loss over long distances, making it ideal for power distribution.
  • **Variable Voltage:** The ability to change voltage levels is particularly useful in residential and industrial applications, where different appliances require varied voltage levels.
When discussing AC circuits, we often come across terms like phase and frequency. These describe the oscillation characteristics. For instance, the **frequency** (measured in Hertz) indicates how often the direction of current changes per second. In most household systems, the frequency is standardized, like 60 Hz in North America. Understanding these properties helps in analyzing circuit behaviors, especially in multi-phase systems where phase sequence is crucial for synchronous operations.
Phase Angles
Phase angles in AC circuits determine a wave's position at any given point in time. They are crucial because they help us understand how different AC quantities relate to each other. A typical example would be voltage waves in a three-phase system.

  • **Understanding Phase Angle:** The phase angle is indicated in degrees or radians and represents where exactly the wave is in its cycle when compared to a reference point, often referred to as the zero-position or starting point.
  • **Determining Order:** When analyzing sets of voltages, like those in the original exercise, determining the correct phase sequence can prevent potential issues in equipment connected to these circuits, like motors which could rotate in the wrong direction if phased incorrectly.
By extracting the phase angle from voltage equations, as solved previously, one can ascertain the correct sequence by simply arranging them from the smallest to the largest angle. This sequence is critical in systems that employ synchronized multi-phase AC inputs.
Voltage Equations
Voltage equations are invaluable for understanding and predicting the behavior of electrical systems. These equations provide the mathematical expressions that describe voltages as they vary with time in AC circuits. Commonly expressed in cosine or sine form, these equations illustrate several key aspects:

  • **Amplitude:** This is the peak value of the voltage. In the equations given, for example, 120 V and 3240 V represent the maximum magnitude of the voltage.
  • **Phase Shift:** Noted in the form of a plus or minus degree notation, this tells us about the shift of the wave compared to a standard reference.
  • **Angular Frequency:** Represented by the Greek letter omega (\( \omega \)), it correlates with the frequency of the AC supply and is crucial for calculating time-based behaviors.
When handling such equations, understanding how these components play together helps in predicting performances and troubleshooting AC circuits, which is essential for efficient design and operation of electrical setups.

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Most popular questions from this chapter

A balanced, three-phase circuit is characterized as follows: Source voltage in the b-phase is \(20 \angle-90^{\circ} \mathrm{V}\) Source phase sequence is acb; Line impedance is \(1+j 3 \Omega / \phi\) Load impedance is \(117-j 99 \Omega / \phi\) a) Draw the single phase equivalent for the a-phase. b) Calculated the a-phase line current. c) Calculated the a-phase line voltage for the three-phase load.

Three balanced three-phase loads are connected in parallel. Load 1 is \(Y\) -connected with an impedance of \(300+j 100 \Omega / \phi\); load 2 is \(\Delta\) -connected with an impedance of \(5400-j 2700 \Omega / \phi\); and load 3 is \(112.32+j 95.04 \mathrm{kVA} .\) The loads are fed from a distribution line with an impedance of \(1+j 10 \Omega / \phi\) The magnitude of the line-to- neutral voltage at the load end of the line is \(7.2 \mathrm{kV}\) a) Calculate the total complex power at the sending end of the line. b) What percentage of the average power at the sending end of the line is delivered to the loads?

The total power delivered to a balanced throe phase load when operating at a line voliage of \(6600 \sqrt{3} \mathrm{V}\) is \(1188 \mathrm{kW}\) at a lagging power factor of \(0.6 .\) The impedance of the distribution line ing the load is \(0.5+j 4 \Omega / \phi\). Under these supply operating conditions, the drop in the magnitude of the line voltage between the sending end and the load end of the line is excessive. To compensate, a bank of A-connected capacitors is placed in parallel with the load. The capacitor bank is designed to furnish \(1920 \mathrm{kVAR}\) of magnetizing reactive power when operated at a line voltage of \(6600 \sqrt{3} \mathrm{V}\) a) What is the magnitude of the voltage at the sending end of the line when the load is operat. ing at a line voltage of \(6600 \sqrt{3} \mathrm{V}\) and the capac. itor bank is disconnected? b) Repeat (a) with the capacitor bank connected c) What is the average power efficiency of the line in (a)? d) What is the average power efficiency in \((b) ?\) e) If the system is operating at a frequency of \(60 \mathrm{Hz}\) what is the size of each capacitor in microfarads?

The time-domain expressions for three line-to-neutral voltages at the terminals of a Y-connected load are $$\begin{array}{l} v_{\mathrm{AN}}=7967 \cos \omega t \mathrm{V} \\ v_{\mathrm{BN}}=7967 \cos \left(\omega t+120^{\circ}\right) \mathrm{V} \\ v_{\mathrm{CN}}=7967 \cos \left(\omega t-120^{\circ}\right) \mathrm{V} \end{array}$$ What are the time-domain expressions for the three line-to-line voltages \(v_{\mathrm{AB}}, v_{\mathrm{BC}},\) and \(v_{\mathrm{CA}} ?\)

The total apparent power supplied in a balanced. three-phase \(Y\) - \(\Delta\) system is 3600 VA. The line volit age is 208 V. If the line impedance is negligible and the power factor angle of the load is \(25^{\circ}\), determine the impedance of the load.

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