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The magnitude of the line voltage at the terminals of a balanced \(Y\) -connected load is 12,800 V. The load impedance is \(216+j 63 \Omega / \phi\). The load is fed from a line that has an impedance of \(0.25+j 2 \Omega / \phi\) a) What is the magnitude of the line current? b) What is the magnitude of the line voltage at the source?

Short Answer

Expert verified
a) 32.73 A, b) 12,865.94 V.

Step by step solution

01

Convert Line Voltage to Phase Voltage

Since the load is Y-connected, we need to convert the line voltage to phase voltage. The relationship between line and phase voltage in a Y-connection is given by:\[ V_{ph} = \frac{V_L}{\sqrt{3}} \]Substitute the line voltage:\[ V_{ph} = \frac{12,800}{\sqrt{3}} \approx 7391.55 \text{ V} \]
02

Calculate the Phase Current

Next, calculate the phase current using Ohm's Law, which states that the current is the voltage divided by the impedance:\[ I_{ph} = \frac{V_{ph}}{Z_{load}} \]Where \( Z_{load} = 216 + j63 \Omega \). The magnitude of the impedance can be calculated as:\[ |Z_{load}| = \sqrt{216^2 + 63^2} \approx 225.84 \Omega \]Thus, the phase current becomes:\[ I_{ph} = \frac{7391.55}{225.84} \approx 32.73 \text{ A} \]
03

Calculate the Line Current

Since the load is balanced and Y-connected, the line current is equal to the phase current:\[ I_L = I_{ph} = 32.73 \text{ A} \]
04

Calculate Voltage Drop Across the Line Impedance

Calculate the voltage drop across the line impedance using Ohm's Law:\[ V_{drop} = I_L \times Z_{line} \]Where \( Z_{line} = 0.25 + j2 \Omega \). Calculate the magnitude of \( Z_{line} \):\[ |Z_{line}| = \sqrt{0.25^2 + 2^2} \approx 2.015 \Omega \]Hence, the magnitude of the voltage drop becomes:\[ V_{drop} = 32.73 \times 2.015 \approx 65.94 \text{ V} \]
05

Calculate the Line Voltage at the Source

Finally, calculate the magnitude of the line voltage at the source by adding the voltage drop to the magnitude of the line voltage at the load:\[ V_{source} = V_{load} + V_{drop} \]Since these are line voltages for a Y-connected system, we use:\[ V_{source} = 12,800 + 65.94 \approx 12,865.94 \text{ V} \]

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Key Concepts

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

Y-connected load
In electrical engineering, a Y-connected load is a common configuration in three-phase circuits. It is also referred to as a "star" connection because the wires and loads connect at a central point, forming a shape similar to the letter "Y." This connection is popular because it allows for two voltages to be utilized. You can access both line-to-line voltages and line-to-neutral (or phase) voltages.
  • In a Y-connected load, the phases are connected to each of the three lines that meet at a central neutral point.
  • Each load impedance (in this exercise, given as \(216 + j63 \Omega \)) is connected in a separate branch
  • The central point (neutral) often remains unconnected or may be connected to a ground reference.
The **primary advantage** of a Y-connection is that it requires fewer conductors compared to other configurations, such as the delta (Δ) configuration. This makes it both cost-effective and efficient for transmitting three-phase power over long distances.
Understanding the Y-connection is crucial, especially when calculating electrical quantities like phase currents and phase voltages, as these differ from line currents and line voltages.
Line voltage and phase voltage
To solve problems involving Y-connected loads, it's essential to understand the relationship between line voltage and phase voltage. This distinction plays a significant role in many calculations.
  • **Line Voltage \(V_L\):** This is the voltage measured across any two lines in the three-phase system. In the exercise provided, the line voltage is given as 12,800 V.
  • **Phase Voltage \(V_{ph}\):** This is the voltage measured between any one line and the neutral point of the Y-connection.
The relationship between these two voltages is
\[ V_{ph} = \frac{V_L}{\sqrt{3}} \] This formula shows that the phase voltage is lower than the line voltage by a factor of \( \sqrt{3} \).
This difference occurs due to the geometrical nature of how the Y-connection distributes voltage among its phases. It is crucial to apply this relationship when moving between line and phase calculations in a Y-connected system.
Impedance in AC circuits
Impedance is a fundamental concept in AC circuits, representing the opposition a circuit offers to the flow of alternating current. In Y-connected loads, understanding impedance is mandatory for calculating currents and voltages accurately.
  • **Impedance \(Z\):** It is a complex quantity, generally represented as \(Z = R + jX\), where \(R\) is the resistance and \(X\) is the reactance.
  • The magnitude of the impedance, essential for calculations, is given by \(|Z| = \sqrt{R^2 + X^2}\).
In the exercise, the load impedance is \(216 + j63 \Omega\) per phase. The magnitude is computed to facilitate the calculation of phase currents using Ohm's Law:
\[ I_{ph} = \frac{V_{ph}}{|Z_{load}|} \] Impedance also affects how voltage drops across lines and loads. In this problem, the line impedance \((0.25 + j2 \Omega)\) influences the total line voltage. This aspect is essential for ensuring that the electrical system functions correctly, allowing engineers to design reliable three-phase systems.

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Most popular questions from this chapter

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?

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)$$\begin{aligned} &v_{\mathrm{a}}=3240 \cos \left(\omega t-26^{\circ}\right) \mathrm{V}\\\ &v_{b}=3240 \cos \left(\omega t+94^{\circ}\right) \mathrm{V}\\\ &v_{\mathrm{c}}=3240 \cos \left(\omega t-146^{\circ}\right) \mathrm{V} \end{aligned}$$

In a balanced three-phase system, the source has an abc sequence, is Y-connected, and \(\mathbf{V}_{\mathrm{an}}=120 / 20^{\circ} \mathrm{V} .\) The source feeds two loads, both of which are \(Y\) -connected. The impedance of load 1 is \(8+j 6 \Omega / \phi\). The complex power for the a \(\cdot\) phase of load 2 is \(600 / 36^{\circ} \mathrm{VA}\). Find the total complex power supplied by the source.

For each set of voltages, state whether or not the voltages form a balanced three-phase set. If the set is balanced, state whether the phase sequence is positive or negative. If the set is not balanced, explain why. a) \(v_{\mathrm{a}}=339 \cos 377 t \mathrm{V}\) \(v_{\mathrm{b}}=339 \cos \left(377 t-120^{\circ}\right) \mathrm{V}\) \(v_{\mathrm{c}}=339 \cos \left(377 t+120^{\circ}\right) \mathrm{V}\) b) \(v_{\mathrm{a}}=622 \sin 377 t \mathrm{V}\) \(v_{b}=622 \sin \left(377 t-240^{\circ}\right) \mathrm{V}\) \(v_{\mathrm{c}}=622 \sin \left(377 t+240^{\circ}\right) \mathrm{V}\) c) \(v_{\mathrm{a}}=933 \sin 377 t \mathrm{V}\) \(v_{b}=933 \sin \left(377 t+240^{\circ}\right) \mathrm{V}\) \(v_{\mathrm{c}}=933 \cos \left(377 t+30^{\circ}\right) \mathrm{V}\) d) \(v_{a}=170 \sin \left(\omega t+60^{\circ}\right) \mathrm{V}\) \(v_{b}=170 \sin \left(\omega t+180^{\prime \prime}\right) V\) \(v_{c}=170 \cos \left(\omega t-150^{\circ}\right) \mathrm{V}\) e) \(v_{a}=339 \cos \left(\omega t+30^{\circ}\right) \mathrm{V}\) \(v_{b}=339 \cos \left(\omega t-90^{\circ}\right) V\) \(v_{c}=393 \cos \left(\omega t+240^{\circ}\right) \mathrm{V}\) f) \(v_{\mathrm{a}}=3394 \sin \left(\omega t+70^{\circ}\right) \mathrm{V}\) \(v_{b}=3394 \cos \left(\omega t-140^{\circ}\right) \mathrm{V}\) \(v_{c}=3394 \cos \left(\omega t+180^{\circ}\right) \mathrm{V}\)

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 \(v_{\mathrm{CA}} ?\)

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