Question

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

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

Step-by-step solution

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:\[ V_{ph} = \frac{12,800}{\sqrt{3}} \approx 7391.55 \text{ V} \]

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:\[ I_{ph} = \frac{7391.55}{225.84} \approx 32.73 \text{ A} \]

Calculate the Line Current

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

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:\[ V_{drop} = 32.73 \times 2.015 \approx 65.94 \text{ V} \]

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:\[ V_{source} = 12,800 + 65.94 \approx 12,865.94 \text{ V} \]

Key concepts

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.