Question

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

Short answer

\( v_{AB} = -13796 \sin(\omega t + 60^\circ) \), \( v_{BC} = -13796 \sin(\omega t) \), \( v_{CA} = 13796 \sin(\omega t - 60^\circ) \).

Step-by-step solution

Convert Line-to-Neutral to Line-to-Line

The line-to-line voltage (e.g., \( v_{AB} \)) in a Y-connected load is found by subtracting one line-to-neutral voltage from another. Specifically, \( v_{AB} = v_{AN} - v_{BN} \), \( v_{BC} = v_{BN} - v_{CN} \), and \( v_{CA} = v_{CN} - v_{AN} \). We proceed to apply this to obtain each line-to-line voltage.

Calculate Line Voltage v_AB

Using \( v_{AB} = v_{AN} - v_{BN} \), we substitute:\[ v_{AB} = 7967 \, \cos \omega t - 7967 \, \cos(\omega t + 120^\circ) \]Apply the identity \( \cos A - \cos B = -2 \, \sin\left(\frac{A+B}{2}\right) \, \sin\left(\frac{A-B}{2}\right) \):\[ v_{AB} = -15934 \, \sin\left(\omega t + 60^\circ\right) \sin(60^\circ) \]Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \),\[ v_{AB} = -15934 \, \sin\left(\omega t + 60^\circ\right) \times \frac{\sqrt{3}}{2} \approx -13796 \, \sin\left(\omega t + 60^\circ\right) \]

Calculate Line Voltage v_BC

Using \( v_{BC} = v_{BN} - v_{CN} \), we substitute:\[ v_{BC} = 7967 \, \cos(\omega t + 120^\circ) - 7967 \, \cos(\omega t - 120^\circ) \]Using the identity \( \cos A - \cos B = -2 \, \sin\left(\frac{A+B}{2}\right) \, \sin\left(\frac{A-B}{2}\right) \):\[ v_{BC} = -15934 \, \sin(\omega t) \, \sin(120^\circ) \]Since \( \sin(120^\circ) = \frac{\sqrt{3}}{2} \),\[ v_{BC} = -15934 \, \sin(\omega t) \times \frac{\sqrt{3}}{2} \approx -13796 \, \sin(\omega t) \]

Calculate Line Voltage v_CA

Using \( v_{CA} = v_{CN} - v_{AN} \), we substitute:\[ v_{CA} = 7967 \, \cos(\omega t - 120^\circ) - 7967 \, \cos(\omega t) \]Applying the identity \( \cos A - \cos B = -2 \, \sin\left(\frac{A+B}{2}\right) \, \sin\left(\frac{A-B}{2}\right) \):\[ v_{CA} = -15934 \, \sin\left(\omega t - 60^\circ\right) \sin(-60^\circ) \]Since \( \sin(-60^\circ) = -\frac{\sqrt{3}}{2} \),\[ v_{CA} = 15934 \, \sin\left(\omega t - 60^\circ\right) \times \frac{\sqrt{3}}{2} \approx 13796 \, \sin\left(\omega t - 60^\circ\right) \]

Key concepts

Y-connected load

In an AC circuit, a **Y-connected load** is a common configuration used for connecting three electrical loads in a symmetrical pattern, resembling the shape of the letter "Y". It's also called a "star configuration."
This configuration involves three components: each component's terminal is connected to a common central point or neutral point. The main advantages of a Y-connected load are:

• It allows the use of a neutral wire, which provides a path for fault currents and can aid in balancing loads.
• It's often used in systems where a mixture of single-phase and three-phase loads is required.

In the specific context of the given exercise, the Y-connected load involves three resistive or inductive elements, where the voltages are denoted as line-to-neutral voltages: \( v_{AN}, v_{BN}, v_{CN} \). Understanding how to use and calculate these voltages is crucial for the analysis of power systems.

Line-to-line voltage

**Line-to-line voltage** in a three-phase Y-connected system refers to the voltage between any two lines, and it is distinct from line-to-neutral voltage, which is the voltage between a line and the neutral point.
The relation between these voltages in a balanced Y system is: the line-to-line voltage is \( \sqrt{3} \) times the line-to-neutral voltage. This calculation can be derived from trigonometric laws.
Here's a breakdown of how to calculate this for each pair, using the exercises' example:

• Voltage \( v_{AB} \) is found by subtracting \( v_{BN} \) from \( v_{AN} \).
• Voltage \( v_{BC} \) involves subtracting \( v_{CN} \) from \( v_{BN} \).
• Voltage \( v_{CA} \) comes from subtracting \( v_{AN} \) from \( v_{CN} \).

These relationships help identify the correct voltages in the system, making them crucial for load and power calculations.

Trigonometric identities

**Trigonometric identities** are mathematical equations that simplify calculations involving trigonometric functions like sine and cosine. In the realm of electrical engineering, especially in AC circuit analysis, these identities are paramount.
For converting line-to-neutral voltages into line-to-line voltages, the identity \( \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \) plays a significant role. It helps in simplifying expressions when dealing with phase angle differences.
For example:

• In calculating \( v_{AB} \), the identity helps in breaking down the cosine terms into a product of sine terms that are more straightforward to work with.
• The same identity applies while calculating both \( v_{BC} \) and \( v_{CA} \), aiding in expressing these voltages more manageably.

Memorizing and understanding these identities can make complex problems more approachable and significantly streamline calculations.

Phase angle difference

**Phase angle difference** refers to the angular displacement between two alternating quantities with the same frequency, such as voltages or currents. In AC circuits, it's vital to understand phase angles as they affect power and energy distribution.
Each of the line-to-neutral voltages in this exercise has its phase angle:

• \( v_{AN} \) serves as the reference phase.
• \( v_{BN} \) differs by +120° from \( v_{AN} \).
• \( v_{CN} \) is -120° relative to \( v_{AN} \).

By using the angle difference, we can analyze the time shifts between various voltages, which is critical when subtracting these voltages to find line-to-line voltages.
Recognizing and applying the concept of phase angles effectively allows engineers to calculate not only voltages but also solves other relevant problems in AC power systems.