Question

A balanced three-phase circuit has the following characteristics: Y-Y connected; The line voltage at the source, \(V_{b b, i}\) \(120 \sqrt{3} / 0^{\circ} \mathrm{V}\) The phase sequence is positive; The line impedance is \(2+j 3 \Omega / \phi\) The load impedance is \(28+j 37 \Omega /\) a) Draw the single phase equivalent circuit for the a-phase b) Calculated the line current in the a-phase c) Calculated the line voltage at the load in the a-phase

Short answer

The line current is \(2.4\angle -53.13^\circ \) A, and the voltage at the load is determined by subtracting the line voltage drop.

Step-by-step solution

Understand the Problem

To solve the problem, we need to draw the single-phase equivalent circuit for a balanced Y-Y connected circuit and find the line current and the line voltage at the load. Given are the source voltage, line and load impedance.

Drawing the Single Phase Equivalent Circuit

For a Y-Y system, the single-phase equivalent circuit for the a-phase shows only one phase of the system. Represent the source as a voltage source with internal impedance, the transmission line with given line impedance, and the load with its impedance.

Identify Source Voltage

The line-to-line voltage is given as \( V_{bb,i} = 120\sqrt{3} / 0^{\circ} \mathrm{V} \). Convert this to line-to-neutral voltage \( V_{an,i} \) using \( V_{an,i} = \frac{120\sqrt{3}}{\sqrt{3}} e^{j0^\circ} = 120 \angle 0^\circ \mathrm{V} \).

Calculate Impedances of Each Component

The line impedance \( Z_l = 2 + j3\, \Omega \) and the load impedance \( Z_{load} = 28 + j37\, \Omega \) are given directly.

Determine Total Impedance

The total impedance in series is the sum of the line impedance and load impedance: \( Z_{total} = Z_l + Z_{load} = (2 + j3) + (28 + j37) = 30 + j40\, \Omega \).

Calculate the Line Current in a-phase

Using Ohm's Law, \( I_a = \frac{V_{an,i}}{Z_{total}} = \frac{120\angle 0^\circ}{30 + j40} \). First, convert \( Z_{total} \) to polar form: \(|Z_{total}| = \sqrt{30^2 + 40^2} = 50\, \Omega \), and \(\theta = \tan^{-1}(\frac{40}{30}) = 53.13^\circ \). Hence, \( Z_{total} = 50\angle 53.13^\circ \). So, \( I_a = \frac{120\angle 0^\circ}{50\angle 53.13^\circ} = 2.4\angle -53.13^\circ \mathrm{A} \).

Calculate the Line Voltage at the Load in a-phase

The voltage drop along the line is \( V_{line} = I_a \times Z_l \). Substitute the values: \( V_{line} = 2.4\angle -53.13^\circ \times (2 + j3) \). Convert \( 2 + j3 \) to polar form: \(|Z_l| = \sqrt{2^2 + 3^2} = \sqrt{13} \), \( \theta = \tan^{-1}(\frac{3}{2}) = 56.31^\circ \), \( Z_l = \sqrt{13}\angle 56.31^\circ \). Thus, \( V_{line} = 2.4\angle -53.13^\circ \times \sqrt{13}\angle 56.31^\circ \). The magnitude is \( 2.4\sqrt{13} \) and angle is \( -53.13^\circ + 56.31^\circ = 3.18^\circ \). Finally, \( V_{load} = V_{an,i} - V_{line} = 120\angle 0^\circ - 8.54\angle 3.18^\circ \). Approximate via vector subtraction.

Key concepts

Impedance calculation

In any electrical circuit, impedance plays a crucial role. Impedance is a measure of resistance that incorporates both resistive and reactive components. In three-phase circuits, calculating the total impedance is essential for understanding how electricity flows through the circuit. This total impedance is determined by adding together the impedances of all individual components involved.

In the specific case we're discussing, each phase in the Y-Y connection features a line impedance of \(2 + j3 \;\Omega\) and a load impedance of \(28 + j37 \;\Omega\). These values are expressed in rectangular form, indicating their real and imaginary parts. To find the total impedance \(Z_{total}\) experienced by a single phase, you sum the line and load impedances as they are connected in series.

• Real part: \(2 + 28 = 30\)
• Imaginary part: \(3 + 37 = 40\)

Thus, the total impedance in rectangular form becomes \(30 + j40 \;\Omega\). To convert this to polar form, which can be more useful for calculations, compute both the magnitude and phase angle. The magnitude is given by \( \sqrt{30^2 + 40^2} = 50 \;\Omega\) and the phase angle by \(\tan^{-1}(\frac{40}{30}) = 53.13^\circ\).

Ohm's Law

Ohm’s Law is fundamental in electrical engineering, representing a simple relationship between voltage, current, and impedance (or resistance). The law is expressed as \(V = I \cdot Z\), where \(V\) is voltage, \(I\) is current, and \(Z\) is impedance. In three-phase circuits, this principle is vital for understanding how the electrical parameters interact within each phase.

For our problem, using Ohm's Law allows us to calculate the line current \(I_a\) in the a-phase. With the known line-to-neutral voltage \(V_{an,i}\) and total impedance \(Z_{total}\), rearranging Ohm’s Law provides the formula \(I = \frac{V}{Z}\). Therefore, the line current \(I_a\) is given by:\[I_a = \frac{120\angle 0^\circ}{50\angle 53.13^\circ}\]
Once the total impedance is in polar form, as \(50\angle 53.13^\circ\), it’s easy to divide this into the given voltage. The resulting current \( I_a \) would be \(2.4\angle -53.13^\circ \;\mathrm{A}\), highlighting how phase angles affect both magnitude and direction in AC circuits.

Y-Y connection

In three-phase systems, the Y-Y (or wye-wye) connection is a common configuration, connecting both the source and the load in a wye formation. This creates a balanced system if each phase has equal impedance, as given in the problem statement.

A Y-Y connection typically involves three phases connected to a common neutral point. This setup has several advantages:

• Reduced insulation requirements due to lower voltage between phases and neutral
• Facilitates the use of a neutral wire for grounding safety
• Simplifies voltage transformations

In a Y-Y system, the line voltage is related to the phase voltage. For example, the line-to-line voltage \( V_{bb,i} \) is \(120 \sqrt{3} / 0^{\circ} \mathrm{V}\). Converting this to the line-to-neutral or phase voltage \(V_{an,i}\) requires dividing by \(\sqrt{3}\), resulting in \(120\angle 0^\circ \;\mathrm{V}\). This conversion is critical in analyzing and solving Y-Y connected circuit problems.

Voltage in a-phase

Understanding the voltage in the a-phase is vital for solving problems in three-phase circuits. The a-phase voltage directly impacts the current flow and, consequently, the performance of the entire circuit. When analyzing three-phase circuits, particularly in a Y-Y connection, it starts with knowing the line-to-neutral voltage.

In this problem, the initial given is a line-to-line voltage \( V_{bb,i} = 120\sqrt{3}/0^\circ \;\mathrm{V}\). Converting this to line-to-neutral voltage helps us understand the voltage across each phase. Using the formula:\[V_{an,i} = \frac{V_{bb,i}}{\sqrt{3}} = 120 \angle 0^\circ \;\mathrm{V}\]With this conversion, you work within the phase domain, which is crucial when analyzing how the voltage drop happens along the line and affects the load. Such insights facilitate precise calculations in electrical engineering and help diagnose performance issues in the circuit.

Line current calculation

Calculating the line current in a Y-Y connected three-phase system using the information given involves understanding both the impedance characteristics and applying Ohm's Law. This is significant because the line current influences the energy flow throughout the circuit and helps maintain a balanced load.

After determining the source voltage as a phase voltage of \(120\angle 0^\circ \;\mathrm{V}\) and the total impedance \(50 \angle 53.13^\circ\), you proceed by applying Ohm’s Law:\[I_a = \frac{120\angle 0^\circ}{50\angle 53.13^\circ} = 2.4\angle -53.13^\circ \;\mathrm{A}\]
This mathematical process translates the electrical parameters to real, applicable outcomes. Note how the negative angle of the current signals a lag between voltage and current typical of an inductive load, where the current lags the voltage. Recognizing such patterns is essential for troubleshooting and optimizing circuit performance.