Question

An abc sequence balanced three-phase \(Y\) -connected source supplies power to a balanced, three-phase \(\Delta\) -connected load with an impedance of \(12+j 9 \Omega / \phi\). The source voltage in the a-phase is \(120 / 80^{\circ} \mathrm{V} .\) The line impedance is \(1+\mathrm{j} 1 \Omega / \phi\) Draw the single phase equivalent circuit for the a-phase and use it to find the current in the a-phase of the load.

Short answer

The a-phase load current is approximately 7.31 A at 42.15°.

Step-by-step solution

Identify System Components

First, recognize that the system consists of a source and load connected by a line impedance. The source is Y-connected and the load is Δ-connected. The load impedance is given as \(12 + j9 \text{ Ω} / \phi\). The line impedance is \(1 + j1 \text{ Ω} / \phi\). The source voltage for the a-phase is \(120 / 80^{\circ} \text{ V}\).

Convert Line to Phase Voltage

In a Y-connected system, the phase voltage \(V_{p}\) is \(V_{a}\) because the system voltage given is the phase voltage (120V magnitude, 80 degrees angle).

Calculate Total Impedance

For the Δ-connected load, convert the given impedance per phase. The load impedance \(Z_{L} = 12 + j9 \text{ Ω}\). Include the line impedance in series: \(Z_{T} = (12 + j9) + (1 + j1) = 13 + j10 \text{ Ω}\).

Use Ohm’s Law for Phase Current

Ohm's Law states \(I = \frac{V}{Z}\). The phase current can be found by dividing the source phase voltage by the total impedance:\[ I_{a} = \frac{120 / 80^{\circ}}{13 + j10} \].

Calculate Load Current

Convert the impedance \(Z_{T}\) into polar form using \(Z_{T} = 16.40 / 37.85^{\circ} \) (calculated using magnitude \( \sqrt{13^2 + 10^2} \) and angle \(\tan^{-1}\frac{10}{13}\)). Compute the current:\[ I_{a} = \frac{120 / 80^{\circ}}{16.40 / 37.85^{\circ}} \]\[ I_{a} = 120 \angle 80^{\circ} - 16.40 \angle 37.85^{\circ} \]\[ I_{a} \approx 7.31 \angle 42.15^{\circ} \text{ A} \].

Interpret the Result

The calculated phase current is \(7.31 \text{ A} \) with an angle of \(42.15^{\circ}\) with respect to the source voltage. This is the current entering the load in the a-phase.

Key concepts

Three-Phase Circuits

Three-phase circuits are systems used extensively in electrical power generation and distribution. They consist of three separate alternating currents that are out of phase by 120 degrees. This design helps optimize power delivery and efficiency.

In a balanced three-phase system, the loads are equally divided among the phases, which results in more stable voltage levels and improved performance. Three-phase systems are common in industrial and commercial settings because they can deliver more power compared to single-phase systems.

• Each phase is represented by a separate circuit with its own voltage and current.
• These circuits are interconnected to provide a consistent power supply.
• Three-phase circuits can be represented in two main configurations: Y (star) and Δ (delta).
• In our exercise, the source is Y-connected, and the load is Δ-connected.

Impedance Calculation

Impedance is a critical factor in analyzing electrical circuits, especially in AC systems. It is defined as the total opposition that a circuit offers to the flow of alternating current, consisting of resistance and reactance.

Calculating total impedance in the circuit is essential for determining how current and voltage will behave. Impedance is typically represented as a complex number, with a real component (resistance) and an imaginary component (reactance), such as in the example of a load impedance of 12 + j9 Ω per phase.

• To calculate the total impedance, sum the line impedance and the load impedance.
• In our exercise, the calculated total impedance is 13 + j10 Ω.
• Impedance can be converted to polar form, which simplifies calculations when dealing with AC voltages and currents.

Ohm's Law

Ohm’s Law is fundamental in electric circuit analysis, providing a straightforward relationship between voltage, current, and impedance. In three-phase systems, Ohm’s Law is applied to find phase currents using the formula: \[ I = \frac{V}{Z} \] where \( I \) is the current, \( V \) is the voltage, and \( Z \) is the impedance.

In our problem, we apply Ohm’s Law to find the current flowing through the a-phase of the load. With a phase voltage of 120V at 80° and a total impedance of 13 + j10 Ω, the phase current can be determined:

• First, convert impedance to polar form (16.40 / 37.85°).
• Then calculate the current using the formula, factoring in both the magnitude and angle of voltage and impedance.
• The resulting current provides insights into the behavior of the circuit.

Delta and Y Connections

Delta (Δ) and Y (star) connections are two methods for configuring the components in a three-phase system. Each has its own characteristics and advantages.

In a Δ-connected system, the ends of each component are connected to form a closed loop. This arrangement is common in motor applications and certain types of loads. It allows for lower line currents and is efficient for certain power levels.
In a Y-connected system, each component is connected to a common point, forming a Y shape. This configuration is preferred for power transmission because it provides a neutral point, which can be earthed.

Understanding these connections is essential in solving exercises involving three-phase systems, like the one we have here:

• Our given system uses a Y-connected source and a Δ-connected load.
• The conversion between Δ and Y connections: Remember that a direct conversion might not be needed if the loads are balanced, but it’s crucial to identify what each setup means for power distribution.
• In Δ connections, each phase sees the full line voltage, whereas in Y connections, each phase sees a voltage that is √3 times smaller.

Knowledge of these connections helps in correctly analyzing and computing values in complex circuits.