Question

A balanced, three-phase circuit is characterized as follows: Source voltage in the b-phase is \(20 \angle-90^{\circ} \mathrm{V}\) Source phase sequence is acb; Line impedance is \(1+j 3 \Omega / \phi\) Load impedance is \(117-j 99 \Omega / \phi\) a) Draw the single phase equivalent for the a-phase. b) Calculated the a-phase line current. c) Calculated the a-phase line voltage for the three-phase load.

Short answer

a-phase line current: 0.13 ∠ 69.72° A, a-phase line voltage: 19.9 ∠ 29° V.

Step-by-step solution

Identify Source Voltage for a-phase

Given that the source sequence is acb, the source voltage for a-phase can be determined. Since we know the b-phase voltage is given as \( 20 \angle -90^{\circ} \mathrm{V} \), and the sequence is acb (anti-clockwise), we apply a phase shift of \(-120^{\circ}\). Thus, the a-phase source voltage is \( V_a = 20 \angle -90^{\circ} + 120^{\circ} = 20 \angle 30^{\circ} \mathrm{V} \).

Calculate Total Impedance for the a-phase

In a single-phase equivalent of a three-phase circuit, we add the line impedance and the load impedance. The line impedance is \( Z_{line} = 1 + j3 \Omega \) and the load impedance is \( Z_{load} = 117 - j99 \Omega \). Therefore, the total impedance \( Z_{tot} \) is computed by adding these two: \( Z_{tot} = (1 + j3) + (117 - j99) = 118 - j96 \Omega \).

Calculate the a-phase Line Current

Using Ohm's Law (\( I = \frac{V}{Z} \)), calculate the line current for the a-phase. The voltage \( V_a = 20 \angle 30^{\circ} \mathrm{V} \) and impedance \( Z_{tot} = 118 - j96 \Omega \) can be used: \[ I_a = \frac{20 \angle 30^{\circ}}{118 - j96} \]. First, compute the magnitude \( |Z_{tot}| = \sqrt{118^2 + (-96)^2} = 153 \Omega \) and the angle \( \theta = \tan^{-1}\left(\frac{-96}{118}\right) \approx -39.72^{\circ} \). Thus, \( I_a = \frac{20}{153} \angle (30^{\circ} + 39.72^{\circ}) \approx 0.13 \angle 69.72^{\circ} \mathrm{A} \).

Calculate the a-phase Line Voltage for the Load

The line voltage across the load for the a-phase can be found using Voltage Divider Rule: \( V_{La} = I_a \cdot Z_{load} \). Substitute \( I_a = 0.13 \angle 69.72^{\circ} \mathrm{A} \) and \( Z_{load} = 117 - j99 \Omega \): \( V_{La} = 0.13 \angle 69.72^{\circ} \mathrm{A} \times (117 - j99) \). The magnitude \( |Z_{load}| = \sqrt{117^2 + (-99)^2} = 153 \Omega \) and the angle \( \theta = \tan^{-1}\left(\frac{-99}{117}\right) \approx -40.72^{\circ} \). Hence \( V_{La} \approx 19.9 \angle (69.72^{\circ} - 40.72^{\circ}) \approx 19.9 \angle 29^{\circ} \mathrm{V} \).

Key concepts

Source voltage calculations

In three-phase circuit analysis, determining the proper source voltage for each phase is crucial. Each phase voltage must be calculated based on the given sequence and known phase voltages. In our example, the b-phase voltage is known as \(20 \angle -90^{\circ}\text{ V}\) and the sequence is acb. This sequence implies an anti-clockwise arrangement. To find the a-phase voltage, we introduce a phase shift of \(-120^{\circ}\) from b-phase. Thus, the calculation is \(V_a = 20 \angle -90^{\circ} + 120^{\circ} = 20 \angle 30^{\circ} \text{ V}\). Understanding these sequence relationships is vital. Adding or subtracting the correct phase angle ensures that everything remains balanced, a key feature in three-phase systems.

Impedance in electric circuits

Impedance plays a critical role in understanding how currents and voltages behave in AC circuits. It represents the total opposition a circuit offers to the flow of alternating current and is comprised of two components: resistance (R) and reactance (X).

• Resistance is the real part and does not change with frequency.
• Reactance is the imaginary part and depends on frequency, indicating how inductors and capacitors oppose AC.

To find the total impedance in a circuit, sum the resistances and reactances of the components.
In our exercise, the line and load impedances are added to form a total impedance \(Z_{tot} = 118 - j96 \Omega\). Understanding impedance is crucial for determining how voltages and currents interact within the circuit.

Ohm's Law

Ohm's Law is a fundamental principle that links current (I), voltage (V), and impedance (Z) in AC circuits. It's expressed as \(I = \frac{V}{Z}\). This formula helps calculate unknown values when two of the parameters are known.
In the given problem, Ohm's Law is used to find the a-phase line current, \(I_a\). With the source voltage \(V_a = 20 \angle 30^{\circ} \text{ V}\) and total impedance \(Z_{tot} = 118 - j96 \Omega\), we use Ohm's Law to compute:
\[ I_a = \frac{20 \angle 30^{\circ}}{118 - j96} \]First, calculate the magnitude \(|Z_{tot}| = 153 \Omega\) and angle \(\theta = \text{atan} \left( \frac{-96}{118} \right) \approx -39.72^{\circ}\). Finally, \(I_a = 0.13 \angle 69.72^{\circ} \text{ A}\). This calculation underscores the relationship between voltage, impedance, and current passing through the circuit.

Voltage divider rule

The Voltage Divider Rule is a handy method for determining the voltage across a particular component in series within a circuit.
It states that the voltage across a load can be found using the current flowing through it and its impedance.
In the context of our exercise, the a-phase line voltage \(V_{La}\) is calculated as follows:
\[ V_{La} = I_a \cdot Z_{load} \]Given the line current \(I_a = 0.13 \angle 69.72^{\circ}\text{ A}\) and load impedance \(Z_{load} = 117 - j99 \Omega\), calculate
\[ |Z_{load}| = \sqrt{117^2 + (-99)^2} = 153 \Omega \]and the angle \(\theta = \tan^{-1}\left(\frac{-99}{117}\right) \approx -40.72^{\circ}\). Finally, compute\( V_{La} \approx 19.9 \angle 29^{\circ} \text{ V}\).The Voltage Divider Rule simplifies understanding of how voltage varies across components in a complex circuit.