Chapter 11: Problem 14
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
Step by step solution
Identify Source Voltage for a-phase
Calculate Total Impedance for the a-phase
Calculate the a-phase Line Current
Calculate the a-phase Line Voltage for the Load
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Source voltage calculations
Impedance in electric circuits
- 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.
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
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
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.