Question

A balance three phase power system, consist of AC genenator, Y connection with V(AB)=480<0°V,50 Hz Which suppling two series loads through a transmission line with (0.5+j0.7)Ω impedance, where:

load 1 is Δ conmection and has impedance is Z₁=(3.0−j6.0)Ω,
load 2 is Y connection and has impedanee is Z₂=(2.0+j1.5)Ω.

Draw a 3 phase circuit of the system

Answer 1

The question involves drawing a balanced three-phase AC circuit that includes an AC generator, transmission line, and two series loads. The transmission line has an impedance of (0.5+j0.7) Ω. Load 1 is in delta with impedance Z₁=(3.0−6.0) Ω, and Load 2 is in wye with impedance Z₂=(2.0+j1.5) Ω.

Step-by-step explanation:

The student's question pertains to a balanced three-phase power system with a transmission line and two different loads connected in delta (Δ) and wye (Y) configurations respectively. In the provided power system, the AC generator with a phase voltage of V(AB) equals 480∠0°V at a frequency of 50 Hz is supplying power to the loads through the transmission line having an impedance of (0.5+j0.7) Ω. Load 1 connected in delta has an impedance of Z₁=(3.0−6.0) Ω, and Load 2 connected in wye has an impedance of Z₂=(2.0+j1.5) Ω.

While specific details of how to draw the circuit are not provided in the answer, the student is expected to draw a three-phase circuit that includes:

• A three-phase generator with a Y-connection.
• A transmission line with each phase having the given impedance.
• Two loads, Load 1 in delta configuration and Load 2 in wye configuration, each with the described impedance values.

For reference, the student can consider how to calculate the voltages, currents, and power for individual circuit elements like resistors, inductors, and capacitors in an AC circuit.