Question

An acb sequence balanced three-phase Y-connected source supplies power to a balanced three- phase A-connected load with an impedance of 12 + 315 Ω/ᵠ. The source voltage in the b-phase is 2302–50°V. The line impedance is 1+j1 Ω/ᵠ. Use the single-phase equivalent circuit for the a- phase to find the current in the a-phase of the load. Express your answer in amperes to three significant figures.

Enter your answer using angle notation. Express argument in degrees.

Answer 1

The current in the a-phase of the load is 1528.298 A at a phase angle of -0.304°.

Step-by-step explanation:

To find the current in the a-phase of the load, we can use the single-phase equivalent circuit for the a-phase. The source voltage in the b-phase is given as 2302-50°V.

We can calculate the total impedance of the load by adding the impedance of the load and the line impedance:

Ztotal = Zload + Zline

Substituting the given values:

Ztotal = (12 + 315i) + (1 + i) = 13 + 316i Ω

Now, we can find the current in the a-phase using Ohm's Law:

Ia = Va / Ztotal

Substituting the given values:

Ia = (2302-50°V) / (13 + 316i Ω)

Calculating the current magnitude and phase angle:

|Ia| = |2302-50°V| / |13 + 316i Ω| = 1528.298 A

∠Ia = ∠(2302-50°V) - ∠(13 + 316i Ω) = -0.304°

Therefore, the current in the a-phase of the load is 1528.298 A at a phase angle of -0.304°.