Final answer:
When the generator has terminal A open and the other two terminals are connected to each other with a short circuit from this connection to the ground, the current into the ground is 350 A and the currents in each phase of the generator are 0 A for phase A, (250ω - 600ω² + 350) cis 90ᵒ A for phase B, and (350 - 600ω + 250ω²) cis 90ᵒ A for phase C.
Step-by-step explanation:
In this scenario, when terminal A of the generator is open and the other two terminals are short-circuited to each other and the ground, we can find the currents in each phase of the generator.
Given the symmetrical components of currents as Ia1=600cis−90ᵒ, Ia2=250 cis 90ᵒ, and Ia0=350 cis 90ᵒA, we can calculate the current into the ground and the currents in each phase.
The current into the ground is the sum of the zero sequence currents: I0 = Ia0 + Ib0 + Ic0. In this case, I0 = 350 cis 90ᵒ + 350 cis 210ᵒ + 350 cis -30ᵒ = 350 cis 270ᵒ = 350 A.
The currents in each phase can be calculated as follows:
Ia = Ia0 + Ia1 + Ia2 = 350 cis 90ᵒ + 600 cis -90ᵒ + 250 cis 90ᵒ = (350 - 600 + 250) cis 90ᵒ = 0 cis 90ᵒ = 0 A
Ib = Ia0 + Ia1ω² + Ia2ω = 350 cis 90ᵒ + 600 cis -90ᵒ (ω²) + 250 cis 90ᵒ ω = (350 - 600ω² + 250ω) cis 90ᵒ = (250ω - 600ω² + 350) cis 90ᵒ A
Ic = Ia0 + Ia1ω + Ia2ω² = 350 cis 90ᵒ + 600 cis -90ᵒ ω + 250 cis 90ᵒ (ω²) = (350 - 600ω + 250ω²) cis 90ᵒ A