Answer:


To find the line currents, we can use the formula:
I_Line = I_Phase / √3
where I_Phase is the current in one phase.

We are given the line voltage V_Line = 400V. Since the system is star connected, the phase voltage is equal to the line voltage. Therefore, V_Phase = V_Line = 400V.

We are also given the impedances of the load in polar form. We can convert them to rectangular form using the following formulas:
Z_R = Z_Magnitude * cos(Z_Angle)
Z_X = Z_Magnitude * sin(Z_Angle)

Using these formulas, we get:
Z_R = 5 * cos(0) = 5
Z_X = 5 * sin(0) = 0

Z_R = 10 * cos(π/2) = 0
Z_X = 10 * sin(π/2) = 10

Z_R = -10 * cos(π/2) = 0
Z_X = -10 * sin(π/2) = -10

Now, we can find the phase currents using the formula:
I_Phase = V_Phase / Z_Phase
where Z_Phase is the impedance of one phase of the load.

For R phase,
Z_Phase_R = Z_R + jZ_X = 5 + j0
I_Phase_R = V_Phase / Z_Phase_R = 400 / (5 + j0) = 80 ∠0° A

For Y phase,
Z_Phase_Y = Z_R + jZ_X = 0 + j10
I_Phase_Y = V_Phase / Z_Phase_Y = 400 / (0 + j10) = 40 ∠-90° A

For B phase,
Z_Phase_B = Z_R + jZ_X = 0 - j10
I_Phase_B = V_Phase / Z_Phase_B = 400 / (0 - j10) = 40 ∠90° A

Finally, we can find the line currents using the formula:
I_Line = I_Phase / √3

For R phase,
I_Line_R = I_Phase_R / √3 = 80 ∠0° / √3 = 46.4 ∠0° A

For Y phase,
I_Line_Y = I_Phase_Y / √3 = 40 ∠-90° / √3 = 23.2 ∠-30° A

For B phase,
I_Line_B = I_Phase_B / √3 = 40 ∠90° / √3 = 23.2 ∠30° A

Therefore, the line currents are:
I_Line_R = 46.4 ∠0° A
I_Line_Y = 23.2 ∠-30° A
I_Line_B = 23.2 ∠30° A


To find the real and reactive power of each phase, we can use the following formulas:
P = V_Phase * I_Phase * cos(θ)
Q = V_Phase * I_Phase * sin(θ)
where θ is the phase angle