Positive Sequence Networks MCQ Quiz - Objective Question with Answer for Positive Sequence Networks - Download Free PDF

X_s = 0.85 pu

X_T1 = X_T2 = 0.157

X_L1 = X_L2 = 0.35 pu

E = 1.5 pu

V = 1.0 pu

P₀ = 1.0 pu

Equivalent circuit of above data can be drawn as

X_T = j0.85 + j0.157 + j(0.35||0.35) + j0.157

X_T = 1.339 pu

As $P_0=\frac{EV}{X}\sin \delta$

P₀ = 1pu

$P=\frac{EV}{X}\sin \delta$

For the minimum value of E, sin δ = 1.

$E_{min}=\frac{P_{0}X}{V}=\frac{1\times 1.339}{1}=1.339pu$

X_s = 0.85 pu

X_T1 = X_T2 = 0.157

X_L1 = X_L2 = 0.35 pu

E = 1.5 pu

V = 1.0 pu

P₀ = 1.0 pu

Equivalent circuit of above data can be drawn as

X_T = j0.85 + j0.157 + j(0.35||0.35) + j0.157

X_T = 1.339 pu

As $P_0=\frac{EV}{X}\sin \delta$

P₀ = 1pu

$P=\frac{EV}{X}\sin \delta$

For the minimum value of E, sin δ = 1.

$E_{min}=\frac{P_{0}X}{V}=\frac{1\times 1.339}{1}=1.339pu$

I_a = -I_c θ, I_b = 0

I_a = 15 ∠0°; I_C = 15 ∠180° = -15

${\mathrm{I}}_{a1}=\frac{1}{3}\left(I_a+\beta I_c+{\beta }^2{I}_b\right)$ = (7.5 + j4.33) A

$I_{a2}=\frac{1}{3}\left(I_a+{\beta }^2{I}_C+\beta I_b\right)$ = (2.5 + j4.33)

$I_{a0}=\left(I_a+I_b+I_C\right)=0$