System Voltage Profile

Considering the following polarities and current orientations:

One gets from applying Kirchhoff's circuit laws :

$\large \displaystyle j \cdot \begin{bmatrix} -95 & 100 & 0 \\ 100 & -190 & 100 \\ 0 & 100 & -95 \end{bmatrix} \cdot \begin{bmatrix} i_1 \\ i_2 \\ i_3 \end{bmatrix} = \begin{bmatrix} 100 \angle \delta \\ 0 \\ -100 \end{bmatrix}$

$\large \displaystyle \begin{bmatrix} i_1 \\ i_2 \\ i_3 \end{bmatrix} = - \frac{j10}{741} \cdot \begin{bmatrix} 322 \angle \delta - 400 \\ 380 \angle \delta - 380 \\ 400 \angle \delta - 322 \end{bmatrix}$

Since:

$\large \displaystyle V_1 = -j100 (i_1 - i_2)$

Then:

$\large \displaystyle V_1 = \frac{1000}{741} \cdot (58 \angle \delta + 20)$

So:

$\large \displaystyle |V_1| = \frac{1000}{741} \cdot \sqrt{ (58 \cos(\delta) + 20)^2 + (58 \sin(\delta))^2 } < 100$

$\large \displaystyle 100 \cdot ( 3764 + 2320 \cos(\delta) ) < 549081$

$\large \displaystyle \cos(\delta) < 0.744314655$

So :

$\large \displaystyle \delta > 41.89^{\circ}$

Or:

$\color{#3D99F6} \boxed{ \large \displaystyle \delta_{min} = 42^{\circ}}$