A two bus power system shown in the figure supplies load of 1.0 + j 0.5 p.u.

The values of V₁ in p.u. and δ₂ respectively are

\(\begin{array}{l} \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{j0.1}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{0.1\angle 90^\circ }\\ 0&1 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} {{V_S}}\\ {{I_S}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_r}}\\ {{I_r}} \end{array}} \right] \end{array}\)

V_S = V₁ ∠0, B = 0.1 ∠90

V_r = 1∠δ₂, A=1∠0

V_S = AV_r + BI_r

\({I_r} = \frac{{{V_S}}}{B} - \frac{A}{B}{V_r} = \frac{{{V_1}\angle 0}}{{0.1\angle 90}}\left( {1\angle {\delta _2}} \right)\)

= 10V₁ ∠-90 -10∠δ₂ – 90

\(\begin{array}{l} I_r^ \star = 10{V_1}\angle 90 - 10\angle 90 - {\delta _2}\\ {S_r} = {P_r} + j{Q_r} = {V_r}I_r^ \star = \left( {1\angle {\delta _2}} \right)\left( {10{V_1}\angle 90 - {\delta _2}} \right) \end{array}\)

= 10 V₁ ∠90 + δ₂ – 10∠90  

= 10 [V₁ cos (90 + δ₂) + jV₁ sin (90 + δ₂)] – j10

= 10 [-V₁ sin δ₂ + jV₁ cos δ₂]-j10

= [-10V₁ sin δ₂] + j[10V₁ cos δ₂ - 10]  

Given S_r = 1 + j0.5

-10V₁ sin δ₂ = +1 ⇒ 10V₁ sin δ₂ = -1

10V₁ cos δ₂ – 10 = 0.5 ⇒ 10V₁ cos δ₂ = 10.5

\( \Rightarrow \tan {\delta _2} = \frac{{ - 1}}{{10.5}}\)

⇒ δ₂ = -5.44

10V₁ sin δ₂ = -1

⇒ V₁ = 1.054