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consider two buses connected by an impedance of (0 + 5j) Ω. The bus ‘1’ voltage is 100 ∠30° V, and bus ‘2’ voltage is 100 ∠0° V. The real and reactive power supplied by bus ‘1’ respectively are
1. 1000 W, 268 VAr
2. -1000 W, -134 VAr
3. 276.9 W, -56.7 VAr
4. -276.9 W, 56.7 VAr

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Correct Answer - Option 1 : 1000 W, 268 VAr

P = Line impedance angle, δ = 30°, α = 0°, A = D = 1.0, β = 5∠90

\(\begin{array}{l} {P_{sending}} = \left| {\frac{D}{B}} \right|{\left| {{N_S}} \right|^2}\cos \left( {β - \alpha } \right) - \frac{{\left| {{V_S}} \right|\left| {{V_r}} \right|}}{{\left| B \right|}}\cos \left( {β + \delta } \right)\\ {Q_{sending}} = \left| {\frac{D}{B}} \right|{\left| {{V_S}} \right|^2}\sin \left( {β - \alpha } \right) - \frac{{\left| {{V_S}} \right|\left| {{V_r}} \right|}}{{\left| B \right|}}\cos \left( {β + \delta } \right)\\ \therefore {P_{sending}} = \left| {\frac{{1.0}}{5}} \right|{100^2}\cos \left( {90 - 0} \right) - \frac{{100 \times 100}}{{\left| 5 \right|}}\cos \left( {90 + 30} \right) \end{array}\)

= 1000 W

\({Q_{sending}} = \left| {\frac{{1.0}}{5}} \right|{100^2}\sin \left( {90 - 0} \right) - \frac{{100 \times 100}}{{\left| 5 \right|}}\sin \left( {90 + 30} \right)\)

= 268 VAR

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