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Question

# The terminal voltage of the generator is held at 1.20 p.u. and the voltage of the infinite bus is 1.0 p.u. The value of steady state power limit of a system consisting of a generator equivalent reactance 0.5 p.u. connected to an infinite bus through a series reactance of 1.0 p.u. is

A
2.512 p.u.
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B
0.591 p.u
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C
1.151 p.u.
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D
1.615 p.u.
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Solution

## The correct option is C 1.151 p.u. I=(Vt−Vj1)=(1.2∠θ−1∠0∘1∠90∘)=(1.2∠(θ−90∘)−1∠−90∘) E∠δ=Vt+j0.5I E∠δ=1.2∠θ+(1.2∠(−90+θ)+j)0.5j =1.2∠θ+1.2×0.5∠θ−0.5 =1.2∠θ+0.6∠θ−0.5 At steady state power, δ=90∘ 1.2cosθ+0.6cosθ−0.5=0 1.8cosθ=0.5 θ=73.87∘ I=1.2∠73.87∘−1j1=1.3316∠30.1∘A E=V+j1.5I =1+j1.5×1.3316∠30.1∘ =1.727∠90∘p.u. Pmax=1.727×11.5=1.151p.u.

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