Discussion
In the given figure, a three-phase static capacitive reactor of reactance 1 pu per phase is connected through a switch at motor bus bar. Calculate the limit of steady state power with and without reactor switch closed. The internal voltage of the generator is 1.2 pu and that of the motor is 1 pu.
Answer: A
The steady state power limit between two sources is \(P_{max} = \dfrac{E_gE_m}{X_{total}}\). Without the reactor the total reactance is the sum of the series parts.
\[ X = 1.0 + 0.1 + 0.25 + 0.1 + 1.0 = 2.45\ \text{pu} \]
\[ P_{max} = \frac{1.2 \times 1}{2.45} = 0.49\ \text{pu} \]
This is option A, and it is the value without the reactor switch closed. The item asks for the limit with and without the reactor, but the options give a single number, so the key is read as the without reactor figure. With the shunt capacitor switched in at the motor bus, the transfer reactance changes and the limit rises above this value.
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