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Question:

*"A synchronous generator is running overexcited with excitation voltage Ef = 1.40 p.u and connected to the network at voltage of 1 p.u. This generator has synchronous reactance of 1.20 p.u., and delivering active power of 0.50 p.u. to the network. In the network, there is 1% increase in real power due to power drop from uncontrolled network connected renewable power source. Therefore, the prime mover (i.e. turbine) input is increased by 1%. The excitation or voltage controller of synchronous generator is not responding to this change.*

**Explain how the reactive power delivery from the synchronous generator will change under this condition**"So, the Q delivered by the generator per phase is:

[itex]Q = \frac{E_fV_t}{X_d}cos(\delta)-\frac{V_{t}^{2}}{X_d}-V_{t}^{2}(\frac{1}{X_d}-\frac{1}{X_q})sin^2(\delta)[/itex] (1)

So according to this equation the Q will not change due to increase in turbine input.

However if we take a look at this equation:

[itex]P = \frac{E_fV_t}{X_d}sin(\delta)-\frac{V_{t}^{2}}{X_d}-V_{t}^{2}(\frac{1}{X_d}-\frac{1}{X_q})sin(2\delta)[/itex] (2)

An increase of P will cause an increase in [itex]V_t[/itex], and an increase of [itex]V_t[/itex] will

**increase the reactive power delivered by the sync. generator**according to equation (1)

Can someone tell me if this is the right way of thinking or if I am missing something please tell me.

Best Regards