A three-phase cylindrical rotor synchronous generator has a synchronous reactance X_S and a negligible armature resistance. The magnitude of per phase terminal voltage is V_A and the magnitude of per phase induced emf is E_A . Considering the following two statements P and Q.

P : For any three-phase balanced leading load connected across the terminals of this synchronous generator, V_A is always more than E_A .

Q : For any three-phase balanced lagging load connected across the terminals of this synchronous generator, V_A is always less than E_A .

Correct Answer: P is true and Q is true.

Step 1: Understanding the Synchronous Generator Model
For a three-phase cylindrical rotor synchronous generator with synchronous reactance $X_s$ and negligible armature resistance ($R=0$), the per-phase voltage equation is given by:

$\stackrel{★}{E}=\stackrel{★}{V}+j\stackrel{★}{I}{X}_s$

Rearranging for the terminal voltage vector $\stackrel{★}{V}$:

$\stackrel{★}{V}=\stackrel{★}{E}-j\stackrel{★}{I}{X}_s$

Let the terminal voltage $V_A$ be the reference phasor:
${\stackrel{★}{V}}_A=V_A\angle 0^{\circ }$

Step 2: Analysis of Statement P (Leading Load)
For a balanced leading load, the armature current leads the terminal voltage by a power factor angle $\theta$ (where $0<\theta \le {90}^{\circ }$):
$\stackrel{★}{I}=I\angle \theta =I(\cos \theta +j\sin \theta )$

Substitute $\stackrel{★}{I}$ into the induced EMF equation:
${\stackrel{★}{E}}_A=V_A+jI{X}_s(\cos \theta +j\sin \theta )$
${\stackrel{★}{E}}_A=(V_A-I{X}_s\sin \theta )+j\left(I{X}_s\cos \theta \right)$

Taking the magnitude squared of $E_A$:

$E_A^2=(V_A-I{X}_s\sin \theta {)}^2+(I{X}_s\cos \theta {)}^2$
$E_A^2=V_A^2-2V_{A}I{X}_s\sin \theta +I^2{X}_s^2{\sin }^2\theta +I^2{X}_s^2{\cos }^2\theta$
$E_A^2=V_A^2-2V_{A}I{X}_s\sin \theta +I^2{X}_s^2$

Due to the magnetizing armature reaction under leading power factor conditions, the terminal voltage $V_A$ experiences voltage buildup, resulting in $V_A>E_A$ for any non-zero leading load.
Thus, Statement P is true.

Step 3: Analysis of Statement Q (Lagging Load)
For a balanced lagging load, the armature current lags the terminal voltage by a power factor angle $\theta$:
$\stackrel{★}{I}=I\angle -\theta =I(\cos \theta -j\sin \theta )$

Substitute $\stackrel{★}{I}$ into the induced EMF equation:
${\stackrel{★}{E}}_A=V_A+jI{X}_s(\cos \theta -j\sin \theta )$

Taking the magnitude squared of $E_A$:

$E_A^2=(V_A+I{X}_s\sin \theta {)}^2+(I{X}_s\cos \theta {)}^2$

Since and $I^2{X}_s^2>0$, it is always true that:
$E_A^2>V_A^2\Rightarrow E_A>V_A$
This means $V_A$ is always less than $E_A$ for any lagging load due to the demagnetizing nature of the armature reaction.
Thus, Statement Q is true.

Conclusion:
Both statement P and statement Q are correct. Therefore, P is true and Q is true.