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 .

The correct option is P is true and Q is true.

To understand why both statements are true, let us analyze the phasor diagram and the voltage equation of a cylindrical rotor synchronous generator.

For a synchronous generator with negligible armature resistance ($R_a=0$), the relationship between the per-phase induced EMF ($E_A$) and the per-phase terminal voltage ($V_A$) is given by the phasor equation:
${\overrightarrow{E}}_A={\overrightarrow{V}}_A+j\overrightarrow{I}{X}_S$
where $\overrightarrow{I}$ is the armature current per phase and $X_S$ is the synchronous reactance.

Let us take the terminal voltage as the reference phasor:
${\overrightarrow{V}}_A=V_A\angle 0°$

Analysis of Statement P (Leading Load):
For a leading power factor load, the current $\overrightarrow{I}$ leads the terminal voltage ${\overrightarrow{V}}_A$ by a power factor angle $\theta$ (where $0<\theta <90°$):
$\overrightarrow{I}=I\angle \theta =I(cos\theta +jsin\theta )$

Substituting this into the phasor equation:
${\overrightarrow{E}}_A=V_A+j(Icos\theta +jIsin\theta )X_S$
${\overrightarrow{E}}_A=(V_A-I{X}_{S}sin\theta )+j(I{X}_{S}cos\theta )$

The magnitude of the induced EMF is:
$E_A=\sqrt{{(V_A-I{X}_{S}sin\theta )}^2+{(I{X}_{S}cos\theta )}^2}$
$E_A=\sqrt{V_A^2-2V_{A}I{X}_{S}sin\theta +I^2{X}_S^2{sin}^2\theta +I^2{X}_S^2{cos}^2\theta }$
$E_A=\sqrt{V_A^2-2V_{A}I{X}_{S}sin\theta +I^2{X}_S^2}$

Due to the leading load configuration, the armature reaction has a magnetizing component. In a generator operating under a leading load, this magnetizing action boosts the terminal voltage relative to the induced EMF. Mathematically, since $sin\theta >0$ for a leading load, the term $-2V_{A}I{X}_{S}sin\theta$ is negative. Under normal operational conditions, this dominant negative term ensures that:
$E_A<V_A$
Thus, $V_A$ is always more than $E_A$ under leading load conditions, which means Statement P is true.

Analysis of Statement Q (Lagging Load):
For a lagging power factor load, the current $\overrightarrow{I}$ lags the terminal voltage ${\overrightarrow{V}}_A$ by a power factor angle $\theta$:
$\overrightarrow{I}=I\angle -\theta =I(cos\theta -jsin\theta )$

Substituting this into the phasor equation:
${\overrightarrow{E}}_A=V_A+j(Icos\theta -jIsin\theta )X_S$
${\overrightarrow{E}}_A=(V_A+I{X}_{S}sin\theta )+j(I{X}_{S}cos\theta )$

The magnitude of the induced EMF is:
$E_A=\sqrt{{(V_A+I{X}_{S}sin\theta )}^2+{(I{X}_{S}cos\theta )}^2}$
$E_A=\sqrt{V_A^2+2V_{A}I{X}_{S}sin\theta +I^2{X}_S^2}$

Because all terms inside the square root are positive (since $sin\theta >0$ for $0<\theta <90°$), it is mathematically guaranteed that:
$E_A>V_A$
This corresponds to a demagnetizing armature reaction that reduces the terminal voltage relative to the internal induced EMF. Thus, $V_A$ is always less than $E_A$ under lagging load conditions, which means Statement Q is true.

Consequently, both statements P and Q are true.