A three-phase cylindrical rotor synchronous generator has a synchronous reactance Xₛ 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${}_A$

Which of the following options is correct?

This question asks us to evaluate two statements about the relationship between the per-phase terminal voltage ($V_A$) and the per-phase induced electromotive force ($E_A$) in a three-phase cylindrical rotor synchronous generator with synchronous reactance ($X_s$) and negligible armature resistance ($R_a\approx 0$).

Understanding the Synchronous Generator Phasor Equation

For a synchronous generator with negligible armature resistance, the relationship between the induced EMF ($E_A$), terminal voltage ($V_A$), armature current ($I_a$), and synchronous reactance ($X_s$) is given by the phasor equation:

$$E_A=V_A+j{I}_a{X}_s$$

Here:

- $E_A$ is the internally generated voltage (induced EMF) per phase.
- $V_A$ is the terminal voltage per phase.
- $I_a$ is the armature current per phase.
- $X_s$ is the synchronous reactance per phase.
- '$j$ ' represents a $90^{\circ }$ leading phase shift.

The term '$j{I}_a{X}_s$ ' represents the voltage drop across the synchronous reactance.

Analyzing Load Conditions

The relationship between $E_A$ and $V_A$ depends significantly on the power factor of the load, which determines whether the armature current $I_a$ leads or lags the terminal voltage $V_A$.Evaluating Statement P

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

Let's analyze this using the phasor equation. For a leading load, $I_a$ leads $V_A$. Let $V_A=|V_A|\angle 0^{\circ }$ and $I_a=|I_a|\angle \varphi$, where $\varphi$ is positive.

The equation is $E_A=V_A+j{I}_a{X}_s$. Substituting the phasors:

$$E_A=|V_A|\angle 0^{\circ }+(|I_a|\angle \varphi )(j{X}_s)$$

$$E_A=|V_A|+|I_a|X_s\angle (\varphi +90^{\circ })$$

The magnitude squared of $E_A$ is:

$$|E_A{|}^2=|V_A{|}^2+(|I_a|X_s{)}^2+2|V_A||I_a|X_s\sin (\varphi )$$

Wait, let's re-calculate the magnitude squared using $E_A=V+jI{X}_s$ where $V=V_A$, $I=I_a$, $V$ is real axis, $I=I\angle \varphi$. $E_A=V+(I\angle \varphi )j{X}_s=V+I{X}_s(\cos (\varphi +90^{\circ })+j\sin (\varphi +90^{\circ }))=V+I{X}_s(-\sin \varphi +j\cos \varphi )$.

$$|E_A{|}^2=(V-I{X}_s\sin \varphi {)}^2+(I{X}_s\cos \varphi {)}^2$$

$$|E_A{|}^2=V^2-2VI{X}_s\sin \varphi +(I{X}_s{)}^2{\sin }^2\varphi +(I{X}_s{)}^2{\cos }^2\varphi$$

$$|E_A{|}^2=V^2+(I{X}_s{)}^2-2VI{X}_s\sin \varphi$$

Since $V,I,X_s$ are positive magnitudes and $\varphi >0$ for a leading load, $\sin \varphi >0$. The term $-2VI{X}_s\sin \varphi$ is negative. This means $|E_A{|}^2$ is less than $V^2+(I{X}_s{)}^2$. It is possible for $|E_A|$ to be less than $|V_A|$ if the term $2VI{X}_s\sin \varphi$ is sufficiently large compared to $(I{X}_s{)}^2$. For instance, with a highly capacitive load (large $\varphi$, approaching $90^{\circ }$), $|E_A|$ can become significantly smaller than $|V_A|$. Therefore, the statement "$V_A$ is always more than $E_A$ " is false.

Evaluating Statement Q

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

For a lagging load, $I_a$ lags $V_A$. Let $V_A=|V_A|\angle 0^{\circ }$ and $I_a=|I_a|\angle -\varphi$, where $\varphi$ is positive.

The equation is $E_A=V_A+j{I}_a{X}_s$. Substituting the phasors:

$$E_A=|V_A|\angle 0^{\circ }+(|I_a|\angle -\varphi )(j{X}_s)$$

$$E_A=|V_A|+|I_a|X_s\angle (-\varphi +90^{\circ })$$

The magnitude squared of $E_A$ is:

$$|E_A{|}^2=(|V_A|+|I_a|X_s\cos (90^{\circ }-\varphi ){)}^2+(|I_a|X_s\sin (90^{\circ }-\varphi ){)}^2$$

$$|E_A{|}^2=(|V_A|+|I_a|X_s\sin \varphi {)}^2+(|I_a|X_s\cos \varphi {)}^2$$

$$|E_A{|}^2=|V_A{|}^2+2|V_A||I_a|X_s\sin \varphi +(|I_a|X_s{)}^2{\sin }^2\varphi +(|I_a|X_s{)}^2{\cos }^2\varphi$$

$$|E_A{|}^2=|V_A{|}^2+(|I_a|X_s{)}^2+2|V_A||I_a|X_s\sin \varphi$$

Since $|V_A|$, $|I_a|$, $X_s$, and $\sin \varphi$ (for $\varphi >0$) are all positive quantities, the term $2|V_A||I_a|X_s\sin \varphi$ is always positive.

Therefore, $|E_A{|}^2$ is strictly greater than $|V_A{|}^2+(|I_a|X_s{)}^2$, which certainly means $|E_A{|}^2>|V_A{|}^2$. Consequently, $|E_A|>|V_A|$.

So, the statement "$V_A$ is always less than $E_A$ " is true.

Conclusion

Based on the analysis:

- Statement P is false.
- Statement Q is true.

This corresponds to option 1.