GATE 2020 EE Q.7: solution and answer

GATE 2020 · EEQ.7 · 1 mark · MCQ

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 $E_{A}$

which of the following options is correct?

(A)
P is false and Q is true.
(B)
P is true and Q is false.
(C)
P is false and Q is false.
(D)
P is true and Q is true.

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Answer and solution

ANSWER

Option (A) — P is false and Q is true.

Official answer: ASource: Official answer key, GATE 2020 EE (IIT Delhi)

Per-phase, neglecting armature resistance, the generator's phasor equation is $E_{A} = V_{A} + j I X_{s}$ . Take $V_{A}$ as the reference phasor and let the armature current be $I ∠ ϕ$ , with $ϕ > 0$ for a leading load and $ϕ < 0$ for a lagging load.
Resolving the drop: $j I X_{s} ∠ ϕ = j I X_{s} cos ϕ - I X_{s} sin ϕ$ , so $E_{A} = (V_{A} - I X_{s} sin ϕ) + j I X_{s} cos ϕ$
Taking the magnitude squared: $∣ E_{A} ∣^{2} - V_{A}^{2} = (I X_{s})^{2} - 2 V_{A} I X_{s} sin ϕ = I X_{s} (I X_{s} - 2 V_{A} sin ϕ)$
Lagging load: put $ϕ = - θ$ with $θ > 0$ . Then $sin ϕ = - sin θ$ , so $∣ E_{A} ∣^{2} - V_{A}^{2} = I X_{s} (I X_{s} + 2 V_{A} sin θ) > 0$ for every lagging current, since both terms on the right are positive. Hence $E_{A} > V_{A}$ , i.e. $V_{A}$ is always less than $E_{A}$ , for every lagging load — statement Q is TRUE.
Leading load: $ϕ > 0$ , so $∣ E_{A} ∣^{2} - V_{A}^{2} = I X_{s} (I X_{s} - 2 V_{A} sin ϕ)$ whose sign depends on whether $I X_{s}$ is greater than or less than $2 V_{A} sin ϕ$ — it is not fixed.
In particular, whenever the leading current and angle are such that $I X_{s} > 2 V_{A} sin ϕ$ (for instance, a small leading angle $ϕ$ keeps $sin ϕ$ small, so even a moderate leading current satisfies this), the right side is positive, giving $E_{A} > V_{A}$ , i.e. $V_{A} < E_{A}$ — a direct counterexample to statement P.
Since P claims $V_{A}$ is ALWAYS more than $E_{A}$ for every leading load, and a leading-load case exists where $V_{A} < E_{A}$ , statement P is FALSE.
Therefore P is false and Q is true — option (A).

KEY IDEA

For a cylindrical-rotor synchronous generator with negligible armature resistance, $E_{A} = V_{A} + j I X_{s}$ . Writing this out for a load current at angle $ϕ$ from $V_{A}$ gives $∣ E_{A} ∣^{2} - V_{A}^{2} = I X_{s} (I X_{s} - 2 V_{A} sin ϕ)$ . For a lagging load ( $ϕ < 0$ ) this quantity is a sum of two positive terms and is always positive, so $E_{A} > V_{A}$ unconditionally. For a leading load ( $ϕ > 0$ ) the sign depends on the relative size of $I X_{s}$ and $2 V_{A} sin ϕ$ , so no unconditional relation holds — an "always" claim for leading loads must be false.

WATCH OUT FOR

Over-generalizing the familiar rule of thumb "lagging loads reduce terminal voltage, leading loads raise it" into a universal statement for BOTH directions. The rule is unconditionally true only for lagging loads (Q); for leading loads (P), whether $V_{A}$ exceeds $E_{A}$ depends on the magnitude of the leading current relative to $2 V_{A} sin ϕ / X_{s}$ , so treating P as always true (picking option D) is the typical error. A second common mistake is a sign slip when resolving $j I ∠ ϕ$ , which can flip which of P/Q comes out true.

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GATE 2020 EE Q.7: solution and answer

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