GATE EE

Power system stability and the equal area criterion

Power system stability is the ability of synchronous machines to stay in step after a disturbance. The swing equation and the equal area criterion let you test, with simple areas on the power-angle curve, whether a load change or a cleared fault leaves the machine in step. Typical numerical problems ask for the critical clearing angle, the maximum swing angle or the critical clearing time from a given Pm, Pmax and H.

4 previous-year questions · 4 solved · 6 practice questions

Types of stability and the classical model

Steady-state stability is the ability to stay in synchronism after a small, slow change, such as a gradual rise in load. Transient stability is the ability to stay in synchronism after a large, sudden disturbance such as a fault, its clearing, or loss of a line or load, judged over roughly the first swing. Dynamic stability (today usually called small-signal stability) concerns small disturbances over a longer period, where the action of controls such as the excitation system and governor decides whether oscillations die out.

Numerical problems use the classical model: the machine is a constant voltage $E^{'}$ behind its transient reactance, the mechanical input power $P_{m}$ stays constant during the swing, losses and damping are neglected, rotor speed stays close to synchronous (so per-unit torque and per-unit power are equal), and a group of machines that swing together is treated as one. All powers are per unit on a stated MVA base, $H$ is on that same base, and $δ$ is the rotor angle with respect to the infinite bus, in electrical radians inside the swing equation unless stated otherwise.

Type	Disturbance	What decides the outcome
Steady-state	Small and gradual	Whether $δ$ stays below $9 0^{\circ}$ on the power-angle curve (positive synchronising coefficient)
Transient	Large and sudden (fault, switching)	Whether the accelerating area can be matched by the decelerating area (equal area criterion)
Dynamic (small-signal)	Small, over a longer time	Damping provided by machines and controls

Swing equation and inertia constant

When the mechanical input $P_{m}$ exceeds the electrical output $P_{e}$ , the surplus accelerates the rotor; when it is less, the rotor slows down. The rotor angle follows the swing equation. With $H$ the inertia constant on the machine's own MVA rating, and $P_{m}$ , $P_{e}$ in per unit on that rating:

SWING EQUATION (PER UNIT)

\frac{H}{π f} \frac{d ^{2} δ}{d t ^{2}} = P_{m} - P_{e} = P_{a}

$δ$ in electrical radians, $t$ in seconds, $f$ the system frequency in Hz, $P_{m}$ and $P_{e}$ in per unit on the same base as $H$ ; $P_{a}$ is the accelerating power. Equivalent form: $\frac{2 H}{ω _{s}} \ddot{δ} = P_{m} - P_{e}$ with $ω_{s} = 2 π f$ .

SWING EQUATION WITH THE ANGLE IN ELECTRICAL DEGREES

\frac{H}{180 f} \frac{d ^{2} δ}{d t ^{2}} = P_{m} - P_{e}

Same equation with $δ$ in electrical degrees (since $π$ rad $= 18 0^{\circ}$ ). Do not mix the two forms.

INERTIA CONSTANT

H = \frac{kinetic energy stored at synchronous speed (MJ)}{machine rating S _{machine} ( MVA )} = \frac{\frac{1}{2} J ω _{s m}^{2}}{S _{machine}}

Unit: seconds (MJ per MVA). In the second form $J$ is in kg m $^{2}$ , $ω_{s m}$ the synchronous mechanical speed in rad/s and $S_{machine}$ in VA. Stored energy is $H S_{machine}$ megajoules when $S$ is in MVA.

ELECTRICAL AND MECHANICAL ANGLE

θ_{e} = \frac{p}{2} θ_{m}, ω_{e} = \frac{p}{2} ω_{m}

$p$ is the number of poles. The swing equation above uses the electrical angle.

CHANGE OF BASE

H_{new} = H_{old} \frac{S _{old}}{S _{new}}

The stored energy in MJ is fixed; only the MVA it is divided by changes.

EQUIVALENT INERTIA OF MACHINES SWINGING TOGETHER

H_{e q} = \frac{H _{1} S _{1} + H _{2} S _{2} + \dots}{S _{base}}

Applies to machines that swing coherently (same $δ$ and speed), each $H_{i}$ on its own rating $S_{i}$ . The stored energies in MJ add.

TWO MACHINES SWINGING AGAINST EACH OTHER

H_{12} = \frac{H _{1} H _{2}}{H _{1} + H _{2}}, P_{m} = \frac{H _{2} P _{m 1} - H _{1} P _{m 2}}{H _{1} + H _{2}}

For two non-coherent machines joined by a lossless link, with $H_{1}$ , $H_{2}$ , $P_{m 1}$ , $P_{m 2}$ on one common base: the relative angle $δ_{12} = δ_{1} - δ_{2}$ obeys $\frac{H _{12}}{π f} \ddot{δ}_{12} = P_{m} - P_{e 1}$ . Obtained by subtracting the two swing equations with $P_{e 2} = - P_{e 1}$ .

ROTOR ACCELERATION FOR CONSTANT SURPLUS POWER

\frac{d ^{2} δ}{d t ^{2}} = \frac{π f}{H} (P_{m} - P_{e}) rad/s^{2}

If $P_{e} = 0$ during a fault and the rotor is at synchronous speed when the fault occurs ( $\dot{δ} = 0$ at $t = 0$ ), then $δ = δ_{0} + \frac{π f P _{m}}{2 H} t^{2}$ because the acceleration is constant.

Power-angle curve and steady-state limit

For a machine connected to an infinite bus through a total reactance $X$ (losses neglected), the electrical power is a sine curve of the rotor angle. The operating point is where the horizontal line $P = P_{m}$ meets the curve.

POWER-ANGLE EQUATION

P_{e} = P_{m a x} sin δ, P_{m a x} = \frac{E V}{X}

$E$ and $V$ are the voltage magnitudes at the two ends, $X$ the total series reactance between them; per unit values give $P_{e}$ in per unit.

INITIAL ANGLE

δ_{0} = sin^{- 1} (\frac{P _{m}}{P _{m a x}})

Of the two angles with this sine, the stable operating point is the one below $9 0^{\circ}$ .

SYNCHRONISING POWER COEFFICIENT

P_{s} = \frac{d P _{e}}{d δ}_{δ_{0}} = P_{m a x} cos δ_{0}

Positive for $δ_{0} < 9 0^{\circ}$ (stable), zero at $9 0^{\circ}$ , negative beyond. Units: per unit power per electrical radian.

FREQUENCY OF SMALL OSCILLATION

ω_{n} = \frac{π f P _{s}}{H} rad/s, f_{n} = \frac{ω _{n}}{2 π} Hz

Undamped natural frequency, from linearising the swing equation about $δ_{0}$ with $Δ δ$ in radians: $\frac{H}{π f} Δ \ddot{δ} = - P_{s} Δ δ$ .

STEADY-STATE STABILITY LIMIT

P_{limit} = P_{m a x} at δ = 9 0^{\circ}

Reached when load is increased slowly: beyond $δ = 9 0^{\circ}$ the coefficient $P_{s}$ becomes negative and the machine falls out of step.

Equal area criterion

Multiply the swing equation by $\dot{δ}$ and integrate: the kinetic energy gained by the rotor is proportional to $\int (P_{m} - P_{e}) d δ$ . The rotor stops swinging forward (speed returns to synchronous) at the angle $δ_{m}$ where the accelerating area (where $P_{m} > P_{e}$ ) equals the decelerating area (where $P_{e} > P_{m}$ ). The speed deviation is largest at $δ_{1}$ , the angle where $P_{m}$ meets the post-disturbance curve at its stable intersection. The machine is stable if $δ_{m}$ is reached before $δ$ passes $δ_{m a x} = π - δ_{1}$ , the far intersection on the same curve.

EQUAL AREA CONDITION

\int_{δ_{0}}^{δ_{m}} (P_{m} - P_{e}) d δ = 0 ⟺ A_{acc} = A_{dec}

$δ_{m}$ is the maximum swing angle. Valid for the classical model (constant $P_{m}$ , no damping).

SUDDEN LOAD (OR $P_M$) CHANGE, SAME CURVE

P_{m 1} (δ_{m} - δ_{0}) + P_{m a x} (cos δ_{m} - cos δ_{0}) = 0

$P_{m}$ jumps from $P_{m 0}$ to $P_{m 1}$ with the network unchanged, $δ_{0} = sin^{- 1} (P_{m 0} / P_{m a x})$ ; angles in radians where they multiply $P_{m 1}$ . Stable only if $δ_{m} \leq π - δ_{1}$ with $δ_{1} = sin^{- 1} (P_{m 1} / P_{m a x})$ .

LIMITING SUDDEN STEP

P_{m 1} (π - δ_{1} - δ_{0}) = P_{m a x} (cos δ_{0} + cos δ_{1}), δ_{1} = sin^{- 1} \frac{P _{m 1}}{P _{m a x}}

The largest $P_{m 1}$ the machine can take suddenly from the starting angle $δ_{0}$ : put $δ_{m} = π - δ_{1}$ in the previous equation. From no load ( $δ_{0} = 0$ ) it becomes $P_{m 1} (π - δ_{1}) = P_{m a x} (1 + cos δ_{1})$ , whose solution is $P_{m 1} \approx 0.725 P_{m a x}$ ( $δ_{1} \approx 46. 4^{\circ}$ ).

POWER-CURVE RATIOS DURING AND AFTER A FAULT

r_{1} = \frac{X _{pre}}{X _{during}}, r_{2} = \frac{X _{pre}}{X _{post}}

With the same $E$ and $V$ , each curve's peak is $E V / X$ , so the during-fault and post-fault peaks are $r_{1} P_{m a x}$ and $r_{2} P_{m a x}$ , where $X$ is the transfer reactance between the machine EMF and the infinite bus in each state.

Procedure for a fault followed by clearing

Draw three curves in per unit: pre-fault $P_{m a x} sin δ$ , during-fault $r_{1} P_{m a x} sin δ$ , post-fault $r_{2} P_{m a x} sin δ$ ( $r_{1} = 0$ for a three-phase fault at the machine terminals).
Find $δ_{0} = sin^{- 1} (P_{m} / P_{m a x})$ on the pre-fault curve.
Find $δ_{m a x} = π - sin^{- 1} (P_{m} / (r_{2} P_{m a x}))$ on the post-fault curve.
Accelerating area $A_{1} = \int_{δ_{0}}^{δ_{c}} (P_{m} - r_{1} P_{m a x} sin δ) d δ$ while the fault lasts.
Decelerating area $A_{2} = \int_{δ_{c}}^{δ_{m a x}} (r_{2} P_{m a x} sin δ - P_{m}) d δ$ after clearing.
The limiting case $A_{1} = A_{2}$ gives the critical clearing angle $δ_{c}$ .

WORKED EXAMPLE

A generator delivers $P_{m} = 0.4$ pu to an infinite bus through a network whose maximum transferable power is $P_{m a x} = 1.0$ pu. The input power is suddenly raised to $P_{m 1} = 0.8$ pu with the network unchanged. Find the maximum rotor angle reached and say whether the machine stays in synchronism.

Solution

Initial angle: $δ_{0} = sin^{- 1} 0.4 = 23.5 8^{\circ} = 0.4115$ rad, so $cos δ_{0} = 0.9165$ .
New equilibrium: $δ_{1} = sin^{- 1} 0.8 = 53.1 3^{\circ} = 0.9273$ rad. The limit of stability is $π - δ_{1} = 2.2143$ rad $= 126.8 7^{\circ}$ .
Equal areas: $0.8 (δ_{m} - 0.4115) + cos δ_{m} - 0.9165 = 0$ , i.e. $0.8 δ_{m} + cos δ_{m} = 1.2457$ .
Try $δ_{m} = 1.6255$ rad: $0.8 (1.6255) + cos (1.6255) = 1.3004 - 0.0547 = 1.2457$ , which satisfies the equation.
$δ_{m} = 1.6255$ rad $\approx 93. 1^{\circ}$ , which is below $126.8 7^{\circ}$ .

Answer: Maximum swing about $93. 1^{\circ}$ ; the machine remains in synchronism.

EXAM TIP

The critical clearing angle from equal areas does not depend on $H$ ; only the clearing time does. A larger $H$ gives a longer critical clearing time for the same angle.

Critical clearing angle and critical clearing time

The critical clearing angle $δ_{c}$ is the largest angle at which the fault can be cleared with the machine still stable; clearing at exactly $δ_{c}$ makes the decelerating area just equal to the accelerating area, with the swing ending at $δ_{m a x}$ . The critical clearing time $t_{c}$ is the time the rotor takes to reach $δ_{c}$ from $δ_{0}$ while the fault is on.

GENERAL CRITICAL CLEARING ANGLE

P_{m} (δ_{m a x} - δ_{0}) + r_{2} P_{m a x} cos δ_{m a x} - r_{1} P_{m a x} cos δ_{0} = (r_{2} - r_{1}) P_{m a x} cos δ_{c}

From $A_{1} = A_{2}$ with during-fault curve $r_{1} P_{m a x} sin δ$ and post-fault curve $r_{2} P_{m a x} sin δ$ , $δ_{m a x} = π - sin^{- 1} (P_{m} / (r_{2} P_{m a x}))$ . Angles in radians in the product with $P_{m}$ . Needs $r_{2} \neq = r_{1}$ .

CRITICAL CLEARING ANGLE, ZERO POWER DURING THE FAULT AND UNCHANGED NETWORK AFTER CLEARING

cos δ_{c} = \frac{P _{m}}{P _{m a x}} (π - 2 δ_{0}) - cos δ_{0}

$r_{1} = 0$ , $r_{2} = 1$ , so $δ_{m a x} = π - δ_{0}$ ; $δ_{0}$ in radians. Typical case: three-phase fault at the machine terminals cleared without a line being lost.

CRITICAL CLEARING TIME, ZERO POWER DURING THE FAULT

t_{c} = \frac{2 H ( δ _{c} - δ _{0} )}{π f P _{m}}

Valid only when $P_{e} = 0$ during the fault ( $r_{1} = 0$ ) and the rotor starts at synchronous speed, so acceleration is constant; angles in radians, $H$ in seconds on the same base as $P_{m}$ . For $r_{1} \neq = 0$ the swing equation has to be integrated step by step.

WORKED EXAMPLE

A generator with $H = 5$ s on its own base, $f = 50$ Hz, supplies $P_{m} = 0.6$ pu to an infinite bus with $P_{m a x} = 1.2$ pu. A three-phase fault at the generator terminals reduces the delivered power to zero. The fault is cleared and the network is restored as before. Find the critical clearing angle and the critical clearing time.

Solution

$δ_{0} = sin^{- 1} (0.6/1.2) = 3 0^{\circ} = 0.5236$ rad; $δ_{m a x} = 15 0^{\circ} = 2.6180$ rad.
$cos δ_{c} = \frac{0.6}{1.2} (2.6180 - 0.5236) + cos 15 0^{\circ} = 0.5 (2.0944) - 0.8660 = 0.1812$ .
$δ_{c} = cos^{- 1} 0.1812 = 79. 6^{\circ} = 1.3886$ rad, so $δ_{c} - δ_{0} = 0.8650$ rad.
$t_{c} = \frac{2 ( 5 ) ( 0.8650 )}{π ( 50 ) ( 0.6 )} = \frac{8.650}{94.25} = 0.09178 = 0.303$ s.

Answer: $δ_{c} \approx 79. 6^{\circ}$ and $t_{c} \approx 0.30$ s.

EXAM TIP

Work in radians in every term that multiplies $P_{m}$ (for example $π - 2 δ_{0}$ ) and in $t_{c}$ . Convert $δ_{c}$ to degrees only for the final answer. A fault with $r_{1} = 0$ means $P_{e} = 0$ during the fault, which is the only case where the closed-form $t_{c}$ applies.

EXAM TIP

Check that the result satisfies $δ_{0} < δ_{c} < δ_{m a x}$ . If it does not (possible when $r_{1} > 0$ or the post-fault curve is weak), the limiting case does not exist in that range: compare the areas directly to decide whether the system is stable for every clearing time or for none.

What improves stability, and exam traps

Measures that improve transient stability

Faster fault clearing, which gives a smaller accelerating area.
Higher $P_{m a x}$ : lower series reactance, extra parallel lines, series capacitors, or higher excitation and voltage.
Fast-acting excitation systems, high-speed automatic reclosing, and single-pole switching and reclosing.
Fast valving of the turbine, which cuts $P_{m}$ for a short time after the fault.
Braking resistors switched in for a short time after a fault to absorb surplus accelerating power.
Larger inertia $H$ (slower rotor acceleration, so longer time available to clear the fault).

Quantity	Depends on $H$ ?	Depends on $P_{m} / P_{m a x}$ ?
Critical clearing angle $δ_{c}$	No	Yes
Critical clearing time $t_{c}$	Yes (grows as $H$ )	Yes
Frequency of small oscillation $ω_{n}$	Yes (falls as $1/ H$ )	Yes (through $cos δ_{0}$ )
Steady-state limit	No	No (it is $P_{m a x}$ )

EXAM TIP

Check the base: if $H$ is given on a machine's own rating but $P_{m}$ is in per unit on a system base, convert $H$ first using $H_{new} = H_{old} S_{old} / S_{new}$ , or express $P_{m}$ on the machine base.

EXAM TIP

If the post-fault curve is lower than the pre-fault curve (a line was tripped, $r_{2} < 1$ ), then $δ_{m a x} = π - sin^{- 1} (P_{m} / (r_{2} P_{m a x}))$ is smaller and $δ_{c}$ decreases. If $P_{m} > r_{2} P_{m a x}$ there is no post-fault equilibrium and the machine is unstable no matter how fast the fault is cleared.

PREVIOUS-YEAR QUESTIONS

4 questions from the EE papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE EE 2025 · GATE EE 2024 · GATE EE 2022 · GATE EE 2018

GATE 2025 EE Q.49

GATE 2025 · EEQ.49 · 2 marks · MCQ

In the system shown below, the generator was initially supplying power to the grid. A temporary LLLG bolted fault occurs at F very close to circuit breaker 1. The circuit breakers open to isolate the line. The fault self-clears. The circuit breakers reclose and restore the line. Which one of the following diagrams best indicates the rotor accelerating and decelerating areas?

GATE 2025 EE question 49: Single-line diagram: Gen connected to a bus, two parallel lines to the Grid bus; — Figure from the GATE 2025 EE paper. It is part of the question.

(A)
Fig. (i)
(B)
Fig. (ii)
(C)
Fig. (iii)
(D)
Fig. (iv)

Answer and solution

GATE 2025 EE Q.49: answer and solution

GATE 2024 EE Q.56

GATE 2024 · EEQ.56 · 2 marks · NAT

The single line diagram of a lossless system is shown in the figure. The system is operating in steady-state at a stable equilibrium point with the power output of the generator being $P_{ma x} sin δ$ , where $δ$ is the load angle and the mechanical power input is $0.5 P_{ma x}$ . A fault occurs on line 2 such that the power output of the generator is less than $0.5 P_{ma x}$ during the fault. After the fault is cleared by opening line 2, the power output of the generator is ${P_{ma x} / 2} sin δ$ . If the critical fault clearing angle is $π /2$ radians, the accelerating area on the power angle curve is ______ times $P_{ma x}$ (rounded off to 2 decimal places).

GATE 2024 EE question 56: Single line diagram: a generator (sine-wave circle symbol) on a bus labelled 'Generator bus', connected through a… — Figure from the GATE 2024 EE paper. It is part of the question.

Answer and solution

GATE 2024 EE Q.56: answer and solution

GATE 2022 EE Q.57

GATE 2022 · EEQ.57 · 2 marks · NAT

A 20 MVA, 11.2 kV, 4-pole, 50 Hz alternator has an inertia constant of 15 MJ/MVA. If the input and output powers of the alternator are 15 MW and 10 MW, respectively, the angular acceleration in mechanical degree/s^2 is __________. (round off to nearest integer)

Answer and solution

GATE 2022 EE Q.57: answer and solution

See the question in the paper

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GATE 2022 EE Q.57: answer and solution

GATE 2018 EE Q.6

GATE 2018 · EEQ.6 · 1 mark · MCQ

Consider a lossy transmission line with $V_{1}$ and $V_{2}$ as the sending and receiving end voltages, respectively. $Z$ and $X$ are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be

(A)
greater than $\frac{V _{1} V _{2}}{X}$
(B)
less than $\frac{V _{1} V _{2}}{X}$
(C)
equal to $\frac{V _{1} V _{2}}{X}$
(D)
equal to $\frac{V _{1} V _{2}}{Z}$

Answer and solution

GATE 2018 EE Q.6: answer and solution

See the question in the paper

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GATE 2018 EE Q.6: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 12 marks · NAT

A generator is connected to an infinite bus. Take per-unit powers on the generator rating; classical model, losses and damping neglected. $P_{m a x} = 1.5$ pu, the machine initially delivers power at a rotor angle $δ_{0} = 3 0^{\circ}$ , $H = 5$ s and $f = 50$ Hz. The natural angular frequency of undamped small rotor oscillations about this operating point, in rad/s, is (round off to two decimals).

Answer and solution

Answer: $6.39$ (accepted range $\pm 0.05$ )

$P_{s} = P_{m a x} cos δ_{0} = 1.5 cos 3 0^{\circ} = 1.299$ pu/rad.
$ω_{n} = π f P_{s} / H = π (50) (1.299) /5 = 40.81$ .
$ω_{n} \approx 6.39$ rad/s.

PRACTICE 22 marks · NAT

A generator with $H = 4$ s and $f = 50$ Hz supplies $P_{m} = 0.9$ pu to an infinite bus. The network has $P_{m a x} = 1.5$ pu before and after a three-phase fault at the generator terminals, which makes the delivered power zero while it lasts. Take $δ_{c} = 71.7 7^{\circ}$ as the critical clearing angle. The critical clearing time in seconds is (round off to three decimals).

Answer and solution

Answer: $0.186$ (accepted range $\pm 0.004$ )

$δ_{0} = sin^{- 1} (0.9/1.5) = sin^{- 1} 0.6 = 36.8 7^{\circ} = 0.6435$ rad.
$δ_{c} = 71.7 7^{\circ} = 1.2526$ rad, so $δ_{c} - δ_{0} = 0.6091$ rad.
$t_{c} = 2 H (δ_{c} - δ_{0}) / (π f P_{m}) = 2 (4) (0.6091) / (π (50) (0.9)) = 4.873/141.37 = 0.03447$ .
$t_{c} \approx 0.186$ s.

PRACTICE 31 mark · MCQ

Two coherent generators are in one station: 200 MVA with $H = 5$ s and 100 MVA with $H = 4$ s, each on its own rating. The equivalent inertia constant on a 100 MVA base is:

(A)
9 s
(B)
14 s
(C)
7 s
(D)
28 s

Answer and solution

Answer: (B)

Stored energies: $5 \times 200 = 1000$ MJ and $4 \times 100 = 400$ MJ, total 1400 MJ.
Divide by the 100 MVA base: $1400/100 = 14$ s.

PRACTICE 41 mark · NAT

A 500 MVA generator has an inertia constant $H = 3.2$ s on its own rating. Its inertia constant on a 100 MVA system base, in seconds, is:

Answer and solution

Answer: $16$ (accepted range $\pm 0.01$ )

Stored energy $= 3.2 \times 500 = 1600$ MJ.
$H_{100} = 1600/100 = 16$ s.

PRACTICE 52 marks · MCQ

A generator delivers $P_{m} = 0.8$ pu to an infinite bus; the network can transfer $P_{m a x} = 1.0$ pu before, and again after, a three-phase fault. During the fault at the generator terminals the electrical power is zero. The critical clearing angle is closest to:

(A)
$58. 2^{\circ}$
(B)
$64. 6^{\circ}$
(C)
$71. 3^{\circ}$
(D)
$79. 6^{\circ}$

Answer and solution

Answer: (B)

$δ_{0} = sin^{- 1} 0.8 = 53.1 3^{\circ} = 0.9273$ rad.
$cos δ_{c} = 0.8 (π - 2 (0.9273)) - cos 53.1 3^{\circ} = 0.8 (1.2870) - 0.6 = 0.4296$ .
$δ_{c} = cos^{- 1} 0.4296 \approx 64. 6^{\circ}$ .

PRACTICE 62 marks · MSQ

For a generator on an infinite bus with $P_{e} = P_{m a x} sin δ$ and constant $P_{m}$ (classical model, losses neglected), which of the following statements is/are correct?

(A)
For a given surplus power, a larger inertia constant gives a smaller rotor acceleration.
(B)
The synchronising power coefficient is zero at $δ = 9 0^{\circ}$ .
(C)
For a fault with zero power transfer during the fault, the critical clearing angle depends on $H$ .
(D)
The steady-state stability limit is reached when $δ = 0^{\circ}$ .

Answer and solution

Answer: (A), (B)

A is true: $\ddot{δ} = π f (P_{m} - P_{e}) / H$ , so acceleration varies as $1/ H$ .
B is true: $P_{s} = P_{m a x} cos δ$ , which is zero at $9 0^{\circ}$ .
C is false: the equal area condition involves only $P_{m}$ , $P_{m a x}$ and the angles, not $H$ .
D is false: the limit occurs at $δ = 9 0^{\circ}$ , where $P_{e} = P_{m a x}$ .

OFFICIAL GATE 2027 SYLLABUS WORDING

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GATE EE

System stability concepts, Equal area criterion

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