GATE EE Power Systems Formula Sheet: Key Formulas and Traps

Q: What is the formula for the critical clearing angle in GATE EE?

For a fault during which the generator delivers no power, with the same network restored after clearing, cosδcr = (π - 2δ₀)sinδ₀ - cosδ₀, with the initial angle in radians inside the bracket. For an initial angle of 20 degrees this gives about 95.97 degrees. If the network changes after clearing, use the full equal area criterion instead.

Q: How do you change a per unit impedance to a new base?

Multiply the old per unit value by the ratio of new to old MVA base, and by the square of old to new voltage base. So 0.12 pu on 40 MVA and 13.2 kV becomes 0.432 pu on 100 MVA and 11 kV. A common slip is to invert one of the two ratios.

Q: What is the fault current for a single line to ground fault?

The three sequence networks are connected in series, so each sequence current is E/(Z₁ + Z₂ + Z₀ + 3Zf) and the fault current is three times that. A neutral impedance also enters the zero sequence network as three times its value. Leaving out either factor of three is the usual error.

Q: What is the condition for economic load dispatch without losses?

Every unit that is not at a limit runs at the same incremental cost, equal to lambda, and the outputs add up to the demand. With a quadratic cost curve each output is lambda minus the linear coefficient, divided by twice the quadratic coefficient. A unit that hits a limit is fixed there and the others share the rest.

Q: How do two generators share a load change through governor droop?

Each unit picks up power in proportion to its rating divided by its per unit droop, because all units see the same frequency change. A 600 MW unit at 5 per cent droop and a 400 MW unit at 4 per cent share a 220 MW rise as 120 MW and 100 MW, with frequency falling 0.5 Hz on 50 Hz.

By MD ANISH AHAMADUpdated 4 Oct 202610 min read

GATE EE Power Systems Formula Sheet: Key Formulas and Traps

The key GATE EE power systems formulas are the line constants and ABCD models, the power-angle equation, per unit base change, the bus admittance matrix, the equal incremental cost rule, governor droop, the sequence-network fault currents, relay settings and the swing equation with the equal area criterion. Each group below gives the formulas with their conditions, one worked example and the trap that costs marks.

In this guide

Key takeaways
The notation this sheet uses
Line constants and the ABCD models of a line
Power transfer, voltage drop and compensation
Per unit, the bus admittance matrix and load flow
Economic dispatch with and without losses
Frequency control and governor droop
Symmetrical components and fault currents
Protection and circuit breakers
Stability, the swing equation and the equal area criterion
Quick revision

Key takeaways

Inductance uses the reduced radius $r' = 0.7788r$ ; capacitance uses the actual radius.
A per unit base change scales with the MVA ratio and with the square of the inverted voltage ratio.
Economic dispatch gives every unit inside its limits the same incremental cost.
Units share a load change in proportion to their rating divided by their per unit droop.
A neutral impedance appears as three times its value, only in the zero sequence network.
The swing equation and the critical clearing angle need angles in electrical radians.

The notation this sheet uses

A per unit value is the actual value divided by a chosen base. You pick a three-phase MVA base and a line voltage base; the impedance and current bases follow.

Line quantities are measured between lines, phase quantities across one phase. In star, $V_L = \sqrt{3}V_{ph}$ , $I_L = I_{ph}$ and three-phase power is $\sqrt{3}V_LI_L$ .

Sequence components split any unbalanced set of three phasors into three balanced sets: positive (subscript 1), negative (2) and zero (0). The operator a rotates a phasor by 120 degrees:

a = 1\angle 120^\circ = -0.5 + j0.866, \qquad a^2 = 1\angle 240^\circ, \qquad 1 + a + a^2 = 0

The load angle $\delta$ is the angle by which the sending voltage, or a generator's internal voltage $E$ , leads the receiving voltage.

Line constants and the ABCD models of a line

Series inductance and shunt capacitance depend on the spacing between phases and the conductor size. The ABCD constants relate sending-end voltage and current to receiving-end values.

Formula	Watch out for
$\displaystyle L = 2\times10^{-7}\ln\frac{\text{GMD}}{\text{GMR}}$ H/m per phase	GMR of a solid conductor is $r' = 0.7788r$
$\displaystyle C = \frac{2\pi\varepsilon_0}{\ln(\text{GMD}/r)}$ F/m to neutral	Uses $r$ , not $r'$
$\text{GMD} = (D_{ab}D_{bc}D_{ca})^{1/3}$	Transposed line
Bundle GMR: $\sqrt{r'd}$ , $(r'd^2)^{1/3}$ , $1.09(r'd^3)^{1/4}$	Bundling lowers $L$ , raises $C$ , reduces corona
Short line: $A = D = 1$ , $B = Z$ , $C = 0$	Up to about 80 km
Nominal $\pi$ : $\displaystyle A = D = 1 + \frac{YZ}{2}$ , $B = Z$ , $\displaystyle C = Y\left(1 + \frac{YZ}{4}\right)$	Nominal T: $\displaystyle B = Z\left(1 + \frac{YZ}{4}\right)$ , $C = Y$
Long line: $A = D = \cosh\gamma l$ , $B = Z_c\sinh\gamma l$ , $\displaystyle C = \frac{\sinh\gamma l}{Z_c}$	$\gamma = \sqrt{zy}$ , $Z_c = \sqrt{z/y}$ per unit length
$\displaystyle \text{SIL} = \frac{V_L^2}{Z_c}$	No net reactive power at SIL
No load: $\displaystyle V_R = \frac{V_S}{A}$	Ferranti effect: $V_R > V_S$

Worked example. For radius 1.2 cm at 3 m equilateral spacing, $r' = 0.7788 \times 0.012 = 0.009346$ m and $L = 2\times10^{-7}\ln(3/0.009346) = 2\times10^{-7}\times5.771 = 1.154$ mH/km. Now a nominal- $\pi$ line with $Z = j100\ \Omega$ and $Y = j0.0004$ S on no load:

A = 1 + \frac{(j0.0004)(j100)}{2} = 1 - 0.02 = 0.98, \qquad V_R = \frac{V_S}{0.98} = 1.0204\,V_S

The trap is the sign: $j \times j = -1$ , so $A$ falls below 1 and the voltage rises about 2 per cent.

Power transfer, voltage drop and compensation

Over a lossless line, real power follows the load angle and reactive power the voltage difference. Compensation changes the reactance or supplies vars locally.

Formula	Watch out for
$\displaystyle P = \frac{V_SV_R}{X}\sin\delta$ , maximum at $\delta = 90^\circ$	Line voltages give three-phase power
$\displaystyle Q_R = \frac{V_SV_R\cos\delta - V_R^2}{X}$	Vars flow from higher voltage to lower
$\displaystyle \Delta V \approx \frac{PR + QX}{V} = I(R\cos\phi + X\sin\phi)$	Greatest at $\tan\phi = X/R$ ; zero for leading $\tan\phi = R/X$
Line loss $\displaystyle \propto \frac{1}{V^2\cos^2\phi}$	Same power delivered
Series capacitor: $\displaystyle P_{\max} = \frac{V^2}{X_L(1 - k)}$ , $\displaystyle k = \frac{X_C}{X_L}$	Risk of subsynchronous resonance
$Q_C = P(\tan\phi_1 - \tan\phi_2)$	Star bank $\displaystyle C = \frac{Q_C}{3\omega V_{ph}^2}$ ; delta bank needs one third of that

Shunt capacitors raise voltage under heavy load; shunt reactors hold it down on light load.

Worked example. With $V_S = V_R = 1$ pu and $X_L = 0.8$ pu, $P_{\max} = 1/0.8 = 1.25$ pu. Half compensation, $k = 0.5$ , gives $P_{\max} = 1/(0.8 \times 0.5) = 2.5$ pu. Separately, raising a 300 kW load from 0.8 to 0.95 lagging needs $Q_C = 300(0.75 - 0.3287) = 126.4$ kvar.

The trap: a delta bank sees line voltage, so it needs one third of the star capacitance.

Per unit, the bus admittance matrix and load flow

Per unit puts every device on one base, the bus admittance matrix $Y_{\text{bus}}$ describes the network, and load flow solves its power equations.

Formula	Watch out for
$\displaystyle Z_{\text{base}} = \frac{V_{\text{base}}^2}{S_{\text{base}}}$ , $\displaystyle I_{\text{base}} = \frac{S_{\text{base}}}{\sqrt{3}V_{\text{base}}}$	Three-phase $S$ , line $V$
$\displaystyle Z_{\text{new}} = Z_{\text{old}}\frac{S_{\text{new}}}{S_{\text{old}}}\left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2$	Voltage ratio inverted and squared
$Y_{ii}$ is the sum of admittances at bus $i$ ; $Y_{ij} = -y_{ij}$	Include shunts and half the line charging
Slack: $\vert V\vert$ , $\delta$ ; PV: $P$ , $\vert V\vert$ ; PQ: $P$ , $Q$	A generator holding voltage is PV
$P_i - jQ_i = V_i^*\sum_k Y_{ik}V_k$	Conjugate on $V_i$
Newton–Raphson polar Jacobian order $2(n - 1) - n_{PV}$	Quadratic convergence, few iterations
DC load flow: $\displaystyle P_{ij} = \frac{\delta_i - \delta_j}{X_{ij}}$	Per unit, angles in radians

One Gauss–Seidel update uses new values as soon as they exist:

V_i \leftarrow \frac{1}{Y_{ii}}\left[\frac{P_i - jQ_i}{V_i^*} - \sum_{k \ne i} Y_{ik}V_k\right]

An off-nominal real tap keeps $Y_{\text{bus}}$ symmetric; a phase-shifting transformer makes $Y_{ij} \ne Y_{ji}$ . With no shunt path to earth it is singular.

Worked example. A reactance of 0.12 pu on 40 MVA and 13.2 kV, moved to 100 MVA and 11 kV, becomes $\displaystyle 0.12 \times \frac{100}{40} \times \left(\frac{13.2}{11}\right)^2 = 0.12 \times 2.5 \times 1.44 = 0.432$ pu. A 14-bus system with 1 slack, 4 PV and 9 PQ buses has a Jacobian of order $2 \times 13 - 4 = 22$ .

Remember: A transformer's per unit impedance is the same from either side when the voltage bases follow its turns ratio, and three-phase and per-phase per unit values are equal.

Economic dispatch with and without losses

Economic dispatch meets the demand at least total fuel cost, so every free unit runs at the same incremental cost.

Formula	Watch out for
$\displaystyle \frac{dC_i}{dP_i} = \lambda$ for every unit inside its limits	A unit at a limit is fixed there
$\displaystyle C_i = a_i + b_iP_i + c_iP_i^2 \Rightarrow P_i = \frac{\lambda - b_i}{2c_i}$	Incremental cost is $b_i + 2c_iP_i$
$\sum_i P_i = P_D$ without losses; $P_D + P_L$ with losses	Losses are extra generation
$\displaystyle L_i\frac{dC_i}{dP_i} = \lambda$ , $\displaystyle L_i = \frac{1}{1 - \partial P_L/\partial P_i}$	Penalty factor above 1 if the unit raises losses
$P_L = \sum_i\sum_j P_iB_{ij}P_j$	Loss formula with B-coefficients

Worked example. Let $IC_1 = 10 + 0.04P_1$ and $IC_2 = 12 + 0.06P_2$ in Rs/MWh, with a 250 MW load. Write each output in terms of $\lambda$ and add:

\frac{\lambda - 10}{0.04} + \frac{\lambda - 12}{0.06} = 250 \;\Rightarrow\; 41.67\lambda = 700 \;\Rightarrow\; \lambda = 16.8

So $P_1 = 6.8/0.04 = 170$ MW and $P_2 = 4.8/0.06 = 80$ MW. If unit 1 has a 150 MW maximum, fix it there; unit 2 then carries 100 MW at $\lambda = 12 + 0.06 \times 100 = 18$ Rs/MWh.

The trap is applying equal $\lambda$ to a unit already at its limit.

The last-minute sheet in the GATE EE 2027 book covers every syllabus topic in this compact form, each result with the conditions under which it holds.

Frequency control and governor droop

Governors and automatic generation control (AGC) tie real power to frequency; the automatic voltage regulator (AVR) ties reactive power to voltage.

Formula	Watch out for
$\displaystyle R = -\frac{\Delta f}{\Delta P}$	Per unit on the unit's own rating
$\displaystyle \Delta P_i = \frac{\Delta f}{f_0} \cdot \frac{P_{\text{rated},i}}{R_i}$	Share follows rating over droop
$\displaystyle \Delta f = -\frac{\Delta P_L}{D + 1/R}$ , $\displaystyle \beta = D + \frac{1}{R}$	Steady state with governors only
AGC adds integral control	It returns $\Delta f$ to zero

Worked example. A 600 MW unit at 5 per cent droop and a 400 MW unit at 4 per cent share a 220 MW rise on 50 Hz. The total stiffness is $\displaystyle \frac{600}{0.05} + \frac{400}{0.04} = 22000$ MW per unit frequency. So $\displaystyle \frac{\Delta f}{f_0} = \frac{220}{22000} = 0.01$ , and frequency falls 0.5 Hz. Unit 1 takes $0.01 \times 12000 = 120$ MW and unit 2 takes 100 MW.

The trap is sharing by rating alone, which would split the 220 MW as 132 and 88.

Symmetrical components and fault currents

Any unbalanced set equals the sum of its three sequence sets. For phase a:

\begin{aligned} V_{a0} &= \tfrac{1}{3}(V_a + V_b + V_c) \\ V_{a1} &= \tfrac{1}{3}(V_a + aV_b + a^2V_c) \\ V_{a2} &= \tfrac{1}{3}(V_a + a^2V_b + aV_c) \end{aligned}

Going back, $V_a = V_{a0} + V_{a1} + V_{a2}$ , $V_b = V_{a0} + a^2V_{a1} + aV_{a2}$ and $V_c = V_{a0} + aV_{a1} + a^2V_{a2}$ .

Formula	Watch out for
Three-phase: $\displaystyle I_f = \frac{E}{Z_1 + Z_f}$ ; $\displaystyle \text{MVA}_{sc} = \frac{\text{MVA}_{\text{base}}}{Z_{1,\text{pu}}}$	Positive sequence only
Line to ground: $\displaystyle I_{a1} = I_{a2} = I_{a0} = \frac{E}{Z_1 + Z_2 + Z_0 + 3Z_f}$ , $I_f = 3I_{a1}$	Two factors of 3
Line to line: $\displaystyle I_{a1} = -I_{a2} = \frac{E}{Z_1 + Z_2 + Z_f}$ , $\displaystyle \vert I_b\vert = \frac{\sqrt{3}E}{\vert Z_1 + Z_2 + Z_f\vert}$	$I_{a0} = 0$
Double line to ground: $\displaystyle I_{a1} = \frac{E}{Z_1 + Z_2 \parallel (Z_0 + 3Z_f)}$	$I_{a2}$ , $I_{a0}$ by current division
$I_n = 3I_{a0}$ ; neutral $Z_n$ enters as $3Z_n$	Zero sequence network only
Line: $Z_1 = Z_2 < Z_0$ ; machine: $Z_0 < Z_2 < Z_1$	Earth return raises a line's $Z_0$
Fault at bus $k$ : $\displaystyle I_f = \frac{V_f}{Z_{kk} + Z_f}$ , $V_i = V_f - Z_{ik}I_f$	Uses $Z_{\text{bus}}$ , not $Y_{\text{bus}}$

Zero sequence current needs a return path: a grounded star passes it, while a delta or ungrounded star blocks it on that side. A star–delta transformer shifts positive and negative sequences by $\pm30^\circ$ .

Worked example. Take $E = 1$ pu, $Z_1 = Z_2 = j0.25$ and $Z_0 = j0.1$ pu. Three-phase: $1/0.25 = 4$ pu. Line to ground: $3/0.6 = 5$ pu. Line to line: $\sqrt{3}/0.5 = 3.46$ pu. With $Z_n = j0.05$ , the zero sequence path becomes $j0.25$ and line to ground falls to $3/0.75 = 4$ pu.

Trap: A neutral impedance is multiplied by three because all three zero sequence currents return through it. It never appears in the positive or negative sequence network.

Protection and circuit breakers

Relays detect the fault and trip in order; the breaker interrupts at a natural current zero.

Formula	Watch out for
IDMT: $\displaystyle t = \frac{0.14\,\text{TMS}}{\text{PSM}^{0.02} - 1}$	IEC standard inverse curve
$\displaystyle \text{PSM} = \frac{I_{\text{fault}}}{\text{CT ratio} \times \text{relay setting}}$	Grade upstream relays slower
$\displaystyle Z_{\text{sec}} = Z_{\text{pri}} \times \frac{\text{CT ratio}}{\text{PT ratio}}$	CT ratio on top
Impedance: circle about origin; reactance: line of constant $X$ ; mho: circle through origin	Mho is directional; reactance ignores arc resistance
Breaking capacity $= \sqrt{3}VI_{\text{break}}$ ; making current $= 2.55 \times I_{\text{break}}$	$2.55 = 1.8 \times \sqrt{2}$
Restriking: $\displaystyle f_n = \frac{1}{2\pi\sqrt{LC}}$ , peak up to $2V_m$ , maximum rate of rise $V_m\omega_n$	Resistance switching damps it

For differential protection of a star–delta transformer, connect the CTs in delta on the star side and in star on the delta side. Distance zone 1 covers about 80 to 90 per cent of the line instantly.

Worked example. A 4000 A fault, a 400/1 CT and a 1 A setting give $\text{PSM} = 4000/400 = 10$ . With TMS 0.2, $t = 0.14 \times 0.2/(10^{0.02} - 1) = 0.028/0.04713 = 0.594$ s. A 20 Ω line seen through a 600/1 CT and a 132 kV/110 V PT measures $20 \times 600/1200 = 10$ Ω at the relay. A 20 kA breaking current means a making current of 51 kA.

The virtual calculator guide shows how to evaluate $10^{0.02}$ quickly.

Stability, the swing equation and the equal area criterion

The rotor accelerates by the difference between mechanical input and electrical output. In per unit, with $\delta$ in electrical radians:

\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e \;\Rightarrow\; \frac{d^2\delta}{dt^2} = \frac{\pi f}{H}(P_m - P_e)

Formula	Watch out for
$\displaystyle H = \frac{\text{stored kinetic energy}}{\text{MVA rating}}$ ; $\displaystyle H_{\text{new}} = H_{\text{old}}\frac{S_{\text{old}}}{S_{\text{new}}}$	Electrical angle $=$ pole pairs $\times$ mechanical angle
$\displaystyle P_{\max} = \frac{EV}{X}$ ; synchronising coefficient $\displaystyle \frac{EV}{X}\cos\delta$	A line outage raises $X$ , lowers $P_{\max}$
$\displaystyle \delta_{\max} = \pi - \sin^{-1}\frac{P_m}{P_{\max,\text{post}}}$	Post-fault curve
$\cos\delta_{cr} = (\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0$	Zero power during fault, network unchanged after
$\displaystyle t_{cr} = \sqrt{\frac{2H(\delta_{cr} - \delta_0)}{\pi f P_m}}$	Angles in radians

The system is stable if the decelerating area can equal the accelerating area before $\delta$ reaches $\delta_{\max}$ .

Worked example. With $H = 4$ MJ/MVA, 50 Hz and $P_a = 0.6$ pu, $\displaystyle \frac{d^2\delta}{dt^2} = \frac{\pi \times 50 \times 0.6}{4} = 23.56$ elec rad/s², or 1350 elec deg/s². For $\delta_0 = 20^\circ$ (0.3491 rad), $\cos\delta_{cr} = 2.4435 \times 0.3420 - 0.9397 = -0.1040$ , so $\delta_{cr} = 95.97^\circ$ and $\delta_{cr} - \delta_0 = 1.3259$ rad. With $H = 5$ and $P_m = 0.8$ pu, $t_{cr} = \sqrt{10 \times 1.3259/(\pi \times 50 \times 0.8)} = 0.325$ s.

In one line: Find the accelerating area while the fault is on, then check that the post-fault curve can return an equal area before the angle reaches its maximum.

Numerical answer questions carry no negative marks under the marking scheme, which is common to every paper. GATE 2027 runs from 6 to 21 February 2027; see the exam dates.

More formula sheets: all of GATE EE · Control Systems · Electrical Machines · Power Electronics

Quick revision

Use $r' = 0.7788r$ for inductance and the true radius for capacitance.
On no load, $V_R = V_S/A$ , and $A < 1$ gives the Ferranti rise.
Change base with the MVA ratio and the inverted voltage ratio squared.
Equal incremental cost holds only for units inside their limits.
Droop sharing follows rating over droop, and AGC removes the frequency error.
Line to ground fault current is $3E/(Z_1 + Z_2 + Z_0 + 3Z_f)$ , and $Z_n$ enters as $3Z_n$ .
Relay ohms are primary ohms times CT ratio over PT ratio.
Keep swing equation angles in electrical radians.

Frequently asked questions

What is the formula for the critical clearing angle in GATE EE?

For a fault during which the generator delivers no power, with the same network restored after clearing, $\cos\delta_{cr} = (\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0$ , with the initial angle in radians inside the bracket. For an initial angle of 20 degrees this gives about 95.97 degrees. If the network changes after clearing, use the full equal area criterion instead.

How do you change a per unit impedance to a new base?

Multiply the old per unit value by the ratio of new to old MVA base, and by the square of old to new voltage base. So 0.12 pu on 40 MVA and 13.2 kV becomes 0.432 pu on 100 MVA and 11 kV. A common slip is to invert one of the two ratios.

What is the fault current for a single line to ground fault?

The three sequence networks are connected in series, so each sequence current is $E/(Z_1 + Z_2 + Z_0 + 3Z_f)$ and the fault current is three times that. A neutral impedance also enters the zero sequence network as three times its value. Leaving out either factor of three is the usual error.

What is the condition for economic load dispatch without losses?

Every unit that is not at a limit runs at the same incremental cost, equal to lambda, and the outputs add up to the demand. With a quadratic cost curve each output is lambda minus the linear coefficient, divided by twice the quadratic coefficient. A unit that hits a limit is fixed there and the others share the rest.

How do two generators share a load change through governor droop?

Each unit picks up power in proportion to its rating divided by its per unit droop, because all units see the same frequency change. A 600 MW unit at 5 per cent droop and a 400 MW unit at 4 per cent share a 220 MW rise as 120 MW and 100 MW, with frequency falling 0.5 Hz on 50 Hz.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.

GATE EE Power Systems Formula Sheet: Key Formulas and Traps

Key takeaways

The notation this sheet uses

Line constants and the ABCD models of a line

Power transfer, voltage drop and compensation

Per unit, the bus admittance matrix and load flow

Economic dispatch with and without losses

Frequency control and governor droop

Symmetrical components and fault currents

Protection and circuit breakers

Stability, the swing equation and the equal area criterion

Quick revision

Frequently asked questions

Sources

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