GATE EE Electrical Machines Formula Sheet: Key Formulas and Traps

Q: What is the formula for the slip of an induction motor?

Slip is s = (Nₛ - N)/Nₛ, where Nₛ = 120f/P is the synchronous speed and N the rotor speed. A 4-pole, 50 Hz motor at 1440 rpm has Nₛ = 1500 rpm and a slip of 0.04. The rotor frequency is then sf = 2 Hz, and the rotor copper loss is the slip times the air-gap power.

Q: At what load is the efficiency of a transformer maximum?

Efficiency is maximum when the copper loss equals the core loss. Copper loss grows as the square of the load, so the maximum falls at the fraction x = √(Pᵢ / Pcu,fl) of full load. A transformer whose full-load copper loss is four times its core loss reaches maximum efficiency at half load, not at full load.

Q: What is the EMF equation of a DC machine?

The generated EMF is E = PPhi Z N/(60A), where P is the number of poles, Phi the flux per pole, Z the number of armature conductors, N the speed in rpm and A the number of parallel paths. Take A = 2 for a wave winding and A = P for a lap winding. In short, E = kPhiω.

Q: How much does star-delta starting reduce the starting torque?

Star-delta starting puts 1/√3 of the line voltage across each winding. Both the starting line current and the starting torque fall to one-third of their direct-on-line values. A motor that draws six times full-load current and develops 1.5 times full-load torque direct on line draws twice full-load current and develops half full-load torque in star.

Q: What is the power angle equation of a synchronous machine?

For a cylindrical-rotor machine with armature resistance neglected, the power per phase is P = (EV/Xₛ)sinδ, where δ is the load angle between the excitation EMF and the terminal voltage. Using line voltages gives the three-phase power directly. The maximum power occurs at δ = 90^∘ and equals EV/Xₛ.

By MD ANISH AHAMADUpdated 4 Oct 20269 min read

GATE EE Electrical Machines Formula Sheet: Key Formulas and Traps

The key GATE EE electrical machines formulas are the transformer EMF, efficiency and regulation equations, the DC machine EMF and speed relations, the induction motor's slip and power split, and the synchronous machine's power-angle equation. Each carries a condition, such as per-phase values, referred values or a load fraction, and most lost marks come from that condition. This sheet takes the machines in learning order, works an example for each and names the trap.

In this guide

Key takeaways
The terms and notation used on this sheet
How the single-phase transformer is modelled and tested
How three-phase connections shift voltage and how the auto-transformer saves copper
How DC machine speed follows back EMF and flux
How slip ties together every induction machine formula
How synchronous machines convert power through the load angle
How losses decide the efficiency of every machine
How to use this sheet before the exam
Quick revision

Key takeaways

Most machine formulas are per phase, so convert line values before you substitute.
A transformer's flux is set by voltage over frequency, and its core losses follow from that flux.
Transformer efficiency peaks when copper loss equals core loss, at the fraction $\sqrt{P_i / P_{cu,fl}}$ of full load.
DC motor speed is proportional to back EMF over flux, not to the supply voltage.
Induction air-gap power splits as $1 : s : (1 - s)$ into itself, rotor copper loss and mechanical power.
Synchronous power is $(EV/X_s)\sin\delta$ , and the angle in it is electrical.

The terms and notation used on this sheet

A per-phase quantity belongs to one winding of a three-phase machine; in star the phase voltage is $V_L/\sqrt{3}$ . A referred value is an impedance moved across a transformer. Moved to the primary it is multiplied by $a^2$ , with $a = N_1/N_2$ , and marked with a prime, as in $R_2'$ .

Slip $s$ is the fraction by which an induction rotor runs behind its rotating field. A per-unit value is the actual value divided by a base; $\epsilon_r$ and $\epsilon_x$ are a transformer's per-unit resistance and reactance. The load fraction $x$ is actual load over full load.

Voltages and currents are RMS; $\Phi_m$ and $B_m$ are peak values. The load angle $\delta$ lies between excitation EMF and terminal voltage, and $\phi$ is the power-factor angle. $P$ means poles in speed formulas and power elsewhere, as in the book.

How the single-phase transformer is modelled and tested

The induced EMF depends on frequency, turns and peak flux. The equivalent circuit puts winding resistance and leakage reactance in series and the core in a shunt branch. The open-circuit test finds the shunt branch; the short-circuit test finds the series branch.

Formula	Watch out for
$E = 4.44\,f N \Phi_m$ ; $\displaystyle \frac{V_1}{V_2} = \frac{N_1}{N_2} = a$ ; $\displaystyle \frac{I_1}{I_2} = \frac{1}{a}$	$\Phi_m$ is peak, $E$ is RMS
$Z_2' = a^2 Z_2$ (referred to the primary)	$a^2$ , not $a$
Open circuit: $\displaystyle \cos\phi_0 = \frac{P_0}{V_0 I_0}$ , $\displaystyle R_c = \frac{V_0}{I_0\cos\phi_0}$ , $\displaystyle X_m = \frac{V_0}{I_0\sin\phi_0}$	LV side, rated voltage; $P_0$ is core loss
Short circuit: $\displaystyle Z = \frac{V_{sc}}{I_{sc}}$ , $\displaystyle R = \frac{P_{sc}}{I_{sc}^2}$ , $X = \sqrt{Z^2 - R^2}$	HV side, rated current; referred to that side
Regulation $\approx \epsilon_r\cos\phi \pm \epsilon_x\sin\phi$	$+$ lagging, $-$ leading; zero when leading with $\tan\phi = R/X$
$\displaystyle \eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2 P_{cu,fl}}$ ; maximum at $x = \sqrt{P_i / P_{cu,fl}}$	Copper loss scales as $x^2$ ; core loss is fixed

All-day efficiency divides 24-hour energy output by output plus all energy lost, which is why distribution transformers are designed for low core loss.

Worked example. A 50 kVA transformer with 400 W core loss and 900 W full-load copper loss, at unity power factor:

x = \sqrt{\tfrac{400}{900}} = 0.667,\quad \eta_{\max} = \frac{33.33}{33.33 + 0.4 + 0.4} = 97.66\ \text{per cent}

For regulation with $\epsilon_r = 0.02$ and $\epsilon_x = 0.05$ at 0.8 lagging: $0.016 + 0.030 = 0.046$ , or 4.6 per cent; at 0.8 leading, $-1.4$ per cent.

Trap: Short-circuit test values are referred to the side the test was done on, usually the high-voltage side. Divide the impedance by $a^2$ before using it with low-voltage currents.

How three-phase connections shift voltage and how the auto-transformer saves copper

Connections change both the line-voltage ratio and the phase angle. The vector group names the shift on a clock: each hour is 30 degrees, reading the low-voltage phasor against the high-voltage one.

Formula	Watch out for
Line ratio: Yy, Dd $\displaystyle \frac{N_1}{N_2}$ ; Yd $\displaystyle \frac{\sqrt{3}\,N_1}{N_2}$ ; Dy $\displaystyle \frac{N_1}{\sqrt{3}\,N_2}$	Turns ratio is the phase ratio
$0^\circ$ : Yy0, Dd0, Dz0; $180^\circ$ : Yy6, Dd6, Dz6; $-30^\circ$ : Yd1, Dy1, Yz1; $+30^\circ$ : Yd11, Dy11, Yz11	Group 1: LV lags; group 11: LV leads
Delta: $I_L = \sqrt{3}\,I_{\text{winding}}$ , displaced $30^\circ$	Magnitude and angle both change
Open delta: $\displaystyle \frac{1}{\sqrt{3}} = 57.7$ per cent of the delta rating	Not two-thirds
Parallel sharing: $\displaystyle S_1 = S\,\frac{Z_2}{Z_1 + Z_2}$	Inverse to impedance, common base
Auto, $\displaystyle a = \frac{V_H}{V_L}$ : transformed $\displaystyle 1 - \frac{1}{a}$ , conducted $\displaystyle \frac{1}{a}$	Copper saving is $\displaystyle \frac{1}{a}$
Reconnected as auto: rating $\displaystyle \times \frac{V_{H,\text{auto}}}{V_{\text{series}}}$	Divide by the series-winding voltage

Parallel transformers need the same voltage ratio, polarity, phase sequence and phase displacement, and equal per-unit impedances to share load well.

Worked example. A 5 kVA, 400/100 V transformer reconnected as a 500/400 V auto-transformer has a 100 V series winding:

S_{\text{auto}} = 5 \times \frac{500}{100} = 25\ \text{kVA}, \quad \text{transformed} = \left(1 - \frac{400}{500}\right) \times 25 = 5\ \text{kVA}

The series winding still carries its rated 50 A, now at 500 V; the other 20 kVA is conducted.

How DC machine speed follows back EMF and flux

Torque in any machine comes from a field and a current interacting. For a singly excited device with a linear magnetic circuit, $\displaystyle T = \frac{1}{2} i^2 \frac{dL}{d\theta}$ . In a DC machine, EMF and torque both scale with flux, and the armature equation links EMF, terminal voltage and current.

Formula	Watch out for
$\displaystyle E = \frac{P\Phi Z N}{60A} = k\Phi\omega$ ; $\displaystyle T = \frac{P\Phi Z I_a}{2\pi A} = k\Phi I_a$	$A = 2$ wave, $A = P$ lap
Generator $V = E - I_a R_a$ ; motor $V = E + I_a R_a$	Generates when $E > V$
$\displaystyle \frac{N_2}{N_1} = \frac{E_2}{E_1}\cdot\frac{\Phi_1}{\Phi_2}$ , with $E = V - I_a R_a$	Scale by $E$ , not $V$
Series motor: $T \propto I_a^2$ , $\displaystyle N \propto \frac{1}{I_a}$	Flux follows current; never unloaded
Armature control below base speed; field weakening above	Constant torque below, constant power above

A self-excited generator needs residual flux, an aiding field connection and field resistance below the critical value.

Worked example. A 250 V separately excited motor, $R_a = 0.4\ \Omega$ , draws 50 A at 1200 rpm. The supply falls to 200 V at constant torque and flux, so $I_a$ stays 50 A:

E_1 = 230\ \text{V},\quad E_2 = 180\ \text{V},\quad N_2 = 1200 \times \frac{180}{230} = 939.13\ \text{rpm}

In one line: Find the new armature current from the load torque law, then the new back EMF, then scale speed by the EMF ratio and inverse flux ratio.

Scaling by supply voltage gives 960 rpm, ignoring the 20 V armature drop.

How slip ties together every induction machine formula

Slip sets the rotor frequency, the rotor EMF and the share of air-gap power lost as heat. Learn the power split first; most results follow from it.

Formula	Watch out for
$\displaystyle N_s = \frac{120f}{P}$ ; $\displaystyle s = \frac{N_s - N}{N_s}$ ; $f_2 = sf$ ; rotor EMF $sE_2$	$P$ is poles, not pole pairs
$P_g : P_{cu2} : P_m = 1 : s : (1 - s)$ ; $\displaystyle T = \frac{P_g}{\omega_s}$	Efficiency cannot exceed $1 - s$
Motoring $0 < s < 1$ ; generating $s < 0$ ; plugging $1 < s < 2$	Generating still needs supply vars
$\displaystyle s_m = \frac{R_2'}{\sqrt{R_1^2 + (X_1 + X_2')^2}} \approx \frac{R_2'}{X_2'}$	$T_{\max} \propto V^2$ , independent of $R_2'$
$\displaystyle \frac{T}{T_{\max}} = \frac{2}{s/s_m + s_m/s}$	$R_1$ neglected
$\displaystyle \frac{T_{st}}{T_{fl}} = \left(\frac{I_{st}}{I_{fl}}\right)^2 s_{fl}$	Starting torque greatest when $R_2' = X_2'$
Star-delta: line current and torque $\displaystyle \frac{1}{3}$ of DOL; auto tap $x$ : both $x^2$	Motor current falls by $x$ only
Single-phase: backward slip $2 - s$ , rotor frequency $(2 - s)f$	Not self-starting

With $V$ per phase, the torque is:

T = \frac{3}{\omega_s}\cdot\frac{V^2\,(R_2'/s)}{(R_1 + R_2'/s)^2 + (X_1 + X_2')^2}

The blocked-rotor test gives $R_{eq}$ and $X_{eq}$ , with $R_2' = R_{eq} - R_1$ ; the no-load test gives the rotational losses and magnetising branch. Constant $V/f$ keeps flux and $T_{\max}$ constant below base speed.

Worked example. A 4-pole, 50 Hz motor at 1440 rpm with 10 kW air-gap power, where $\omega_s = 157.08$ rad/s:

s = \frac{1500 - 1440}{1500} = 0.04,\quad P_{cu2} = 400\ \text{W},\quad P_m = 9.6\ \text{kW},\quad T = \frac{10000}{157.08} = 63.66\ \text{N m}

If it takes six times full-load current direct on line, $T_{st}/T_{fl} = 36 \times 0.04 = 1.44$ ; in star-delta, $0.48$ .

Remember: Torque is air-gap power over synchronous speed, or mechanical power over rotor speed. Mixing the two pairs gives a wrong torque.

How synchronous machines convert power through the load angle

The load angle sets the real power; the excitation sets the reactive power.

Formula	Watch out for
$E = 4.44\,f\Phi N K_w$ per phase; M-G set $\displaystyle \frac{f_1}{P_1} = \frac{f_2}{P_2}$	$K_w$ is the winding factor
Generator $E = V + jX_s I$ ; motor $V = E + jX_s I$	Sign of $jX_s I$ flips
$\displaystyle P = \frac{EV}{X_s}\sin\delta$ per phase; line values give three-phase	Maximum at $\delta = 90^\circ$
$\displaystyle Q = \frac{EV\cos\delta - V^2}{X_s}$ (generator)	Over-excited supplies lagging vars
Salient: $\displaystyle P = \frac{EV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q} - \frac{1}{X_d}\right)\sin 2\delta$	Reluctance term needs no excitation
Regulation $\displaystyle \frac{E - V}{V}$ ; $\displaystyle Z_s = \frac{V_{oc}}{I_{sc}}$ at equal field current; $\displaystyle \text{SCR} = \frac{1}{X_d}$ (pu)	EMF method pessimistic, MMF optimistic
$\displaystyle \theta_e = \frac{P}{2}\,\theta_m$	$\delta$ is electrical

On an infinite bus at constant power, more excitation lowers $\delta$ and swings generator current from leading to lagging. A motor's V-curve has least current at unity power factor. Synchronise by matching voltage, frequency, phase sequence and phase; start a motor on its damper winding.

Worked example. $E = 13.2$ kV, $V = 11$ kV (line), $X_s = 8\ \Omega$ per phase, $\delta = 30^\circ$ :

P = \frac{13.2 \times 11 \times 10^6}{8}\sin 30^\circ = 18.15 \times 0.5 = 9.075\ \text{MW}

The trap is the angle: a 4-pole rotor that slips back 2 mechanical degrees has a 4 electrical degree load angle.

How losses decide the efficiency of every machine

Variable losses are copper losses, growing as current squared. Constant losses are core loss at fixed voltage and frequency, friction and windage at fixed speed, and a DC shunt field's copper loss.

Formula	Watch out for
$\displaystyle \eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{\text{input} - \text{losses}}{\text{input}}$	First form for generators, second for motors
$\eta_{\max}$ when variable loss $=$ constant loss; DC: $I_a = \sqrt{P_{\text{const}} / R_a}$	Same rule as the transformer
$P_h \propto f B_m^n$ ( $n \approx 1.6$ ); $P_e \propto f^2 B_m^2 t^2$	$t$ is lamination thickness
$\displaystyle B_m \propto \frac{V}{f}$ ; at constant $V/f$ : $P_h \propto f$ , $P_e \propto f^2$	At fixed $V$ , higher $f$ lowers $B_m$

Worked example. 100 W hysteresis and 50 W eddy loss at 50 Hz, run at 60 Hz with voltage raised in proportion:

P_h = 100 \times 1.2 = 120\ \text{W},\quad P_e = 50 \times 1.2^2 = 72\ \text{W},\quad \text{total} = 192\ \text{W}

The trap is scaling both losses alike. For a DC shunt motor with 500 W constant loss and $R_a = 0.5\ \Omega$ , maximum efficiency comes at $\sqrt{1000} = 31.62$ A.

How to use this sheet before the exam

A numerical answer type question has no options, so practise roots and sines on the GATE virtual calculator. MCQ, MSQ and NAT marking is in the exam pattern and marking scheme guide, which is common to every paper. Check what you may carry in the admit card and exam-day rules, and plan around the GATE 2027 exam dates.

The GATE EE 2027 book has a last-minute sheet covering every syllabus topic with its conditions. It runs to 643 pages, with 914 questions with worked solutions and 10 full mock tests, built on all 5 GATE EE papers from 2022 to 2026 counted question by question.

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

Quick revision

$E = 4.44\,f N \Phi_m$ ; flux follows $V/f$ ; referred impedance is $a^2 Z$ .
Transformer efficiency peaks at $x = \sqrt{P_i / P_{cu,fl}}$ .
Regulation is $\epsilon_r\cos\phi \pm \epsilon_x\sin\phi$ , minus for leading loads.
An auto-transformer's rating is the two-winding rating times high voltage over series-winding voltage.
DC motor speed is $E/\Phi$ ; find the current from the load torque law first.
Induction power splits $1 : s : (1 - s)$ ; star-delta cuts line current and torque to one-third.
Synchronous power is $(EV/X_s)\sin\delta$ , with $\delta$ electrical.
Maximum efficiency comes where variable loss equals constant loss.

Frequently asked questions

What is the formula for the slip of an induction motor?

Slip is $s = (N_s - N)/N_s$ , where $N_s = 120f/P$ is the synchronous speed and $N$ the rotor speed. A 4-pole, 50 Hz motor at 1440 rpm has $N_s = 1500$ rpm and a slip of 0.04. The rotor frequency is then $sf = 2$ Hz, and the rotor copper loss is the slip times the air-gap power.

At what load is the efficiency of a transformer maximum?

Efficiency is maximum when the copper loss equals the core loss. Copper loss grows as the square of the load, so the maximum falls at the fraction $x = \sqrt{P_i / P_{cu,fl}}$ of full load. A transformer whose full-load copper loss is four times its core loss reaches maximum efficiency at half load, not at full load.

What is the EMF equation of a DC machine?

The generated EMF is $E = P\Phi Z N/(60A)$ , where $P$ is the number of poles, $\Phi$ the flux per pole, $Z$ the number of armature conductors, $N$ the speed in rpm and $A$ the number of parallel paths. Take $A = 2$ for a wave winding and $A = P$ for a lap winding. In short, $E = k\Phi\omega$ .

How much does star-delta starting reduce the starting torque?

Star-delta starting puts $1/\sqrt{3}$ of the line voltage across each winding. Both the starting line current and the starting torque fall to one-third of their direct-on-line values. A motor that draws six times full-load current and develops 1.5 times full-load torque direct on line draws twice full-load current and develops half full-load torque in star.

What is the power angle equation of a synchronous machine?

For a cylindrical-rotor machine with armature resistance neglected, the power per phase is $P = (EV/X_s)\sin\delta$ , where $\delta$ is the load angle between the excitation EMF and the terminal voltage. Using line voltages gives the three-phase power directly. The maximum power occurs at $\delta = 90^\circ$ and equals $EV/X_s$ .

Sources

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