GATE EE Formula Sheet 2027: Key Formulas by Subject

By MD ANISH AHAMADUpdated 4 Oct 202610 min read

This GATE EE formula sheet collects the most-used formulas of all ten technical sections of the 2027 syllabus, plus the Engineering Mathematics and General Aptitude essentials. Each formula sits beside the condition or trap that costs marks. It is a free selection from the last-minute sheet of the GATE EE 2027 book, so use it to check your recall, then practise each formula on previous-year questions.

In this guide

Key takeaways
How to read this sheet
Engineering Mathematics
Electric Circuits
Electromagnetic Fields
Signals and Systems
Electrical Machines
Power Systems
Control Systems
Electrical and Electronic Measurements
Analog and Digital Electronics
Power Electronics
General Aptitude essentials
How to use the sheet in the exam
Subject-wise formula sheets
Quick revision

Key takeaways

A formula without its condition is the commonest wrong answer: a linear element, a lossless line, continuous conduction, a stable closed loop.
Phasors and meter readings are RMS; $V_m$ is always a peak, and three-phase voltages are line-to-line unless stated.
Per-unit impedances must be on one common base before you add them.
Many traps are factor errors: $\sqrt{3}$ between line and phase, $3Z_n$ in the zero-sequence network, $2\pi$ between $f$ and $\omega$ .
You cannot carry paper into the exam hall, so these formulas must be in memory.

How to read this sheet

A few terms are used throughout, so fix them first. A phasor is the complex number that stands for a sinusoid; on this sheet its size is the RMS value, which is the peak divided by $\sqrt{2}$ for a sine. A per-unit (pu) value is a quantity divided by a chosen base, so that 0.1 pu means one tenth of the base. Line quantities are measured between two lines, phase quantities across one winding.

Notation: $j = \sqrt{-1}$ ; $\omega = 2\pi f$ ; $\ln$ is the natural logarithm; $\displaystyle R_1 \parallel R_2 = \frac{R_1 R_2}{R_1 + R_2}$ ; $\delta$ in a machine or stability formula is the electrical load angle; $D$ is a converter's duty ratio and $\alpha$ a thyristor's firing angle.

Engineering Mathematics

Formula	Watch out for
$\sum \lambda_i = \operatorname{tr} A$ ; $\prod \lambda_i = \det A$ ; $A^k$ has $\lambda^k$ , $A^{-1}$ has $1/\lambda$ , $A + cI$ has $\lambda + c$	The eigenvector stays the same; only the eigenvalue changes
$Ax = b$ : unique iff $\operatorname{rank} A = \operatorname{rank}[A \mid b] = n$ ; none iff $\operatorname{rank} A < \operatorname{rank}[A \mid b]$	Infinitely many when the ranks are equal but less than $n$
Cayley–Hamilton, $2 \times 2$ : $A^2 - (\operatorname{tr} A)\,A + (\det A)\,I = 0$	Gives $A^{-1}$ without inverting
$y' + P y = Q$ : integrating factor $e^{\int P\,dx}$	Put the equation in this form first
$\oint_C f(z)\,dz = 2\pi j \sum \text{Res}$ (inside $C$ ); $\displaystyle \oint_C \frac{f(z)}{z - a}\,dz = 2\pi j f(a)$	Poles outside $C$ contribute nothing
Two-variable extremum: $rt - s^2 > 0$ with $r > 0$ minimum, $r < 0$ maximum; $rt - s^2 < 0$ saddle	$rt - s^2 = 0$ gives no conclusion
Binomial mean $np$ , variance $np(1 - p)$ ; Poisson mean $=$ variance $= \lambda$ ; $\displaystyle Z = \frac{X - \mu}{\sigma}$	Sample variance divides by $n - 1$

Electric Circuits

The first-order response is the result used most often. Any voltage or current in a circuit with one capacitor or one inductor moves from its starting value to its final value exponentially:

x(t) = x(\infty) + \big[x(0^+) - x(\infty)\big]\,e^{-t/\tau}, \qquad \tau = R_{\text{th}}C \;\text{ or }\; L/R_{\text{th}}

Here $R_{\text{th}}$ is the Thevenin resistance seen by the capacitor or inductor, not the resistance in series with the source.

Formula	Watch out for
Max power: $R_L = R_{\text{th}}$ , $\displaystyle P_{\max} = \frac{V_{\text{th}}^2}{4R_{\text{th}}}$ ; AC: $Z_L = Z_{\text{th}}^*$	Resistive load only: $R_L = \lvert Z_{\text{th}} \rvert$
Series resonance: $\displaystyle \omega_0 = \frac{1}{\sqrt{LC}}$ , $\displaystyle Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ , bandwidth $\displaystyle = \frac{R}{L} = \frac{\omega_0}{Q}$	Parallel RLC: $Q = R\sqrt{C/L}$ , the inverse form
$P = \sqrt{3}\,V_L I_L \cos\phi$	$\phi$ is the angle of the phase impedance
Two wattmeters: $P = W_1 + W_2$ , $\displaystyle \tan\phi = \frac{\sqrt{3}\,(W_1 - W_2)}{W_1 + W_2}$	Below 0.5 power factor one reading is negative
PF correction: $Q_C = P(\tan\phi_1 - \tan\phi_2)$ ; $\displaystyle C = \frac{Q_C}{\omega V^2}$	$\omega$ , not $f$
Reciprocal: $z_{12} = z_{21}$ , $AD - BC = 1$ ; cascade: ABCD matrices multiply	Symmetric: $A = D$
Balanced star–delta: $Z_Y = Z_\Delta / 3$	Unbalanced star, no neutral: find the neutral shift first

Electromagnetic Fields

Point charge $\displaystyle E = \frac{Q}{4\pi\varepsilon_0 r^2}$ ; infinite line $\displaystyle E = \frac{\rho_L}{2\pi\varepsilon_0 \rho}$ ; infinite plane $\displaystyle E = \frac{\rho_s}{2\varepsilon_0}$ , independent of distance
Capacitance: parallel plate $\displaystyle C = \frac{\varepsilon A}{d}$ ; coaxial $\displaystyle C = \frac{2\pi\varepsilon l}{\ln(b/a)}$ ; layered dielectrics stacked between the plates are in series, side by side in parallel
Infinite wire $\displaystyle H = \frac{I}{2\pi\rho}$ ; centre of a loop $\displaystyle H = \frac{I}{2a}$ ; long solenoid $\displaystyle H = \frac{NI}{l}$
Faraday: $\displaystyle e = -N\frac{d\Phi}{dt}$ ; motional $e = Blv$ ; sinusoidal flux $E_{\text{rms}} = 4.44\,f N \Phi_m$
Magnetic circuit: $\displaystyle \mathcal{R} = \frac{l}{\mu A}$ , $\text{mmf} = NI = \Phi\mathcal{R}$ , $\displaystyle L = \frac{N^2}{\mathcal{R}}$ ; an air gap usually dominates the total reluctance
Parallel wires: force $\displaystyle \frac{\mu_0 I_1 I_2}{2\pi d}$ per metre, attractive for currents in the same direction

Signals and Systems

Formula	Watch out for
LTI causal iff $h(t) = 0$ for $t < 0$ ; BIBO stable iff $\int \lvert h(t) \rvert\,dt < \infty$	$y = x + 1$ is not linear
Discrete convolution of lengths $N_1$ , $N_2$ gives $N_1 + N_2 - 1$ samples	Output starts at the sum of the start times
$\cos(\Omega_0 n)$ periodic iff $\displaystyle \frac{\Omega_0}{2\pi} = \frac{m}{N}$ is rational	Period $N$ in lowest terms
Sampling: $f_s \ge 2f_m$ ; a tone $f_0$ aliases to $\lvert f_0 - k f_s \rvert$	$x^2(t)$ has twice the bandwidth
$\displaystyle e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}$ ; $x(t - t_0) \leftrightarrow e^{-j\omega t_0}X(\omega)$	$a > 0$ for the transform to exist
Final value: $x(\infty) = \lim_{s \to 0} sX(s)$	Only if every pole of $sX(s)$ is in the open left half-plane
$\displaystyle a^n u[n] \leftrightarrow \frac{z}{z - a}$ , $\lvert z \rvert > \lvert a \rvert$	Stable iff the ROC contains $\lvert z \rvert = 1$
RMS of a sum of different frequencies: $\sqrt{V_0^2 + V_{1}^2 + V_{2}^2 + \cdots}$	Same frequency: add phasors first

Remember: Average the square first, then take the root. A half-wave rectified sine has average $V_m/\pi$ but RMS $V_m/2$ .

Electrical Machines

Formula	Watch out for
Transformer: $E = 4.44\,f N \Phi_m$ ; maximum efficiency at load fraction $x = \sqrt{P_i / P_{\text{cu,fl}}}$	It occurs where copper loss equals core loss
Regulation $\approx \varepsilon_r \cos\phi \pm \varepsilon_x \sin\phi$	$+$ for lagging, $-$ for leading power factor
DC machine: $\displaystyle E = \frac{P\Phi Z N}{60A}$ ; motor $V = E + I_a R_a$ ; $\displaystyle N \propto \frac{V - I_a R_a}{\Phi}$	$A = 2$ for wave, $A = P$ for lap
Induction: $\displaystyle N_s = \frac{120f}{P}$ , $\displaystyle s = \frac{N_s - N}{N_s}$ ; $P_g : P_{\text{cu2}} : P_m = 1 : s : (1 - s)$	Efficiency cannot exceed $1 - s$
$\displaystyle s_m \approx \frac{R_2'}{X_2'}$ ; $T_{\max} \propto V^2$	$T_{\max}$ is independent of $R_2'$
Star–delta starting: line current and torque $\displaystyle \frac{1}{3}$ of direct-on-line	Auto-transformer tap $x$ gives $x^2$
Synchronous: $\displaystyle P = \frac{EV}{X_s}\sin\delta$ ; salient adds $\displaystyle \frac{V^2}{2}\left(\frac{1}{X_q} - \frac{1}{X_d}\right)\sin 2\delta$	Reluctance power exists even with zero excitation

For example, a transformer with core loss 1.5 kW and full-load copper loss 6 kW reaches maximum efficiency at $x = \sqrt{1.5/6} = 0.5$ , that is, at half load.

The free sheet stops at the most-used rows. The GATE EE 2027 book has the full last-minute sheet: every topic of the 2027 syllabus, each entry printed with its condition, and recomputed numerical examples.

Power Systems

Formula	Watch out for
$\displaystyle Z_{\text{base}} = \frac{V_{\text{base}}^2}{S_{\text{base}}}$ ; $\displaystyle Z_{\text{new}} = Z_{\text{old}}\,\frac{S_{\text{new}}}{S_{\text{old}}}\left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2$	Convert everything to one base before adding
Lossless transfer: $\displaystyle P = \frac{V_S V_R}{X}\sin\delta$	Maximum at $\delta = 90^\circ$
$\displaystyle L = 2 \times 10^{-7} \ln\frac{\text{GMD}}{\text{GMR}}$ H/m; solid conductor $\text{GMR} = 0.7788\,r$	Capacitance uses $r$ , not $0.7788\,r$
Nominal $\pi$ : $\displaystyle A = D = 1 + \frac{YZ}{2}$ , $B = Z$ , $\displaystyle C = Y\left(1 + \frac{YZ}{4}\right)$ ; $\displaystyle \text{SIL} = \frac{V_L^2}{Z_c}$	Short line: $A = D = 1$ , $C = 0$
Economic dispatch: $\displaystyle \frac{dC_i}{dP_i} = \lambda$ ; with losses, penalty factor $\displaystyle L_i = \frac{1}{1 - \partial P_L / \partial P_i}$	A unit at its limit is fixed there
Line-to-ground: $\displaystyle I_{a1} = I_{a2} = I_{a0} = \frac{E}{Z_1 + Z_2 + Z_0 + 3Z_f}$ , $I_f = 3I_{a1}$	Neutral impedance enters as $3Z_n$
Three-phase fault $\displaystyle I_f = \frac{E}{Z_1 + Z_f}$ ; line-to-line $\displaystyle \lvert I_b \rvert = \frac{\sqrt{3}\,E}{\lvert Z_1 + Z_2 + Z_f \rvert}$	No zero sequence in a line-to-line fault
Swing: $\displaystyle \frac{d^2\delta}{dt^2} = \frac{\pi f\,(P_m - P_e)}{H}$	$\delta$ in electrical radians
Isolated area: $\displaystyle \Delta f = -\frac{\Delta P_L}{D + 1/R}$	Units share load in proportion to rating over droop

For a change of base, a reactance of 0.15 pu on 25 MVA becomes $\displaystyle 0.15 \times \frac{100}{25} = 0.6$ pu on 100 MVA at the same voltage.

Trap: The base-change formula multiplies by the new MVA over the old, but by the old kV over the new, squared. Swapping either ratio is the classic slip.

Control Systems

The standard second-order system $\displaystyle \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$ carries most time-response questions. For $0 < \zeta < 1$ :

M_p = e^{-\pi\zeta/\sqrt{1 - \zeta^2}}, \qquad t_p = \frac{\pi}{\omega_d}, \qquad t_s = \frac{4}{\zeta\omega_n}\ (2\%), \qquad \omega_d = \omega_n\sqrt{1 - \zeta^2}

With $\zeta = 0.6$ the overshoot is $e^{-2.356} \approx 0.095$ , about 9.5 per cent.

Closed loop $\displaystyle T = \frac{G}{1 + GH}$ ; Mason $\displaystyle T = \frac{\sum_k P_k \Delta_k}{\Delta}$
Steady-state error (unity feedback): step $\displaystyle \frac{1}{1 + K_p}$ , ramp $\displaystyle \frac{1}{K_v}$ , parabola $\displaystyle \frac{1}{K_a}$ , with $K_v = \lim_{s \to 0} sG(s)$
Routh: sign changes in the first column $=$ right-half-plane roots; cubic $s^3 + a_2 s^2 + a_1 s + a_0$ is stable iff all coefficients are positive and $a_2 a_1 > a_0$
Nyquist: $Z = N + P$ , with $N$ the clockwise encirclements of $-1 + j0$ ; stable iff $Z = 0$
Root locus: asymptote angles $\displaystyle \frac{(2q + 1)\,180^\circ}{n - m}$ , centroid $\displaystyle \frac{\sum p_i - \sum z_i}{n - m}$
Lead compensator: $\displaystyle \sin\phi_m = \frac{a - 1}{a + 1}$ at $\displaystyle \omega_m = \frac{1}{T\sqrt{a}}$
State space: $G(s) = C(sI - A)^{-1}B + D$ ; $\Phi(t) = \mathcal{L}^{-1}\big[(sI - A)^{-1}\big]$

Electrical and Electronic Measurements

Ammeter shunt $\displaystyle R_{sh} = \frac{R_m}{m - 1}$ ; voltmeter multiplier $R_s = (m - 1)\,R_m$
AC bridge balance $Z_1 Z_4 = Z_2 Z_3$ : magnitudes and angles both balance; Maxwell $L_1 = R_2 R_3 C_4$ ; Schering $\displaystyle C_1 = \frac{C_2 R_4}{R_3}$
Limiting error: sums add absolute errors; products and quotients add relative errors; $x^n$ has $n$ times the relative error
A rectifier meter reads $1.11 \times$ the average, so it is correct only on a sine
Oscilloscope rise time $\displaystyle t_r \approx \frac{0.35}{\text{bandwidth}}$ ; a $\displaystyle 3\tfrac{1}{2}$ -digit meter counts to 1999
Never open-circuit a loaded current transformer's secondary

Analog and Digital Electronics

Formula	Watch out for
Half-wave $\displaystyle V_{dc} = \frac{V_m}{\pi}$ ; full-wave $\displaystyle \frac{2V_m}{\pi}$ ; ripple factor 1.21 and 0.482	Centre-tap PIV $2V_m$ , bridge PIV $V_m$
$\displaystyle g_m = \frac{I_C}{V_T}$ , $\displaystyle r_\pi = \frac{\beta}{g_m}$ ; CE gain $-g_m (R_C \parallel R_L)$	$V_T \approx 26$ mV at room temperature
MOSFET saturation: $\displaystyle I_D = \frac{1}{2}\mu C_{ox}\frac{W}{L}(V_{GS} - V_t)^2$	Valid only for $V_{DS} \ge V_{GS} - V_t$
Inverting $\displaystyle -\frac{R_f}{R_1}$ ; non-inverting $\displaystyle 1 + \frac{R_f}{R_1}$ ; $\displaystyle A_f = \frac{A}{1 + A\beta}$	Virtual short needs negative feedback
Wien bridge $\displaystyle f = \frac{1}{2\pi RC}$ , gain $\ge 3$ ; RC phase shift $\displaystyle f = \frac{1}{2\pi RC\sqrt{6}}$ , $\lvert A \rvert \ge 29$	Barkhausen: loop gain 1 at $0^\circ$
555 astable $\displaystyle f = \frac{1.44}{(R_A + 2R_B)\,C}$ ; slew limit $\displaystyle f_{\max} = \frac{SR}{2\pi V_m}$	Duty cycle is above 50 per cent
ADC: LSB $\displaystyle = \frac{V_{FS}}{2^n}$ ; $\text{SNR} \approx 6.02n + 1.76$ dB; flash needs $2^n - 1$ comparators	Quantisation error is $\displaystyle \pm\frac{1}{2}$ LSB
JK: $Q^+ = J\overline{Q} + \overline{K}Q$ ; mod- $N$ counter needs $\lceil \log_2 N \rceil$ flip-flops	Ring counter mod $n$ , Johnson mod $2n$

Power Electronics

Formula	Watch out for
Buck $D$ ; boost $\displaystyle \frac{1}{1 - D}$ ; buck–boost $\displaystyle -\frac{D}{1 - D}$	Continuous conduction only
Buck ripple $\displaystyle \Delta I_L = \frac{V_o (1 - D)}{Lf}$	Peak to peak, not peak
Single-phase full converter $\displaystyle \frac{2V_m}{\pi}\cos\alpha$ ; semi-converter $\displaystyle \frac{V_m}{\pi}(1 + \cos\alpha)$	A semi-converter cannot invert
Three-phase full converter $\displaystyle \frac{3V_{mL}}{\pi}\cos\alpha$ ; overlap subtracts $\displaystyle \frac{3\omega L_s I_d}{\pi}$	$V_{mL}$ is the line peak
$\displaystyle \text{pf} = \frac{I_{s1}}{I_s}\cos\phi_1$ ; distortion factor $\displaystyle = \frac{1}{\sqrt{1 + \text{THD}^2}}$	Harmonics add current, not power
Single-phase bridge: $I_{s1} = 0.90\,I_d$ , $\text{pf} = 0.90\cos\alpha$ ; three-phase: $I_{s1} = 0.78\,I_d$ , $\text{pf} = 0.955\cos\alpha$	Constant, ripple-free $I_d$ assumed
SPWM full bridge: fundamental peak $m_a V_{dc}$ for $m_a \le 1$	$m_a > 1$ brings back low-order harmonics
Commutation: $\displaystyle t_c = \frac{C V_s}{I_o}$ must exceed $t_q$	$t_c$ is the circuit's, $t_q$ the thyristor's

In one line: Every DC–DC ratio comes from volt-second balance on the inductor, so derive it rather than memorise it if you are unsure.

General Aptitude essentials

Successive percentage changes multiply: $+a\%$ then $+b\%$ is a factor $\displaystyle \left(1 + \frac{a}{100}\right)\left(1 + \frac{b}{100}\right)$
$\displaystyle {}^nP_r = \frac{n!}{(n - r)!}$ ; $\displaystyle {}^nC_r = \frac{n!}{r!\,(n - r)!}$
AP sum $\displaystyle S_n = \frac{n}{2}\big(2a + (n - 1)d\big)$ ; GP sum $\displaystyle \frac{a(r^n - 1)}{r - 1}$
Unit digits repeat with cycle length 4 for 2, 3, 7 and 8, and 2 for 4 and 9
Two curves meet where their difference has real roots: check $b^2 - 4ac$
Circular segment area $\displaystyle \frac{1}{2}r^2(\theta - \sin\theta)$ , $\theta$ in radians

The General Aptitude preparation guide covers how to take these 15 marks.

How to use the sheet in the exam

Calculators of any kind, papers and data handbooks are banned in the hall; the exam-day rules guide lists what you may and may not take in. Rough work goes on the scribble pad. Do the arithmetic on the on-screen calculator, and practise with it beforehand using the virtual calculator guide.

NAT and MSQ answers carry no negative marking, while a wrong MCQ answer loses marks; the marking scheme, common to every GATE paper, sets out each type. The book also adds 914 questions with worked solutions and 10 full mock tests to practise these formulas on.

Subject-wise formula sheets

Each of these goes deeper on one part of the GATE EE paper, with a worked example and the trap for every formula:

Quick revision

Label every number as RMS, peak or average, and as line or phase, before you use it.
Put every per-unit impedance on one base: new MVA over old, old kV over new squared.
A first-order circuit needs only $x(0^+)$ , $x(\infty)$ and $\tau$ with the Thevenin resistance.
In induction machines the power split is $1 : s : (1 - s)$ , and $T_{\max}$ does not depend on rotor resistance.
A line-to-ground fault connects all three sequence networks in series, and $I_f = 3I_{a1}$ .
Second-order overshoot depends only on $\zeta$ ; settling time depends on $\zeta\omega_n$ .
Converter ratios hold only in continuous conduction; check it before you use them.
Use the final value theorem only when every pole of $sX(s)$ is in the left half-plane.

Frequently asked questions

Can I take a formula sheet into the GATE EE exam?

No. The GATE brochure bans papers, loose sheets, data handbooks, tables and calculators of any kind inside the hall. You get an on-screen virtual calculator and a scribble pad from the invigilator. A useful habit is to write the few formulas you are most likely to forget on the scribble pad in the first five minutes. Confirm the current rules at gate2027.iitm.ac.in.

What is the slip at maximum torque of an induction motor?

The slip at maximum torque is the referred rotor resistance divided by the magnitude of the stator resistance plus the total leakage reactance, which is close to the rotor resistance over the rotor reactance when stator resistance is neglected. Maximum torque varies as the square of the supply voltage and does not depend on rotor resistance. Extra rotor resistance only moves the slip at which it occurs, and starting torque is greatest when rotor resistance equals rotor reactance.

How do you calculate a single line-to-ground fault current?

Connect the positive, negative and zero sequence networks in series. Each sequence current equals the prefault voltage divided by the sum of the three sequence impedances plus three times the fault impedance, and the fault current is three times one sequence current. A neutral impedance enters the zero sequence network three times over. Put every impedance on a common per unit base first.

What are the voltage ratios of buck, boost and buck-boost converters?

In continuous conduction the buck gives D times the input, the boost gives the input divided by one minus D, and the buck-boost gives minus D over one minus D, where D is the duty ratio. All three come from volt-second balance on the inductor. Below the critical inductance the current falls to zero each cycle and these ratios no longer hold.

How should I revise formulas before GATE EE?

Read a compact formula sheet once a day in the final two weeks and once more on the exam morning. For every formula, know the condition under which it holds and the trap built around it. Label every number as RMS, peak or average and as line or phase, carry full precision, and round only the final numerical answer.

Sources

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