GATE Network Theory Formula Sheet for ECE and EE, with Traps

Q: What is the most important network theory formula for GATE?

No single formula carries the section, but two families recur in the counted papers. Two-port parameters appeared in all five counted GATE EC papers, one question each. First-order transients appeared in all five counted GATE EE papers from 2022 to 2026. Learn the general first-order response and the reciprocity and symmetry conditions until you can write them without thinking.

Q: How do you find the Thevenin resistance with a dependent source?

Never switch off a dependent source. Switch off only the independent sources, then apply a test source of 1 V or 1 A at the terminals and take the ratio of voltage to current. Alternatively, find the open-circuit voltage and the short-circuit current with all sources active and divide the first by the second.

Q: What is the condition for maximum power transfer in an AC circuit?

The load impedance must be the complex conjugate of the Thevenin impedance, ZL = Zₜₕ^*. The maximum power is then | Vₜₕ |²/(4Rₜₕ) with the rms Thevenin voltage. If the load can only be a pure resistance, choose it equal to the magnitude of the Thevenin impedance instead. The power delivered is then less than the conjugate-match maximum.

Q: What is the difference between the Q factor of series and parallel resonance?

For a series RLC circuit, Q = ω₀ L/R, so a smaller resistance gives a sharper peak. For a parallel RLC circuit, Q = R/(ω₀ L), the reciprocal form, so a larger resistance gives a sharper peak. In both cases the bandwidth is the resonant frequency divided by Q. Using the series form on a parallel circuit is a common error.

Q: Are three-phase circuits in the GATE ECE syllabus?

The GATE 2027 EC syllabus does not name three-phase circuits in its networks section. The GATE 2027 EE syllabus does, as balanced three-phase circuits, alongside star-delta transformation and complex power. ECE candidates can skip the three-phase table on this sheet, while EE candidates should treat it as core. Check the current syllabus PDF on the official site before deciding.

By MD ANISH AHAMADUpdated 9 Oct 20269 min read

GATE Network Theory Formula Sheet for ECE and EE, with Traps

Network theory is one subject in both papers: Networks, Signals and Systems for ECE, Electric circuits for EE. The formulas you need are Thevenin and Norton, maximum power transfer, the first-order response, the RLC damping cases, phasor power, resonance, the two-port sets, three-phase relations and the Laplace element models. This sheet teaches them in learning order, each with a worked example and the trap that costs marks.

In this guide

Key takeaways
The terms this sheet uses
Source transformation and star–delta conversion
Thevenin, Norton and maximum power transfer
Superposition, and what it cannot add
First-order RC and RL transients
Second-order RLC: the three damping cases
Sinusoidal steady state and AC power
Series and parallel resonance
Two-port parameters: Z, Y, h and ABCD
Three-phase circuits (EE syllabus)
Laplace-domain elements with initial conditions
Where this sheet fits
Quick revision

Key takeaways

A dependent source is never switched off, whether you are finding a Thevenin resistance or applying superposition.
Superposition adds voltages and currents, never powers.
Every first-order circuit follows one formula built from the initial value, the final value and the time constant.
A series RLC circuit is damped more by a larger R, and a parallel one by a smaller R.
Average power uses the impedance angle, and AC meters read rms, not peak.
The h-parameter reciprocity condition carries a minus sign, and the Z, Y and ABCD conditions do not.

The terms this sheet uses

A linear network is one built from R, L, C, coupled coils and linear sources, so doubling every source doubles every response. A dependent source is one whose value is set by a voltage or current elsewhere in the circuit.

The time constant $\tau$ sets how quickly a first-order response settles. A phasor is the complex rms value of a sinusoid at one frequency. A two-port is a network with an input pair and an output pair of terminals, described by four parameters.

Signs follow the passive convention throughout.

Source transformation and star–delta conversion

Both tools redraw a circuit without changing what the rest of it sees, so you can shrink it before writing equations.

Rule	Formula	Watch out for
Source transformation	$V_s$ in series with $R$ $\Leftrightarrow$ $I_s = V_s/R$ in parallel with $R$	The arrow points to the source's $+$ terminal
Delta to star	$\displaystyle R_a = \dfrac{R_{ab}R_{ca}}{R_{ab} + R_{bc} + R_{ca}}$	Numerator: the two delta arms touching node $a$
Star to delta	$\displaystyle R_{ab} = \dfrac{R_aR_b + R_bR_c + R_cR_a}{R_c}$	Divide by the star arm opposite
Balanced	$R_Y = R_\Delta/3$	Not $3R_\Delta$

Worked example. A delta with $R_{ab} = 10\ \Omega$ , $R_{bc} = 20\ \Omega$ and $R_{ca} = 30\ \Omega$ becomes a star:

\begin{aligned} R_a &= \frac{10 \times 30}{60} = 5\ \Omega, \qquad R_b = \frac{10 \times 20}{60} = 3.33\ \Omega, \qquad R_c = \frac{20 \times 30}{60} = 10\ \Omega \\ R_{ab} &= \frac{5(3.33) + 3.33(10) + 10(5)}{10} = 10\ \Omega \quad \text{(check)} \end{aligned}

The trap is an ideal source. An ideal voltage source has no series R, so it cannot be transformed on its own.

Thevenin, Norton and maximum power transfer

Any linear two-terminal network looks, from its terminals, like one source with one impedance. Find the open-circuit voltage, then the impedance seen at the terminals.

Result	Formula	Watch out for
Thevenin, Norton	$V_{th} = V_{oc}$ , $\; Z_{th} = V_{oc}/I_{sc}$ , $\; I_N = V_{th}/Z_{th}$	With independent sources only: short $V$ , open $I$
Maximum power, DC	$R_L = R_{th}$ , $\displaystyle \; P_{max} = \dfrac{V_{th}^2}{4R_{th}}$	Efficiency is then 50 per cent
Maximum power, AC	$Z_L = Z_{th}^{*}$ , $\displaystyle \; P_{max} = \dfrac{\vert V_{th} \vert^2}{4R_{th}}$	$V_{th}$ in rms
Resistive load only	$R_L = \vert Z_{th} \vert$	Delivers less than $P_{max}$

Worked example. A 24 V source with 6 Ω in series feeds 12 Ω across the terminals. Then $V_{th} = 24 \times 12/18 = 16$ V, $R_{th} = 6 \parallel 12 = 4\ \Omega$ , $I_N = 4$ A, and a 4 Ω load takes $P_{max} = 256/16 = 16$ W.

Worked example (AC). With $V_{th} = 10$ V rms and $Z_{th} = 3 + j4\ \Omega$ :

\begin{aligned} Z_L = 3 - j4\ \Omega:&\quad P_{max} = \frac{10^2}{4 \times 3} = 8.33\ \text{W} \\ R_L = \vert Z_{th} \vert = 5\ \Omega:&\quad P = \frac{10^2 \times 5}{(3+5)^2 + 4^2} = 6.25\ \text{W} \end{aligned}

Remember: a dependent source stays on. Apply a 1 V or 1 A test source at the terminals and take the ratio, or divide the open-circuit voltage by the short-circuit current.

In the ECE book's count, dc network analysis appears in four of the five counted EC papers. The 2019 paper, the one counted paper set by IIT Madras, asked no Thevenin, Norton or maximum power question. One paper is a signal, not a rule.

Superposition, and what it cannot add

In a linear circuit, the response to several independent sources is the sum of the responses to each one alone. Switch the others off: voltage sources become shorts, current sources become opens.

Rule	Formula	Watch out for
Superposition	$i = i_1 + i_2 + \dots + i_n$	Linear elements only
Power	$P = R(i_1 + i_2)^2 \ne Ri_1^2 + Ri_2^2$	Power is not linear in current
Different frequencies	$P = \sum_k R\,I_k^2$ (rms)	Phasors of different $\omega$ never add

Worked example. Two sources drive 2 A and 1 A, in the same direction, through 5 Ω. The true power is $5 \times 3^2 = 45$ W. Adding the separate powers gives $20 + 5 = 25$ W, which is wrong.

The trap is the last row in reverse. Powers from sources at different frequencies do add, because the cross terms average to zero.

First-order RC and RL transients

A circuit with one capacitor or one inductor moves from its initial value to its final value along one exponential. Three numbers fix the whole response.

Result	Formula	Watch out for
General response	$x(t) = x(\infty) + [x(0^+) - x(\infty)]\,e^{-t/\tau}$	Valid for any voltage or current in the circuit
Time constant	$\tau = R_{th}C$ or $\tau = L/R_{th}$	$R_{th}$ seen by the C or L, sources off
Continuity	$v_C(0^+) = v_C(0^-)$ , $\; i_L(0^+) = i_L(0^-)$	Other variables may jump
DC steady state	C open circuit, L short circuit	Gives $x(\infty)$
Progress	63.2, 86.5, 95.0 per cent after $\tau$ , $2\tau$ , $3\tau$	Half-way at $\tau \ln 2$

Worked example. A capacitor of 50 µF holds 2 V and charges towards 10 V through 2 kΩ:

\begin{aligned} \tau &= 2000 \times 50 \times 10^{-6} = 0.1\ \text{s} \\ v_C(t) &= 10 + (2 - 10)\,e^{-t/0.1} = 10 - 8e^{-10t}\ \text{V} \\ v_C(0.1) &= 10 - 8e^{-1} = 7.06\ \text{V} \end{aligned}

In one line: find the value just after switching, the value long after, and the resistance the element sees, then write the one formula.

The trap is a capacitor loop or an inductor cut-set created by the switch. Capacitor voltage can then jump, so conserve charge or flux instead.

Second-order RLC: the three damping cases

With both L and C, the response depends on how the damping $\alpha$ compares with the undamped natural frequency $\omega_0$ . Their ratio is the damping ratio $\zeta$ .

Quantity	Series RLC	Parallel RLC
Characteristic equation	$\displaystyle s^2 + \dfrac{R}{L}s + \dfrac{1}{LC} = 0$	$\displaystyle s^2 + \dfrac{1}{RC}s + \dfrac{1}{LC} = 0$
Damping	$\displaystyle \alpha = \dfrac{R}{2L}$	$\displaystyle \alpha = \dfrac{1}{2RC}$
Natural frequency	$\omega_0 = 1/\sqrt{LC}$	$\omega_0 = 1/\sqrt{LC}$
Damping ratio	$\displaystyle \zeta = \dfrac{R}{2}\sqrt{C/L}$	$\displaystyle \zeta = \dfrac{1}{2R}\sqrt{L/C}$

Case	Condition	Natural response
Overdamped	$\alpha > \omega_0$	$A_1e^{s_1t} + A_2e^{s_2t}$ , $\; s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}$
Critically damped	$\alpha = \omega_0$	$(A_1 + A_2t)\,e^{-\alpha t}$
Underdamped	$\alpha < \omega_0$	$e^{-\alpha t}(A_1\cos\omega_dt + A_2\sin\omega_dt)$ , $\; \omega_d = \sqrt{\omega_0^2 - \alpha^2}$

Worked example. A series circuit with $R = 2\ \Omega$ , $L = 1$ H and $C = 0.2$ F:

\begin{aligned} \alpha &= \frac{2}{2 \times 1} = 1\ \text{s}^{-1}, \qquad \omega_0 = \frac{1}{\sqrt{0.2}} = 2.236\ \text{rad/s}, \qquad \zeta = 0.447 \\ \omega_d &= \sqrt{5 - 1} = 2\ \text{rad/s}, \qquad s = -1 \pm j2 \quad \text{(underdamped)} \\ R_{crit} &= 2\sqrt{L/C} = 2\sqrt{5} = 4.47\ \Omega \end{aligned}

The trap is the parallel circuit. There a larger R means less damping, the opposite of the series case.

Sinusoidal steady state and AC power

At one frequency every element becomes an impedance, and phasors solve the circuit like a resistive one.

Quantity	Formula	Watch out for
Impedance	$Z_R = R$ , $\; Z_L = j\omega L$ , $\displaystyle \; Z_C = \dfrac{1}{j\omega C} = \dfrac{-j}{\omega C}$	$\omega = 2\pi f$
RMS	$V = V_m/\sqrt{2}$	$V_m\cos\omega t$ is a peak value
Average power	$P = VI\cos\phi = I^2R$ (W)	$\phi = \angle Z$
Reactive power	$Q = VI\sin\phi = I^2X$ (var)	Positive for inductive loads
Complex power	$S = VI^{*} = P + jQ$ , $\; \vert S \vert = VI$ (VA)	Conjugate the current
Power factor	$\cos\phi$	Lagging when current lags voltage

Worked example. The source $v = 100\sqrt{2}\cos(1000t)$ V feeds 30 Ω in series with 40 mH:

\begin{aligned} X_L &= 1000 \times 0.04 = 40\ \Omega, \qquad Z = 30 + j40 = 50\angle 53.13^\circ\ \Omega \\ I &= 100/50 = 2\ \text{A rms} \\ P &= 2^2 \times 30 = 120\ \text{W}, \qquad Q = 2^2 \times 40 = 160\ \text{var}, \qquad S = 200\ \text{VA}, \qquad \text{pf} = 0.6\ \text{lagging} \end{aligned}

The trap is the meter. The ECE book flags rms meter readings as the recent framing of phasor questions, which appear in all five counted EC papers. Using the peak 141.4 V would double every power.

Series and parallel resonance

Resonance is the frequency at which the inductive and capacitive reactances cancel. The impedance becomes purely resistive.

Quantity	Series RLC	Parallel RLC
Resonant frequency	$\displaystyle f_0 = \dfrac{1}{2\pi\sqrt{LC}}$	$\displaystyle f_0 = \dfrac{1}{2\pi\sqrt{LC}}$
Impedance at $f_0$	$Z = R$ , minimum	$Z = R$ , maximum
Quality factor	$\displaystyle Q = \dfrac{\omega_0L}{R} = \dfrac{1}{R}\sqrt{L/C}$	$\displaystyle Q = \dfrac{R}{\omega_0L} = R\sqrt{C/L}$
Bandwidth	$\displaystyle \dfrac{\omega_0}{Q} = \dfrac{R}{L}$ rad/s	$\displaystyle \dfrac{\omega_0}{Q} = \dfrac{1}{RC}$ rad/s
Magnification	$V_L = V_C = QV$	Circulating current $= Q \times$ supply current
Band edges	$\omega_1\omega_2 = \omega_0^2$	$\omega_1\omega_2 = \omega_0^2$

Worked example. A series circuit with $R = 10\ \Omega$ , $L = 10$ mH and $C = 1$ µF:

\begin{aligned} \omega_0 &= \frac{1}{\sqrt{10^{-2} \times 10^{-6}}} = 10^4\ \text{rad/s}, \qquad f_0 = 1591.5\ \text{Hz} \\ Q &= \frac{10^4 \times 10^{-2}}{10} = 10, \qquad \text{BW} = \frac{10^4}{10} = 1000\ \text{rad/s} = 159.2\ \text{Hz} \end{aligned}

Trap: put the same three parts in parallel and the series formula still gives Q of 10. The parallel formula gives 0.1, the reciprocal, and only that one is right.

Two-port parameters: Z, Y, h and ABCD

Each parameter set writes two port variables in terms of the other two. Both currents enter the network.

Set	Equations	Reciprocal	Symmetric
Z	$V_1 = z_{11}I_1 + z_{12}I_2$ , $\; V_2 = z_{21}I_1 + z_{22}I_2$	$z_{12} = z_{21}$	$z_{11} = z_{22}$
Y	$I_1 = y_{11}V_1 + y_{12}V_2$ , $\; I_2 = y_{21}V_1 + y_{22}V_2$	$y_{12} = y_{21}$	$y_{11} = y_{22}$
h	$V_1 = h_{11}I_1 + h_{12}V_2$ , $\; I_2 = h_{21}I_1 + h_{22}V_2$	$h_{12} = -h_{21}$	$h_{11}h_{22} - h_{12}h_{21} = 1$
ABCD	$V_1 = AV_2 - BI_2$ , $\; I_1 = CV_2 - DI_2$	$AD - BC = 1$	$A = D$

Connection	Rule
Series–series	Z matrices add
Parallel–parallel	Y matrices add
Series–parallel	h matrices add
Cascade	ABCD matrices multiply, in order
Conversion	$\displaystyle A = \dfrac{z_{11}}{z_{21}}$ , $\displaystyle \; B = \dfrac{\Delta z}{z_{21}}$ , $\displaystyle \; C = \dfrac{1}{z_{21}}$ , $\displaystyle \; D = \dfrac{z_{22}}{z_{21}}$

Worked example. A T-network has 2 Ω in each series arm and 4 Ω in the shunt arm:

\begin{aligned} z_{11} &= z_{22} = 2 + 4 = 6\ \Omega, \qquad z_{12} = z_{21} = 4\ \Omega, \qquad \Delta z = 36 - 16 = 20 \\ A &= \tfrac{6}{4} = 1.5, \qquad B = \tfrac{20}{4} = 5\ \Omega, \qquad C = \tfrac{1}{4} = 0.25\ \text{S}, \qquad D = 1.5 \\ AD - BC &= 2.25 - 1.25 = 1 \quad \text{(reciprocal)}, \qquad A = D \quad \text{(symmetric)} \end{aligned}

Cascading a 2 Ω series arm, a 0.25 S shunt and another 2 Ω arm gives the same matrix.

The trap is the h set. Its reciprocity carries a minus sign, and its symmetry condition is a determinant, not $h_{11} = h_{22}$ . The ECE book counts a two-port in all five EC papers, one each, with the parameter set varying between y, z and ABCD.

Three-phase circuits (EE syllabus)

The GATE 2027 EE syllabus names balanced three-phase circuits; the EC syllabus does not. Solve a balanced system on one phase of its star equivalent.

Quantity	Star	Delta
Voltage	$V_L = \sqrt{3}\,V_{ph}$	$V_L = V_{ph}$
Current	$I_L = I_{ph}$	$I_L = \sqrt{3}\,I_{ph}$
Shift (abc)	$V_{ab}$ leads $V_{an}$ by $30^\circ$	Line current lags phase current by $30^\circ$
Power	$P = \sqrt{3}\,V_LI_L\cos\phi = 3V_{ph}I_{ph}\cos\phi$	Same
Reactive, apparent	$Q = \sqrt{3}\,V_LI_L\sin\phi$ , $\; S = \sqrt{3}\,V_LI_L$	Same
Equivalence	$Z_Y = Z_\Delta/3$	Balanced loads only

Worked example. A 400 V supply feeds $Z = 8 + j6\ \Omega$ per phase:

\begin{aligned} \text{Star: } V_{ph} &= \frac{400}{\sqrt{3}} = 230.9\ \text{V}, \quad I_L = \frac{230.9}{10} = 23.09\ \text{A}, \quad P = 3(23.09)^2(8) = 12.8\ \text{kW} \\ \text{Delta: } I_{ph} &= \frac{400}{10} = 40\ \text{A}, \quad I_L = 69.28\ \text{A}, \quad P = 3(40)^2(8) = 38.4\ \text{kW} \end{aligned}

The trap is the angle. In $\cos\phi$ , $\phi$ is the angle of the phase impedance, not the angle between line voltage and line current.

Laplace-domain elements with initial conditions

The Laplace model turns every element into an impedance and every stored energy into a source, so one algebraic solution covers the whole response.

Element	Series model	Parallel model
R	$R$	$R$
L	$sL$ with a source $L\,i(0^-)$ aiding the initial current	$sL$ with a current source $i(0^-)/s$
C	$\displaystyle \dfrac{1}{sC}$ with a source $v(0^-)/s$ matching the stored polarity	$\displaystyle \dfrac{1}{sC}$ with a current source $C\,v(0^-)$
Final value	$\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)$	Poles of $sF(s)$ in the open left half-plane

Worked example. A 0.5 F capacitor charged to 4 V discharges through 2 Ω:

\begin{aligned} I(s) &= \frac{4/s}{2 + \dfrac{1}{0.5s}} = \frac{4}{2s + 2} = \frac{2}{s + 1} \\ i(t) &= 2e^{-t}\ \text{A}, \qquad \text{consistent with } \tau = RC = 1\ \text{s and } i(0^+) = 4/2 = 2\ \text{A} \end{aligned}

The trap is the source value. The series capacitor source is $v(0^-)/s$ , a step in the s-domain, not the bare number $v(0^-)$ .

Where this sheet fits

The GATE ECE formula sheet and the GATE EE formula sheet cover every section. Transforms are in the ECE signals and systems sheet, time-response specifications in the EE control systems sheet. For decimal answers, read the virtual calculator guide.

The GATE ECE 2027 book has 925 questions with worked solutions and 10 full mock tests. The GATE EE 2027 book has 914 and 10. In the EE count, first-order transients are the only Electric circuits topic present in all five papers from 2022 to 2026.

Quick revision

Thevenin: open-circuit voltage, then a test source for the resistance when a dependent source is present.
DC maximum power at $R_L = R_{th}$ ; AC at $Z_L = Z_{th}^{*}$ ; resistive only at $R_L = \vert Z_{th} \vert$ .
Superpose currents, then square; never add powers at one frequency.
First order: $x(\infty) + [x(0^+) - x(\infty)]e^{-t/\tau}$ with $R_{th}$ in $\tau$ .
Series $\alpha = R/(2L)$ , parallel $\alpha = 1/(2RC)$ ; compare with $\omega_0 = 1/\sqrt{LC}$ .
$P = VI\cos\phi$ with rms values and the impedance angle.
Series $Q = \omega_0L/R$ , parallel $Q = R/(\omega_0L)$ , bandwidth $\omega_0/Q$ .
Reciprocal: $z_{12} = z_{21}$ , $y_{12} = y_{21}$ , $h_{12} = -h_{21}$ , $AD - BC = 1$ ; cascade multiplies ABCD.

Frequently asked questions

What is the most important network theory formula for GATE?

No single formula carries the section, but two families recur in the counted papers. Two-port parameters appeared in all five counted GATE EC papers, one question each. First-order transients appeared in all five counted GATE EE papers from 2022 to 2026. Learn the general first-order response and the reciprocity and symmetry conditions until you can write them without thinking.

How do you find the Thevenin resistance with a dependent source?

Never switch off a dependent source. Switch off only the independent sources, then apply a test source of 1 V or 1 A at the terminals and take the ratio of voltage to current. Alternatively, find the open-circuit voltage and the short-circuit current with all sources active and divide the first by the second.

What is the condition for maximum power transfer in an AC circuit?

The load impedance must be the complex conjugate of the Thevenin impedance, $Z_L = Z_{th}^{*}$ . The maximum power is then $\vert V_{th} \vert^{2}/(4R_{th})$ with the rms Thevenin voltage. If the load can only be a pure resistance, choose it equal to the magnitude of the Thevenin impedance instead. The power delivered is then less than the conjugate-match maximum.

What is the difference between the Q factor of series and parallel resonance?

For a series RLC circuit, $Q = \omega_0 L/R$ , so a smaller resistance gives a sharper peak. For a parallel RLC circuit, $Q = R/(\omega_0 L)$ , the reciprocal form, so a larger resistance gives a sharper peak. In both cases the bandwidth is the resonant frequency divided by Q. Using the series form on a parallel circuit is a common error.

Are three-phase circuits in the GATE ECE syllabus?

The GATE 2027 EC syllabus does not name three-phase circuits in its networks section. The GATE 2027 EE syllabus does, as balanced three-phase circuits, alongside star-delta transformation and complex power. ECE candidates can skip the three-phase table on this sheet, while EE candidates should treat it as core. Check the current syllabus PDF on the official site before deciding.

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 Network Theory Formula Sheet for ECE and EE, with Traps

Key takeaways

The terms this sheet uses

Source transformation and star–delta conversion

Thevenin, Norton and maximum power transfer

Superposition, and what it cannot add

First-order RC and RL transients

Second-order RLC: the three damping cases

Sinusoidal steady state and AC power

Series and parallel resonance

Two-port parameters: Z, Y, h and ABCD

Three-phase circuits (EE syllabus)

Laplace-domain elements with initial conditions

Where this sheet fits

Quick revision

Frequently asked questions

Sources

Keep reading