GATE EC · EE · IN

Sinusoidal steady state, AC power and resonance

Sinusoidal steady-state analysis replaces the differential equations of an RLC circuit with complex algebra: every source and response becomes a phasor and every element an impedance, so nodal, mesh and the network theorems apply unchanged. On that base sit rms values, the AC power quantities (active, reactive, apparent and complex power, power factor and its correction), maximum power transfer with complex impedances, series and parallel resonance with $Q$ and bandwidth, locus diagrams, and simple RLC filters. Most problems here reduce to a few lines of phasor arithmetic once the right convention (rms phasors, passive sign convention, $S = V I^{*}$ ) and the right formula are chosen.

34 previous-year questions · 34 solved · 6 practice questions

Periodic waveforms: average, rms, form and crest factor

For a periodic signal of period $T$ , the average (dc) value is the mean over one period and the rms (effective) value is the square root of the mean of the square. The rms value is the dc value that would dissipate the same average power in a resistor, which is why meter readings, nameplate ratings and every AC power formula use rms. For waveforms whose full-cycle average is zero (sine, symmetric square, symmetric triangle), the average used in the form factor is the half-cycle average.

AVERAGE VALUE

V_{avg} = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} v (t) d t

Any periodic waveform; integrate over any one full period.

RMS VALUE

V_{rms} = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} v^{2} (t) d t

Any periodic waveform. Square first, then average, then take the root.

SINUSOID

V_{rms} = \frac{V _{m}}{2} \approx 0.707 V_{m}, V_{avg, half cycle} = \frac{2 V _{m}}{π} \approx 0.637 V_{m}

$v = V_{m} sin (ω t + θ)$ . The full-cycle average is zero; the rms value does not depend on the phase $θ$ .

FORM FACTOR AND CREST (PEAK) FACTOR

k_{f} = \frac{V _{rms}}{V _{avg}}, k_{c} = \frac{V _{peak}}{V _{rms}}

$V_{avg}$ is the half-cycle average for a waveform whose full-cycle average is zero; $V_{peak}$ is the largest absolute value. For a sine, $k_{f} = π / (22) \approx 1.11$ and $k_{c} = 2 \approx 1.414$ .

Waveform (peak $V_{m}$ )	Average	RMS	Form factor	Crest factor
Sine $V_{m} sin ω t$	$0$ (full cycle); $2 V_{m} / π$ (half cycle)	$V_{m} / 2$	$1.11$	$2 = 1.414$
Full-wave rectified sine	$2 V_{m} / π$	$V_{m} / 2$	$1.11$	$1.414$
Half-wave rectified sine	$V_{m} / π$	$V_{m} /2$	$π /2 = 1.571$	$2$
Symmetric square, $\pm V_{m}$	$0$ (full); $V_{m}$ (half)	$V_{m}$	$1$	$1$
Symmetric triangle, $\pm V_{m}$	$0$ (full); $V_{m} /2$ (half)	$V_{m} / 3$	$2/ 3 = 1.155$	$3 = 1.732$
Sawtooth, $0$ to $V_{m}$	$V_{m} /2$	$V_{m} / 3$	$1.155$	$1.732$
Rectangular pulse train: $V_{m}$ for a fraction $D$ of each period, $0$ otherwise ( $0 < D \leq 1$ )	$D V_{m}$	$V_{m} D$	$1/ D$	$1/ D$

RMS OF A SUM OF DIFFERENT FREQUENCIES

V_{rms} = V_{0}^{2} + n \sum V_{n, rms}^{2} = V_{0}^{2} + n \sum \frac{V _{mn}^{2}}{2}

Only when the components have different frequencies (dc counts as frequency zero): products of different frequencies average to zero over the common period. Add same-frequency terms as phasors first.

EXAM TIP

Trap: $10 sin ω t + 10 sin (ω t + 6 0^{\circ})$ is ONE sinusoid of amplitude $∣10∠ 0^{\circ} + 10∠6 0^{\circ} ∣ = 17.32$ , so its rms value is $12.25$ , not $50 + 50 = 10$ . Squares of rms values add only for different frequencies, or for same-frequency terms exactly $9 0^{\circ}$ apart (such as $sin ω t$ and $cos ω t$ ).

WORKED EXAMPLE

Find the rms value of $i (t) = 5 + 10 sin (100 π t) + 10 sin (100 π t + 6 0^{\circ}) + 4 cos (300 π t)$ A.

Solution

The two $100 π$ terms have the same frequency, so add them as peak phasors: $10∠ 0^{\circ} + 10∠6 0^{\circ} = 15 + j 8.66 = 17.32∠3 0^{\circ}$ . Their rms value is $17.32/ 2 = 12.25$ A, whose square is $150$ .
The $300 π$ term has rms value $4/ 2 = 2.83$ A, square $8$ . The dc term contributes $5^{2} = 25$ .
Different frequencies add in squares: $I_{rms} = 25 + 150 + 8 = 183$ .

Answer: $I_{rms} = 183 \approx 13.53$ A

Phasors, impedance and admittance

In a linear, time-invariant circuit driven by sinusoids of one frequency $ω$ , once the transient (natural) response has died out, every voltage and current is a sinusoid of that same frequency; only amplitude and phase differ. A phasor stores those two numbers as one complex number. Convention in these notes: phasors are rms and referred to cosine, so $v (t) = 2 ∣ V ∣ cos (ω t + θ) \leftrightarrow V = ∣ V ∣∠ θ$ . Convert a sine with $sin x = cos (x - 9 0^{\circ})$ , or take sine as the reference for every source in the problem; mixing references is the error, not the choice. If a problem uses peak phasors, every power expression below that is written in phasor magnitudes (such as $∣ V ∣∣ I ∣ cos ϕ$ , $∣ I ∣^{2} R$ , $V I^{*}$ and $∣ V_{T h} ∣^{2} / (4 R_{T h})$ ) gains a factor $\frac{1}{2}$ .

Because $d / d t$ becomes multiplication by $j ω$ , each element obeys $V = Z I$ with a complex impedance, and KCL, KVL, nodal and mesh analysis, source transformation, superposition, Thevenin and Norton all carry over unchanged with complex arithmetic.

ELEMENT IMPEDANCES

Z_{R} = R, Z_{L} = j ω L, Z_{C} = \frac{1}{j ω C} = - \frac{j}{ω C}

Inductor: voltage leads current by $9 0^{\circ}$ . Capacitor: current leads voltage by $9 0^{\circ}$ . Resistor: in phase.

IMPEDANCE

Z = R + j X = ∣ Z ∣∠ ϕ, ∣ Z ∣ = R^{2} + X^{2}, ϕ = tan^{- 1} \frac{X}{R}

$X > 0$ : inductive, current lags. $X < 0$ : capacitive, current leads. $ϕ$ is the angle by which voltage leads current. The arctangent form assumes $R > 0$ (a passive impedance); for $R = 0$ , $ϕ = + 9 0^{\circ}$ or $- 9 0^{\circ}$ with the sign of $X$ .

ADMITTANCE

Y = \frac{1}{Z} = G + j B, G = \frac{R}{R ^{2} + X ^{2}}, B = \frac{- X}{R ^{2} + X ^{2}}

$G = 1/ R$ only when $X = 0$ . Susceptances: capacitor $B_{C} = ω C > 0$ , inductor $B_{L} = - 1/ (ω L) < 0$ .

COMBINATIONS AND DIVIDERS

Z_{series} = k \sum Z_{k}, Y_{parallel} = k \sum Y_{k}, V_{k} = V \frac{Z _{k}}{\sum Z}, I_{1} = I \frac{Z _{2}}{Z _{1} + Z _{2}}

Same as dc, with complex numbers: the voltage divider is for elements in series, the current divider for two parallel branches. Magnitudes add only when the phasors are in phase.

Phasor-domain analysis, in order

Check that every source has the same $ω$ . Choose one reference (cos or sin) and one scale (rms or peak), and write each source as a phasor.
Replace R, L, C by $R$ , $j ω L$ , $1/ (j ω C)$ ; a mutual inductance by $j ω M$ , signed by the dot convention (a current entering the dotted terminal of one coil induces a voltage in the other coil that is positive at that coil's dotted terminal).
Solve with nodal or mesh analysis, Thevenin/Norton, source transformation or superposition, in complex algebra.
Convert back: the rms phasor $∣ V ∣∠ θ$ gives $v (t) = 2 ∣ V ∣ cos (ω t + θ)$ .
If the sources have different frequencies, solve one circuit per frequency (the impedances change with $ω$ ) and add the responses in the time domain. Never add phasors of different frequencies.

A phasor diagram draws these complex numbers to scale. Take the quantity common to all elements as the reference: current in a series circuit, voltage in a parallel one. In series RLC the drops $V_{R}$ (in phase with $I$ ), $V_{L}$ ( $9 0^{\circ}$ ahead of $I$ ) and $V_{C}$ ( $9 0^{\circ}$ behind $I$ ) add as vectors, $∣ V ∣ = V_{R}^{2} + (V_{L} - V_{C})^{2}$ , so a single element voltage can exceed the supply voltage.

EXAM TIP

Meter readings are magnitudes. In a series RL circuit with $V_{R} = 60$ V and $V_{L} = 80$ V, the supply is $6 0^{2} + 8 0^{2} = 100$ V, not $140$ V. KVL holds for phasors, not for voltmeter readings.

AC power: active, reactive, apparent and complex

Convention: passive sign convention (current enters the $+$ terminal of the element), so a positive result is power absorbed; phasors are rms. For a load with $V = ∣ V ∣∠ θ_{v}$ and $I = ∣ I ∣∠ θ_{i}$ , the power-factor angle is $ϕ = θ_{v} - θ_{i}$ , which equals the angle of the load impedance.

INSTANTANEOUS POWER

p (t) = v (t) i (t) = ∣ V ∣∣ I ∣ cos ϕ + ∣ V ∣∣ I ∣ cos (2 ω t + θ_{v} + θ_{i})

With $v = 2 ∣ V ∣ cos (ω t + θ_{v})$ and $i = 2 ∣ I ∣ cos (ω t + θ_{i})$ : a constant plus a term at $2 ω$ whose average is zero.

ACTIVE, REACTIVE AND APPARENT POWER

P = ∣ V ∣∣ I ∣ cos ϕ (W), Q = ∣ V ∣∣ I ∣ sin ϕ (var), ∣ S ∣ = ∣ V ∣∣ I ∣ = P^{2} + Q^{2} (VA)

rms values. $P$ is the average of $p (t)$ . $Q$ is the amplitude of the part of $p (t)$ that flows back and forth between the source and the fields of L and C, with zero average.

COMPLEX POWER

S = V I^{*} = P + j Q = ∣ I ∣^{2} Z = \frac{∣ V ∣ ^{2}}{Z ^{*}}

rms phasors, power absorbed by the load. With peak phasors, $S = \frac{1}{2} V_{m} I_{m}^{*}$ . The conjugate goes on the current; writing $V^{*} I$ instead flips the sign of $Q$ .

POWER IN EACH ELEMENT

P_{R} = ∣ I ∣^{2} R = \frac{∣ V _{R} ∣ ^{2}}{R}, Q_{L} = ∣ I ∣^{2} X_{L} = \frac{∣ V _{L} ∣ ^{2}}{X _{L}} > 0, Q_{C} = - ∣ I ∣^{2} X_{C} = - \frac{∣ V _{C} ∣ ^{2}}{X _{C}} < 0

$X_{L} = ω L$ and $X_{C} = 1/ (ω C)$ are positive magnitudes. An inductor absorbs reactive power, a capacitor delivers it, and neither absorbs average power.

POWER FACTOR

pf = cos ϕ = \frac{P}{∣ S ∣} = \frac{R}{∣ Z ∣}

Lagging when current lags voltage ( $ϕ > 0$ , $Q > 0$ , inductive); leading when current leads ( $ϕ < 0$ , $Q < 0$ , capacitive); unity at $ϕ = 0$ . Always state lagging or leading with the number.

CONSERVATION OF COMPLEX POWER

all elements \sum S_{k} = 0 \Rightarrow \sum P_{delivered} = \sum P_{absorbed}, \sum Q_{delivered} = \sum Q_{absorbed}

Every element taken with the passive sign convention. $P$ and $Q$ of loads add whether they are in series or parallel; apparent powers $∣ S ∣$ do NOT add in general (only when every load has the same power-factor angle).

Power-factor correction. A lagging load draws more current than its real power needs, which raises line $I^{2} R$ loss and the kVA rating the supply must provide. A capacitor connected in parallel (so the load voltage is unchanged) supplies part of the load's reactive power: $P$ is unchanged, the reactive power drawn from the supply falls, and so does the supply current.

SHUNT CAPACITOR FOR PF CORRECTION

Q_{C} = P (tan ϕ_{1} - tan ϕ_{2}), C = \frac{P ( tan ϕ _{1} - tan ϕ _{2} )}{ω ∣ V ∣ ^{2}}

$ϕ_{1}$ , $ϕ_{2}$ = pf angles before and after (lagging taken positive), $∣ V ∣$ = rms voltage across the capacitor, $P$ in W. For a final leading pf take $ϕ_{2}$ negative.

WORKED EXAMPLE

A single-phase load takes 10 kW at 0.8 pf lagging from a 230 V (rms), 50 Hz supply. Find the shunt capacitance that raises the supply pf to 0.95 lagging, and the supply current before and after.

Solution

$tan ϕ_{1} = tan (cos^{- 1} 0.8) = 0.75$ , so $Q_{1} = 10000 \times 0.75 = 7500$ var.
$tan ϕ_{2} = tan (cos^{- 1} 0.95) = 0.3287$ , so $Q_{2} = 10000 \times 0.3287 = 3287$ var.
The capacitor must supply $Q_{C} = 7500 - 3287 = 4213$ var.
$C = \frac{Q _{C}}{ω ∣ V ∣ ^{2}} = \frac{4213}{( 2 π \times 50 ) ( 23 0 ^{2} )} = \frac{4213}{1.662 \times 1 0 ^{7}} = 253.5 μ F$ .
Supply current before: $∣ S_{1} ∣/∣ V ∣ = (10 000/0.8) /230 = 54.3$ A. After: $(10 000/0.95) /230 = 45.8$ A. $P$ is the same 10 kW in both cases.

Answer: $C \approx 253.5 μ F$ ; the supply current falls from 54.3 A to 45.8 A.

Non-sinusoidal periodic waveforms. Expand $v$ and $i$ into dc plus harmonics. The product of two sinusoids of different frequencies averages to zero over the common period, so average power comes only from voltage and current terms of the same frequency.

AVERAGE POWER WITH HARMONICS

P = V_{0} I_{0} + n = 1 \sum \infty V_{n} I_{n} cos ϕ_{n}

$V_{n}$ , $I_{n}$ = rms values of the $n$ -th harmonic; $ϕ_{n}$ = angle by which $V_{n}$ leads $I_{n}$ . A harmonic present in only one of $v$ , $i$ contributes nothing. Power factor $= P / (V_{rms} I_{rms})$ using the full rms values. If $v$ is a pure sinusoid and only $i$ is distorted, this becomes $pf = (I_{1} / I_{rms}) cos ϕ_{1}$ : distortion factor times displacement factor.

WORKED EXAMPLE

A two-terminal network (passive sign convention) has $v (t) = 100 + 1002 sin ω t + 502 sin 3 ω t$ V and $i (t) = 5 + 102 sin (ω t - 6 0^{\circ}) + 22 sin 5 ω t$ A. Find the average power and the power factor.

Solution

dc: $100 \times 5 = 500$ W.
Fundamental: rms values $100$ V and $10$ A, voltage leading current by $6 0^{\circ}$ : $100 \times 10 \times cos 6 0^{\circ} = 500$ W.
The 3rd harmonic is only in $v$ and the 5th only in $i$ , so neither contributes. $P = 1000$ W.
$V_{rms} = 10 0^{2} + 10 0^{2} + 5 0^{2} = 150$ V; $I_{rms} = 5^{2} + 1 0^{2} + 2^{2} = 129 = 11.36$ A.
$pf = \frac{1000}{150 \times 11.36} = 0.587$ .

Answer: $P = 1000$ W; pf $\approx 0.587$

EXAM TIP

Quick checks: a pure L or C absorbs $P = 0$ . For loads in parallel, add the $P$ 's and the $Q$ 's separately and only then take $∣ S ∣ = P^{2} + Q^{2}$ . A capacitor rated $Q_{C}$ at voltage $V$ and frequency $f$ delivers $Q_{C} (V^{'} / V)^{2} (f^{'} / f)$ at voltage $V^{'}$ and frequency $f^{'}$ , because its reactive power is $∣ V ∣^{2} ω C$ .

Maximum power transfer in AC

Replace the network seen by the load with its Thevenin equivalent: $V_{T h}$ (rms) in series with $Z_{T h} = R_{T h} + j X_{T h}$ . $V_{T h}$ is the open-circuit voltage at the load terminals; $Z_{T h} = V_{oc} / I_{sc}$ , or the ratio $V / I$ at the terminals when a test source is applied with every independent source set to zero (the test-source method is needed when dependent sources are present). For a load $Z_{L} = R_{L} + j X_{L}$ the load current is $V_{T h} / (Z_{T h} + Z_{L})$ , and what maximises the load power depends on which parts of $Z_{L}$ are allowed to vary.

LOAD POWER

P_{L} = \frac{∣ V _{T h} ∣ ^{2} R _{L}}{( R _{T h} + R _{L} ) ^{2} + ( X _{T h} + X _{L} ) ^{2}}

$V_{T h}$ rms. Start every restricted-load problem from this expression.

What the load may vary	Condition for maximum	Maximum load power
$R_{L}$ and $X_{L}$ both free	$Z_{L} = Z_{T h}^{*}$ : $R_{L} = R_{T h}$ , $X_{L} = - X_{T h}$	$\frac{∣ V _{T h} ∣ ^{2}}{4 R _{T h}}$
Purely resistive load ( $X_{L} = 0$ )	$R_{L} = ∣ Z_{T h} ∣ = R_{T h}^{2} + X_{T h}^{2}$	Substitute into $P_{L}$
$R_{L}$ free, $X_{L}$ fixed	$R_{L} = R_{T h}^{2} + (X_{T h} + X_{L})^{2}$	Substitute into $P_{L}$
$X_{L}$ free, $R_{L}$ fixed	$X_{L} = - X_{T h}$	$\frac{∣ V _{T h} ∣ ^{2} R _{L}}{( R _{T h} + R _{L} ) ^{2}}$
Angle of $Z_{L}$ fixed, $∣ Z_{L} ∣$ free	$∣ Z_{L} ∣ = ∣ Z_{T h} ∣$	Substitute into $P_{L}$

CONJUGATE MATCH

Z_{L} = Z_{T h}^{*} \Rightarrow P_{L, m a x} = \frac{∣ V _{T h} ∣ ^{2}}{4 R _{T h}}

rms $V_{T h}$ (with a peak value $V_{T h, m}$ it is $V_{T h, m}^{2} / (8 R_{T h})$ ). The Thevenin source then delivers twice this power, so the efficiency measured against the Thevenin equivalent is 50%.

PURELY RESISTIVE LOAD

R_{L} = ∣ Z_{T h} ∣ = R_{T h}^{2} + X_{T h}^{2}

When the load cannot have reactance. Gives less power than the conjugate match unless $X_{T h} = 0$ .

WORKED EXAMPLE

A network has $V_{T h} = 50∠ 0^{\circ}$ V (rms) and $Z_{T h} = 3 + j 4 Ω$ at the operating frequency. Find the maximum load power (a) for an arbitrary load impedance and (b) for a purely resistive load.

Solution

(a) Conjugate match, $Z_{L} = 3 - j 4 Ω$ : $P_{m a x} = \frac{5 0 ^{2}}{4 \times 3} = 208.3$ W.
(b) $R_{L} = ∣ Z_{T h} ∣ = 5 Ω$ . Current $∣ I ∣ = \frac{50}{∣8 + j 4∣} = \frac{50}{80} = 5.59$ A.
$P = ∣ I ∣^{2} R_{L} = \frac{2500}{80} \times 5 = 156.25$ W.

Answer: (a) 208.3 W with $Z_{L} = 3 - j 4 Ω$ ; (b) 156.25 W with $R_{L} = 5 Ω$

EXAM TIP

With a purely resistive load, do not set $R_{L} = R_{T h}$ when $X_{T h} \neq = 0$ ; the optimum is $R_{L} = ∣ Z_{T h} ∣$ . And a conjugate match holds at one frequency only, because the load reactance must cancel $X_{T h}$ at that $ω$ .

EXAM TIP

Two more traps. (1) If the load is fixed and the source resistance $R_{T h}$ is what can be adjusted, $P_{L}$ is largest at the smallest possible $R_{T h}$ (zero if allowed), not at $R_{T h} = R_{L}$ : every matching condition above assumes a fixed source and an adjustable load. (2) An ideal transformer of turns ratio $N_{1} : N_{2}$ , with the network on the $N_{1}$ side, presents the load to the network as $(N_{1} / N_{2})^{2} Z_{L}$ . It scales $∣ Z_{L} ∣$ but not its angle, so it is the 'angle fixed, $∣ Z_{L} ∣$ free' case: for a resistive load choose $(N_{1} / N_{2})^{2} R_{L} = ∣ Z_{T h} ∣$ , which is $R_{T h}$ when $Z_{T h}$ is resistive.

Resonance: series, parallel and the practical tank

A circuit containing both L and C is resonant at the frequency where its input impedance (equivalently admittance) is purely real, so the source voltage and current are in phase (unity pf). The net reactive power absorbed, $2 ω (\overset{ˉ}{W}_{m} - \overset{ˉ}{W}_{e})$ with $\overset{ˉ}{W}_{m}$ and $\overset{ˉ}{W}_{e}$ the time-averaged energies stored in the inductors and in the capacitors, is then zero: the reactive powers of L and C cancel, and energy swaps between L and C instead of between the circuit and the source. In this section $Q$ denotes the quality factor, not reactive power.

RESONANT FREQUENCY

ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C}

Ideal series RLC and ideal parallel RLC (frequency-independent R). The value of R does not shift it.

Series RLC, $Z = R + j (ω L - 1/ (ω C))$ : at $ω_{0}$ , $∣ Z ∣ = R$ is minimum and the current $V / R$ is maximum. Below $ω_{0}$ the circuit is capacitive (current leads); above $ω_{0}$ it is inductive (current lags).

QUALITY FACTOR, SERIES RLC

Q = \frac{ω _{0} L}{R} = \frac{1}{ω _{0} C R} = \frac{1}{R} \frac{L}{C}

$R$ is the total series resistance, including the coil's own resistance and the source's internal resistance.

QUALITY FACTOR, GENERAL DEFINITION

Q = 2 π \frac{maximum energy stored}{energy dissipated per cycle}

Evaluated at resonance, counting the total energy stored in L and C; it reduces to the series and parallel formulas.

VOLTAGE MAGNIFICATION (SERIES)

∣ V_{L} ∣ = ∣ V_{C} ∣ = Q ∣ V ∣ at ω_{0}

The L and C voltages are $18 0^{\circ}$ apart and cancel, but each is $Q$ times the supply voltage.

HALF-POWER FREQUENCIES (SERIES)

ω_{1, 2} = \mp \frac{R}{2 L} + (\frac{R}{2 L})^{2} + \frac{1}{L C}, ω_{0} = ω_{1} ω_{2}

Constant-amplitude voltage source. At $ω_{1}$ and $ω_{2}$ : $∣ I ∣ = I_{m a x} / 2$ , $P = P_{m a x} /2$ , $∣ Z ∣ = 2 R$ , impedance angle $- 4 5^{\circ}$ at $ω_{1}$ (capacitive) and $+ 4 5^{\circ}$ at $ω_{2}$ (inductive). $ω_{0}$ is their exact geometric mean.

BANDWIDTH (SERIES)

BW = ω_{2} - ω_{1} = \frac{R}{L} rad/s = \frac{ω _{0}}{Q}, in Hz: \frac{R}{2 π L} = \frac{f _{0}}{Q}

Exact for any $Q$ . For high $Q$ (roughly $Q \geq 10$ ), $ω_{1, 2} \approx ω_{0} \mp BW /2$ ; the arithmetic symmetry about $ω_{0}$ is only approximate, the geometric one is exact.

FREQUENCIES OF PEAK CAPACITOR AND INDUCTOR VOLTAGE (SERIES)

ω_{C, m a x} = ω_{0} 1 - \frac{1}{2 Q ^{2}}, ω_{L, m a x} = \frac{ω _{0}}{1 - \frac{1}{2 Q ^{2}}}

Constant-amplitude source, $Q > 1/ 2$ . $V_{C}$ peaks slightly below $ω_{0}$ and $V_{L}$ slightly above, both at the same height $∣ V_{C} ∣_{m a x} = ∣ V_{L} ∣_{m a x} = Q ∣ V ∣/ 1 - 1/ (4 Q^{2})$ , a little above $Q ∣ V ∣$ . The current and $V_{R}$ peak exactly at $ω_{0}$ .

WORKED EXAMPLE

A series RLC circuit has $R = 10 Ω$ , $L = 0.1$ H, $C = 10 μ F$ and is fed from a 50 V (rms) variable-frequency source. Find $ω_{0}$ , $Q$ , the bandwidth, the half-power frequencies, and the current and capacitor voltage at resonance.

Solution

$ω_{0} = 1/ 0.1 \times 1 0^{- 5} = 1/ 1 0^{- 6} = 1000$ rad/s ( $f_{0} = 159.2$ Hz).
$Q = ω_{0} L / R = 100/10 = 10$ ; $BW = R / L = 100$ rad/s, which equals $ω_{0} / Q$ .
$ω_{1, 2} = \mp 50 + 5 0^{2} + 1 0^{6} = \mp 50 + 1001.25$ , so $ω_{1} = 951.25$ rad/s and $ω_{2} = 1051.25$ rad/s. Check: $ω_{1} ω_{2} = 1 0^{6} = ω_{0}^{2}$ .
At resonance $I = 50/10 = 5$ A and $P = 5^{2} \times 10 = 250$ W.
$∣ V_{C} ∣ = ∣ V_{L} ∣ = Q V = 500$ V. Check: $I / (ω_{0} C) = 5/ (1000 \times 1 0^{- 5}) = 500$ V.

Answer: $ω_{0} = 1000$ rad/s, $Q = 10$ , BW $= 100$ rad/s, $ω_{1} = 951.25$ rad/s, $ω_{2} = 1051.25$ rad/s, $I = 5$ A, $∣ V_{C} ∣ = 500$ V

Parallel RLC (ideal R, L and C in parallel, $Y = G + j (ω C - 1/ (ω L))$ ) is the dual: at $ω_{0}$ the admittance is minimum, the impedance is maximum and equal to $R$ , and the current drawn from a voltage source is minimum ( $V / R$ ); fed from a current source, the voltage is maximum. Below $ω_{0}$ the circuit is inductive and above it capacitive, the opposite of series.

QUALITY FACTOR AND BANDWIDTH, PARALLEL RLC

Q = \frac{R}{ω _{0} L} = ω_{0} C R = R \frac{C}{L}, BW = \frac{1}{R C} rad/s = \frac{ω _{0}}{Q}

Ideal parallel RLC. Current magnification: at $ω_{0}$ , $∣ I_{L} ∣ = ∣ I_{C} ∣ = Q ∣ I_{source} ∣$ . A current source with internal resistance $R_{s}$ puts $R_{s}$ in parallel with $R$ : use $R ∥ R_{s}$ in $Q$ and BW.

HALF-POWER FREQUENCIES (PARALLEL)

ω_{1, 2} = \mp \frac{1}{2 R C} + (\frac{1}{2 R C})^{2} + \frac{1}{L C}

Ideal parallel RLC driven by a constant-amplitude current source, so $∣ V ∣ = ∣ V ∣_{m a x} / 2$ there; again $ω_{0} = ω_{1} ω_{2}$ .

SERIES-PARALLEL EQUIVALENT AT ONE FREQUENCY

R_{p} = \frac{R _{s}^{2} + X _{s}^{2}}{R _{s}} = R_{s} (1 + Q^{2}), X_{p} = \frac{R _{s}^{2} + X _{s}^{2}}{X _{s}} = X_{s} (1 + \frac{1}{Q ^{2}}), Q = \frac{∣ X _{s} ∣}{R _{s}} = \frac{R _{p}}{∣ X _{p} ∣}

A series $R_{s} + j X_{s}$ and a parallel combination of $R_{p}$ and $j X_{p}$ have the same impedance at the frequency where the values are computed, and only there. For $Q ≫ 1$ , $R_{p} \approx Q^{2} R_{s}$ and $X_{p} \approx X_{s}$ , which is where the tank result $Z_{D} \approx Q^{2} R$ below comes from.

Practical tank circuit. A real coil has series resistance, so the usual tank is a coil ( $R$ in series with $L$ ) in parallel with $C$ . Setting the imaginary part of $Y = j ω C + \frac{1}{R + j ω L}$ to zero gives the unity-pf frequency and the purely resistive impedance there.

PRACTICAL TANK (COIL R, L IN PARALLEL WITH C)

ω_{0} = \frac{1}{L C} - \frac{R ^{2}}{L ^{2}}, Z_{D} = \frac{L}{C R}

$ω_{0}$ = frequency of unity pf; $Z_{D}$ = dynamic impedance (purely resistive) at that frequency. A real $ω_{0}$ exists only if $R < L / C$ . Exactly, $Z_{D} = R (1 + Q^{2})$ with the coil $Q = ω_{0} L / R$ taken at this $ω_{0}$ ; for $Q ≫ 1$ , $ω_{0} \approx 1/ L C$ and $Z_{D} \approx Q^{2} R$ .

Property	Series RLC	Parallel RLC (ideal)
$∣ Z ∣$ at $ω_{0}$	minimum, $= R$	maximum, $= R$
Current from a voltage source at $ω_{0}$	maximum, $V / R$	minimum, $V / R$
Magnification at $ω_{0}$	voltage: $∣ V_{L} ∣ = ∣ V_{C} ∣ = Q V$	current: $∣ I_{L} ∣ = ∣ I_{C} ∣ = Q I$
$Q$	$ω_{0} L / R$	$R / (ω_{0} L)$
Bandwidth (rad/s)	$R / L$	$1/ (R C)$
Below $ω_{0}$	capacitive (leading)	inductive (lagging)
Above $ω_{0}$	inductive (lagging)	capacitive (leading)
Larger $R$	lower $Q$ , wider BW	higher $Q$ , narrower BW

Selectivity is the ability to pass a narrow band around $ω_{0}$ and reject nearby frequencies. It rises with $Q$ , since $BW = ω_{0} / Q$ .

Locus diagrams trace the tip of a phasor as one parameter varies. The impedance of a series circuit with one varying element moves along a straight line; the admittance, and so the current for a fixed voltage, is the inverse of a line that misses the origin, which is a circle through the origin. In a parallel circuit whose varying element is a branch on its own, the roles swap: the admittance, and so the current, moves along a straight line.

Standard loci (supply voltage V at angle 0°, held fixed)

Series R-L, $R$ varied from $0$ to $\infty$ , $X_{L}$ fixed: $I$ traces a semicircle of diameter $V / X_{L}$ in the lagging (fourth) quadrant, from $(V / X_{L}) ∠ - 9 0^{\circ}$ at $R = 0$ to the origin.
Series R-C, $R$ varied: the same kind of semicircle, diameter $V / X_{C}$ , in the leading (first) quadrant, from $(V / X_{C}) ∠9 0^{\circ}$ at $R = 0$ to the origin.
Series R-L, $X_{L}$ varied from $0$ to $\infty$ , $R$ fixed: semicircle of diameter $V / R$ , from $(V / R) ∠ 0^{\circ}$ to the origin, lagging.
Series RLC, $ω$ varied from $0$ to $\infty$ : $Z$ moves up the vertical line $Re Z = R$ , and $I$ traces the full circle of diameter $V / R$ through the origin: the leading half below $ω_{0}$ , $(V / R) ∠ 0^{\circ}$ at $ω_{0}$ , the lagging half above, with the half-power points at $\pm 4 5^{\circ}$ on it.
R-L branch in parallel with a variable capacitor: $I = I_{R L} + j ω C V$ , so the supply current moves along a vertical straight line, starting from the fixed branch current $I_{R L}$ at $C = 0$ and rising as $C$ increases. It is smallest, and in phase with $V$ , where that line crosses the real axis.
On the $R$ -varied locus of a series circuit, the active power $∣ V ∣∣ I ∣ cos ϕ$ is largest at the point with the largest component in phase with $V$ , where $R = X$ and $ϕ = \pm 4 5^{\circ}$ : maximum power transfer ( $R_{L} = ∣ Z_{T h} ∣$ with a purely reactive $Z_{T h}$ ) seen geometrically.

EXAM TIP

For the practical tank, the unity-pf frequency, the frequency of maximum impedance and $1/ L C$ are three slightly different frequencies that coincide only as $R \to 0$ . A question that says 'in phase' or 'unity power factor' wants $ω_{0} = 1/ (L C) - R^{2} / L^{2}$ .

EXAM TIP

Increasing $R$ has opposite effects: it lowers $Q$ in a series circuit but raises $Q$ in an ideal parallel circuit. Remember $Q_{series} = X / R$ and $Q_{parallel} = R / X$ , with $X$ the reactance of L (or C) at $ω_{0}$ .

Passive filters with R, L and C

A filter's transfer function is $H (j ω) = V_{out} / V_{in}$ , with the output open-circuited unless a load is stated. The cut-off (corner, $- 3$ dB, half-power) frequency is where $∣ H ∣$ falls to $1/ 2$ of its pass-band value: the dc gain for a low-pass, the high-frequency gain for a high-pass, the peak for a band-pass. (A second-order low-pass with $Q > 1/ 2$ peaks above its dc gain; its cut-off is still measured from the dc gain.) Identify the type from the limits: as $ω \to 0$ a capacitor is an open circuit and an inductor a short circuit; as $ω \to \infty$ the reverse.

FIRST-ORDER LOW-PASS

H (j ω) = \frac{1}{1 + j ω / ω _{c}}, ω_{c} = \frac{1}{R C} (series R, output across C) or \frac{R}{L} (series L, output across R)

At $ω_{c}$ : $∣ H ∣ = 1/ 2$ ( $- 3$ dB), phase $- 4 5^{\circ}$ . Asymptotic roll-off $- 20$ dB/decade above $ω_{c}$ .

FIRST-ORDER HIGH-PASS

H (j ω) = \frac{j ω / ω _{c}}{1 + j ω / ω _{c}}, ω_{c} = \frac{1}{R C} (series C, output across R) or \frac{R}{L} (series R, output across L)

At $ω_{c}$ : $∣ H ∣ = 1/ 2$ , phase $+ 4 5^{\circ}$ . Asymptotic slope $+ 20$ dB/decade below $ω_{c}$ .

Second order: series RLC driven by $V_{in}$ , with $ω_{0} = 1/ L C$ , $Q = ω_{0} L / R$ and $s = j ω$ . The element the output is taken across decides the filter type.

Output across	$H (s)$	Type	Key values
$R$	$\frac{( R / L ) s}{s ^{2} + ( R / L ) s + 1/ ( L C )}$	Band-pass	centre $ω_{0}$ , $∣ H (j ω_{0}) ∣ = 1$ , BW $= R / L$
$C$	$\frac{1/ ( L C )}{s ^{2} + ( R / L ) s + 1/ ( L C )}$	Low-pass	$∣ H (0) ∣ = 1$ , $∣ H (j ω_{0}) ∣ = Q$
$L$	$\frac{s ^{2}}{s ^{2} + ( R / L ) s + 1/ ( L C )}$	High-pass	$∣ H (j \infty) ∣ = 1$ , $∣ H (j ω_{0}) ∣ = Q$
$L$ and $C$ together	$\frac{s ^{2} + 1/ ( L C )}{s ^{2} + ( R / L ) s + 1/ ( L C )}$	Band-stop (notch)	zero output at $ω_{0}$ , stop-band width $R / L$

STANDARD SECOND-ORDER FORMS

H_{L P} (s) H_{H P} (s) H_{B P} (s) H_{B S} (s) = \frac{ω _{0}^{2}}{s ^{2} + ( ω _{0} / Q ) s + ω _{0}^{2}} = \frac{s ^{2}}{s ^{2} + ( ω _{0} / Q ) s + ω _{0}^{2}} = \frac{( ω _{0} / Q ) s}{s ^{2} + ( ω _{0} / Q ) s + ω _{0}^{2}} = \frac{s ^{2} + ω _{0}^{2}}{s ^{2} + ( ω _{0} / Q ) s + ω _{0}^{2}}

Match the denominator to read $ω_{0}$ and $Q$ directly: series RLC has $ω_{0} / Q = R / L$ , parallel RLC has $ω_{0} / Q = 1/ (R C)$ . Damping ratio $ζ = 1/ (2 Q)$ . The band-pass and band-stop bandwidth is $ω_{0} / Q$ . Asymptotic slopes: low-pass $- 40$ dB/decade above $ω_{0}$ , high-pass $+ 40$ dB/decade below it, band-pass $\pm 20$ dB/decade on its two skirts.

CUT-OFF OF THE SECOND-ORDER LOW-PASS

(\frac{ω _{c}}{ω _{0}})^{2} = (1 - \frac{1}{2 Q ^{2}}) + (1 - \frac{1}{2 Q ^{2}})^{2} + 1

$- 3$ dB point relative to the dc gain. $Q = 1/ 2$ (maximally flat) gives $ω_{c} = ω_{0}$ . For the matching high-pass, $ω_{c, H P} = ω_{0}^{2} / ω_{c, L P}$ .

WORKED EXAMPLE

An RC high-pass filter (series C, output across R, no load) has $R = 2 k Ω$ and $C = 0.1 μ F$ . Find the cut-off frequency, and the gain magnitude and phase at twice that frequency.

Solution

$ω_{c} = 1/ (R C) = 1/ (2000 \times 1 0^{- 7}) = 5000$ rad/s, so $f_{c} = 5000/ (2 π) = 795.8$ Hz.
At $ω = 2 ω_{c}$ : $H = \frac{j 2}{1 + j 2}$ , $∣ H ∣ = \frac{2}{5} = 0.894$ (about $- 0.97$ dB).
Phase $= 9 0^{\circ} - tan^{- 1} 2 = 9 0^{\circ} - 63.4 3^{\circ} = 26.5 7^{\circ}$ ; the output leads the input.

Answer: $f_{c} \approx 795.8$ Hz; at $2 f_{c}$ , $∣ H ∣ \approx 0.894$ and phase $\approx + 26. 6^{\circ}$

EXAM TIP

Loading changes a filter. A resistor $R_{load}$ across C in an RC low-pass makes the dc gain $R_{load} / (R + R_{load})$ and the cut-off $ω_{c} = 1/ (C (R ∥ R_{load}))$ . Take the Thevenin equivalent at the capacitor terminals instead of using the unloaded formula.

EXAM TIP

Identify a filter quickly: open every C and short every L to get the $ω \to 0$ output, then short every C and open every L for $ω \to \infty$ . Output only at dc: low-pass. Only at high frequency: high-pass. Zero at both ends: band-pass. Non-zero at both ends with a null at one frequency: band-stop (notch).

PREVIOUS-YEAR QUESTIONS

34 questions from the EC, EE and IN 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 ECE 2026 · GATE ECE 2025 · GATE ECE 2023 · GATE ECE 2022 · GATE ECE 2020 · GATE ECE 2018 · GATE EE 2026 · GATE EE 2022 · GATE EE 2020 · GATE EE 2018 · GATE IN 2026 · GATE IN 2025 · GATE IN 2024 · GATE IN 2023 · GATE IN 2022 · GATE IN 2020 · GATE IN 2018

GATE 2026 EC Q.31

GATE 2026 · ECQ.31 · 1 mark · NAT

Consider the circuit shown in the Figure, where the input $v_{i} (t)$ is in Volt.

The average power (in mW) dissipated in the load resistance of $1 k Ω$ at the resonant frequency is ____. (rounded off to two decimal places)

$v_{i} (t) = 12 sin (ω t)$

GATE 2026 EC question 31: An ideal sinusoidal voltage source v_i(t) (circle with a sine symbol) sits in the left vertical branch. — Figure from the GATE 2026 EC paper. It is part of the question.

Answer and solution

GATE 2026 EC Q.31: answer and solution

GATE 2026 EC Q.39

GATE 2026 · ECQ.39 · 2 marks · MCQ

In the given circuit, $L = 1 μ H$ and $C = 1 μ F$ . The phasor diagram for $I_{C}$ and $I_{L}$ is also shown. Assume that the phase $(θ_{1} + θ_{2})$ is $9 0^{\circ}$ at a frequency of 159.15 kHz.

Among the following options, what is the nearest integer value of $R_{C} \times R_{L}$ ?

GATE 2026 EC question 39: Left: a voltage V is applied across two terminals, + at the top and - at the bottom. — Figure from the GATE 2026 EC paper. It is part of the question.

(A)
0
(B)
1
(C)
2
(D)
10

Answer and solution

GATE 2026 EC Q.39: answer and solution

GATE 2026 EE Q.26

GATE 2026 · EEQ.26 · 1 mark · MCQ

A single-phase voltage source $v_{s} = 325 sin (2 π 50 t)$ V delivers a current, $i = 12 sin (2 π 50 t) + 9 sin (2 π 150 t)$ A to a load.

The load power factor, correct up to two decimal places, is:

(A)
1.00
(B)
0.80
(C)
0.65
(D)
0.57

Answer and solution

GATE 2026 EE Q.26: answer and solution

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GATE 2026 EE Q.26: answer and solution

GATE 2026 EE Q.42

GATE 2026 · EEQ.42 · 2 marks · MCQ

An electrical component has voltage drop $v = V_{m} sin (ω t)$ , when the current through it is $i = I_{m} sin (ω t - θ)$ .

What is the average power dissipated over a half cycle corresponding to $ω$ ?

(A)
$0$
(B)
$V_{m} I_{m} cos θ$
(C)
$\frac{V _{m} I _{m}}{2} cos θ$
(D)
$\frac{V _{m} I _{m}}{4} cos θ$

Answer and solution

GATE 2026 EE Q.42: answer and solution

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GATE 2026 EE Q.42: answer and solution

GATE 2026 IN Q.16

GATE 2026 · INQ.16 · 1 mark · MCQ

The RLC filter shown below acts as a ________.

GATE 2026 IN question 16: Two-port RLC network. — Figure from the GATE 2026 IN paper. It is part of the question.

(A)
low pass filter
(B)
high pass filter
(C)
band pass filter
(D)
band stop filter

Answer and solution

GATE 2026 IN Q.16: answer and solution

GATE 2026 IN Q.33

GATE 2026 · INQ.33 · 1 mark · NAT

In a series $R L$ circuit shown in the figure, the current through the resistor lags the AC source voltage by $4 5^{\circ}$ . Let $R = 100 π Ω$ and $L = 2 H$ .

The frequency of the source is ____ Hz. (rounded off to the nearest integer)

GATE 2026 IN question 33: A single-loop series circuit. — Figure from the GATE 2026 IN paper. It is part of the question.

Answer and solution

GATE 2026 IN Q.33: answer and solution

GATE 2025 EC Q.20

GATE 2025 · ECQ.20 · 1 mark · MCQ

Let $i_{C}$ , $i_{L}$ , and $i_{R}$ be the currents flowing through the capacitor, inductor, and resistor, respectively, in the circuit given below. The AC admittances are given in Siemens (S).

Which one of the following is TRUE?

GATE 2025 EC question 20: A voltage source 1 ∠ 90° V on the left (+ terminal at top, − at bottom) connects between a top rail and a bottom rail. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
$i_{C} = 0.25∠18 0^{\circ} A, i_{L} = 0.1∠ 0^{\circ} A, i_{R} = 0.2∠9 0^{\circ} A$
(B)
$i_{C} = 4∠18 0^{\circ} A, i_{L} = 10∠ 0^{\circ} A, i_{R} = 5∠9 0^{\circ} A$
(C)
$i_{C} = 0.25∠27 0^{\circ} A, i_{L} = 0.1∠9 0^{\circ} A, i_{R} = 0.2∠9 0^{\circ} A$
(D)
$i_{C} = 4∠9 0^{\circ} A, i_{L} = 10∠27 0^{\circ} A, i_{R} = 5∠ 0^{\circ} A$

Answer and solution

GATE 2025 EC Q.20: answer and solution

GATE 2025 EC Q.39

GATE 2025 · ECQ.39 · 2 marks · MCQ

In the circuit below, $M_{1}$ is an ideal AC voltmeter and $M_{2}$ is an ideal AC ammeter. The source voltage (in Volts) is $v_{s} (t) = 100 cos (200 t)$ .

What should be the value of the variable capacitor $C$ such that the RMS readings on $M_{1}$ and $M_{2}$ are 25 V and 5 A, respectively?

GATE 2025 EC question 39: A sinusoidal voltage source v_s(t) (+ terminal at top, - at bottom) is connected to the left-hand input terminals of a… — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
25 μF
(B)
4 μF
(C)
0.25 μF
(D)
Insufficient information to find $C$

Answer and solution

GATE 2025 EC Q.39: answer and solution

GATE 2025 IN Q.25

GATE 2025 · INQ.25 · 1 mark · MCQ

Three parallel admittances $Y_{a} = - j 0.2 S$ , $Y_{b} = 0.3 S$ , and $Y_{c} = j 0.4 S$ connected in parallel with a voltage source $V_{s} = 10∠4 5^{\circ} V$ , draw a total current $I_{s}$ from the source. The currents flowing through each of these admittances are $I_{a}$ , $I_{b}$ , and $I_{c}$ , respectively. Let $I = I_{b} + I_{c}$ . The phase relation between $I$ and $I_{s}$ is

(A)
$I$ leads $I_{s}$ by $19.4 4^{\circ}$
(B)
$I$ lags $I_{s}$ by $19.4 4^{\circ}$
(C)
$I$ leads $I_{s}$ by $33.6 9^{\circ}$
(D)
$I$ lags $I_{s}$ by $33.6 9^{\circ}$

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GATE 2025 IN Q.25: answer and solution

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GATE 2025 IN Q.65

GATE 2025 · INQ.65 · 2 marks · NAT

For the circuit shown in the figure, the active power supplied by the source is ________ W (rounded off to one decimal place).

GATE 2025 IN question 65: A sinusoidal voltage source 100 ∠ 0° V sits on the left vertical side. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

GATE 2025 IN Q.65: answer and solution

GATE 2024 IN Q.30

GATE 2024 · INQ.30 · 1 mark · NAT

The voltage applied and the current drawn by a circuit are

v (t) = 95 + 200 cos (120 π t) + 90 cos (360 π t - 6 0^{\circ}) V

i (t) = 4 cos (120 π t - 6 0^{\circ}) + 1.5 cos (240 π t - 7 5^{\circ}) A

The average power absorbed by the circuit is ________ W (rounded off to nearest integer).

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GATE 2024 IN Q.30: answer and solution

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GATE 2023 EC Q.18

GATE 2023 · ECQ.18 · 1 mark · MCQ

A series $R L C$ circuit has a quality factor $Q$ of 1000 at a center frequency of $1 0^{6}$ rad/s. The possible values of $R$ , $L$ and $C$ are

(A)
$R = 1 Ω$ , $L = 1 μ H$ and $C = 1 μ F$
(B)
$R = 0.1 Ω$ , $L = 1 μ H$ and $C = 1 μ F$
(C)
$R = 0.01 Ω$ , $L = 1 μ H$ and $C = 1 μ F$
(D)
$R = 0.001 Ω$ , $L = 1 μ H$ and $C = 1 μ F$

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GATE 2023 EC Q.18: answer and solution

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GATE 2023 IN Q.46

GATE 2023 · INQ.46 · 2 marks · NAT

In the circuit shown, $ω = 100 π$ rad/s, $R_{1} = R_{2} = 2.2 Ω$ and $L = 7$ mH. The capacitance $C$ for which $Y_{in}$ is purely real is ________ mF (rounded off to two decimal places).

GATE 2023 IN question 46: An AC voltage source V_0 cos (ω t) sits on the left vertical wire between a top rail and a bottom rail. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

GATE 2023 IN Q.46: answer and solution

GATE 2023 IN Q.65

GATE 2023 · INQ.65 · 2 marks · NAT

The voltage source Vs = $102 sin (20000 π t)$ V has an internal resistance of 50 $Ω$ . The RMS value of the current through R is ______ mA (rounded off to one decimal place).

GATE 2023 IN question 65: A single series loop. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

GATE 2023 IN Q.65: answer and solution

GATE 2022 EC Q.38

GATE 2022 · ECQ.38 · 2 marks · MCQ

For the circuit shown, the locus of the impedance $Z (j ω)$ is plotted as $ω$ increases from zero to infinity. The values of $R_{1}$ and $R_{2}$ are:

GATE 2022 EC question 38: Left: a two-terminal network seen from the input, labelled Z(j ω) with an arrow pointing into it. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
$R_{1} = 2 k Ω, R_{2} = 3 k Ω$
(B)
$R_{1} = 5 k Ω, R_{2} = 2 k Ω$
(C)
$R_{1} = 5 k Ω, R_{2} = 2.5 k Ω$
(D)
$R_{1} = 2 k Ω, R_{2} = 5 k Ω$

Answer and solution

GATE 2022 EC Q.38: answer and solution

GATE 2022 EC Q.39

GATE 2022 · ECQ.39 · 2 marks · MCQ

Consider the circuit shown in the figure with input $V (t)$ in volts. The sinusoidal steady state current $I (t)$ flowing through the circuit is shown graphically (where $t$ is in seconds). The circuit element $Z$ can be ________.

GATE 2022 EC question 39: Left: a single series loop. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
a capacitor of 1 F
(B)
an inductor of 1 H
(C)
a capacitor of $3$ F
(D)
an inductor of $3$ H

Answer and solution

GATE 2022 EC Q.39: answer and solution

GATE 2022 EE Q.27

GATE 2022 · EEQ.27 · 1 mark · NAT

An inductor having a $Q$ -factor of 60 is connected in series with a capacitor having a $Q$ -factor of 240. The overall $Q$ -factor of the circuit is ________. (round off to nearest integer)

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GATE 2022 EE Q.27: answer and solution

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GATE 2022 EE Q.28

GATE 2022 · EEQ.28 · 1 mark · NAT

The network shown below has a resonant frequency of 150 kHz and a bandwidth of 600 Hz. The $Q$ -factor of the network is ________. (round off to nearest integer)

GATE 2022 EE question 28: A two-terminal network with input terminals at top-left and bottom-left (open circles). — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

GATE 2022 EE Q.28: answer and solution

GATE 2022 IN Q.33

GATE 2022 · INQ.33 · 1 mark · NAT

In the bandpass filter circuit shown, $R_{0} = 50 Ω$ , $L_{0} = 1 mH$ , $C_{0} = 10 nF$ . The Q factor of the filter is ________ (round off to two decimal places)

GATE 2022 IN question 33: Two-port circuit, input terminals on the left, output terminals on the right. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

GATE 2022 IN Q.33: answer and solution

GATE 2022 IN Q.52

GATE 2022 · INQ.52 · 2 marks · NAT

In the circuit shown, the load is driven by a sinusoidal ac voltage source $V_{1} = 100∠ 0^{\circ} V$ at 50 Hz. Given $R_{1} = 20 Ω$ , $C_{1} = (\frac{1000}{π}) μ F$ , $L_{1} = (\frac{20}{π}) mH$ and $R_{2} = 4 Ω$ , the power factor is ________ (round off to one decimal place)

GATE 2022 IN question 52: A single-loop circuit. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

GATE 2022 IN Q.52: answer and solution

GATE 2022 IN Q.60

GATE 2022 · INQ.60 · 2 marks · NAT

A 1 kHz sine-wave generator having an internal resistance of 50 Ω generates an open-circuit voltage of 10 V $_{p-p}$ . When a capacitor is connected across the output terminals, the voltage drops to 8 V $_{p-p}$ . The capacitance of the capacitor (in microfarads) is _______ (round off to two decimal places)

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GATE 2020 EC Q.16

GATE 2020 · ECQ.16 · 1 mark · NAT

The current in the RL-circuit shown below is $i (t) = 10 cos (5 t - π /4) A$ . The value of the inductor (rounded off to two decimal places) is ______ H.

GATE 2020 EC question 16: A single series loop. — Figure from the GATE 2020 EC paper. It is part of the question.

Answer and solution

GATE 2020 EC Q.16: answer and solution

GATE 2020 EC Q.28

GATE 2020 · ECQ.28 · 2 marks · MCQ

The current I in the given network is

GATE 2020 EC question 28: Three vertical branches connected between a top rail T and a bottom rail B. — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
0 A.
(B)
2.38∠−96.37° A.
(C)
2.38∠143.63° A.
(D)
2.38∠−23.63° A.

Answer and solution

GATE 2020 EC Q.28: answer and solution

GATE 2020 EE Q.18

GATE 2020 · EEQ.18 · 1 mark · NAT

Currents through ammeters A2 and A3 in the figure are $1∠1 0^{\circ}$ and $1∠7 0^{\circ}$ , respectively. The reading of the ammeter A1 (rounded off to 3 decimal places) is ______ A.

GATE 2020 EE question 18: A single wire carries current I_1 (arrow pointing right) into ammeter A1. — Figure from the GATE 2020 EE paper. It is part of the question.

Answer and solution

GATE 2020 EE Q.18: answer and solution

GATE 2020 EE Q.46

GATE 2020 · EEQ.46 · 2 marks · NAT

A cylindrical rotor synchronous generator with constant real power output and constant terminal voltage is supplying 100 A current to a 0.9 lagging power factor load. An ideal reactor is now connected in parallel with the load, as a result of which the total lagging reactive power requirement of the load is twice the previous value while the real power remains unchanged. The armature current is now ________ A (rounded off to 2 decimal places).

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GATE 2020 IN Q.33

GATE 2020 · INQ.33 · 2 marks · NAT

The rms value of the phasor current $I$ in the circuit shown (in amperes) is ____

GATE 2020 IN question 33: A single-loop series circuit. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

GATE 2020 IN Q.33: answer and solution

GATE 2018 EC Q.35

GATE 2018 · ECQ.35 · 2 marks · MCQ

For the circuit given in the figure, the voltage $V_{C}$ (in volts) across the capacitor is

GATE 2018 EC question 35: A single series loop. — Figure from the GATE 2018 EC paper. It is part of the question.

(A)
$1.25 2 sin (5 t - 0.25 π)$
(B)
$1.25 2 sin (5 t - 0.125 π)$
(C)
$2.5 2 sin (5 t - 0.25 π)$
(D)
$2.5 2 sin (5 t - 0.125 π)$

Answer and solution

GATE 2018 EC Q.35: answer and solution

GATE 2018 EE Q.8

GATE 2018 · EEQ.8 · 1 mark · MCQ

In the figure, the voltages are $v_{1} (t) = 100 cos (ω t)$ , $v_{2} (t) = 100 cos (ω t + π /18)$ and $v_{3} (t) = 100 cos (ω t + π /36)$ . The circuit is in sinusoidal steady state, and $R << ω L$ . $P_{1}$ , $P_{2}$ and $P_{3}$ are the average power outputs. Which one of the following statements is true?

GATE 2018 EE question 8: Three ideal voltage sources share a common bottom (ground) wire. — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
$P_{1} = P_{2} = P_{3} = 0$
(B)
$P_{1} < 0, P_{2} > 0, P_{3} > 0$
(C)
$P_{1} < 0, P_{2} > 0, P_{3} < 0$
(D)
$P_{1} > 0, P_{2} < 0, P_{3} > 0$

Answer and solution

GATE 2018 EE Q.8: answer and solution

GATE 2018 EE Q.32

GATE 2018 · EEQ.32 · 2 marks · MCQ

The equivalent impedance $Z_{e q}$ for the infinite ladder circuit shown in the figure is

GATE 2018 EE question 32: An infinite ladder network between two input terminals (top-left and bottom-left dots). — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
$j 12 Ω$
(B)
$- j 12 Ω$
(C)
$j 13 Ω$
(D)
$13 Ω$

Answer and solution

GATE 2018 EE Q.32: answer and solution

GATE 2018 EE Q.48

GATE 2018 · EEQ.48 · 2 marks · NAT

The voltage across the circuit in the figure, and the current through it, are given by the following expressions:

$v (t) = 5 - 10 cos (ω t + 6 0^{\circ}) V$

$i (t) = 5 + X cos (ω t) A$

where $ω = 100 π$ radian/s. If the average power delivered to the circuit is zero, then the value of $X$ (in Ampere) is ______ (up to 2 decimal places).

GATE 2018 EE question 48: A two-terminal box labelled "Electrical Circuit". — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

GATE 2018 EE Q.48: answer and solution

GATE 2018 EE Q.55

GATE 2018 · EEQ.55 · 2 marks · NAT

The voltage $v (t)$ across the terminals $a$ and $b$ as shown in the figure, is a sinusoidal voltage having a frequency $ω = 100$ radian/s. When the inductor current $i (t)$ is in phase with the voltage $v (t)$ , the magnitude of the impedance $Z$ (in $Ω$ ) seen between the terminals $a$ and $b$ is ________ (up to 2 decimal places).

GATE 2018 EE question 55: Terminals a (top, marked +) and b (bottom, marked -), with v(t) applied across them and an arrow labelled Z pointing into… — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

GATE 2018 EE Q.55: answer and solution

GATE 2018 IN Q.6

GATE 2018 · INQ.6 · 1 mark · MCQ

A series R-C circuit is excited by a $1∠0 V$ sinusoidal ac voltage source. The locus diagram of the phasor current $\underline{I} = (x + j y) A$ , when $C$ is varied, while keeping $R$ fixed, is

GATE 2018 IN question 6: Each option is a sketch on complex-plane axes: horizontal axis labelled x (arrows both ways), vertical axis labelled +jy… — Figure from the GATE 2018 IN paper. It is part of the question.

(A)
Figure — semicircle in the first quadrant: starts at the origin, bulges upward (+jy), ends on the positive x-axis.
(B)
Figure — semicircle in the fourth quadrant: starts at the origin, bulges downward (−jy), ends on the positive x-axis.
(C)
Figure — semicircle in the second quadrant: from a point on the negative x-axis, bulges upward (+jy), ends at the origin.
(D)
Figure — semicircle in the third quadrant: from a point on the negative x-axis, bulges downward (−jy), ends at the origin.

Answer and solution

GATE 2018 IN Q.6: answer and solution

GATE 2018 IN Q.8

GATE 2018 · INQ.8 · 1 mark · NAT

A coil having an impedance of $(10 + j 100) Ω$ is connected in parallel to a variable capacitor as shown in figure. Keeping the excitation frequency unchanged, the value of the capacitor is changed so that parallel resonance occurs. The impedance across terminals p-q at resonance (in $Ω$ ) is ______.

GATE 2018 IN question 8: Two open terminals, p at the top left and q at the bottom left. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

GATE 2018 IN Q.8: answer and solution

GATE 2018 IN Q.31

GATE 2018 · INQ.31 · 2 marks · NAT

In the figure, an RLC load is supplied by a 230 V, 50 Hz single phase source. The magnitude of the reactive power (in VAr) supplied by the source is ________.

GATE 2018 IN question 31: A 230 V, 50 Hz sinusoidal AC source sits on the left. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

GATE 2018 IN Q.31: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 11 mark · NAT

The rms value (in V) of $v (t) = 6 + 82 sin (314 t) + 42 cos (942 t)$ V is ________ (round off to two decimal places).

Answer and solution

Answer: $10.77$ (accepted range $\pm 0.02$ )

The three components have different frequencies ( $0$ , $314$ and $942$ rad/s), so their mean squares add.
Their rms values are $6$ V, $82 / 2 = 8$ V and $4$ V.
$V_{rms} = 6^{2} + 8^{2} + 4^{2} = 116 = 10.77$ V.

PRACTICE 22 marks · MSQ

A load connected to a 250 V (rms) single-phase supply absorbs complex power $S = 3 + j 4$ kVA, where $S = V I^{*}$ with rms phasors and the passive sign convention. Which of the following statements is/are correct?

(A)
The load power factor is 0.6 lagging.
(B)
The magnitude of the load current is 20 A.
(C)
The load impedance is $7.5 - j 10 Ω$ .
(D)
A capacitor connected in parallel with the load and supplying 4 kvar reduces the supply current to 12 A.

Answer and solution

Answer: (A), (B), (D)

$∣ S ∣ = 3^{2} + 4^{2} = 5$ kVA; pf $= P /∣ S ∣ = 3/5 = 0.6$ . $Q > 0$ means the load absorbs reactive power: inductive, lagging. (A) is correct.
$∣ I ∣ = ∣ S ∣/∣ V ∣ = 5000/250 = 20$ A. (B) is correct.
$Z = ∣ V ∣^{2} / S^{*} = 62 500/ (3000 - j 4000) = 7.5 + j 10 Ω$ , inductive as expected. (C) is wrong.
With the capacitor, the supply provides $S = 3 + j 0$ kVA, so $∣ I ∣ = 3000/250 = 12$ A. (D) is correct.

PRACTICE 32 marks · NAT

A source network has the Thevenin equivalent $V_{T h} = 20∠ 0^{\circ}$ V (rms) in series with $Z_{T h} = 6 + j 8 Ω$ at the operating frequency. A purely resistive load $R_{L}$ is connected and adjusted for maximum power transfer. The maximum power (in W) delivered to $R_{L}$ is ________ (round off to two decimal places).

Answer and solution

Answer: $12.5$ (accepted range $\pm 0.05$ )

For a purely resistive load, $R_{L} = ∣ Z_{T h} ∣ = 6^{2} + 8^{2} = 10 Ω$ .
$∣ I ∣ = \frac{20}{∣16 + j 8∣} = \frac{20}{320} = 1.118$ A.
$P = ∣ I ∣^{2} R_{L} = \frac{400}{320} \times 10 = 12.5$ W. (A conjugate-matched load would receive $2 0^{2} / (4 \times 6) = 16.67$ W; setting $R_{L} = 6 Ω$ gives only $400 \times 6/ (144 + 64) = 11.54$ W.)

PRACTICE 42 marks · MCQ

A coil of resistance $20 Ω$ and inductance $0.2$ H is connected in parallel with a $2 μ F$ capacitor. The angular frequency at which the combination draws a current in phase with the applied sinusoidal voltage, and the input impedance of the combination at that frequency, are respectively

(A)
$1578$ rad/s and $5000 Ω$
(B)
$1581$ rad/s and $5000 Ω$
(C)
$1578$ rad/s and $20 Ω$
(D)
$1581$ rad/s and $20 Ω$

Answer and solution

Answer: (A)

Unity pf for a lossy coil in parallel with C: $ω_{0} = \frac{1}{L C} - \frac{R ^{2}}{L ^{2}} = \frac{1}{0.4 \times 1 0 ^{- 6}} - \frac{400}{0.04} = 2.5 \times 1 0^{6} - 1 0^{4} = 2.49 \times 1 0^{6} = 1578$ rad/s.
Dynamic impedance: $Z_{D} = \frac{L}{C R} = \frac{0.2}{2 \times 1 0 ^{- 6} \times 20} = 5000 Ω$ , purely resistive.
$1/ L C = 1581$ rad/s ignores the coil resistance, and $20 Ω$ is what a series circuit would present at resonance. Option (A).

PRACTICE 51 mark · MCQ

A series RLC circuit has $R = 20 Ω$ , $L = 50$ mH and $C = 0.5 μ F$ . Its quality factor $Q$ at resonance and its half-power bandwidth are

(A)
$Q = 15.8$ , bandwidth $= 400$ rad/s
(B)
$Q = 15.8$ , bandwidth $= 200$ rad/s
(C)
$Q = 31.6$ , bandwidth $= 400$ rad/s
(D)
$Q = 0.063$ , bandwidth $= 400$ rad/s

Answer and solution

Answer: (A)

$ω_{0} = 1/ L C = 1/ 0.05 \times 0.5 \times 1 0^{- 6} = 1/ 2.5 \times 1 0^{- 8} = 6324.6$ rad/s.
$Q = ω_{0} L / R = (6324.6 \times 0.05) /20 = 316.2/20 = 15.8$ .
$BW = R / L = 20/0.05 = 400$ rad/s. Check: $ω_{0} / Q = 6324.6/15.81 = 400$ rad/s. Option (A).

PRACTICE 62 marks · MSQ

A series RLC circuit with $R = 20 Ω$ , $L = 10$ mH and $C = 1 μ F$ is driven by a sinusoidal voltage $v_{in}$ , and the output $v_{out}$ is taken across the capacitor with no load. For $H (j ω) = V_{out} / V_{in}$ , which of the following statements is/are correct?

(A)
The circuit is a low-pass filter.
(B)
$∣ H (j ω) ∣ = 5$ at $ω = 1 0^{4}$ rad/s.
(C)
$∣ H (j ω) ∣ \to 1$ as $ω \to \infty$ .
(D)
At $ω = 1 0^{4}$ rad/s the output lags the input by $9 0^{\circ}$ .

Answer and solution

Answer: (A), (B), (D)

$H (s) = \frac{1/ ( L C )}{s ^{2} + ( R / L ) s + 1/ ( L C )}$ : $∣ H (0) ∣ = 1$ and $∣ H ∣ \to 0$ as $ω \to \infty$ , so it is a low-pass filter. (A) is correct and (C) is wrong.
$ω_{0} = 1/ L C = 1/ 1 0^{- 2} \times 1 0^{- 6} = 1 0^{4}$ rad/s and $Q = ω_{0} L / R = (1 0^{4} \times 0.01) /20 = 5$ .
At $s = j ω_{0}$ the terms $- ω_{0}^{2}$ and $1/ (L C)$ cancel: $H (j ω_{0}) = \frac{1 0 ^{8}}{j 1 0 ^{4} \times 2000} = \frac{1 0 ^{8}}{j 2 \times 1 0 ^{7}} = - j 5$ .
So $∣ H (j ω_{0}) ∣ = 5 = Q$ , and the angle is $- 9 0^{\circ}$ : the output lags the input by $9 0^{\circ}$ . (B) and (D) are correct.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

Sinusoidal steady state analysis: phasors, complex power, maximum power transfer

GATE EE

sinusoidal steady-state analysis, resonance … complex power and power factor in AC circuits

GATE IN

Peak, average and rms values of AC quantities; apparent, active and reactive powers, power factor; phasor analysis, impedance and admittance; series and parallel resonance, locus diagrams, realization of basic filters with R, L, and C elements

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