Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).
AC impedance & phasors
Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z
Method
Convert the frequency: ω = 2πf.
Find each reactance: XL = ωL and XC = 1/(ωC).
Add impedances like resistors, with Z = R + jX.
Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.
AC power & power factor
Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)
LC resonance, Q factor & bandwidth
Natural frequency
ω0 = 1/√(LC), f0 = 1/(2π√(LC))
Energy
W = ½CV0² = ½Cv² + ½Li²
Peak current
Imax = V0√(C/L)
Quality factor
Q = 2π·(energy stored)/(energy lost per cycle)
Series RLC
Q = (1/R)√(L/C), BW = R/(2πL)
Parallel RLC
Q = R√(C/L), BW = 1/(2πRC)
Bandwidth
BW = f2 − f1 = f0/Q, f0 = √(f1f2)
Half-power frequencies
f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q))
Tank with coil resistance r
RD = L/(C·r) = Q²·r
Decay of the amplitude
e−Rt/(2L)
Method
Find the resonance frequency first: ω₀ = 1/√(LC) and f₀ = ω₀/(2π).
Decide whether the circuit is series or parallel: series Q = (1/R)√(L/C), parallel Q = R√(C/L).
Use BW = f₀/Q for the bandwidth, or the shortcuts R/(2πL) (series) and 1/(2πRC) (parallel).
For energy questions, use that the total energy ½CV₀² stays constant in an ideal LC circuit and swings between C and L every quarter period.
Three-phase power
Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η
RC & RL transients
Time constant
τ = RC | τ = L/R
Response
x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
Capacitor at DC
open circuit; vC cannot jump
Inductor at DC
short circuit; iL cannot jump
Settled
t ≈ 5τ
Method
Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.
Thévenin & Norton
Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)
Method
Remove the load and mark the two terminals a and b.
Find the open-circuit voltage between a and b: this is Vth.
Switch off all independent sources and find the resistance seen from a–b: this is Rth.
Put the load back on the equivalent circuit: Vth in series with Rth.
For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.
2Practice problems
1.Complex numbersMedium
Find the argument θ of z = −3 − j5 in degrees, as the principal value between −180° and 180°.
z = −3 − j5
θ = °
2.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁶ for z = −√3 + j.
z = −√3 + jn = 6
Re(z⁶) =
3.AC impedance & phasorsMedium
A resistor and a capacitor are connected in parallel across a 100 V (rms), 100 Hz supply. Find the rms current I.
V = 100 V (rms)f = 100 HzR = 60 ΩC = 22 µF
I = A
4.AC impedance & phasorsHard
A series RLC circuit has R = 2 Ω, L = 2 mH and C = 10 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.
R = 2 ΩL = 2 mHC = 10 µF
B = Hz
5.AC power & power factorMedium
Two loads are connected in parallel to a 240 V single-phase supply: a 28 kW load at a power factor of 0.6 lagging and a 4 kW heater (unity power factor). Find the total current I drawn from the supply.
A 230 V (rms), 1000 Hz source has an internal impedance Zs = 8 + j40 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
Vs = 230 V (rms)f = 1000 HzZs = 8 + j40 Ω
Pmax = W
7.LC resonance, Q factor & bandwidthMedium
A series RLC circuit resonates at 1 kHz with Q = 2. Find the lower half-power frequency f1 exactly, in Hz.
f0 = 1 kHzQ = 2
f1 = Hz
8.LC resonance, Q factor & bandwidthHard
The half-power frequencies of a resonance curve are 10 kHz and 12 kHz. Find the resonance frequency as their geometric mean, and then the quality factor Q.
f1 = 10 kHzf2 = 12 kHz
Q =
9.Three-phase powerMedium
A balanced three-phase load draws a line current of 120 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.95 lagging. Find the total real power P drawn by the load in kW.
VL = 690 VIL = 120 Acos φ = 0.95
P = kW
10.Three-phase powerHard
A balanced three-phase load takes 80 kW at a power factor of 0.72 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 80 kWpf1 = 0.72pf2 = 0.9VL = 400 Vf = 60 Hz
Cph = µF
11.RC & RL transientsMedium
The inductor carries no current before the switch closes at t = 0. Find the energy W(∞) stored in the inductor once the circuit has settled, in mJ.
Vs = 20 VR = 5 ΩL = 68 mH
W(∞) = mJ
12.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 20 V with the polarity shown. How long does it take for vC to reach 14 V? Find t.
Vs = 5 VR = 10 kΩC = 100 µFvC(0) = 20 V
t = ms
13.Thévenin & NortonMedium
Find the Thévenin voltage Vth seen from terminals a–b.
V1 = 6 VR1 = 20 ΩR2 = 8 ΩR3 = 6 Ω
Vth = V
14.Thévenin & NortonHard
A load RL = 12 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.
Find the phase ∠G(jω) of G(s) = 3(s + 13)/(s(s + 1)) at ω = 5 rad/s, in degrees.
G(s) = 3(s + 13)/(s(s + 1))ω = 5 rad/s
∠G(jω) = °
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 4)(s + 10)). Find the gain K that gives a gain margin of 15 dB.
L(s) = K/(s(s + 4)(s + 10))GM = 15 dB
K =
3Solutions
1.Complex numbers
Answer: θ = −121 °
Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.
Quadrant
a = −3, b = −5 → III
Reference angle
α = atan(5/3) = 59.04°
Argument
θ = −(180° − α) = −121°
Answer
θ = −121°
In polar form z = 5.831∠−121°.
2.Complex numbers
Answer: Re(z⁶) = −64
De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.
Polar form
z = −√3 + j = 2∠150°
De Moivre
z⁶ = (2)⁶∠900° = 64∠180°
Rectangular form
z⁶ = 64·(cos 180° + j sin 180°) = −64
Re(z⁶) = −64
Answer
Re(z⁶) = −64
3.AC impedance & phasors
Answer: I = 2.165 A
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
Reactance
XC = 1/(ωC) = 1/(2π·100·22·10⁻⁶) = 72.34 Ω
Branch currents
Both branches are connected straight across the supply, so each sees the full voltage V.
IR = V/R = 100/60 = 1.667 A
IX = V/XC = 100/72.34 = 1.382 A
Total current
The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.
I = √(IR² + IX²) = √(1.667² + 1.382²) = 2.165 A
Answer
I = 2.165 A
4.AC impedance & phasors
Answer: B = 159.2 Hz
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
Resonant frequency
f0 = 1/(2π·√(0.002·10·10⁻⁶)) = 1125 Hz
Quality factor
Q compares the reactance at resonance with the resistance: Q = ω₀L/R.
AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.
Reactive power of each load
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Q1 = 28·tan(arccos 0.6) = 37.33 kvar
Q2 = 0 = 0 kvar
Totals
P = 28 + 4 = 32 kW
Q = 37.33 + 0 = 37.33 kvar
S = √(32² + 37.33²) = 49.17 kVA
Total current
I = S/V = 49171/240 = 204.9 A
Answer
I = 204.9 A
6.AC power & power factor
Answer: Pmax = 1653 W
AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.
Conjugate match
The load reactance cancels the source reactance, and the load resistance equals the source resistance.
ZL = Zs* = 8 − j40 Ω
Maximum power
I = Vs/(2Rs) = 230/16 = 14.38 A
Pmax = I²·RL = Vs²/(4Rs) = 230²/(4·8) = 1653 W
Answer
Pmax = 1653 W
7.LC resonance, Q factor & bandwidth
Answer: f1 = 780.8 Hz
Q factor and bandwidth: in a resonant circuit Q = 2π·(energy stored)/(energy lost per cycle). For series RLC Q = ω0L/R = (1/R)√(L/C); for parallel RLC Q = R/(ω0L) = R√(C/L). The bandwidth between the half-power frequencies f1 and f2 (where the power is half and the current or voltage 1/√2 of its peak) is BW = f2 − f1 = f0/Q, which is R/(2πL) in series and 1/(2πRC) in parallel. The resonance frequency is the geometric mean f0 = √(f1f2), and f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q)). Losses also make the oscillation decay: in series RLC the amplitude falls as e−Rt/(2L).
Square-root term
√(1 + 1/(4·2²)) = 1.031
Lower half-power frequency
f1 = 1000·(1.031 − 0.25) = 780.8 Hz = 780.8 Hz
Answer
f1 = 780.8 Hz
The upper one is f₂ = 1.281 kHz; check: f₂ − f₁ = 500 Hz = f₀/Q.
8.LC resonance, Q factor & bandwidth
Answer: Q = 5.477
Q factor and bandwidth: in a resonant circuit Q = 2π·(energy stored)/(energy lost per cycle). For series RLC Q = ω0L/R = (1/R)√(L/C); for parallel RLC Q = R/(ω0L) = R√(C/L). The bandwidth between the half-power frequencies f1 and f2 (where the power is half and the current or voltage 1/√2 of its peak) is BW = f2 − f1 = f0/Q, which is R/(2πL) in series and 1/(2πRC) in parallel. The resonance frequency is the geometric mean f0 = √(f1f2), and f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q)). Losses also make the oscillation decay: in series RLC the amplitude falls as e−Rt/(2L).
Resonance frequency
f0 = √(10000·12000) = 10954 Hz
Bandwidth
BW = 12000 − 10000 = 2000 Hz
Quality factor
Q = 10954/2000 = 5.477
Answer
Q = 5.477
9.Three-phase power
Answer: P = 136.2 kW
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.
Apparent power
S = √3·VL·IL = √3·690·120 = 143.4 kVA
Real power
P = S·cos φ = 143.4·0.95 = 136.2 kW
Answer
P = 136.2 kW
10.Three-phase power
Answer: Cph = 636 µF
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.
Reactive power before and after
The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.
Q1 = 80·tan(acos 0.72) = 80·0.9639 = 77.11 kvar
Q2 = 80·tan(acos 0.9) = 80·0.4843 = 38.75 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 38.36 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.
V = 400/√3 = 230.9 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 12787/(377·230.9²) = 636 µF
Answer
Cph = 636 µF
11.RC & RL transients
Answer: W(∞) = 544 mJ
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Time constant and final current
Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.
τ = L/R = 68 mH / 5 Ω = 13.6 ms
i(∞) = Vs/R = 20/5 = 4 A
Stored energy
W(∞) = ½·L·i(∞)² = ½·68·4² = 544 mJ
Answer
W(∞) = 544 mJ
12.RC & RL transients
Answer: t = 510.8 ms
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Initial value, final value, time constant
The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.
t = τ·ln((20 − 5)/(14 − 5)) = 1000·ln(1.667) = 510.8 ms
Answer
t = 510.8 ms
13.Thévenin & Norton
Answer: Vth = 1.714 V
Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.
Find Rth with every source switched off
Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.
R1 ∥ R2 = 20·8/(20+8) = 5.714 Ω
Rth = (R1 ∥ R2) + R3 = 5.714 + 6 = 11.71 Ω
Find Vth, the open-circuit voltage
With a–b open, no current flows through R₃, so there is no drop across it and V_b = V_A. R₁ and R₂ form a voltage divider.
Vth = V1·R2/(R1+R2) = 6·8/(20+8) = 1.714 V
14.Thévenin & Norton
Answer: PL = 31.31 W
Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.
Find Rth with every source switched off
Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.
R1 ∥ R2 = 4·8/(4+8) = 2.667 Ω
Rth = (R1 ∥ R2) + R3 = 2.667 + 20 = 22.67 Ω
Find Vth, the open-circuit voltage
With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.
VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 12 + (−3)·2.667 = 4 V
Vth = VB = VA + Is·R3 = 4 + (−3)·20 = −56 V
Connect the load to the Thévenin equivalent
The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.
IL = Vth/(Rth + RL) = −56/(22.67 + 12) = −1.615 A
PL = IL²·RL = −1.615²·12 = 31.31 W
15.Bode plots & margins
Answer: ∠G(jω) = −147.7 °
Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.
Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.