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 = −4 − j8 in degrees, as the principal value between −180° and 180°.
z = −4 − j8
θ = °
2.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z⁶ for z = −2 − j2.
z = −2 − j2n = 6
Im(z⁶) =
3.AC impedance & phasorsMedium
A series RLC circuit with C = 0.47 µF must resonate at 200 Hz. Find the inductance L needed.
f0 = 200 HzC = 0.47 µF
L = mH
4.AC impedance & phasorsHard
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 100 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.
V = 100 V (rms)f = 200 HzR1 = 40 ΩR2 = 60 ΩC = 22 µF
|Z| = Ω
5.AC power & power factorMedium
A single-phase motor takes 4 kW at a power factor of 0.65 lagging from a 230 V supply. Find the line current I.
P = 4 kWpf = 0.65V = 230 V
I = A
6.AC power & power factorHard
A single-phase load takes 17 kW at a power factor of 0.75 lagging from a 230 V, 60 Hz supply. A 150 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.
P = 17 kWpf1 = 0.75V = 230 Vf = 60 HzC = 150 µF
cos φ =
7.LC resonance, Q factor & bandwidthMedium
A series RLC circuit has R = 47 Ω, L = 10 mH and C = 22 nF. Find its bandwidth BW in Hz.
R = 47 ΩL = 10 mHC = 22 nF
BW = Hz
8.LC resonance, Q factor & bandwidthHard
The half-power frequencies of a resonance curve are 900 kHz and 1.125 MHz. Find the resonance frequency as their geometric mean, and then the quality factor Q.
f1 = 900 kHzf2 = 1.125 MHz
Q =
9.Three-phase powerMedium
A three-phase induction motor delivers 75 kW of mechanical output power at full load. At this load its efficiency is 92% and its power factor is 0.88 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.
Pout = 75 kWη = 92%cos φ = 0.88VL = 480 V
IL = A
10.Three-phase powerHard
A balanced three-phase load takes 120 kW at a power factor of 0.72 lagging from a 415 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 120 kWpf1 = 0.72pf2 = 0.92VL = 415 Vf = 50 Hz
Cph = µF
11.RC & RL transientsMedium
The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 80% of its final value? Find t.
Vs = 20 VR = 33 ΩL = 47 mH
t = ms
12.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 8 V with the polarity shown. How long does it take for vC to reach 14.4 V? Find t.
Vs = 24 VR = 6.8 kΩC = 47 µFvC(0) = 8 V
t = ms
13.Thévenin & NortonMedium
Find the Thévenin voltage Vth seen from terminals a–b.
V1 = 6 VR1 = 2 ΩR2 = 3 ΩR3 = 15 Ω
Vth = V
14.Thévenin & NortonHard
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
V1 = 5 VIs = 1.5 AR1 = 15 ΩR2 = 10 ΩR3 = 12 Ω
IN = A
15.Bode plots & marginsMedium
Find the exact magnitude |G(jω)| of G(s) = 108/((s + 3)(s + 12)) at ω = 5 rad/s, in dB.
G(s) = 108/((s + 3)(s + 12))ω = 5 rad/s
|G(jω)| = dB
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 250/(s(s + 5)(s + 9)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 250/(s(s + 5)(s + 9))
ω180 = rad/s
3Solutions
1.Complex numbers
Answer: θ = −116.6 °
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 = −4, b = −8 → III
Reference angle
α = atan(8/4) = 63.43°
Argument
θ = −(180° − α) = −116.6°
Answer
θ = −116.6°
In polar form z = 8.944∠−116.6°.
2.Complex numbers
Answer: Im(z⁶) = −512
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 = −2 − j2 = 2√2∠−135°
De Moivre
z⁶ = (2√2)⁶∠−810° = 512∠−90°
Rectangular form
z⁶ = 512·(cos −90° + j sin −90°) = −j512
Im(z⁶) = −512
Answer
Im(z⁶) = −512
3.AC impedance & phasors
Answer: L = 1347 mH
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|.
Resonance condition
At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).
ω₀ = 2π·200 = 1257 rad/s
L = 1/(ω₀²·C) = 1/(1257²·0.47·10⁻⁶) = 1347 mH
Answer
L = 1347 mH
4.AC impedance & phasors
Answer: |Z| = 61.96 Ω
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/(2π·200·22·10⁻⁶) = 36.17 Ω
Parallel combination
Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.
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.
Line current
Only the real part of the power does work, so divide by the power factor.
I = P/(V·cos φ) = 4000/(230·0.65) = 26.76 A
Answer
I = 26.76 A
6.AC power & power factor
Answer: cos φ = 0.8169
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 before and after
The real power P does not change; only the reactive power drawn from the supply does.
Q1 = 17·tan(arccos 0.75) = 14.99 kvar
QC = V²·2πf·C = 230²·2π·60·150·10⁻⁶ = 2991 var = 2.991 kvar
Q2 = 14.99 − 2.991 = 12 kvar
New apparent power
S2 = √(17² + 12²) = 20.81 kVA
New power factor
cos φ = P/S2 = 17/20.81 = 0.8169
Answer
cos φ = 0.8169
7.LC resonance, Q factor & bandwidth
Answer: BW = 748 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).
Bandwidth in rad/s
R/L = 47/0.01 = 4700 rad/s
Bandwidth in Hz
BW = 4700/(2π) = 748 Hz = 748 Hz
Answer
BW = 748 Hz
With f₀ = 10.73 kHz this gives Q = 14.34.
8.LC resonance, Q factor & bandwidth
Answer: Q = 4.472
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 = √(900000·1.125 × 10⁶) = 1.006 × 10⁶ Hz
Bandwidth
BW = 1.125 × 10⁶ − 900000 = 225000 Hz
Quality factor
Q = 1.006 × 10⁶/225000 = 4.472
Answer
Q = 4.472
9.Three-phase power
Answer: IL = 111.4 A
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.
Electrical input power
Pin = Pout/η = 75/0.92 = 81.52 kW
Line current
IL = Pin/(√3·VL·cos φ) = 81522/(√3·480·0.88) = 111.4 A
Answer
IL = 111.4 A
10.Three-phase power
Answer: Cph = 397.6 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 = 120·tan(acos 0.72) = 120·0.9639 = 115.7 kvar
Q2 = 120·tan(acos 0.92) = 120·0.426 = 51.12 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 64.54 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.
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 = 47 mH / 33 Ω = 1.424 ms
i(∞) = Vs/R = 20/33 = 0.6061 A
Solve for the time
0.8 = 1 − e−t/τ → t = −τ·ln(1 − 0.8)
t = −1.424·ln(0.2) = 2.292 ms
Answer
t = 2.292 ms
12.RC & RL transients
Answer: t = 163.3 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((8 − 24)/(14.4 − 24)) = 319.6·ln(1.667) = 163.3 ms
Answer
t = 163.3 ms
13.Thévenin & Norton
Answer: Vth = 3.6 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 = 2·3/(2+3) = 1.2 Ω
Rth = (R1 ∥ R2) + R3 = 1.2 + 15 = 16.2 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) = 6·3/(2+3) = 3.6 V
14.Thévenin & Norton
Answer: IN = 1.611 A
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 = 15·10/(15+10) = 6 Ω
Rth = (R1 ∥ R2) + R3 = 6 + 12 = 18 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) = 2 + 1.5·6 = 11 V
Vth = VB = VA + Is·R3 = 11 + 1.5·12 = 29 V
Norton equivalent
A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.
IN = Vth/Rth = 29/18 = 1.611 A
15.Bode plots & margins
Answer: |G(jω)| = 3.075 dB
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.