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
The number z = 6·ej1.5 has its angle in radians. Find its imaginary part b.
r = 6θ = 1.5 rad
b =
2.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 8∠180°. Click it.
w = 8∠180°n = 3
1A (2∠60°)
2B (2∠−120°)
3C (3.6∠120°)
4D (2∠90°)
3.AC impedance & phasorsMedium
A series RLC circuit is connected to a 12 V (rms), 100 Hz supply. Find the rms current I.
V = 12 V (rms)f = 100 HzR = 25 ΩL = 250 mHC = 100 µF
I = A
4.AC impedance & phasorsHard
A series RLC circuit is driven by a 10 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.
V = 10 V (rms)R = 10 ΩL = 5 mHC = 0.1 µF
VL = V
5.AC power & power factorMedium
A load with impedance Z = 3 + j4 Ω is connected to a 120 V (rms) single-phase supply. Find the reactive power Q it absorbs.
V = 120 VZ = 3 + j4 Ω
Q = var
6.AC power & power factorHard
A 200 kVA single-phase transformer supplies a load of 93 kW at a power factor of 0.7 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
Smax = 200 kVAP1 = 93 kWpf1 = 0.7
ΔP = kW
7.LC resonance, Q factor & bandwidthMedium
A series RLC circuit has R = 22 Ω, L = 1 mH and C = 22 nF. Find its bandwidth BW in kHz.
R = 22 ΩL = 1 mHC = 22 nF
BW = kHz
8.LC resonance, Q factor & bandwidthHard
The half-power frequencies of a resonance curve are 10 kHz and 11 kHz. Find the resonance frequency as their geometric mean, and then the quality factor Q.
f1 = 10 kHzf2 = 11 kHz
Q =
9.Three-phase powerMedium
A three-phase induction motor delivers 22 kW of mechanical output power at full load. At this load its efficiency is 94% and its power factor is 0.8 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.
Pout = 22 kWη = 94%cos φ = 0.8VL = 690 V
IL = A
10.Three-phase powerHard
A balanced three-phase load takes 50 kW at a power factor of 0.78 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 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 50 kWpf1 = 0.78pf2 = 0.9VL = 415 Vf = 50 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 = 9 VR = 50 ΩL = 100 mH
W(∞) = mJ
12.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 20 V with the polarity shown. Find the capacitor voltage vC at t = 50 ms.
Vs = 12 VR = 2.2 kΩC = 10 µFvC(0) = 20 Vt = 50 ms
vC = V
13.Thévenin & NortonMedium
Find the Thévenin resistance Rth seen from terminals a–b.
V1 = 30 VR1 = 15 ΩR2 = 10 ΩR3 = 10 Ω
Rth = Ω
14.Thévenin & NortonHard
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
V1 = 9 VIs = 4 AR1 = 25 ΩR2 = 15 ΩR3 = 4 Ω
IN = A
15.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 47(s + 11)/(s(s + 4)) at ω = 20 rad/s, in degrees.
G(s) = 47(s + 11)/(s(s + 4))ω = 20 rad/s
∠G(jω) = °
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 16/(s(s + 1)(s + 9)). Find the gain margin GM in dB.
L(s) = 16/(s(s + 1)(s + 9))
GM = dB
3Solutions
1.Complex numbers
Answer: b = 5.985
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.
Euler's formula
z = 6·(cos 1.5 + j sin 1.5)
Components (radians)
a = 6·cos 1.5 = 0.4244
b = 6·sin 1.5 = 5.985
b = 5.985
Answer
b = 5.985
2.Complex numbers
Answer: A (2∠60°)
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.
Modulus and arguments of the roots
|z| = 2
θ = 60°, 180°, −60°
The point
A = 2∠60°
Answer
A (2∠60°)
3.AC impedance & phasors
Answer: I = 0.0837 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|.
Reactances
ω = 2πf = 2π·100 = 628.3 rad/s
XL = ωL = 628.3·0.25 = 157.1 Ω
XC = 1/(ωC) = 1/(628.3·100·10⁻⁶) = 15.92 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
The same current flows through every element of a series circuit.
I = V/|Z| = 12/143.4 = 0.0837 A
Answer
I = 0.0837 A
4.AC impedance & phasors
Answer: VL = 223.6 V
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
At resonance the reactances cancel, so the impedance is just R.
ω₀ = 1/√(LC) = 1/√(0.005·0.1·10⁻⁶) = 44721 rad/s
I = V/R = 10/10 = 1 A
Voltage across the inductor
XL = ω₀L = 44721·0.005 = 223.6 Ω
VL = I·XL = 1·223.6 = 223.6 V
Answer
VL = 223.6 V
This is Q = 22.36 times the supply voltage.
5.AC power & power factor
Answer: Q = 2304 var
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.
Current
|Z| = √(3² + 4²) = 5 Ω
I = V/|Z| = 120/5 = 24 A
Reactive power
Q = I²·X = 24²·4 = 2304 var
Answer
Q = 2304 var
6.AC power & power factor
Answer: ΔP = 83.06 kW
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 the existing load
Q1 = 93·tan(arccos 0.7) = 94.88 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(200² − 94.88²) = 176.1 kW
Extra real power
ΔP = 176.1 − 93 = 83.06 kW
Answer
ΔP = 83.06 kW
7.LC resonance, Q factor & bandwidth
Answer: BW = 3.501 kHz
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 = 22/0.001 = 22000 rad/s
Bandwidth in Hz
BW = 22000/(2π) = 3501 Hz = 3.501 kHz
Answer
BW = 3.501 kHz
With f₀ = 33.93 kHz this gives Q = 9.691.
8.LC resonance, Q factor & bandwidth
Answer: Q = 10.49
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 = √(10000·11000) = 10488 Hz
Bandwidth
BW = 11000 − 10000 = 1000 Hz
Quality factor
Q = 10488/1000 = 10.49
Answer
Q = 10.49
9.Three-phase power
Answer: IL = 24.48 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 = 22/0.94 = 23.4 kW
Line current
IL = Pin/(√3·VL·cos φ) = 23404/(√3·690·0.8) = 24.48 A
Answer
IL = 24.48 A
10.Three-phase power
Answer: Cph = 293.8 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 = 50·tan(acos 0.78) = 50·0.8023 = 40.11 kvar
Q2 = 50·tan(acos 0.9) = 50·0.4843 = 24.22 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 15.9 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.
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 = 100 mH / 50 Ω = 2 ms
i(∞) = Vs/R = 9/50 = 0.18 A
Stored energy
W(∞) = ½·L·i(∞)² = ½·100·0.18² = 1.62 mJ
Answer
W(∞) = 1.62 mJ
12.RC & RL transients
Answer: vC = 12.82 V
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.
vC(0) = 20 V, vC(∞) = Vs = 12 V
τ = R·C = 2.2 kΩ · 10 µF = 22 ms
vC(t) = 12 + (20 − 12)·e−t/τ
Voltage at the given time
vC(50) = 12 + (8)·e−50/22 = 12.82 V
Answer
vC = 12.82 V
13.Thévenin & Norton
Answer: Rth = 16 Ω
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 + 10 = 16 Ω
14.Thévenin & Norton
Answer: IN = 4.252 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 = 25·15/(25+15) = 9.375 Ω
Rth = (R1 ∥ R2) + R3 = 9.375 + 4 = 13.38 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) = 3.375 + 4·9.375 = 40.88 V
Vth = VB = VA + Is·R3 = 40.88 + 4·4 = 56.87 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 = 56.87/13.38 = 4.252 A
15.Bode plots & margins
Answer: ∠G(jω) = −107.5 °
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.