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
Write z = 12∠−20° in rectangular form a + jb and give the real part a.
r = 12θ = −20°
a =
2.Complex numbersHard
The roots of zⁿ = w with n = 4 and w = 16∠90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the imaginary part of z₁.
w = 16∠90°n = 4k = 1
Im(z₁) =
3.AC impedance & phasorsMedium
A resistor and a capacitor are connected in parallel across a 120 V (rms), 200 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
V = 120 V (rms)f = 200 HzR = 80 ΩC = 10 µF
φ = °
4.AC impedance & phasorsHard
A series RLC circuit has R = 5 Ω, L = 100 mH and C = 0.22 µF. Find its quality factor Q at resonance.
R = 5 ΩL = 100 mHC = 0.22 µF
Q =
5.AC power & power factorMedium
An inductive load takes 9 A and 918 W from a 120 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.
V = 120 VI = 9 AP = 918 W
R = Ω
6.AC power & power factorHard
A 24 V (rms), 60 Hz source has an internal impedance Zs = 8 + j5 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
Vs = 24 V (rms)f = 60 HzZs = 8 + j5 Ω
Pmax = W
7.LC resonance, Q factor & bandwidthMedium
In an ideal LC circuit with L = 2.2 mH and C = 10 nF, the capacitor is fully charged at t = 0. How long does it take until all the energy is in the inductor? Give t in µs.
L = 2.2 mHC = 10 nF
t = µs
8.LC resonance, Q factor & bandwidthHard
A charged 47 nF capacitor discharges through a 10 mH coil with a total series resistance of 10 Ω, so the oscillation slowly dies out. Its amplitude decays as e−Rt/(2L). How many oscillation cycles n pass before the amplitude has halved? (The losses are small, so the frequency is f₀.)
L = 10 mHC = 47 nFR = 10 Ω
n =
9.Three-phase powerMedium
A balanced Y-connected load with an impedance of 9 + j12 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 400 V. Find the total real power P absorbed by the load in kW.
VL = 400 VZ = 9 + j12 Ω per phase
P = kW
10.Three-phase powerHard
A balanced three-phase load takes 200 kW at a power factor of 0.7 lagging from a 480 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 200 kWpf1 = 0.7pf2 = 0.98VL = 480 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 at t = 5 ms, in mJ.
Vs = 20 VR = 15 ΩL = 150 mHt = 5 ms
W = mJ
12.RC & RL transientsHard
The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage vC at t = 300 ms.
Vs = 20 VR1 = 2.2 kΩR2 = 4.7 kΩC = 47 µFt = 300 ms
vC = V
13.Thévenin & NortonMedium
A load RL = 5 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.
V1 = 6 VR1 = 2 ΩR2 = 20 ΩR3 = 5 ΩRL = 5 Ω
IL = A
14.Thévenin & NortonHard
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
V1 = 18 VIs = 4 AR1 = 6 ΩR2 = 3 ΩR3 = 30 Ω
IN = A
15.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 25(s + 13)/(s(s + 9)) at ω = 8 rad/s, in degrees.
G(s) = 25(s + 13)/(s(s + 9))ω = 8 rad/s
∠G(jω) = °
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 22/(s(s + 3)(s + 4)). Find the gain margin GM in dB.
L(s) = 22/(s(s + 3)(s + 4))
GM = dB
3Solutions
1.Complex numbers
Answer: a = 11.28
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.
Components
a = 12·cos(−20°) = 11.28
b = 12·sin(−20°) = −4.104
Rectangular form
z = 11.28 − j4.104
a = 11.28
Answer
a = 11.28
2.Complex numbers
Answer: Im(z₁) = 1.848
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.
The 4 roots lie on the circle of radius 2, 90° apart.
3.AC impedance & phasors
Answer: φ = −45.15 °
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·200·10·10⁻⁶) = 79.58 Ω
Branch currents
Both branches are connected straight across the supply, so each sees the full voltage V.
IR = V/R = 120/80 = 1.5 A
IX = V/XC = 120/79.58 = 1.508 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) = √(1.5² + 1.508²) = 2.127 A
Phase angle
The capacitor current leads V, so the total current leads and the angle is negative.
φ = −arctan(IX/IR) = −arctan(1.508/1.5) = −45.15°
Answer
φ = −45.15°
4.AC impedance & phasors
Answer: Q = 134.8
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·√(0.1·0.22·10⁻⁶)) = 1073 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.
Resistance from the real power
R = P/I² = 918/9² = 11.33 Ω
Answer
R = 11.33 Ω
6.AC power & power factor
Answer: Pmax = 18 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 − j5 Ω
Maximum power
I = Vs/(2Rs) = 24/16 = 1.5 A
Pmax = I²·RL = Vs²/(4Rs) = 24²/(4·8) = 18 W
Answer
Pmax = 18 W
7.LC resonance, Q factor & bandwidth
Answer: t = 7.368 µs
LC resonance: a charged capacitor connected to an inductor drives a current that charges it the other way: the energy swings between the electric field of C (½Cv²) and the magnetic field of L (½Li²) at the natural frequency f0 = 1/(2π√(LC)). Without resistance the total energy ½CV0² stays constant, so the peak current is Imax = V0√(C/L). One period is T = 1/f0; the energy moves from C to L in a quarter of it.
Period
T = 2π√(0.0022·1 × 10⁻⁸) = 29.47 µs
Quarter period
t = T/4 = 7.368 µs
Answer
t = 7.368 µs
8.LC resonance, Q factor & bandwidth
Answer: n = 10.18
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).
Half-life of the amplitude
t = 2·0.01·ln 2/10 = 0.001386 s
Number of cycles
n = t·f0 = 0.001386·7341 = 10.18
Answer
n = 10.18
Shortcut: n = Q·ln 2/π with Q = 46.13.
9.Three-phase power
Answer: P = 6.4 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.
Phase voltage
Vph = 400/√3 = 230.9 V
Phase current
|Z| = √(9² + 12²) = 15 Ω
Iph = 230.9/15 = 15.4 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·15.4²·9 = 6.4 kW
Answer
P = 6.4 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
10.Three-phase power
Answer: Cph = 752.6 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 = 200·tan(acos 0.7) = 200·1.02 = 204 kvar
Q2 = 200·tan(acos 0.98) = 200·0.2031 = 40.61 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 163.4 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 = 150 mH / 15 Ω = 10 ms
i(∞) = Vs/R = 20/15 = 1.333 A
Current at the given time
i(t) = 1.333·(1 − e−5/10) = 0.5246 A
Stored energy
W = ½·L·i² = ½·150·0.5246² = 20.64 mJ
Answer
W = 20.64 mJ
12.RC & RL transients
Answer: vC = 3.503 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 capacitor voltage
In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.
vC(0) = 20·4.7/(2.2 + 4.7) = 13.62 V
Time constant after the switch opens
τ = R2·C = 4.7·47 = 220.9 ms
Discharge
vC(t) = 13.62·e−300/220.9 = 3.503 V
Answer
vC = 3.503 V
13.Thévenin & Norton
Answer: IL = 0.4615 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 = 2·20/(2+20) = 1.818 Ω
Rth = (R1 ∥ R2) + R3 = 1.818 + 5 = 6.818 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) = 6·20/(2+20) = 5.455 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) = 5.455/(6.818 + 5) = 0.4615 A
14.Thévenin & Norton
Answer: IN = −3.813 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 = 6·3/(6+3) = 2 Ω
Rth = (R1 ∥ R2) + R3 = 2 + 30 = 32 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) = 6 + (−4)·2 = −2 V
Vth = VB = VA + Is·R3 = −2 + (−4)·30 = −122 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 = −122/32 = −3.813 A
15.Bode plots & margins
Answer: ∠G(jω) = −100 °
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.