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·e−j0.5 has its angle in radians. Find its imaginary part b.
r = 6θ = −0.5 rad
b =
2.Complex numbersHard
The roots of zⁿ = w with n = 6 and w = 64∠180° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 5. Find the argument of z₂ in degrees, between −180° and 180°.
w = 64∠180°n = 6k = 2
θ₂ = °
3.AC impedance & phasorsMedium
A series RLC circuit is connected to a 10 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
V = 10 V (rms)f = 1000 HzR = 80 ΩL = 20 mHC = 100 µF
φ = °
4.AC impedance & phasorsHard
A series RLC circuit is connected to a 50 V (rms), 100 Hz supply. Find the rms current I.
V = 50 V (rms)f = 100 HzR = 15 ΩL = 250 mHC = 4.7 µF
I = A
5.AC power & power factorMedium
Two loads are connected in parallel to a 400 V single-phase supply: a 4 kW load at a power factor of 0.85 lagging and a 5 kW load at a power factor of 0.7 leading. Find the overall power factor cos φ seen by the supply.
V = 400 VP1 = 4 kW, pf1 = 0.85 laggingP2 = 5 kW, pf2 = 0.7 leading
cos φ =
6.AC power & power factorHard
A single-phase load takes 9 kW at a power factor of 0.6 lagging from a 240 V, 50 Hz supply. A 220 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.
P = 9 kWpf1 = 0.6V = 240 Vf = 50 HzC = 220 µF
cos φ =
7.LC resonance, Q factor & bandwidthMedium
A parallel RLC circuit has R = 22 kΩ, L = 1 mH and C = 1 nF. Find its quality factor Q.
R = 22 kΩL = 1 mHC = 1 nF
Q =
8.LC resonance, Q factor & bandwidthHard
The half-power frequencies of a resonance curve are 2 kHz and 3 kHz. Find the resonance frequency as their geometric mean, and then the quality factor Q.
f1 = 2 kHzf2 = 3 kHz
Q =
9.Three-phase powerMedium
A balanced three-phase load draws a line current of 150 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.8 lagging. Find the total real power P drawn by the load in kW.
VL = 415 VIL = 150 Acos φ = 0.8
P = kW
10.Three-phase powerHard
A balanced three-phase load takes 80 kW at a power factor of 0.7 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 80 kWpf1 = 0.7pf2 = 0.95VL = 415 Vf = 50 Hz
QC = kvar
11.RC & RL transientsMedium
The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the capacitor voltage vC at t = 30 ms.
Vs = 10 VR1 = 4.7 kΩR2 = 1 kΩC = 100 µFt = 30 ms
vC = V
12.RC & RL transientsHard
A timer starts charging an uncharged capacitor through a resistor from a 5 V supply at t = 0. It must trigger when the capacitor voltage reaches 3.16 V, exactly 25 ms after the start. With C = 2.2 µF, find the resistance R needed.
Vs = 5 VVth = 3.16 VT = 25 msC = 2.2 µF
R = kΩ
13.Thévenin & NortonMedium
Find the Norton current IN: the current that flows from a to b if the terminals are shorted.
V1 = 18 VR1 = 6 ΩR2 = 10 ΩR3 = 8 Ω
IN = A
14.Thévenin & NortonHard
Find the Thévenin voltage Vth seen from terminals a–b.
V1 = 18 VIs = 2 AR1 = 4 ΩR2 = 8 ΩR3 = 6 Ω
Vth = V
15.Bode plots & marginsMedium
In the series RLC band-pass filter shown the output is taken across R = 50 Ω, with L = 50 mH and C = 0.1 µF. Find the centre (resonant) frequency f0 in Hz.
R = 50 ΩL = 50 mHC = 0.1 µF
f0 = Hz
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 2)(s + 7)). Find the gain K that gives a gain margin of 8 dB.
L(s) = K/(s(s + 2)(s + 7))GM = 8 dB
K =
3Solutions
1.Complex numbers
Answer: b = −2.877
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 −0.5 + j sin −0.5)
Components (radians)
a = 6·cos −0.5 = 5.265
b = 6·sin −0.5 = −2.877
b = −2.877
Answer
b = −2.877
2.Complex numbers
Answer: θ₂ = 150 °
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 of the roots
641/6 = 2
Argument of the root
θ₂ = (180° + 360°·2)/6 = 150°
Argument
θ₂ = 150°
Answer
θ₂ = 150°
The 6 roots lie on the circle of radius 2, 60° apart.
3.AC impedance & phasors
Answer: φ = 57.19 °
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·1000 = 6283 rad/s
XL = ωL = 6283·0.02 = 125.7 Ω
XC = 1/(ωC) = 1/(6283·100·10⁻⁶) = 1.592 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
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/(628.3·4.7·10⁻⁶) = 338.6 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| = 50/182.2 = 0.2745 A
Answer
I = 0.2745 A
5.AC power & power factor
Answer: cos φ = 0.9601
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 = 4·tan(arccos 0.85) = 2.479 kvar
Q2 = −5·tan(arccos 0.7) = −5.101 kvar
Totals
P = 4 + 5 = 9 kW
Q = 2.479 + (−5.101) = −2.622 kvar
S = √(9² + (−2.622)²) = 9.374 kVA
Overall power factor
cos φ = P/S = 9/9.374 = 0.9601 leading
Answer
cos φ = 0.9601
6.AC power & power factor
Answer: cos φ = 0.7466
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 = 9·tan(arccos 0.6) = 12 kvar
QC = V²·2πf·C = 240²·2π·50·220·10⁻⁶ = 3981 var = 3.981 kvar
Q2 = 12 − 3.981 = 8.019 kvar
New apparent power
S2 = √(9² + 8.019²) = 12.05 kVA
New power factor
cos φ = P/S2 = 9/12.05 = 0.7466
Answer
cos φ = 0.7466
7.LC resonance, Q factor & bandwidth
Answer: Q = 22
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
ω₀ = 1/√(LC) = 1000000 rad/s
Quality factor
Q = R/(ω₀L) = 22000/(1000000·0.001) = 22
Answer
Q = 22
The bandwidth is 1/(2πRC) = 7.234 kHz.
8.LC resonance, Q factor & bandwidth
Answer: Q = 2.449
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 = √(2000·3000) = 2449 Hz
Bandwidth
BW = 3000 − 2000 = 1000 Hz
Quality factor
Q = 2449/1000 = 2.449
Answer
Q = 2.449
9.Three-phase power
Answer: P = 86.26 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 = √3·415·150 = 107.8 kVA
Real power
P = S·cos φ = 107.8·0.8 = 86.26 kW
Answer
P = 86.26 kW
10.Three-phase power
Answer: QC = 55.32 kvar
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 0.7) = 80·1.02 = 81.62 kvar
Q2 = 80·tan(acos 0.95) = 80·0.3287 = 26.29 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 55.32 kvar
Answer
QC = 55.32 kvar
11.RC & RL transients
Answer: vC = 0.5351 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.
Thévenin equivalent seen by the capacitor
Short the source to find Rth; open the capacitor to find the voltage it will settle at.
Rth = R1 ∥ R2 = 4.7·1/(4.7 + 1) = 0.8246 kΩ
vC(∞) = 10·1/(4.7 + 1) = 1.754 V
Time constant
τ = Rth·C = 0.8246·100 = 82.46 ms
Charging curve
vC(t) = 1.754·(1 − e−30/82.46) = 0.5351 V
Answer
vC = 0.5351 V
12.RC & RL transients
Answer: R = 11.37 kΩ
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.
Required time constant
3.16 = 5·(1 − e−25/τ) → τ = 25/ln(5/(5 − 3.16))
τ = 25/0.9997 = 25.01 ms
Resistance
kΩ times µF gives milliseconds.
R = τ/C = 25.01/2.2 = 11.37 kΩ
Answer
R = 11.37 kΩ
13.Thévenin & Norton
Answer: IN = 0.9574 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 = 6·10/(6+10) = 3.75 Ω
Rth = (R1 ∥ R2) + R3 = 3.75 + 8 = 11.75 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) = 18·10/(6+10) = 11.25 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 = 11.25/11.75 = 0.9574 A
14.Thévenin & Norton
Answer: Vth = 29.33 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 = 4·8/(4+8) = 2.667 Ω
Rth = (R1 ∥ R2) + R3 = 2.667 + 6 = 8.667 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 + 2·2.667 = 17.33 V
Vth = VB = VA + Is·R3 = 17.33 + 2·6 = 29.33 V
15.Bode plots & margins
Answer: f0 = 2251 Hz
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.
Resonant frequency
ω₀ = 1/√(LC) = 1/√(0.05·1 × 10⁻⁷) = 14142 rad/s
f₀ = ω₀/(2π) = 2251 Hz
Answer
f0 = 2251 Hz
16.Bode plots & margins
Answer: K = 50.16
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.