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 = −8 − j7 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −8 − j7
θ = rad
2.Complex numbersHard
Find the imaginary part of (2 − j)³.
z = 2 − jn = 3
Im(z³) =
3.AC impedance & phasorsMedium
A resistor and an inductor are connected in parallel across a 50 V (rms), 60 Hz supply. Find the rms current I.
V = 50 V (rms)f = 60 HzR = 15 ΩL = 100 mH
I = A
4.AC impedance & phasorsHard
A series RLC circuit is connected to a 120 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.
V = 120 V (rms)f = 500 HzR = 15 ΩL = 50 mHC = 2.2 µF
f0 = Hz
5.AC power & power factorMedium
A single-phase motor takes 15 kW at a power factor of 0.75 lagging from a 230 V supply. Find the line current I.
P = 15 kWpf = 0.75V = 230 V
I = A
6.AC power & power factorHard
A single-phase load takes 45 kW at a power factor of 0.75 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.92. Find the capacitance C needed.
P = 45 kWpf1 = 0.75pf2 = 0.92V = 240 Vf = 60 Hz
C = µF
7.LC resonance, Q factor & bandwidthMedium
A series RLC circuit resonates at 20 kHz with Q = 5. Find the lower half-power frequency f1 exactly, in kHz.
f0 = 20 kHzQ = 5
f1 = kHz
8.LC resonance, Q factor & bandwidthHard
The half-power frequencies of a resonance curve are 100 kHz and 150 kHz. Find the resonance frequency as their geometric mean, and then the quality factor Q.
f1 = 100 kHzf2 = 150 kHz
Q =
9.Three-phase powerMedium
A balanced three-phase load draws a line current of 30 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.
VL = 400 VIL = 30 Acos φ = 0.82
P = kW
10.Three-phase powerHard
A balanced three-phase load takes 300 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 = 300 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. How long after the switch closes does the current reach 80% of its final value? Find t.
Vs = 24 VR = 15 ΩL = 68 mH
t = ms
12.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 12 V with the polarity shown. Find the capacitor voltage vC at t = 30 ms.
Vs = 20 VR = 4.7 kΩC = 22 µFvC(0) = 12 Vt = 30 ms
vC = V
13.Thévenin & NortonMedium
Find the Thévenin resistance Rth seen from terminals a–b.
V1 = 15 VR1 = 15 ΩR2 = 3 ΩR3 = 20 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 = 5 AR1 = 15 ΩR2 = 30 ΩR3 = 25 Ω
IN = A
15.Bode plots & marginsMedium
Find the phase ∠G(jω) of G(s) = 45(s + 19)/(s(s + 18)) at ω = 0.5 rad/s, in degrees.
G(s) = 45(s + 19)/(s(s + 18))ω = 0.5 rad/s
∠G(jω) = °
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 5)(s + 8)). Find the gain K that gives a gain margin of 8 dB.
L(s) = K/(s(s + 5)(s + 8))GM = 8 dB
K =
3Solutions
1.Complex numbers
Answer: θ = −2.423 rad
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.
Modulus
r = √(64 + 49) = 10.63
Argument in radians
θ = −138.8° = −138.8·π/180 = −2.423 rad
Exponential form
z = 10.63·e−j2.423
Answer
θ = −2.423 rad
2.Complex numbers
Answer: Im(z³) = −11
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.
Powers step by step
z² = (2 − j)(2 − j) = 3 − j4
z³ = (3 − j4)(2 − j) = 2 − j11
Check with De Moivre
z = 2.236∠−26.57° → z³ = 11.18∠−79.7°
Im(z³) = −11
Answer
Im(z³) = −11
3.AC impedance & phasors
Answer: I = 3.588 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
XL = ωL = 2π·60·0.1 = 37.7 Ω
Branch currents
Both branches are connected straight across the supply, so each sees the full voltage V.
IR = V/R = 50/15 = 3.333 A
IX = V/XL = 50/37.7 = 1.326 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) = √(3.333² + 1.326²) = 3.588 A
Answer
I = 3.588 A
4.AC impedance & phasors
Answer: f0 = 479.9 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|.
Resonance
At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.
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) = 15000/(230·0.75) = 86.96 A
Answer
I = 86.96 A
6.AC power & power factor
Answer: C = 944.8 µF
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 = P·tan φ1 = 45·tan(41.41°) = 39.69 kvar
Q2 = P·tan φ2 = 45·tan(23.07°) = 19.17 kvar
QC = Q1 − Q2 = 20.52 kvar
Capacitance
A capacitor across the supply delivers Q = V²·ωC.
C = QC/(2πf·V²) = 20516/(2π·60·240²) = 944.8 µF
Answer
C = 944.8 µF
7.LC resonance, Q factor & bandwidth
Answer: f1 = 18.1 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).
Square-root term
√(1 + 1/(4·5²)) = 1.005
Lower half-power frequency
f1 = 20000·(1.005 − 0.1) = 18100 Hz = 18.1 kHz
Answer
f1 = 18.1 kHz
The upper one is f₂ = 22.1 kHz; check: f₂ − f₁ = 4 kHz = f₀/Q.
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 = √(100000·150000) = 122474 Hz
Bandwidth
BW = 150000 − 100000 = 50000 Hz
Quality factor
Q = 122474/50000 = 2.449
Answer
Q = 2.449
9.Three-phase power
Answer: P = 17.04 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 = √3·400·30 = 20.78 kVA
Real power
P = S·cos φ = 20.78·0.82 = 17.04 kW
Answer
P = 17.04 kW
10.Three-phase power
Answer: Cph = 1763 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 = 300·tan(acos 0.78) = 300·0.8023 = 240.7 kvar
Q2 = 300·tan(acos 0.9) = 300·0.4843 = 145.3 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 95.39 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 = 68 mH / 15 Ω = 4.533 ms
i(∞) = Vs/R = 24/15 = 1.6 A
Solve for the time
0.8 = 1 − e−t/τ → t = −τ·ln(1 − 0.8)
t = −4.533·ln(0.2) = 7.296 ms
Answer
t = 7.296 ms
12.RC & RL transients
Answer: vC = 14.01 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) = 12 V, vC(∞) = Vs = 20 V
τ = R·C = 4.7 kΩ · 22 µF = 103.4 ms
vC(t) = 20 + (12 − 20)·e−t/τ
Voltage at the given time
vC(30) = 20 + (−8)·e−30/103.4 = 14.01 V
Answer
vC = 14.01 V
13.Thévenin & Norton
Answer: Rth = 22.5 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 = 15·3/(15+3) = 2.5 Ω
Rth = (R1 ∥ R2) + R3 = 2.5 + 20 = 22.5 Ω
14.Thévenin & Norton
Answer: IN = 5.171 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 = 15·30/(15+30) = 10 Ω
Rth = (R1 ∥ R2) + R3 = 10 + 25 = 35 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 + 5·10 = 56 V
Vth = VB = VA + Is·R3 = 56 + 5·25 = 181 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 = 181/35 = 5.171 A
15.Bode plots & margins
Answer: ∠G(jω) = −90.08 °
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.