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 = −9 − j7 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −9 − j7
θ = rad
2.Complex numbersHard
Find the real part of (2 − j3)³.
z = 2 − j3n = 3
Re(z³) =
3.AC impedance & phasorsMedium
A series RLC circuit is connected to a 100 V (rms), 500 Hz supply. Find the power factor cos φ of the circuit.
V = 100 V (rms)f = 500 HzR = 80 ΩL = 250 mHC = 2.2 µF
cos φ =
4.AC impedance & phasorsHard
A series RLC circuit is connected to a 50 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.
V = 50 V (rms)f = 500 HzR = 100 ΩL = 50 mHC = 4.7 µF
f0 = Hz
5.AC power & power factorMedium
An inductive load takes 20 A and 4080 W from a 240 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find X.
V = 240 VI = 20 AP = 4080 W
X = Ω
6.AC power & power factorHard
A single-phase load takes 10 kW at a power factor of 0.6 lagging from a 240 V, 60 Hz supply. A 330 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.
P = 10 kWpf1 = 0.6V = 240 Vf = 60 HzC = 330 µF
cos φ =
7.LC resonance, Q factor & bandwidthMedium
A 1 nF capacitor charged to 12 V oscillates with an ideal 1 mH inductor. Find the energy WL stored in the inductor at the moment the capacitor voltage is 9 V, in nJ.
C = 1 nFV0 = 12 VvC = 9 V
WL = nJ
8.LC resonance, Q factor & bandwidthHard
Design a series RLC filter with a total resistance of 10 Ω, a resonance frequency of 20 kHz and a bandwidth of 500 Hz. Find the capacitance C in nF.
R = 10 Ωf0 = 20 kHzBW = 500 Hz
C = nF
9.Three-phase powerMedium
A balanced three-phase load draws a line current of 120 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.
VL = 690 VIL = 120 Acos φ = 0.85
P = kW
10.Three-phase powerHard
A balanced three-phase load takes 150 kW at a power factor of 0.7 lagging from a 690 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 = 150 kWpf1 = 0.7pf2 = 0.92VL = 690 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 = 10 ΩL = 47 mH
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 = 25 ms.
Vs = 6 VR1 = 6.8 kΩR2 = 1 kΩC = 22 µFt = 25 ms
vC = V
13.Thévenin & NortonMedium
A load RL = 6 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.
V1 = 10 VR1 = 3 ΩR2 = 12 ΩR3 = 4 ΩRL = 6 Ω
IL = A
14.Thévenin & NortonHard
A load RL = 6 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.
The RC high-pass filter shown has R = 4.7 kΩ, C = 22 nF. A sinusoidal input of 770 Hz is applied. Find the gain |H| = |vout/vin| in dB.
R = 4.7 kΩC = 22 nFf = 770 Hz
|H| = dB
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 4/(s(s + 1)(s + 2)). Find the gain margin GM in dB.
L(s) = 4/(s(s + 1)(s + 2))
GM = dB
3Solutions
1.Complex numbers
Answer: θ = −2.481 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 = √(81 + 49) = 11.4
Argument in radians
θ = −142.1° = −142.1·π/180 = −2.481 rad
Exponential form
z = 11.4·e−j2.481
Answer
θ = −2.481 rad
2.Complex numbers
Answer: Re(z³) = −46
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 − j3)(2 − j3) = −5 − j12
z³ = (−5 − j12)(2 − j3) = −46 − j9
Check with De Moivre
z = 3.606∠−56.31° → z³ = 46.87∠−168.9°
Re(z³) = −46
Answer
Re(z³) = −46
3.AC impedance & phasors
Answer: cos φ = 0.1239
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
XL = 785.4 Ω, XC = 144.7 Ω
Impedance
|Z| = √(80² + (640.7)²) = 645.7 Ω
Power factor
cos φ = R/|Z| = 80/645.7 = 0.1239
The circuit is inductive, so the power factor is lagging.
Answer
cos φ = 0.1239
4.AC impedance & phasors
Answer: f0 = 328.3 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.
Resistance from the real power
R = P/I² = 4080/20² = 10.2 Ω
Reactance from the impedance
|Z| = V/I = 240/20 = 12 Ω
X = √(|Z|² − R²) = √(12² − 10.2²) = 6.321 Ω
Answer
X = 6.321 Ω
6.AC power & power factor
Answer: cos φ = 0.8511
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 = 10·tan(arccos 0.6) = 13.33 kvar
QC = V²·2πf·C = 240²·2π·60·330·10⁻⁶ = 7166 var = 7.166 kvar
Q2 = 13.33 − 7.166 = 6.167 kvar
New apparent power
S2 = √(10² + 6.167²) = 11.75 kVA
New power factor
cos φ = P/S2 = 10/11.75 = 0.8511
Answer
cos φ = 0.8511
7.LC resonance, Q factor & bandwidth
Answer: WL = 31.5 nJ
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.
Total energy
½·1 × 10⁻⁹·12² = 7.2 × 10⁻⁸ J
Energy left in the capacitor
½·1 × 10⁻⁹·9² = 4.05 × 10⁻⁸ J
Energy in the inductor
WL = 3.15 × 10⁻⁸ J = 31.5 nJ
Answer
WL = 31.5 nJ
8.LC resonance, Q factor & bandwidth
Answer: C = 19.89 nF
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).
Inductance from the bandwidth
L = R/(2π·BW) = 10/(2π·500) = 0.003183 H
Capacitance from the resonance frequency
C = 1/((2π·20000)²·0.003183) = 1.989 × 10⁻⁸ F = 19.89 nF
Answer
C = 19.89 nF
Q = f₀/BW = 40.
9.Three-phase power
Answer: P = 121.9 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 = √3·690·120 = 143.4 kVA
Real power
P = S·cos φ = 143.4·0.85 = 121.9 kW
Answer
P = 121.9 kW
10.Three-phase power
Answer: Cph = 198.6 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 = 150·tan(acos 0.7) = 150·1.02 = 153 kvar
Q2 = 150·tan(acos 0.92) = 150·0.426 = 63.9 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 89.13 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 / 10 Ω = 4.7 ms
i(∞) = Vs/R = 9/10 = 0.9 A
Stored energy
W(∞) = ½·L·i(∞)² = ½·47·0.9² = 19.04 mJ
Answer
W(∞) = 19.04 mJ
12.RC & RL transients
Answer: vC = 0.2469 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) = 6·1/(6.8 + 1) = 0.7692 V
Time constant after the switch opens
τ = R2·C = 1·22 = 22 ms
Discharge
vC(t) = 0.7692·e−25/22 = 0.2469 V
Answer
vC = 0.2469 V
13.Thévenin & Norton
Answer: IL = 0.6452 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 = 3·12/(3+12) = 2.4 Ω
Rth = (R1 ∥ R2) + R3 = 2.4 + 4 = 6.4 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) = 10·12/(3+12) = 8 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) = 8/(6.4 + 6) = 0.6452 A
14.Thévenin & Norton
Answer: PL = 1.116 W
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 = 25·2/(25+2) = 1.852 Ω
Rth = (R1 ∥ R2) + R3 = 1.852 + 4 = 5.852 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) = 0.7407 + (−1)·1.852 = −1.111 V
Vth = VB = VA + Is·R3 = −1.111 + (−1)·4 = −5.111 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) = −5.111/(5.852 + 6) = −0.4312 A
PL = IL²·RL = −0.4312²·6 = 1.116 W
15.Bode plots & margins
Answer: |H| = −6.986 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.
Corner frequency
ωc = 1/(RC) = 1/(4700·2.2 × 10⁻⁸) = 9671 rad/s
fc = 1539 Hz → f/fc = 770/1539 = 0.5003
Gain
|H| = 0.5003/√(1 + 0.5003²) = 0.4474
20·log₁₀(0.4474) = −6.986 dB
Answer
|H| = −6.986 dB
16.Bode plots & margins
Answer: GM = 3.522 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.