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
Given z₁ = 3∠70° and z₂ = 4∠−140°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 3∠70°z₂ = 4∠−140°
θ = °
2.Complex numbersHard
One root of z³ + 7z² + 9z − 17 = 0 is z₁ = −4 + j. The coefficients are real. Find the real root r.
z₁ = −4 + j
r =
3.AC impedance & phasorsMedium
A series RLC circuit with L = 1 mH must resonate at 5000 Hz. Find the capacitance C needed.
f0 = 5000 HzL = 1 mH
C = µF
4.AC impedance & phasorsHard
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 24 V (rms), 200 Hz supply. Find the rms current I.
V = 24 V (rms)f = 200 HzR1 = 25 ΩR2 = 80 ΩC = 22 µF
I = A
5.AC power & power factorMedium
Two loads are connected in parallel to a 230 V single-phase supply: a 25 kW load at a power factor of 0.8 lagging and a 27 kW load at a power factor of 0.8 lagging. Find the overall power factor cos φ seen by the supply.
A single-phase load takes 18 kW at a power factor of 0.65 lagging from a 240 V, 60 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 18 kWpf1 = 0.65V = 240 Vf = 60 HzC = 100 µF
I = A
7.LC resonance, Q factor & bandwidthMedium
In an ideal LC circuit with L = 0.1 µH and C = 2.2 µF, 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 = 0.1 µHC = 2.2 µF
t = µs
8.LC resonance, Q factor & bandwidthHard
Design a series RLC filter with a total resistance of 100 Ω, a resonance frequency of 500 kHz and a bandwidth of 10 kHz. Find the capacitance C in nF.
R = 100 Ωf0 = 500 kHzBW = 10 kHz
C = nF
9.Three-phase powerMedium
A balanced Y-connected load with an impedance of 3 + j4 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 = 3 + j4 Ω per phase
P = kW
10.Three-phase powerHard
A balanced three-phase load takes 80 kW at a power factor of 0.72 lagging from a 400 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 Δ. Find the capacitance Cph of each capacitor in µF.
P = 80 kWpf1 = 0.72pf2 = 0.9VL = 400 Vf = 50 Hz
Cph = µF
11.RC & RL transientsMedium
The inductor carries no current before the switch closes at t = 0. Find the current i at t = 1 ms.
Vs = 5 VR = 47 ΩL = 10 mHt = 1 ms
i = A
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 = 100 ms.
Vs = 20 VR1 = 10 kΩR2 = 4.7 kΩC = 68 µFt = 100 ms
vC = V
13.Thévenin & NortonMedium
A load RL = 2 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.
V1 = 30 VR1 = 10 ΩR2 = 12 ΩR3 = 3 ΩRL = 2 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 = 10 VIs = 0.5 AR1 = 8 ΩR2 = 30 ΩR3 = 30 Ω
IN = A
15.Bode plots & marginsMedium
In the series RLC band-pass filter shown the output is taken across R = 20 Ω, with L = 5 mH and C = 0.1 µF. Find the −3 dB bandwidth BW in Hz.
R = 20 ΩL = 5 mHC = 0.1 µF
BW = Hz
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = K/(s(s + 3)(s + 9)). Find the gain K that gives a gain margin of 6 dB.
L(s) = K/(s(s + 3)(s + 9))GM = 6 dB
K =
3Solutions
1.Complex numbers
Answer: θ = −150 °
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 and argument of w
|w| = 3 / 4 = 0.75
θ = 70° − (−140)° = 210° → −150°
Argument
θ = −150°
Answer
θ = −150°
2.Complex numbers
Answer: r = 1
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.
Check with the sum of the roots: z₁ + z₂ + r = −8 + 1 = −7 = −(7).
3.AC impedance & phasors
Answer: C = 1.013 µF
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 condition
At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).
ω₀ = 2π·5000 = 31416 rad/s
C = 1/(ω₀²·L) = 1/(31416²·0.001) = 1.013 µF
Answer
C = 1.013 µF
4.AC impedance & phasors
Answer: I = 0.4909 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
XC = 1/(2π·200·22·10⁻⁶) = 36.17 Ω
Parallel combination
Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.
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 = 25·tan(arccos 0.8) = 18.75 kvar
Q2 = 27·tan(arccos 0.8) = 20.25 kvar
Totals
P = 25 + 27 = 52 kW
Q = 18.75 + 20.25 = 39 kvar
S = √(52² + 39²) = 65 kVA
Overall power factor
cos φ = P/S = 52/65 = 0.8 lagging
Answer
cos φ = 0.8
6.AC power & power factor
Answer: I = 108.7 A
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 = 18·tan(arccos 0.65) = 21.04 kvar
QC = V²·2πf·C = 240²·2π·60·100·10⁻⁶ = 2171 var = 2.171 kvar
Q2 = 21.04 − 2.171 = 18.87 kvar
New apparent power
S2 = √(18² + 18.87²) = 26.08 kVA
New line current
I = S2/V = 26080/240 = 108.7 A
Answer
I = 108.7 A
7.LC resonance, Q factor & bandwidth
Answer: t = 0.7368 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√(1 × 10⁻⁷·2.2 × 10⁻⁶) = 2.947 µs
Quarter period
t = T/4 = 0.7368 µs
Answer
t = 0.7368 µs
8.LC resonance, Q factor & bandwidth
Answer: C = 0.06366 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) = 100/(2π·10000) = 0.001592 H
Capacitance from the resonance frequency
C = 1/((2π·500000)²·0.001592) = 6.366 × 10⁻¹¹ F = 0.06366 nF
Answer
C = 0.06366 nF
Q = f₀/BW = 50.
9.Three-phase power
Answer: P = 19.2 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| = √(3² + 4²) = 5 Ω
Iph = 230.9/5 = 46.19 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·46.19²·3 = 19.2 kW
Answer
P = 19.2 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
10.Three-phase power
Answer: Cph = 254.4 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 = 80·tan(acos 0.72) = 80·0.9639 = 77.11 kvar
Q2 = 80·tan(acos 0.9) = 80·0.4843 = 38.75 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 38.36 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.
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/(10 + 4.7) = 6.395 V
Time constant after the switch opens
τ = R2·C = 4.7·68 = 319.6 ms
Discharge
vC(t) = 6.395·e−100/319.6 = 4.677 V
Answer
vC = 4.677 V
13.Thévenin & Norton
Answer: IL = 1.565 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 = 10·12/(10+12) = 5.455 Ω
Rth = (R1 ∥ R2) + R3 = 5.455 + 3 = 8.455 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) = 30·12/(10+12) = 16.36 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) = 16.36/(8.455 + 2) = 1.565 A
14.Thévenin & Norton
Answer: IN = 0.7174 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 = 8·30/(8+30) = 6.316 Ω
Rth = (R1 ∥ R2) + R3 = 6.316 + 30 = 36.32 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) = 7.895 + 0.5·6.316 = 11.05 V
Vth = VB = VA + Is·R3 = 11.05 + 0.5·30 = 26.05 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 = 26.05/36.32 = 0.7174 A
15.Bode plots & margins
Answer: BW = 636.6 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/√(0.005·1 × 10⁻⁷) = 44721 rad/s
f₀ = ω₀/(2π) = 7118 Hz
Bandwidth
BW = R/L = 20/0.005 = 4000 rad/s = 636.6 Hz
Answer
BW = 636.6 Hz
16.Bode plots & margins
Answer: K = 162.4
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.