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 = 3∠110° in rectangular form a + jb and give the real part a.
r = 3θ = 110°
a =
2.Complex numbersHard
Solve z² − 3z + 18.25 = 0. The roots are a ± jb; find b > 0.
p = −3q = 18.25
b =
3.AC impedance & phasorsMedium
A resistor and an inductor are connected in parallel across a 120 V (rms), 60 Hz supply. Find the rms current I.
V = 120 V (rms)f = 60 HzR = 20 ΩL = 100 mH
I = A
4.AC impedance & phasorsHard
A series RLC circuit is connected to a 12 V (rms), 500 Hz supply. Find the rms current I.
V = 12 V (rms)f = 500 HzR = 40 ΩL = 200 mHC = 47 µF
I = A
5.AC power & power factorMedium
A load with impedance Z = 12 + j5 Ω is connected to a 120 V (rms) single-phase supply. Find the reactive power Q it absorbs.
V = 120 VZ = 12 + j5 Ω
Q = var
6.AC power & power factorHard
A 100 kVA single-phase transformer supplies a load of 38 kW at a power factor of 0.75 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
Smax = 100 kVAP1 = 38 kWpf1 = 0.75
ΔP = kW
7.LC resonance, Q factor & bandwidthMedium
A series RLC circuit has R = 5 Ω, L = 470 µH and C = 2.2 µF. Find its quality factor Q.
R = 5 ΩL = 470 µHC = 2.2 µF
Q =
8.LC resonance, Q factor & bandwidthHard
A parallel tank circuit is made of a 10 nF capacitor and a 1 mH coil with a winding resistance of 1 Ω. At resonance it behaves like a resistor, its dynamic resistance RD = L/(C·r). Find RD in kΩ.
L = 1 mHC = 10 nFr = 1 Ω
RD = kΩ
9.Three-phase powerMedium
A three-phase induction motor delivers 11 kW of mechanical output power at full load. At this load its efficiency is 90% and its power factor is 0.85 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.
Pout = 11 kWη = 90%cos φ = 0.85VL = 690 V
IL = A
10.Three-phase powerHard
A balanced three-phase load takes 200 kW at a power factor of 0.75 lagging from a 480 V (line-to-line), 60 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 = 200 kWpf1 = 0.75pf2 = 0.95VL = 480 Vf = 60 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 time constant τ of the circuit after the switch closes.
Vs = 30 VR1 = 3.3 kΩR2 = 2.2 kΩC = 68 µF
τ = ms
12.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 5 V with the polarity shown. How long does it take for vC to reach 8 V? Find t.
Vs = 10 VR = 3.3 kΩC = 22 µFvC(0) = 5 V
t = ms
13.Thévenin & NortonMedium
A load RL = 10 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.
V1 = 18 VR1 = 15 ΩR2 = 5 ΩR3 = 20 ΩRL = 10 Ω
IL = A
14.Thévenin & NortonHard
A load RL = 10 Ω is connected across a–b. Use a Thévenin equivalent to find the power PL delivered to the load.
The RC low-pass filter shown has R = 47 kΩ, C = 100 nF. A sinusoidal input of 140 Hz is applied. Find the phase shift φ of vout relative to vin, in degrees.
R = 47 kΩC = 100 nFf = 140 Hz
φ = °
16.Bode plots & marginsHard
A unity-feedback loop has the open-loop transfer function L(s) = 316/(s(s + 5)(s + 9)). Find the phase crossover frequency ω180, where ∠L(jω) = −180°.
L(s) = 316/(s(s + 5)(s + 9))
ω180 = rad/s
3Solutions
1.Complex numbers
Answer: a = −1.026
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 = 3·cos(110°) = −1.026
b = 3·sin(110°) = 2.819
Rectangular form
z = −1.026 + j2.819
a = −1.026
Answer
a = −1.026
2.Complex numbers
Answer: b = 4
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.
Discriminant
p² − 4q = 9 − 73 = −64 < 0
Quadratic formula
z = (3 ± j√64)/2 = 1.5 + j4, 1.5 − j4
Answer
b = 4
Answer
b = 4
3.AC impedance & phasors
Answer: I = 6.792 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 = 120/20 = 6 A
IX = V/XL = 120/37.7 = 3.183 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) = √(6² + 3.183²) = 6.792 A
Answer
I = 6.792 A
4.AC impedance & phasors
Answer: I = 0.01927 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|.
Reactances
ω = 2πf = 2π·500 = 3142 rad/s
XL = ωL = 3142·0.2 = 628.3 Ω
XC = 1/(ωC) = 1/(3142·47·10⁻⁶) = 6.773 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| = 12/622.8 = 0.01927 A
Answer
I = 0.01927 A
5.AC power & power factor
Answer: Q = 426 var
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.
Current
|Z| = √(12² + 5²) = 13 Ω
I = V/|Z| = 120/13 = 9.231 A
Reactive power
Q = I²·X = 9.231²·5 = 426 var
Answer
Q = 426 var
6.AC power & power factor
Answer: ΔP = 56.22 kW
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 the existing load
Q1 = 38·tan(arccos 0.75) = 33.51 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(100² − 33.51²) = 94.22 kW
Extra real power
ΔP = 94.22 − 38 = 56.22 kW
Answer
ΔP = 56.22 kW
7.LC resonance, Q factor & bandwidth
Answer: Q = 2.923
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).
Characteristic impedance
√(L/C) = √(0.00047/2.2 × 10⁻⁶) = 14.62 Ω
Quality factor
Q = 14.62/5 = 2.923
Answer
Q = 2.923
The resonance frequency is 4.949 kHz, so BW = f₀/Q = 1.693 kHz.
8.LC resonance, Q factor & bandwidth
Answer: RD = 100 kΩ
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).
Dynamic resistance
RD = 0.001/(1 × 10⁻⁸·1) = 100000 Ω = 100 kΩ
Check with Q
Q = (1/r)·√(L/C) = 316.2, Q²·r = 100000 Ω
Answer
RD = 100 kΩ
9.Three-phase power
Answer: IL = 12.03 A
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.
Electrical input power
Pin = Pout/η = 11/0.9 = 12.22 kW
Line current
IL = Pin/(√3·VL·cos φ) = 12222/(√3·690·0.85) = 12.03 A
Answer
IL = 12.03 A
10.Three-phase power
Answer: QC = 110.6 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 = 200·tan(acos 0.75) = 200·0.8819 = 176.4 kvar
Q2 = 200·tan(acos 0.95) = 200·0.3287 = 65.74 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 110.6 kvar
Answer
QC = 110.6 kvar
11.RC & RL transients
Answer: τ = 89.76 ms
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 = 3.3·2.2/(3.3 + 2.2) = 1.32 kΩ
vC(∞) = 30·2.2/(3.3 + 2.2) = 12 V
Time constant
τ = Rth·C = 1.32·68 = 89.76 ms
Answer
τ = 89.76 ms
12.RC & RL transients
Answer: t = 66.52 ms
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.
t = τ·ln((5 − 10)/(8 − 10)) = 72.6·ln(2.5) = 66.52 ms
Answer
t = 66.52 ms
13.Thévenin & Norton
Answer: IL = 0.1333 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 = 15·5/(15+5) = 3.75 Ω
Rth = (R1 ∥ R2) + R3 = 3.75 + 20 = 23.75 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) = 18·5/(15+5) = 4.5 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) = 4.5/(23.75 + 10) = 0.1333 A
14.Thévenin & Norton
Answer: PL = 12.35 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 = 6·4/(6+4) = 2.4 Ω
Rth = (R1 ∥ R2) + R3 = 2.4 + 20 = 22.4 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) = 2.4 + 1.5·2.4 = 6 V
Vth = VB = VA + Is·R3 = 6 + 1.5·20 = 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) = 36/(22.4 + 10) = 1.111 A
PL = IL²·RL = 1.111²·10 = 12.35 W
15.Bode plots & margins
Answer: φ = −76.4 °
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.
Corner frequency
ωc = 1/(RC) = 1/(47000·1 × 10⁻⁷) = 212.8 rad/s
fc = 33.86 Hz → f/fc = 140/33.86 = 4.134
Phase
φ = − atan(4.134) = −76.4°
Answer
φ = −76.4°
16.Bode plots & margins
Answer: ω180 = 6.708 rad/s
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.