⚡ AC Theory & Electromagnetism — HN Engineering Science
How alternating voltages combine, and how resistors, inductors and capacitors behave together in AC circuits. Work through each section, try the interactive tools, then test yourself in the game and practice questions.
3Topics
6Worked Examples
3Live Tools
12Practice Qs
📌 Where this fits: AC theory builds directly on the DC circuit theorems you have already covered — the same ideas of voltage, current and Ohm's Law still apply, but now everything varies sinusoidally with time, so we need phasors to add and compare signals, and reactance to describe how inductors and capacitors resist changing current.
🔀AC Waveforms & PhasorsCombining signals
🔌Series RLCImpedance & power
📡Parallel RLCAdmittance & resonance
⚙️ApplicationsReal-world uses
🎮Quick GameTest your recall
📝Practice12 questions
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1 · AC Waveforms & Phasor Combination
Adding two sinusoidal signals of the same frequency using the phasor method
📜 Rules Banner
v(t) = Vm sin(ωt ± φ)
Vm = peak (maximum) voltage (V) · ω = angular frequency (rad/s) · φ = phase angle relative to the reference (rad)
ω = 2πf
f = supply frequency (Hz) · ω = angular frequency (rad/s)
Phasor form: Vm∠φ
A phasor is a rotating line whose length is the peak value and whose angle is the phase — it freezes a sine wave into a single vector so two signals of the same frequency can be added geometrically
x = Vm cosφ, y = Vm sinφ
x = horizontal (in-phase) component · y = vertical (quadrature) component — used to convert a phasor to rectangular form before adding
VR = √(xtotal² + ytotal²), φR = tan⁻&sup9;(ytotal / xtotal)
VR = resultant peak value · φR = resultant phase angle, found after adding the rectangular components separately
Amplitude (Vm)The peak height of the sine wave above zero, in volts.
PhasorA vector representing a sinusoid's magnitude and phase at a single instant, used for adding AC quantities.
Leading / laggingA signal with a more positive phase angle "leads" — it reaches its peak earlier in time than one with a smaller or negative angle.
ResultantThe single sine wave produced when two or more sinusoids of the same frequency are added together.
Phasor addition (top) and the corresponding time-domain waveforms (bottom) — the resultant wave has its own peak value and phase shift
📝Worked Example 1 — Northgate Automation, sensor conditioning line (50 Hz)
Two conditioned sensor signals on the same 50 Hz line are added by a summing amplifier: v₁(t) = 6.0 sin(100πt + π/3) V and v₂(t) = 4.0 sin(100πt − π/8) V. Find v_total(t).
Step 1Convert v1 to rectangular: x1 = 6.0cos(60°) = 3.000, y1 = 6.0sin(60°) = 5.196
Why? — splitting each phasor into in-phase (x) and quadrature (y) parts lets us add like quantities separately.
Step 2Convert v2 to rectangular: x2 = 4.0cos(−22.5°) = 3.696, y2 = 4.0sin(−22.5°) = −1.531
Step 3Add components: x_total = 3.000 + 3.696 = 6.696, y_total = 5.196 + (−1.531) = 3.665
Step 4Resultant magnitude: V_R = √(6.696² + 3.665²) = √58.27 = 7.63 V
Step 5Resultant phase: φ_R = tan⁻&sup9;(3.665 / 6.696) = 28.7° = 0.501 rad
Resultv_total(t) = 7.63 sin(100πt + 0.501) V — peak 7.63 V, phase +28.7°, leading v1
📝Worked Example 2 — audio-frequency test signal, step-through
A test rig combines two 1000 rad/s signals: v₁(t) = 3.0 sin(1000t + π/4) V and v₂(t) = 2.0 sin(1000t − π/3) V.
Step 1x1 = 3.0cos(45°) = 2.121, y1 = 3.0sin(45°) = 2.121
Step 2x2 = 2.0cos(−60°) = 1.000, y2 = 2.0sin(−60°) = −1.732
Step 3x_total = 2.121 + 1.000 = 3.121, y_total = 2.121 + (−1.732) = 0.389
Step 4V_R = √(3.121² + 0.389²) = 3.15 V
Step 5φ_R = tan⁻&sup9;(0.389 / 3.121) = 7.1° = 0.124 rad
Resultv_total(t) = 3.15 sin(1000t + 0.124) V
✅ Verifying against software: plotting v1(t), v2(t) and their point-by-point sum in Desmos, MATLAB or Python and reading off the peak and the time of that peak should match the phasor-addition result above (allowing for small rounding). A mismatch usually means a sign error in the phase angle or working in degrees where the software expects radians.
🎯Try it yourself — Phasor Addition Calculator
Enter two AC signals of the same frequency and see the resultant, plus a live plot of all three waveforms
🔌
2 · Series RLC Circuits
Reactance, impedance, current, phase angle, power factor and the power triangle
📜 Rules Banner
XL = 2πfL
XL = inductive reactance (Ω) · f = supply frequency (Hz) · L = inductance (H)
XC = 1 / (2πfC)
XC = capacitive reactance (Ω) · C = capacitance (F)
X = XL − XC
X = net reactance (Ω) — positive means the circuit is net inductive, negative means net capacitive
Z = √(R² + X²)
Z = impedance (Ω) · R = resistance (Ω)
I = V / Z, φ = tan⁻&sup9;(X / R)
I = rms current (A) · V = rms supply voltage (V) · φ = phase angle between V and I
P = VI cosφ, Q = VI sinφ, S = VI
P = true power (W) · Q = reactive power (VAr) · S = apparent power (VA) · cosφ = power factor
fr = 1 / (2π√(LC))
fr = resonant frequency (Hz), the frequency point at which XL = XC
Q = XL / R (at fr), BW = fr / Q
Q = quality factor (how sharp the resonance peak is) · BW = bandwidth (Hz), the range of frequencies passed at half power
fL = fr − BW/2, fH = fr + BW/2
fL, fH = lower and upper −3dB (half-power) cut-off frequencies either side of fr
ReactanceThe opposition an inductor or capacitor offers to AC current, in ohms — frequency-dependent, unlike resistance.
Impedance (Z)The combined opposition of resistance and reactance in an AC circuit.
Power factorcosφ — the fraction of apparent power that does useful work; ranges from 0 to 1.
ResonanceThe frequency fr where XL and XC cancel, leaving the circuit purely resistive.
Dynamic impedance (Zd)The circuit's impedance at resonance — in a series circuit this is at its minimum and equal to R.
Quality factor (Q)The sharpness of the resonance peak; also the voltage magnification — how many times larger VL and VC are than the supply voltage at resonance.
IEC-style series RLC circuit diagram, labelled with V_R, V_L and V_C
📝Worked Example — Northgate Automation control panel supply
A control panel branch is supplied at 230 V rms, 50 Hz and contains R = 18 Ω, L = 95 mH, C = 65 µF in series. Find Z, I, φ, power factor, P, Q, S and the resonant frequency.
X_LX_L = 2π(50)(0.095) = 29.85 Ω
X_CX_C = 1 / (2π(50)(0.000065)) = 48.97 Ω
Net XX = 29.85 − 48.97 = −19.13 Ω (negative ⇒ net capacitive)
ZZ = √(18² + (−19.13)²) = √689.8 = 26.26 Ω
CurrentI = 230 / 26.26 = 8.76 A rms
Phase φφ = tan⁻&sup9;(−19.13 / 18) = −46.7° (current leads voltage ⇒ capacitive circuit)
PFcosφ = cos(46.7°) = 0.685 leading
P, Q, SP = 230(8.76)(0.685) = 1380 W · Q = 230(8.76)(sin46.7°) = 1467 VAr (capacitive) · S = 230(8.76) = 2014 VA
f_rf_r = 1 / (2π√(0.095 × 0.000065)) = 64.0 Hz
🔄Resonance in this circuit — the Acceptor Circuit
A series RLC circuit is called an acceptor circuit: at f_r its impedance is at its minimum (Zd = R), so it easily "accepts" current at that one frequency while opposing others more strongly. Using the same R, L, C as above:
X_L at f_rX_L = 2π(64.0)(0.095) = 38.2 Ω (this equals X_C at resonance, by definition)
Z_dZ_d = R = 18 Ω (the dynamic impedance — the circuit's minimum possible impedance)
I at f_rI = 230 / 18 = 12.78 A (the maximum current this circuit can draw from a 230 V supply, at any frequency)
Q factorQ = X_L / R = 38.2 / 18 = 2.12
V_L, V_C at f_rV_L = I × X_L = 12.78 × 38.2 = 488.5 V — over twice the 230 V supply, but V_L and V_C are equal and opposite so they cancel, leaving V_R = V_S
BandwidthBW = f_r / Q = 64.0 / 2.12 = 30.2 Hz
f_L, f_Hf_L = 64.0 − 15.1 = 48.9 Hz, f_H = 64.0 + 15.1 = 79.1 Hz
⚠️ Why this matters for design: the low dynamic impedance at resonance means current can rise sharply, and the reactive voltages V_L and V_C can each exceed the supply voltage several times over (here 488 V from a 230 V supply). A series-resonant filter or supply branch must be rated for these internal voltages and currents, not just the applied supply voltage.
DiagramImpedance triangle
DiagramPower triangle
✅ At resonance (f = f_r): X_L = X_C so the net reactance X = 0, impedance falls to its minimum value Z = Z_d = R, and current rises to its maximum I = V/R for that supply voltage. Below f_r the circuit is net capacitive (X_C dominates, leading power factor); above f_r it is net inductive (X_L dominates, lagging power factor).
🎯Try it yourself — Series RLC Calculator
Enter your own R, L, C, supply voltage and frequency
📡
3 · Parallel RLC Circuits & Resonance
Admittance, branch currents, total current, and how resonance behaves as the dual of the series case
📜 Rules Banner
G = 1 / R, BL = 1 / XL, BC = 1 / XC
G = conductance (S) · BL, BC = inductive / capacitive susceptance (S) — the parallel-circuit equivalents of reactance
B = BC − BL
B = net susceptance (S) — positive means net capacitive, negative means net inductive (opposite sign convention to series X)
Y = √(G² + B²)
Y = admittance (S, siemens) — the parallel-circuit equivalent of impedance
IT = V × Y
IT = total rms line current (A), the phasor sum of IR, IL and IC · V = rms supply voltage (V), common to every branch
fr = 1 / (2π√(LC))
Same formula as the series case for an ideal (resistanceless-branch) parallel LC combination
Zd = R, Q = R / XL (at fr)
Zd = dynamic impedance, the circuit's maximum impedance, reached at resonance · Q here is the inverse of the series formula — it measures current magnification
Imag = Q × IT
Imag = current magnification — how much larger the circulating tank current (IL or IC) is than the supply current at resonance
Admittance (Y)How easily a circuit admits current — the reciprocal of impedance, measured in siemens (S).
Susceptance (B)The reciprocal of reactance — the parallel-circuit measure of how an L or C branch opposes current.
Dual behaviourA parallel RLC circuit is the mirror image of a series one: impedance is maximum and line current is minimum at resonance, the opposite of the series case.
Tank circuitThe L–C pair on their own: at resonance a large current circulates between them (IL = −IC) while drawing almost nothing from the supply.
IEC-style parallel RLC circuit diagram, labelled with I_R, I_L, I_C and I_T
📝Worked Example — same Northgate Automation branch, reconfigured in parallel
The same components (R = 18 Ω, L = 95 mH, C = 65 µF) are reconnected in parallel across the same 230 V, 50 Hz supply, using the reactances already found in Section 2 (X_L = 29.85 Ω, X_C = 48.97 Ω).
GG = 1 / 18 = 0.0556 S
B_LB_L = 1 / 29.85 = 0.0335 S
B_CB_C = 1 / 48.97 = 0.0204 S
Net BB = 0.0204 − 0.0335 = −0.0131 S (negative ⇒ net inductive)
YY = √(0.0556² + 0.0131²) = 0.0571 S
Total currentI = 230 × 0.0571 = 13.13 A rms
Resonancef_r is unchanged at 64.0 Hz — it depends only on L and C, not on the arrangement
⚠︉ Compare the two arrangements: at 50 Hz the series version drew 8.76 A while the parallel version draws 13.13 A from the same supply and same components — the arrangement changes how the circuit behaves even though nothing else changed. At resonance, though, the series circuit's current would peak while the parallel circuit's line current would dip to its minimum (ideally just G × V, since the reactive branches cancel).
🔄Resonance in this circuit — the Rejecter Circuit
A parallel RLC circuit is called a rejecter circuit (or anti-resonant circuit): at f_r its impedance is at its maximum, so it strongly opposes — "rejects" — current at that one frequency, unlike the series acceptor circuit.
X_L at f_rX_L = 2π(64.0)(0.095) = 38.2 Ω (equal to X_C at resonance)
SusceptanceAt f_r, B_L = B_C so the net susceptance B = 0 and Y falls to its minimum, Y = G
Z_dZ_d = R = 18 Ω (the dynamic impedance — here it is the circuit's maximum possible impedance)
I_T at f_rI_T = V / R = 230 / 18 = 12.78 A (the minimum line current this circuit can draw, in phase with V_S)
Q factorQ = R / X_L = 18 / 38.2 = 0.47
Current magnificationI_mag = Q × I_T = 0.47 × 12.78 = 6.0 A circulating in the L–C tank
BandwidthBW = f_r / Q = 64.0 / 0.47 = 136.0 Hz (a low Q gives a wide, unselective bandwidth here)
💡 Reading the Q factor: Q < 1 here because R (18 Ω) is small next to X_L (38.2 Ω) — a lot of current leaks away through the resistor branch instead of circulating in the tank, so the resonance is broad and the current magnification is modest. A parallel tank designed for sharp tuning (e.g. a radio selector) uses a much larger R relative to X_L so that Q ≫ 1 and I_mag is many times I_T.
🎯Try it yourself — Parallel RLC Calculator
Enter your own R, L, C, supply voltage and frequency
⚙️
Applications
Where AC phasor addition and RLC analysis are used in real engineering practice
🔌 Power factor correction on an industrial supply
A factory's motors and transformers present an inductive load, giving a lagging power factor and pushing up the apparent power (and the supplier's kVA charge) for the same useful work. Engineers add capacitor banks in parallel to supply the reactive power locally, cancelling much of the inductive B and pulling the overall power factor closer to 1.
Series/parallel RLC analysis tells the engineer exactly how much capacitance to add to bring Q close to zero without over-correcting into a leading (capacitive) power factor.
🔊 Radio and audio tuning circuits
A radio receiver's tuning stage is a resonant RLC circuit. Turning the tuning dial changes C (or L), which shifts f_r = 1/(2π√(LC)) so that only the wanted station's frequency produces a large current (acceptor/series) or high impedance (rejecter/parallel), while other frequencies are rejected.
The same resonance mathematics used in Section 3 selects which frequency the circuit responds to.
📡 Signal summing in instrumentation
Where several AC sensor outputs of the same frequency (e.g. from a rotating machine's vibration sensors) are combined by an analogue summing stage, phasor addition predicts the resultant amplitude and phase — useful for checking that signals reinforce rather than partially cancel due to phase misalignment.
This is exactly the Section 1 method: convert each signal to rectangular form, add, convert back to polar.
⚡ Electric motor supply branch design
A motor branch behaves as a series R-L circuit (plus any correction capacitance). Calculating Z, I and φ lets an engineer size the supply cable, the protective device and any power-factor correction capacitor correctly for the actual current the motor will draw, not just its rated power.
Undersizing based on P alone (ignoring Q and S) is a common real-world design error this analysis avoids.
🎮
Quick-Fire Game
Randomised questions on phasors, series and parallel RLC circuits — keep your streak going
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Practice Questions
12 questions covering phasor addition, series RLC and parallel RLC — enter your answer and check
Question 1 · Phasor addition
Two signals on the same line: v1 = 5.0∠20° V and v2 = 3.0∠−50° V. Find the resultant peak value V_R (V), to 2 d.p.
Show working
x_total = 5cos20° + 3cos(−50°) = 4.698 + 1.928 = 6.627
y_total = 5sin20° + 3sin(−50°) = 1.710 − 2.298 = −0.588
V_R = √(6.627² + 0.588²) = 6.65 V
y_total = 5sin20° + 3sin(−50°) = 1.710 − 2.298 = −0.588
V_R = √(6.627² + 0.588²) = 6.65 V
Question 2 · Phasor addition
Using the same two signals as Question 1, find the resultant phase angle φ_R in degrees (to 1 d.p., include the minus sign if negative).
Show working
φ_R = tan⁻&sup9;(−0.588 / 6.627) = −5.1°
Question 3 · Inductive reactance
A 200 mH inductor is supplied at 50 Hz. Find X_L (Ω), to 1 d.p.
Show working
X_L = 2π(50)(0.2) = 62.8 Ω
Question 4 · Capacitive reactance
A 40 µF capacitor is supplied at 50 Hz. Find X_C (Ω), to 1 d.p.
Show working
X_C = 1 / (2π(50)(0.00004)) = 79.6 Ω
Question 5 · Series impedance
A series circuit has R = 25 Ω and net reactance X = −17 Ω. Find the impedance Z (Ω), to 1 d.p.
Show working
Z = √(25² + (−17)²) = √914 = 30.2 Ω
Question 6 · Series current
The circuit in Question 5 is supplied at 230 V rms. Find the current I (A), to 1 d.p.
Show working
I = 230 / 30.2 = 7.6 A
Question 7 · Phase angle & circuit type
For the circuit in Questions 5–6 (R = 25 Ω, X = −17 Ω), find the phase angle φ in degrees (to 1 d.p., include the sign).
Show working
φ = tan⁻&sup9;(−17 / 25) = −34.2° — negative X means the circuit is net capacitive, current leads voltage
Question 8 · Power factor
For φ = −34.2°, find the power factor cosφ, to 2 d.p.
Show working
cos(34.2°) = 0.83 (leading, since the circuit is capacitive)
Question 9 · True power
Using V = 230 V, I = 7.6 A and PF = 0.83 from Questions 6 and 8, find the true power P (W), to the nearest watt.
Show working
P = 230 × 7.6 × 0.83 ≈ 1447 W
Question 10 · Resonant frequency
For L = 200 mH and C = 40 µF, find the resonant frequency f_r (Hz), to 1 d.p.
Show working
f_r = 1 / (2π√(0.2 × 0.00004)) = 56.3 Hz
Question 11 · Parallel net susceptance
A parallel branch has X_L = 40 Ω and X_C = 25 Ω. Find the net susceptance B (S), to 3 d.p. (state as a positive number; the branch is net capacitive).
Show working
B_L = 1/40 = 0.025 S, B_C = 1/25 = 0.04 S ⇒ B = B_C − B_L = 0.015 S
Question 12 · Parallel total current
The parallel branch from Question 11 also has R = 30 Ω and is supplied at 230 V rms. Find the total line current I (A), to 1 d.p.
Show working
G = 1/30 = 0.0333 S ⇒ Y = √(0.0333² + 0.015²) = 0.0366 S ⇒ I = 230 × 0.0366 = 8.4 A