Unit 2: Fundamentals of A.C. circuits
I. Foundations of Alternating-Current Circuits
Alternating-current circuits use voltages and currents whose magnitudes and directions vary periodically with time. Unless stated otherwise, steady-state sinusoidal operation, linear components, constant frequency, and balanced three-phase supplies are assumed.
A. Alternating current and voltage
Alternating current and voltage reverse direction periodically and are commonly represented by sinusoidal waveforms.
- Instantaneous equations: A sinusoidal voltage and current are written as:
v(t) = Vₘ sin(ωt + θᵥ)
i(t) = Iₘ sin(ωt + θᵢ)v(t)andi(t)are instantaneous voltage and current.VₘandIₘare peak values.ωis angular frequency in radians per second.θᵥandθᵢare initial phase angles.
- Frequency and period:
f = 1/T
ω = 2πffis frequency in hertz, andTis period in seconds.- A 50 Hz supply has
T = 1/50 = 0.02 s.
- Defining properties: Sinusoidal AC is periodic, characterized by magnitude, frequency, and phase, and permits phasor-based steady-state analysis.
II. Sinusoidal Quantities and Notation
Sinusoidal signals are described using instantaneous values, representative magnitudes, and angular position.
A. Concept of notations (i, v, I, V)
Letter case distinguishes time-varying quantities from constant-valued AC measures.
- Lowercase notation:
iori(t)andvorv(t)denote instantaneous values at a specified time. - Uppercase notation:
IandVnormally denote RMS values or their complex phasors, according to context. - Peak notation:
Iₘ,Vₘ, or sometimesImax,Vmax, denote maximum magnitudes. - Phasor notation: A phasor may be shown as
V = V∠θᵥ, whereVis RMS magnitude andθᵥis phase angle.
B. Amplitude
Amplitude is the maximum displacement of an alternating quantity from zero.
- Peak value: For
v(t) = Vₘ sin(ωt), the amplitude isVₘvolts. - Peak-to-peak value:
Vpp = 2VₘVppis the difference between positive and negative peaks.- Physical meaning: Amplitude indicates electrical stress; component voltage and current ratings must accommodate peak values, not merely averages.
C. Phase
Phase specifies a sinusoid’s angular position relative to a chosen time origin.
- Phase angle: In
v(t) = Vₘ sin(ωt + θ),θis the phase att = 0. - Leading phase: A positive
θshifts the waveform earlier in time. - Lagging phase: A negative
θshifts it later. - Time-angle conversion:
θ = 360°(Δt/T)Δtis the time displacement andTis the period.
D. Phase difference
Phase difference measures the angular displacement between sinusoids having the same frequency.
- Calculation:
φ = θᵥ − θᵢφis the voltage phase relative to current.- Interpretation:
φ > 0: voltage leads current, as in an inductive circuit.φ < 0: current leads voltage, as in a capacitive circuit.
- Special cases: Signals are in phase at
0°, in quadrature at90°, and in opposition at180°.
- Interpretation:
E. RMS value of an AC signal
The root-mean-square value equals the DC value producing the same heating effect in a resistor.
- General definition:
Xrms = √[(1/T)∫₀ᵀ x²(t) dt]x(t)is the periodic signal,Tits period, andXrmsits RMS value.- Sinusoidal result:
Vrms = Vₘ/√2
Irms = Iₘ/√2- Example: A sinusoid with
Vₘ = 325 VhasVrms ≈ 230 V.
F. Average value of an AC signal
The average value is the arithmetic mean of instantaneous values over a selected interval.
- Complete cycle: A symmetrical sinusoid has zero average because positive and negative half-cycles cancel.
- Half-cycle or rectified average:
Vavg = 2Vₘ/π ≈ 0.637VₘVavgis the mean magnitude over a half-cycle.- Distinction: Average value describes net level, whereas RMS value describes power-producing capability.
III. Impedance and Phasors
Impedance extends resistance to AC circuits by combining opposition in magnitude with phase displacement.
A. Complex representation of impedance
Complex impedance is the phasor ratio of voltage to current.
- Definition:
Z = V/I = R + jX = |Z|∠φZis impedance in ohms,Ris resistance,Xis reactance, andj = √−1.- Magnitude and angle:
|Z| = √(R² + X²)
φ = tan⁻¹(X/R)- Component impedances:
ZR = R
ZL = jωL
ZC = 1/(jωC) = −j/(ωC)Lis inductance in henries andCis capacitance in farads.- Inductive reactance is positive; capacitive reactance is negative.
IV. Steady-State Series-Circuit Analysis
In sinusoidal steady state, phasors convert differential circuit relationships into algebraic impedance equations.
A. Steady-state analysis of RL circuits
A series RL circuit causes current to lag the applied voltage.
- Impedance:
Z = R + jωL
|Z| = √[R² + (ωL)²]
φ = tan⁻¹(ωL/R)- Current:
I = V/Z, so the current phase is−φwhen voltage is the reference. - Voltage relation:
VR = IRis in phase with current, whileVL = IωLleads current by90°.
B. Steady-state analysis of RC circuits
A series RC circuit causes current to lead the applied voltage.
- Impedance:
Z = R − j/(ωC)
|Z| = √[R² + (1/ωC)²]
φ = −tan⁻¹[1/(ωCR)]- Current:
I = V/Z; because the impedance angle is negative, current leads voltage by|φ|. - Voltage relation:
VRis in phase with current, whereasVC = I/(ωC)lags current by90°.
C. Steady-state analysis of series RLC circuits
A series RLC circuit contains resistance and opposing inductive and capacitive reactances.
- Impedance:
Z = R + j(ωL − 1/ωC)- Current and phase:
I = V/|Z|
φ = tan⁻¹[(ωL − 1/ωC)/R]- Operating character:
ωL > 1/(ωC): net inductive; current lags.ωL < 1/(ωC): net capacitive; current leads.- Equal reactances: purely resistive behavior.
D. Resonance in series RLC circuit
Series resonance occurs when inductive and capacitive reactances are equal.
- Resonant condition and frequency:
ω₀L = 1/(ω₀C)
f₀ = 1/(2π√LC)ω₀is resonant angular frequency andf₀is resonant frequency.- At resonance:
Z = R, current is maximum, phase angle is zero, and power factor is unity. - Selectivity:
- At resonance:
Q = ω₀L/R
BW = f₀/QQis quality factor andBWis bandwidth between half-power frequencies.
V. Power and Power Factor
AC power depends on RMS voltage, RMS current, and their phase difference.
A. Power factor and power calculation in RL circuits
An RL circuit has a lagging power factor because current lags voltage.
- Power factor:
pf = cosφ = R/|Z|- Power quantities:
P = VI cosφ
Q = VI sinφ
S = VIPis real power in watts,Qis positive inductive reactive power in vars, andSis apparent power in volt-amperes.- Power triangle:
S² = P² + Q².
- Power triangle:
B. Power factor and power calculation in RC circuits
An RC circuit has a leading power factor because current leads voltage.
- Power factor:
pf = cosφ = R/|Z|; “leading” must accompany the numerical value. - Power: Real power remains
P = VI cosφ. - Reactive power:
Q = VI sinφis negative under the standard sign convention becauseφ < 0. - Apparent power:
S = VIuses RMS values and is always non-negative.
C. Power factor and power calculation in RLC circuits
An RLC circuit may have lagging, leading, or unity power factor.
- Phase and power factor:
φ = tan⁻¹[(ωL − 1/ωC)/R]
pf = cosφ- Classification: Positive net reactance gives lagging power factor; negative net reactance gives leading power factor.
- Power calculation:
P = VI cosφ,Q = VI sinφ, andS = VI. - Resonance: At
φ = 0,pf = 1,Q = 0, andP = VI.
VI. Three-Phase Supply and Interconnection
A three-phase system uses three equal-frequency sinusoidal quantities separated by 120°, enabling nearly constant power transfer and efficient generation and transmission.
A. Three-phase circuits
Three-phase circuits may be balanced or unbalanced and connected in star or delta form.
- Balanced system: The three phase voltages have equal RMS magnitudes and
120°phase displacement. - Phase sequence: The order in which voltages reach positive maxima, commonly R-Y-B or A-B-C, determines motor rotation.
- Balanced power:
P = √3 VL IL cosφVLandILare line voltage and line current;φis the load power-factor angle.
B. Numbering and interconnection of three phases
Three-phase windings require consistent terminal identification and polarity before interconnection.
- Terminal numbering: Corresponding winding starts and finishes may be marked
U1-U2,V1-V2, andW1-W2. - Star interconnection: Three similar ends are joined to form the neutral; the remaining ends connect to line conductors.
- Delta interconnection: The finish of each phase joins the start of the next, forming a closed loop.
- Requirement: Incorrect polarity or phase sequence can produce excessive circulating current or reversed rotation.
C. Delta or mesh connection
A delta connection joins three phase impedances end-to-end in a closed triangular mesh.
- Conductors: A delta system normally uses three line wires and has no neutral point.
- Voltage exposure: Every phase impedance is directly connected across a pair of line conductors.
- Continuity: The closed mesh can permit operation with reduced capacity after one branch is removed, although currents and power become altered.
- Application: Delta is common where loads require full line voltage, including many three-phase motors.
VII. Star and Delta Line–Phase Relations
Line quantities are measured in external conductors, while phase quantities apply to individual source windings or load impedances.
A. Relations between line and phase voltages and currents in star networks
In a balanced star network, each line conductor carries its phase current, while line voltage is the phasor difference of two phase voltages.
- Relations:
VL = √3 Vph
IL = IphVphandIphare phase voltage and phase current.- Angular relation: Each line voltage leads its corresponding phase voltage by
30°for positive phase sequence. - Neutral current: Balanced phase currents sum vectorially to zero, so no neutral current flows.
- Phase impedance:
Zph = Vph/Iph.
- Angular relation: Each line voltage leads its corresponding phase voltage by
B. Relations between line and phase voltages and currents in delta networks
In a balanced delta network, phase voltage equals line voltage, while line current is the phasor difference of adjacent phase currents.
- Relations:
VL = Vph
IL = √3 Iph- Angular relation: Line current is displaced by
30°from the corresponding phase current; the lead or lag description depends on the adopted current directions and phase sequence. - Phase impedance:
Zph = Vph/Iph. - Comparison with star: For the same
VLand phase impedance, a delta-connected load draws three times the line current and consumes three times the power of a star-connected load.
Did this save you a night before the exam?
LPU Notes is free, and it stays free. Ads cover part of the server bill. The rest comes out of a student's own pocket: the domain, the storage, and keeping the site up through the weeks everyone needs it at once.
The payment button didn't load. An ad blocker or a filtered network is the usual reason. to try again.
Nothing here is ever locked, and nothing unlocks. Chip in only if it was worth it. What it pays for →