KCL, KVL, Ohm's law, and the passive sign convention — the bookkeeping that turns any resistive circuit into a solvable set of linear equations.
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B. Series/parallel equivalent circuits
Series-Parallel Equivalents and Dividers
Collapse resistor networks with series/parallel rules, read off voltages and currents with the divider formulas, and convert sources and delta-wye to finish stubborn topologies.
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Reduce any linear two-terminal network to a single source and resistance, swap between Thevenin and Norton forms, and apply superposition and maximum power transfer.
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D. Node and loop analysis
Nodal and Mesh Analysis
Two systematic methods that turn any resistive or AC circuit into a small linear system you can solve fast under exam pressure.
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E. Waveform analysis
Waveform Analysis: RMS and Average
Period, frequency, average, and RMS for the common waveforms — plus first-order RC/RL transients — so you can read any signal the exam throws at you.
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F. Phasors and impedance
Phasors, Impedance, and AC Power
Phasor analysis, R/L/C impedance, series-parallel combination, and the P/Q/S power triangle — the core AC toolkit for the FE-ECE exam.
Phasors turn calculus into algebra: a sinusoid at a fixed frequency becomes a single complex number, and every derivative in a circuit's differential equations becomes a multiplication by jω. That is the whole reason AC analysis is tractable, and the FE Reference Handbook's AC Circuits section gives you the phasor transforms, the impedance table, and the complex-power definitions. The points are lost in the bookkeeping — degrees versus radians, peak versus RMS, the sign of reactive power — so this concept builds the fluency to move between time domain, phasor domain, and the power triangle without slipping.
Phasor representation
A sinusoid x(t)=Xmaxcos(ωt+ϕ) maps to the phasor X=X∠ϕ, a complex number carrying magnitude and phase at the single frequency ω. The FE handbook lets the magnitude be either peak or RMS depending on context — for power work, use RMS magnitudes. Convert freely between polar X∠ϕ and rectangular Xcosϕ+jXsinϕ: addition is easy in rectangular form, multiplication and division easy in polar form.
x(t)=Xmaxcos(ωt+ϕ)⟷X=X∠ϕ
Impedance of R, L, and C
Impedance Z=V/I is the phasor generalization of resistance, a complex number Z=R+jX with resistance R and reactance X
Series and parallel impedance
Impedances combine exactly like resistances. In series they add; in parallel the reciprocals add (admittances add). Admittance Y=1/Z=G+jB has conductance G and susceptance B in siemens, and is the natural quantity for parallel branches and nodal analysis. For two impedances in parallel, the product-over-sum shortcut applies, just as with resistors.
Zseries=k∑Zk,Zpar1=k∑Zk1=k∑Yk
AC power: real, reactive, apparent
With RMS phasors, real power P (watts) is the average power actually consumed, reactive power Q (vars) is the power that sloshes in and out of inductors and capacitors, and apparent power S (volt-amperes) is their magnitude. The angle θ is measured from voltage to current and equals the impedance angle. Inductive (lagging) loads absorb positive Q; capacitive (leading) loads supply it, giving negative Q
Complex power and the power triangle
Complex power packages everything: S=VI∗=P+jQ, where I∗ is the conjugate of the current phasor. Its magnitude is S=P2+Q2
Power-factor correction
A poor (low) power factor means large current for the same real power, so utilities penalize it and engineers correct it. Adding a capacitor in parallel supplies leading reactive power that cancels part of an inductive load's lagging Q, raising the power factor toward unity without changing P. The required capacitor reactive power is the difference between the original and target Q, both computed at the same real power.
QC=P(tanθ1−tanθ2)
Exam strategy
Fix your magnitude convention first: use RMS phasors so power formulas need no factor of 21. Build Z=R+jX in rectangular form, convert to polar to get I=V/Z
Key equations
Phasor transformXmaxcos(ωt+ϕ)↔X∠ϕ
Maps a sinusoid at frequency ω
Worked examples
Series RL load: current and power
Problem. A series circuit of R=30Ω and an inductor with reactance XL=40Ω is driven by 120V
Common pitfalls
•Mixing peak and RMS phasor magnitudes. With peak magnitudes, power formulas need a factor of 21 (P=21VmaxImaxcosθ
References
NCEES FE Reference Handbook — Electrical and Computer Engineering: AC Circuits (Phasor Transforms, Impedance, Complex Power) — Source of the impedance table, phasor transforms, and the P/Q/S complex-power definitions.
NCEES FE Reference Handbook — Mathematics: Complex Numbers / Euler's Identity — Polar-rectangular conversion and complex arithmetic underlying phasor analysis.
Hayt, Kemmerly & Durbin, Engineering Circuit Analysis — Sinusoidal steady-state analysis, impedance, and AC power.
in ohms. A resistor is purely real,
ZR=R
. An inductor has positive reactance
ZL=jωL
— current lags voltage by
90∘
. A capacitor has negative reactance
ZC=1/(jωC)=−j/(ωC)
— current leads voltage by
90∘
. The mnemonic ELI the ICE man captures it: in an inductor (L) voltage E leads current I; in a capacitor (C) current I leads voltage E.
ZR=R,ZL=jωL,ZC=jωC1=−ωCj
.
P=VrmsIrmscosθ,Q=VrmsIrmssinθ,S=VrmsIrms
and its angle is the impedance angle
θ
, forming the right-triangle relationship below. The power factor is
pf=cosθ=P/S
: lagging for inductive loads (current lags voltage), leading for capacitive. Equivalent forms
S=Irms2Z=Vrms2/Z∗
let you compute power directly from impedance.
S=VI∗=P+jQ,S=P2+Q2,pf=cosθ=SP
, and read
θ
straight off the impedance angle. For power, anchor on the triangle
S=P2+Q2
and remember the sign: inductive/lagging gives
+Q
, capacitive/leading gives
−Q
. Compute
P
from
I2R
as an independent check on
Scosθ
. For correction problems, work entirely with
Q
at constant
P
. Sanity-check that
S≥P
always, and that pf lies between 0 and 1.
to a complex phasor. Magnitude is peak or RMS by context; use RMS for power.
ImpedanceZ=IV=R+jX
Ratio of voltage to current phasors. R = resistance, X = reactance, both in ohms.
Inductor / capacitor reactanceXL=ωL,XC=ωC1
ZL=+jXL (current lags 90∘); ZC=−jXC
AdmittanceY=Z1=G+jB
Reciprocal of impedance. G = conductance, B = susceptance, in siemens. Adds in parallel.
Series / parallel impedanceZs=∑Zk,Zp=Z1+Z2Z1Z2
Impedances add in series; product-over-sum for two in parallel (or sum admittances).
Real powerP=VrmsIrmscosθ=Irms2R
Average power dissipated (W). θ = angle from voltage to current = impedance angle.
Reactive powerQ=VrmsIrmssinθ=Irms2X
Reactive power (var). Positive for inductive (lagging), negative for capacitive (leading) loads.
Apparent powerS=VrmsIrms=P2+Q2
Magnitude of complex power (VA). Always S≥P.
Complex powerS=VI∗=P+jQ
I∗ = conjugate of current phasor. Real part is P, imaginary part is Q.
Power factorpf=cosθ=SP
Lagging (inductive) when current lags voltage; leading (capacitive) when it leads. Dimensionless, 0 to 1.
Power from impedanceS=Irms2Z=Z∗Vrms2
Direct complex power from load impedance; convenient when only Z and one of V or I is known.
Power-factor correctionQC=P(tanθ1−tanθ2)
Capacitor reactive power (kvar) to move from θ1 to a better θ2 at fixed real power P.
RMS at
0∘
. Find the current, power factor, and the real, reactive, and apparent power.
Solution. Impedance: Z=30+j40=50∠53.1∘Ω.
Current: I=120/50=2.40A at −53.1∘ (lagging).
Power factor: cos53.1∘=0.600 lagging.
S=VI=120(2.40)=288VA; P=Scosθ=288(0.6)=173W; Q=Ssinθ=288(0.8)=230var (lagging).
Sanity: P=I2R=2.402(30)=173W and Q=I2X=2.402(40)=230var — both match the triangle results.
Z=30+j40=50∠53.1∘Ω
Parallel R and C
Problem. A 50Ω resistor is in parallel with a capacitor whose reactance is 50Ω, driven by 100V RMS. Find the equivalent impedance, the total current, and the real and reactive power.
Solution. Admittances: YR=1/50=0.0200S, YC=+j/50=j0.0200S (capacitor susceptance is positive).
Y=0.0200+j0.0200=0.02828∠45∘S, so Z=1/Y=35.4∠−45∘Ω.
Current: I=VY=100(0.02828)=2.83A at +45∘ (leading).
S=VI=100(2.83)=283VA; P=283cos45∘=200W; Q=−283sin45∘=−200var (capacitive, supplied).
Sanity: P=V2/R=1002/50=200W and ∣Q∣=V2/XC=1002/50=200var — both confirm.
Z=Y1=35.4∠−45∘Ω
Power-factor correction
Problem. A single-phase load draws 10kW at 0.70 power factor lagging from a 480V source. Find the apparent power and reactive power, then the capacitor kvar needed to correct the power factor to 0.95 lagging.
Solution. S1=P/pf=10/0.70=14.3kVA.
θ1=cos−10.70=45.6∘, so Q1=Ptanθ1=10tan45.6∘=10.2kvar (lagging).
Target: θ2=cos−10.95=18.2∘, Q2=10tan18.2∘=3.29kvar.
Capacitor: QC=Q1−Q2=10.2−3.29=6.92kvar.
Sanity: real power is unchanged at 10kW; new apparent power is 10/0.95=10.5kVA, below the original 14.3kVA, so current (and losses) drop as expected.
QC=P(tanθ1−tanθ2)
); with RMS they do not. Choose one convention and stay in it.
•Capacitor reactance sign: ZC=−j/(ωC) is negative imaginary. Writing +jXC flips current from leading to lagging and breaks the power factor.
•Calling the power factor leading or lagging backwards. Lagging = current lags voltage = inductive = +Q; leading = capacitive = −Q. Tie it to the sign of Q, not guesswork.
•Using θ as the source-voltage angle instead of the angle from voltage to current. θ in the power formulas is the impedance angle, ∠Z.
•Adding impedances in parallel directly. Impedances add in series; in parallel add the admittances (or use product-over-sum), then invert.
•Forgetting ω=2πf when computing reactance. XL=ωL=2πfL, not fL; a factor of 2π error scales every reactance.
•In power-factor correction, subtracting apparent powers. Work with reactive power Q at constant P — S values do not subtract because their angles differ.