Basic Electrical Engineering

6% of exam

Charge, current and voltage laws, DC and AC circuits, reactance and impedance, three-phase power, and measuring devices.

4 concepts

Exam areas

B. Current and voltage laws

Electrical Fundamentals: Ohm's and Kirchhoff's Laws

Charge, current, voltage, resistance, and power, then the two conservation laws (KVL, KCL) and the divider shortcuts that solve almost every resistive circuit on the exam.

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C. AC and DC circuits

AC and DC Circuit Analysis

Phasors, reactance, and impedance turn AC steady-state into the same algebra as DC, then the power triangle and series resonance close out the AC and DC circuits topic.

DC analysis asks only for steady values; AC analysis adds time-varying sinusoids, and the trick that tames them is the phasor. By representing a sinusoid of fixed frequency as a single complex number (magnitude and phase), every capacitor and inductor becomes a complex 'resistance' called impedance, and Ohm's law, KVL, KCL, and the divider rules all carry over unchanged — now with complex arithmetic. The FE Other Disciplines exam leans on this heavily: impedance of RLC combinations, the real/reactive/apparent power triangle, power factor, and the one special frequency (resonance) where an RLC circuit looks purely resistive. The handbook's Electrical and Computer Engineering chapter lists every relation here; the skill is moving fluently between rectangular and polar form.

Sinusoids, RMS, and phasors

A sinusoid

v(t)=V_{\max}\cos(\omega t+\phi)

has angular frequency

\omega=2\pi f

. The value that matters for power and for meter readings is the RMS (effective) value: for any sinusoid

V_{\text{rms}}=V_{\max}/\sqrt{2}

. The phasor strips away the

\cos(\omega t)

carrier and keeps the magnitude and phase, conventionally as an RMS magnitude:

\mathbf{V}=V_{\text{rms}}\angle\phi

. All phasors in one problem rotate at the same

\omega

, so you can do algebra on the frozen snapshots.

v(t)=V_{\max}\cos(\omega t+\phi) \;\longleftrightarrow\; \mathbf{V}=V_{\text{rms}}\angle\phi, \qquad V_{\text{rms}}=\frac{V_{\max}}{\sqrt{2}}

Reactance and impedance

Impedance

\mathbf{Z}=R+jX

is the ratio of phasor voltage to phasor current, in ohms. A resistor has

\mathbf{Z}=R

(current in phase with voltage). An inductor opposes change in current and its reactance rises with frequency,

X_L=\omega L

, with current lagging voltage by

90^\circ

. A capacitor opposes change in voltage and its reactance falls with frequency,

X_C=1/(\omega C)

, with current leading voltage by

90^\circ

. The memory hook is ELI the ICE man: in an inductor (L) voltage E leads current I; in a capacitor (C) current I leads voltage E.

\mathbf{Z}_R=R, \qquad \mathbf{Z}_L=j\omega L, \qquad \mathbf{Z}_C=\frac{1}{j\omega C}=-\frac{j}{\omega C}

Combining impedances and Ohm's law for AC

Impedances combine exactly like resistances: in series they add, in parallel the reciprocals add. Once you have the total impedance, the phasor current is the source voltage divided by it, and voltage/current dividers work with impedances in place of resistances. Do series additions in rectangular form (

R+jX

) and divisions in polar form (

|Z|\angle\theta

) — converting between the two is the bulk of the arithmetic.

\mathbf{Z}_{\text{ser}}=\sum_k \mathbf{Z}_k, \quad \mathbf{Z}_{\text{par}}=\left(\sum_k\frac{1}{\mathbf{Z}_k}\right)^{-1}, \quad \mathbf{I}=\frac{\mathbf{V}}{\mathbf{Z}}

The series RLC circuit

A series RLC branch has total impedance

\mathbf{Z}=R+j(X_L-X_C)

. The sign of the net reactance tells you everything: if

X_L>X_C

the circuit is net inductive (current lags, lagging power factor); if

X_C>X_L

it is net capacitive (current leads). The impedance magnitude and angle follow directly, and the current magnitude is

V_{\text{rms}}/|\mathbf{Z}|

\mathbf{Z}=R+j(X_L-X_C), \qquad |\mathbf{Z}|=\sqrt{R^2+(X_L-X_C)^2}, \qquad \theta=\tan^{-1}\frac{X_L-X_C}{R}

Complex power and the power triangle

Real power

P

(watts) does the actual work; reactive power

Q

(vars) sloshes energy in and out of the fields and does no net work; apparent power

S

(volt-amperes) is what the conductors and transformers must be sized for. They form a right triangle with

\theta

, the angle by which voltage leads current, equal to the impedance angle. Complex power is

\mathbf{S}=\mathbf{V}\mathbf{I}^{*}=P+jQ

. The power factor

\cos\theta

is lagging when the load is inductive (current lags), leading when capacitive.

\mathbf{S}=\mathbf{V}\mathbf{I}^{*}=P+jQ, \quad P=VI\cos\theta, \quad Q=VI\sin\theta, \quad S=\sqrt{P^2+Q^2}

Power factor correction

Most industrial loads are inductive, drawing a lagging current that inflates

S

and the line current without adding useful power. Adding a parallel capacitor supplies the reactive power locally, shrinking

Q

from

Q_1

to a target

Q_2

and raising the power factor. The needed capacitor reactive power is

Q_C=Q_1-Q_2=P(\tan\theta_1-\tan\theta_2)

, and the capacitance follows from

Q_C=V^2/X_C=\omega C V^2

. Real power

P

is unchanged — only the reactive bookkeeping moves.

Q_C = P(\tan\theta_1-\tan\theta_2), \qquad C=\frac{Q_C}{\omega V^2}

Series resonance

As frequency varies,

X_L

rises and

X_C

falls; at one frequency they cancel and the series impedance collapses to just

R

. There the circuit is purely resistive, the current peaks, voltage and current are in phase, and the power factor is unity. The resonant frequency and the sharpness (quality factor

Q

, the ratio of resonant center frequency to bandwidth,

Q=\omega_0/\text{BW}

) are standard exam targets. Note: at resonance the individual

X_L

and

X_C

are large and equal but opposite, so the voltages across L and C can far exceed the source voltage.

\omega_0=\frac{1}{\sqrt{LC}}, \qquad Q=\frac{\omega_0 L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}, \qquad \text{BW}=\frac{\omega_0}{Q}

Exam strategy

Fix your reference: take the source voltage as

\angle 0^\circ

and let the impedance angle set the current's phase. Add impedances in rectangular form, then convert to polar to divide for current. For power problems, the fastest route is often

P=I^2R

(real power lives only in resistance) and

Q=I^2X

, then

S=\sqrt{P^2+Q^2}

— no angle needed. Watch the lead/lag language: lagging means inductive (

+Q

, current behind voltage). At resonance set

X_L=X_C

immediately and the impedance is just

R

. Keep

\omega=2\pi f

straight from

f

— mixing rad/s and Hz is the classic slip.

Key equations

RMS value of a sinusoid

V_{\text{rms}}=\frac{V_{\max}}{\sqrt{2}}, \qquad \omega=2\pi f

Effective value used in all power and meter calculations; $f$ in Hz, $\omega$ in rad/s.

Inductive reactance

X_L=\omega L

Ohms; rises with frequency. Inductor current lags voltage by $90^\circ$ . $L$ in henrys.

Capacitive reactance

X_C=\frac{1}{\omega C}

Ohms; falls with frequency. Capacitor current leads voltage by $90^\circ$ . $C$ in farads.

Impedance

\mathbf{Z}=R+jX=|\mathbf{Z}|\angle\theta, \quad |\mathbf{Z}|=\sqrt{R^2+X^2}, \quad \theta=\tan^{-1}\frac{X}{R}

Complex ratio of phasor voltage to phasor current; $X=X_L-X_C$ for a series branch.

AC Ohm's law

\mathbf{I}=\frac{\mathbf{V}}{\mathbf{Z}}

Phasor current from phasor voltage and impedance; magnitudes divide, angles subtract.

Complex power

\mathbf{S}=\mathbf{V}\mathbf{I}^{*}=P+jQ

Uses conjugate of current phasor; $P$ in W, $Q$ in var, $S=|\mathbf{S}|$ in VA.

Real, reactive, apparent power

P=V_{\text{rms}}I_{\text{rms}}\cos\theta, \quad Q=V_{\text{rms}}I_{\text{rms}}\sin\theta, \quad S=V_{\text{rms}}I_{\text{rms}}

$\theta$ = angle by which voltage leads current = impedance angle.

Power in terms of R and X

P=I_{\text{rms}}^2 R, \qquad Q=I_{\text{rms}}^2 X

Real power dissipates only in resistance; reactive power only in reactance.

Power factor

pf=\cos\theta=\frac{P}{S}

Lagging (inductive, current lags) or leading (capacitive, current leads).

Power factor correction

Q_C=P(\tan\theta_1-\tan\theta_2), \qquad C=\frac{Q_C}{\omega V^2}

Parallel capacitor sizing to raise pf from $\cos\theta_1$ to $\cos\theta_2$ at fixed $P$ , $V$ . The FE Reference Handbook 10.6 prints the power triangle ( $P$ , $Q$ , pf) but not this correction formula.

Resonant frequency

\omega_0=\frac{1}{\sqrt{LC}}, \qquad f_0=\frac{1}{2\pi\sqrt{LC}}

Where $X_L=X_C$ ; series impedance equals $R$ and current peaks.

Quality factor and bandwidth (series)

Q=\frac{\omega_0 L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}, \qquad \text{BW}=\frac{\omega_0}{Q}

Sharpness of resonance; bandwidth in rad/s between half-power points.

Worked examples

Series RLC: impedance, current, and power

Problem. A $120\,\text{V}$ rms, $60\,\text{Hz}$ source drives a series circuit with $R=10\,\Omega$ , $L=20\,\text{mH}$ , and $C=100\,\mu\text{F}$ . Find the impedance, the current, the real power, and the power factor (state lead or lag).

Solution.

\omega=2\pi(60)=377\,\text{rad/s}

X_L=\omega L=377(0.020)=7.54\,\Omega

;

X_C=\dfrac{1}{\omega C}=\dfrac{1}{377(100\times10^{-6})}=26.5\,\Omega

\mathbf{Z}=10+j(7.54-26.5)=10-j19.0\,\Omega

, so

|\mathbf{Z}|=\sqrt{10^2+19.0^2}=21.5\,\Omega

\theta=-62.2^\circ

I=\dfrac{120}{21.5}=5.59\,\text{A}

P=I^2R=(5.59)^2(10)=313\,\text{W}

pf=\cos(62.2^\circ)=0.466

leading (net capacitive, since

X_C>X_L

). Sanity check:

S=VI=120(5.59)=671\,\text{VA}

, and

P=S\,pf=671(0.466)=313\,\text{W}

. Consistent.

\mathbf{Z}=R+j(X_L-X_C)=10-j19.0=21.5\angle{-62.2^\circ}\,\Omega

Power factor correction

Problem. A $240\,\text{V}$ rms, $60\,\text{Hz}$ single-phase load draws $10\,\text{kW}$ at $0.70$ power factor lagging. Find the parallel capacitance needed to raise the power factor to $0.95$ lagging.

Solution.

\theta_1=\cos^{-1}0.70=45.6^\circ

\theta_2=\cos^{-1}0.95=18.2^\circ

Q_1=P\tan\theta_1=10000(1.020)=10.2\,\text{kvar}

;

Q_2=P\tan\theta_2=10000(0.329)=3.29\,\text{kvar}

Q_C=Q_1-Q_2=6.92\,\text{kvar}

C=\dfrac{Q_C}{\omega V^2}=\dfrac{6915}{377(240)^2}=3.18\times10^{-4}\,\text{F}=318\,\mu\text{F}

. Sanity check:

P

is unchanged; the new line current drops from

S_1/V=14.3\,\text{kVA}/240=59.5\,\text{A}

S_2/V=10.5\,\text{kVA}/240=43.9\,\text{A}

— exactly the point of correction.

C=\frac{P(\tan\theta_1-\tan\theta_2)}{\omega V^2}=\frac{6915}{377(240)^2}=318\,\mu\text{F}

Series resonance

Problem. A series RLC circuit has $R=5\,\Omega$ , $L=2\,\text{mH}$ , and $C=8\,\mu\text{F}$ . Find the resonant frequency in Hz, the quality factor, and the current if a $10\,\text{V}$ rms source is applied at resonance.

Solution.

\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{(2\times10^{-3})(8\times10^{-6})}}=7906\,\text{rad/s}

, so

f_0=\dfrac{7906}{2\pi}=1.26\,\text{kHz}

Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}=\dfrac{1}{5}\sqrt{\dfrac{2\times10^{-3}}{8\times10^{-6}}}=\dfrac{1}{5}(15.81)=3.16

. At resonance

\mathbf{Z}=R=5\,\Omega

, so

I=\dfrac{10}{5}=2.00\,\text{A}

. Sanity check:

X_L=\omega_0 L=7906(0.002)=15.8\,\Omega

equals

X_C=1/(\omega_0 C)=15.8\,\Omega

— they cancel, confirming resonance.

f_0=\frac{1}{2\pi\sqrt{LC}}=\frac{1}{2\pi\sqrt{(2\times10^{-3})(8\times10^{-6})}}=1.26\,\text{kHz}

Common pitfalls

•Mixing $f$ and $\omega$ : reactance uses $\omega=2\pi f$ , not $f$ . Plugging $60$ instead of $377$ into $X_L=\omega L$ understates the reactance by a factor of $2\pi$ .
•Capacitor reactance sign: $X_C$ contributes $-j/(\omega C)$ to impedance. Forgetting the minus makes the circuit look inductive and flips the power factor from leading to lagging.
•Lead/lag confusion: lagging power factor means the current lags the voltage (inductive, $+Q$ ). 'Leading' is capacitive. Reversing them gives the wrong sign of reactive power.
•Using peak instead of RMS for power: $P=V_{\text{rms}}I_{\text{rms}}\cos\theta$ needs RMS values. Using $V_{\max}$ overstates power by a factor of 2 (per the $V_{\max}I_{\max}/2$ form).
•Real power in reactance: $P=I^2R$ uses only the resistance. Reactive elements store and return energy — they dissipate no average power, so never put $X$ into the $P$ formula.
•Conjugate in complex power: $\mathbf{S}=\mathbf{V}\mathbf{I}^{*}$ uses the conjugate of the current. Using $\mathbf{V}\mathbf{I}$ flips the sign of $Q$ and mislabels inductive vs capacitive.
•Resonance reactances aren't zero: at resonance $X_L=X_C$ individually can be large; only their difference cancels. The L and C voltages can exceed the source voltage by the factor $Q$ .

References

NCEES FE Reference Handbook — Electrical and Computer Engineering (AC Circuits, Impedance)
NCEES FE Reference Handbook — Electrical and Computer Engineering (AC Power, Complex Power)
NCEES FE Reference Handbook — Electrical and Computer Engineering (Resonance)

D. Measuring devices

Electrical Measuring Devices

How voltmeters, ammeters, ohmmeters, and wattmeters are built from one meter movement, why their resistance matters, and how loading error and instrument transformers extend their range.

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E. Three-phase power

Three-Phase Power

Wye vs delta connections, the line/phase and √3 relationships, balanced three-phase power, and motor efficiency — the bookkeeping behind nearly every three-phase exam problem.

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