Power Systems

8% of exam

Power theory, transmission & distribution, transformers, motors and generators.

4 concepts

Exam areas

A. Power theory

AC Power and Three-Phase Systems

Real, reactive, apparent, and complex power; power factor and correction; wye/delta and the √3 relations that drive nearly every power-systems question on the FE.

Power is where circuit theory meets the grid, and it is one of the densest scoring areas on the FE Electrical and Computer exam: a single fluent idea — the power triangle — unlocks single-phase AC power, power-factor correction, and balanced three-phase systems all at once. The trap is that the handbook gives you compact formulas (“AC Power” and “Balanced Three-Phase Systems” in the Electrical and Computer Engineering chapter) but not the bookkeeping discipline that keeps line and phase quantities straight. That discipline is what this concept builds: when you can write

S=P+jQ

in your sleep and know that

S=\sqrt{3}\,V_{LL}I_L

already counts all three phases, these questions become near-automatic.

Instantaneous and average power

For a sinusoidal voltage

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

across an element carrying

i(t)=I_{max}\cos(\omega t-\theta)

, the instantaneous power

p(t)=v(t)i(t)

pulsates at twice the line frequency. Averaging over a cycle leaves only the real (average) power, which depends on the cosine of the angle

\theta

between voltage and current. Using RMS magnitudes

V=V_{max}/\sqrt{2}

and

I=I_{max}/\sqrt{2}

collapses the messy time function into one number. This is why power engineers work in RMS phasors: the constant term is the useful power, and the pulsating term averages to zero.

p(t)=V_{max}I_{max}\cos(\omega t)\cos(\omega t-\theta),\qquad P=\tfrac{1}{2}V_{max}I_{max}\cos\theta = V_{rms}I_{rms}\cos\theta

Real, reactive, and apparent power

Three quantities describe AC power, and the FE expects you to move between them instantly. Real power

P

(watts) does the actual work; reactive power

Q

(vars) shuttles energy back and forth into magnetic and electric fields without net work; apparent power

S

(volt-amperes) is the RMS product

VI

that the conductors and transformers must actually be sized for. The defining angle

\theta

is measured from

V

I

and equals the angle of the load impedance. For an inductive (lagging) load

\theta>0

Q>0

; for a capacitive (leading) load

\theta<0

Q<0

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

Complex power and the power triangle

Bundle real and reactive power into one complex number and the algebra gets easy: complex power is

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

, where

\mathbf{I}^{*}

is the conjugate of the phasor current. The magnitude

|\mathbf{S}|=S

is the apparent power, and

P

Q

S

form a right triangle whose angle is the impedance angle

\theta

. The power triangle is the single most useful picture in this topic: every power-factor and correction problem is just moving along the

Q

leg while holding

P

fixed.

\mathbf{S}=\mathbf{V}\mathbf{I}^{*}=P+jQ,\qquad S=\sqrt{P^{2}+Q^{2}}

Power factor and its correction

Power factor is

pf=\cos\theta=P/S

— the fraction of apparent power that becomes real power. A low (lagging) power factor means a plant draws more current than its real load requires, oversizing conductors and incurring utility penalties. Correction adds shunt capacitors that supply leading vars locally, shrinking the

Q

the source must deliver while leaving

P

untouched. Size the capacitor bank from the change in reactive power:

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

, moving from the old angle

\theta_1

to the target

\theta_2

pf=\cos\theta=\frac{P}{S},\qquad Q_C=P\left(\tan\theta_1-\tan\theta_2\right)

Three-phase wye and delta relations

Three-phase supplies deliver constant total power and use copper efficiently. Everything reduces to keeping line quantities (between two conductors, or through a line) distinct from phase quantities (across or through one load element). In a wye (Y) the line current equals the phase current while the line-to-line voltage is

\sqrt{3}

larger than the line-to-neutral voltage and leads it by

30^\circ

. In a delta (

\Delta

) the phase voltage equals the line voltage while the line current is

\sqrt{3}

times the phase current. A balanced delta load converts to an equivalent wye by

Z_Y=Z_\Delta/3

\text{Wye: } V_{LL}=\sqrt{3}\,V_{LN},\; I_L=I_\phi \qquad \text{Delta: } V_{LL}=V_\phi,\; I_L=\sqrt{3}\,I_\phi

Balanced three-phase power

For any balanced load — wye or delta — the total three-phase power is the same when written in line quantities, and this is the workhorse formula of the entire topic. The angle

\theta

is the load impedance angle (between phase voltage and phase current), never the

30^\circ

wye–delta shift. Because the

\sqrt{3}

form already sums all three phases, you must not multiply it again by 3; the equivalent phase form is

S=3V_\phi I_\phi

P=\sqrt{3}\,V_{LL}I_L\cos\theta,\quad Q=\sqrt{3}\,V_{LL}I_L\sin\theta,\quad S=\sqrt{3}\,V_{LL}I_L=3V_\phi I_\phi

Exam strategy

Anchor every problem with the power triangle and

S=\sqrt{3}\,V_{LL}I_L

. Read a stated three-phase voltage as line-to-line by default (

480\,\text{V}

means

V_{LN}=277\,\text{V}

). For a rated load, get

S=P/pf

first, then

I_L=S/(\sqrt{3}V_{LL})

; only afterward chase connection-specific phase quantities. For impedance loads, convert delta to wye with

Z_Y=Z_\Delta/3

, work one phase against neutral using

V_{LN}=V_{LL}/\sqrt{3}

, and multiply per-phase power by three. For correction, hold

P

fixed and slide down the

Q

leg with

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

Key equations

Real (average) power

P=V_{rms}I_{rms}\cos\theta=\tfrac{1}{2}V_{max}I_{max}\cos\theta

Average power in watts; $\theta$ is the angle from $\mathbf{V}$ to $\mathbf{I}$ (the impedance angle). Use RMS values unless peak values are explicitly given.

Reactive power

Q=V_{rms}I_{rms}\sin\theta

In vars. Positive (absorbed) for inductive/lagging loads, negative for capacitive/leading loads.

Complex and apparent power

\mathbf{S}=\mathbf{V}\mathbf{I}^{*}=P+jQ,\qquad S=|\mathbf{S}|=V_{rms}I_{rms}

$\mathbf{I}^{*}$ is the conjugate of the phasor current; $S$ in VA is what conductors and transformers are sized to.

Power triangle / power factor

S=\sqrt{P^{2}+Q^{2}},\qquad pf=\cos\theta=\frac{P}{S}

Lagging pf = inductive ( $I$ lags $V$ ); leading pf = capacitive ( $I$ leads $V$ ).

Power factor correction

Q_C=P\left(\tan\theta_1-\tan\theta_2\right)

Capacitor reactive power (var) to move from initial angle $\theta_1$ to target $\theta_2$ at constant real power $P$ . The FE Reference Handbook 10.6 prints the power triangle ( $P$ , $Q$ , pf) but not this correction formula.

Wye line-phase relations

V_{LL}=\sqrt{3}\,V_{LN}\angle{+30^\circ},\qquad I_L=I_\phi

Star connection; line-to-line voltage leads line-to-neutral by $30^\circ$ (positive/ABC sequence).

Delta line-phase relations

V_{LL}=V_\phi,\qquad I_L=\sqrt{3}\,I_\phi\angle{-30^\circ}

Delta connection; line current is $\sqrt{3}$ times phase current and lags it by $30^\circ$ (positive sequence).

Balanced three-phase power

S=\sqrt{3}\,V_{LL}I_L=3V_\phi I_\phi

Same total from line or phase quantities; $P=S\cos\theta$ , $Q=S\sin\theta$ with $\theta$ the load impedance angle.

Delta-to-wye impedance

Z_Y=\frac{Z_\Delta}{3}

Balanced loads only; lets you reduce a delta load to per-phase wye analysis.

Per-phase voltage

V_{LN}=\frac{V_{LL}}{\sqrt{3}}

$480\,\text{V}\to277\,\text{V}$ ; $208\,\text{V}\to120\,\text{V}$ ; $13.8\,\text{kV}\to7.97\,\text{kV}$ . Used for single-phase (per-phase) equivalents.

Resistive power

P=\frac{V_{rms}^{2}}{R}=I_{rms}^{2}R

For a pure resistor $\theta=0$ , so $Q=0$ and $S=P$ ; useful as a sanity check.

Worked examples

Three-phase motor line current and reactive power

Problem. A three-phase motor draws $75\,\text{kW}$ at $0.85$ power factor lagging from a $480\,\text{V}$ system. Find the line current and the reactive power.

Solution. Apparent power:

S=P/pf=75/0.85=88.2\,\text{kVA}

. Line current:

I_L=S/(\sqrt{3}V_{LL})=88{,}235/(\sqrt{3}\cdot480)=106\,\text{A}

. Angle:

\theta=\cos^{-1}0.85=31.8^\circ

, so

Q=S\sin\theta=88.2\sin31.8^\circ=46.5\,\text{kvar}

(absorbed, lagging). Sanity check:

P=S\cos\theta=88.2(0.85)=75\,\text{kW}

recovers the given load. Answers:

I_L\approx106\,\text{A}

Q\approx46.5\,\text{kvar}

I_L=\frac{75{,}000}{0.85\,\sqrt{3}\,(480)}\approx106\ \text{A}

Power-factor correction capacitor bank

Problem. The same $75\,\text{kW}$ , $0.85$ lagging load is to be corrected to $0.95$ lagging. Find the required capacitor reactive power and the new line current at $480\,\text{V}$ .

Solution. Old angle

\theta_1=\cos^{-1}0.85=31.79^\circ

\tan\theta_1=0.620

. New angle

\theta_2=\cos^{-1}0.95=18.19^\circ

\tan\theta_2=0.329

Q_C=P(\tan\theta_1-\tan\theta_2)=75(0.620-0.329)=21.8\,\text{kvar}

. New apparent power:

S'=P/pf'=75/0.95=78.9\,\text{kVA}

, so

I_L'=78{,}947/(\sqrt{3}\cdot480)=95.0\,\text{A}

. Sanity check: current drops from

106\,\text{A}

95.0\,\text{A}

— about

10\%

less, as expected for raising pf. Answers:

Q_C\approx21.8\,\text{kvar}

I_L'\approx95.0\,\text{A}

Q_C=P\left(\tan\theta_1-\tan\theta_2\right)=75(0.620-0.329)\approx21.8\ \text{kvar}

Balanced delta load

Problem. A balanced delta load of $Z=20+j15\,\Omega$ per phase is connected to a $240\,\text{V}$ three-phase source. Find the phase current, line current, and total real and reactive power.

Solution. Impedance magnitude and angle:

|Z|=\sqrt{20^{2}+15^{2}}=25\,\Omega

\theta=\tan^{-1}(15/20)=36.9^\circ

. Delta phase voltage equals line voltage:

V_\phi=240\,\text{V}

, so

I_\phi=240/25=9.60\,\text{A}

and

I_L=\sqrt{3}(9.60)=16.6\,\text{A}

. Real power:

P=3V_\phi I_\phi\cos\theta=3(240)(9.60)\cos36.9^\circ=5.53\,\text{kW}

. Reactive power:

Q=3(240)(9.60)\sin36.9^\circ=4.15\,\text{kvar}

. Sanity check:

S=\sqrt{3}(240)(16.6)=6.91\,\text{kVA}

and

\sqrt{5.53^{2}+4.15^{2}}=6.91\,\text{kVA}

— consistent. Answers:

I_\phi=9.60\,\text{A}

I_L\approx16.6\,\text{A}

P\approx5.53\,\text{kW}

Q\approx4.15\,\text{kvar}

Common pitfalls

•Applying $\sqrt{3}$ twice: $S=\sqrt{3}\,V_{LL}I_L$ already counts all three phases. The factor-of-3 alternative uses phase quantities, $S=3V_\phi I_\phi$ — never combine both.
•Using the $30^\circ$ wye–delta shift as the power-factor angle. The $\theta$ in $P=S\cos\theta$ is the load impedance angle only.
•Forgetting that stated three-phase voltages are line-to-line: $480\,\text{V}$ gives $V_{LN}=277\,\text{V}$ in per-phase work, not $480\,\text{V}$ .
•Sign of $Q$ : inductive (lagging) loads absorb positive vars; capacitors supply negative vars. Mixing the sign breaks correction sizing.
•For a delta load, $V_{LL}/Z$ gives PHASE current; multiply by $\sqrt{3}$ to get line current — not the reverse.
•Treating $Z_\Delta$ and $Z_Y$ as interchangeable: a delta bank rewired to wye draws one-third the power at the same line voltage ( $Z_Y=Z_\Delta/3$ ).
•Computing power factor as $Q/S$ or forgetting it must be labeled leading or lagging — $pf=\cos\theta=P/S$ , and the magnitude alone is ambiguous.

References

NCEES FE Reference Handbook — Electrical and Computer Engineering: AC Power (Complex Power)
NCEES FE Reference Handbook — Electrical and Computer Engineering: Balanced Three-Phase (3-φ) Systems
Irwin & Nilsson, Electric Circuits — AC steady-state power and three-phase chapters for background derivations.

B. Transmission and distribution

Transmission and Distribution

Model a line as series R and X, compute voltage drop, regulation, and I²R loss, work the per-unit system, and size feeder power-factor correction.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

C. Transformers

Transformers

Ideal turns ratio and impedance reflection, the equivalent circuit, regulation and efficiency, three-phase connections, and the autotransformer kVA advantage.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

D. Motors and generators

Motors and Generators

DC machine EMF and torque, induction-motor slip and synchronous speed, the synchronous machine power equation, and machine efficiency and power flow.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders