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PE Electrical and Computer: Power formula sheet

91 key equations from 9 study topics, each with what it is for and where to look it up: the PE Electrical and Computer: Power Reference Handbook section, or the code or textbook when the handbook does not print it. No signup. These are the equations from the free chapters of our PE Electrical and Computer: Power study handbook; the full handbook covers every topic at this depth, with worked examples and the traps that cost points.

91

Equations

9

Study topics

10

Concepts

Circuit Analysis
14% of the exam

Three-phase, symmetrical components, per-unit, phasors, 1-φ & DC.

Three-Phase Circuits and Power

Wye and delta relationships, balanced three-phase power, and the √3 conventions that underpin nearly every exam calculation.

Wye relationships
VLL=3 VLN∠+30∘,IL=IϕV_{LL} = \sqrt{3}\,V_{LN}\angle{+30^\circ}, \quad I_L = I_\phi
Line-line voltage leads line-neutral by 30° (ABC sequence).
Delta relationships
VLL=Vϕ,IL=3 Iϕ∠−30∘V_{LL} = V_\phi, \quad I_L = \sqrt{3}\,I_\phi\angle{-30^\circ}
Line current lags its phase current by 30° (ABC sequence).
Three-phase apparent power
S=3 VLLIL=3 VϕIϕS = \sqrt{3}\,V_{LL} I_L = 3\,V_\phi I_\phi
Same total from line or phase quantities.
Real and reactive power
P=3 VLLILcos⁡θ,Q=3 VLLILsin⁡θP = \sqrt{3}\,V_{LL}I_L\cos\theta, \quad Q = \sqrt{3}\,V_{LL}I_L\sin\theta
θ = load impedance angle = angle between phase voltage and phase current.
Power triangle
S=P2+Q2,pf=cos⁡θ=P/SS = \sqrt{P^2 + Q^2}, \quad pf = \cos\theta = P/S
Lagging = inductive (Q > 0 absorbed).
Delta→wye impedance
ZY=ZΔ/3Z_Y = Z_\Delta / 3
Balanced loads only.
Per-phase voltage
VLN=VLL/3V_{LN} = V_{LL}/\sqrt{3}
480 V system → 277 V; 208 V → 120 V; 13.8 kV → 7.97 kV.

Where it lives: NCEES PE Power Reference Handbook §3.1.1 — 3-Phase Circuits · NCEES PE Power Reference Handbook §1.3.1 — Algebra of Complex Numbers · Grainger & Stevenson, Power System Analysis

Read the Circuit Analysis chapter

The Per-Unit System

Normalizing voltage, current, power, and impedance to common bases — the change-of-base formula and why transformers vanish from the math.

Base current (3-φ)
Ibase=Sbase3 VLL,baseI_{base} = \dfrac{S_{base}}{\sqrt{3}\,V_{LL,base}}
S in VA, V in volts → A (or MVA and kV → kA).
Base impedance
Zbase=VLL,base2Sbase=(kV)2MVA ΩZ_{base} = \dfrac{V_{LL,base}^2}{S_{base}} = \dfrac{(\text{kV})^2}{\text{MVA}}\ \Omega
The kV²/MVA shortcut works directly in those units.
Per-unit value
Xpu=Xactual/XbaseX_{pu} = X_{actual}/X_{base}
Dimensionless; ×100 gives percent.
Change of base
Zpunew=Zpuold(VoldVnew)2SnewSoldZ_{pu}^{new} = Z_{pu}^{old}\left(\dfrac{V_{old}}{V_{new}}\right)^2 \dfrac{S_{new}}{S_{old}}
Old voltage base in the numerator; impedance grows with a larger MVA base.
Per-unit Ohm's law
Vpu=IpuZpu,Spu=VpuIpu∗V_{pu} = I_{pu} Z_{pu}, \qquad S_{pu} = V_{pu} I_{pu}^*
No √3 or 3 factors in per-unit.
Three-phase fault current
If,pu=VpuZpu  ⇒  If=If,pu×IbaseI_{f,pu} = \dfrac{V_{pu}}{Z_{pu}} \;\Rightarrow\; I_f = I_{f,pu} \times I_{base}
The most common per-unit application; see the fault-analysis concept.

Where it lives: NCEES PE Power Reference Handbook §3.1.3 — Per Unit System · Grainger & Stevenson, Power System Analysis · IEEE Std 399 (Brown Book)

Read the Circuit Analysis chapter

Rotating Machines
7% of the exam

Machine types & applications, motor starting, slip, torque.

Induction Machines: Equivalent Circuit, Power Flow, and Torque

Master slip, the per-phase equivalent circuit, the 1 : s : (1−s) power-flow ladder, and the torque-slip curve behind every induction-motor exam question.

Slip
s=ns−nnss = \frac{n_s - n}{n_s}
Per-unit. ns=120f/Pn_s = 120f/P.
Rotor speed
n=(1−s) nsωm=(1−s) ωsn = (1-s)\,n_s \qquad \omega_m = (1-s)\,\omega_s
Rotor frequency
fr=sff_r = s f
Rotor currents at slip frequency; standstill rotor reactance XR0X_{R0} becomes sXR0sX_{R0} when running.
Air-gap power
PAG=Pin−PSCL−Pcore=3 (I2′)2 R2′sP_{AG} = P_{in} - P_{SCL} - P_{core} = 3\,(I_2')^2\,\frac{R_2'}{s}
PSCL=3I12R1P_{SCL} = 3I_1^2R_1 = stator copper loss.
Rotor copper loss
PRCL=s PAGP_{RCL} = s\,P_{AG}
Converted (developed) power
Pconv=(1−s) PAGP_{conv} = (1-s)\,P_{AG}
Power-flow split
PAG:PRCL:Pconv=1:s:(1−s)P_{AG} : P_{RCL} : P_{conv} = 1 : s : (1-s)
Given any one of the three and the slip, find the other two; PinP_{in} and PoutP_{out} also need the stator and rotational losses.
Developed torque
Tdev=PAGωs=PconvωmT_{dev} = \frac{P_{AG}}{\omega_s} = \frac{P_{conv}}{\omega_m}
The PAG/ωsP_{AG}/\omega_s form is valid even at standstill.
Shaft output and torque
Pout=Pconv−Pfw−PstrayTout=PoutωmP_{out} = P_{conv} - P_{fw} - P_{stray} \qquad T_{out} = \frac{P_{out}}{\omega_m}
PfwP_{fw} = friction and windage; PstrayP_{stray} = stray load loss.
Thevenin voltage
VTH=Vϕ XmR12+(X1+Xm)2≈Vϕ XmX1+XmV_{TH} = V_\phi\,\frac{X_m}{\sqrt{R_1^2 + (X_1+X_m)^2}} \approx V_\phi\,\frac{X_m}{X_1+X_m}
Per-phase stator reduction seen from the air gap; typically about 0.95Vϕ0.95V_\phi.
Torque-slip equation
Tdev=3VTH2 (R2′/s)ωs[(RTH+R2′/s)2+(XTH+X2′)2]T_{dev} = \frac{3V_{TH}^2\,(R_2'/s)}{\omega_s\left[(R_{TH} + R_2'/s)^2 + (X_{TH}+X_2')^2\right]}
All quantities per phase, referred to the stator; ωs\omega_s in mechanical rad/s. The handbook (§4.2.4) prints the approximate form, with V1V_1, R1R_1, X1X_1 in place of VTHV_{TH}, RTHR_{TH}, XTHX_{TH} (assumes ZM≫Z1Z_M \gg Z_1). Use it when a problem gives the raw circuit.
Slip at maximum torque
smaxT=R2′RTH2+(XTH+X2′)2s_{maxT} = \frac{R_2'}{\sqrt{R_{TH}^2 + (X_{TH}+X_2')^2}}
Proportional to rotor resistance — the wound-rotor control handle. The handbook prints the approximate form R2/smaxT=R12+(X1+X2)2R_2/s_{maxT} = \sqrt{R_1^2 + (X_1+X_2)^2}.
Breakdown (pull-out) torque
Tmax=3VTH22 ωs[RTH+RTH2+(XTH+X2′)2]T_{max} = \frac{3V_{TH}^2}{2\,\omega_s\left[R_{TH} + \sqrt{R_{TH}^2+(X_{TH}+X_2')^2}\right]}
Independent of R2′R_2'. With the handbook's approximate circuit, put V1V_1, R1R_1 and X1X_1 in place of the Thevenin values.
Voltage sensitivity
T∝V2T \propto V^2
At any fixed slip. Current at fixed slip scales as VV.

Where it lives: NCEES PE Power Reference Handbook §4.2 — Induction Machines · NCEES PE Power Reference Handbook §4.2.5 — 3-Phase Induction Motor Power Flow · NEMA MG-1 — Motors and Generators · IEEE Std 112 — Test Procedure for Polyphase Induction Motors and Generators

Read the Rotating Machines chapter

Electric Power Devices
11% of the exam

Transformers, capacitors, energy storage, PV & wind, testing.

Transformers: Model, Tests, Regulation, and Efficiency

From the ideal transformer to the full equivalent circuit: OC/SC tests, voltage regulation, efficiency, all-day efficiency, taps, and parallel operation.

Turns ratio
a=N1N2=V1V2=I2I1a = \dfrac{N_1}{N_2} = \dfrac{V_1}{V_2} = \dfrac{I_2}{I_1}
Impedance refers by a².
OC test (shunt branch)
Rc=Voc2Poc,Xm=Voc2(VocIoc)2−Poc2R_c = \dfrac{V_{oc}^2}{P_{oc}}, \quad X_m = \dfrac{V_{oc}^2}{\sqrt{(V_{oc}I_{oc})^2 - P_{oc}^2}}
Rated voltage, open secondary; gives core loss.
SC test (series branch)
Zeq=VscIsc,Req=PscIsc2,Xeq=Zeq2−Req2Z_{eq} = \dfrac{V_{sc}}{I_{sc}}, \quad R_{eq} = \dfrac{P_{sc}}{I_{sc}^2}, \quad X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2}
Rated current, shorted secondary; gives copper loss.
Voltage regulation (approx.)
%VR≈I(Reqcos⁡θ+Xeqsin⁡θ)Vrated×100%\%VR \approx \dfrac{I(R_{eq}\cos\theta + X_{eq}\sin\theta)}{V_{rated}} \times 100\%
θ > 0 lagging; sinθ term negative for leading loads.
Efficiency
η=xScos⁡θxScos⁡θ+Pcore+x2Pcu\eta = \dfrac{xS\cos\theta}{xS\cos\theta + P_{core} + x^2 P_{cu}}
x = per-unit loading; S = rated VA.
Maximum-efficiency load
x=Pcore/Pcu,ratedx = \sqrt{P_{core}/P_{cu,rated}}
Copper loss equals core loss at peak efficiency.
All-day efficiency
ηday=kWhoutkWhout+24 Pcore+∑xi2Pcu ti\eta_{day} = \dfrac{\text{kWh}_{out}}{\text{kWh}_{out} + 24\,P_{core} + \sum x_i^2 P_{cu}\,t_i}
Core loss accrues all 24 h regardless of load.
Parallel load sharing
SASB=Zpu,BZpu,A  (on a common base)\dfrac{S_A}{S_B} = \dfrac{Z_{pu,B}}{Z_{pu,A}} \;\text{(on a common base)}
Load divides inversely with per-unit impedance.

Where it lives: NCEES PE Power Reference Handbook §4.3.1 — Transformers · IEEE Std C57.12.00 · IEEE Std C57.12.90 · Chapman, Electric Machinery Fundamentals

Read the Electric Power Devices chapter

Transmission & Distribution
12% of the exam

Voltage drop & regulation, PF correction, faults, power flow, stability.

Fault Current Analysis

Bolted three-phase faults in per-unit, the MVA method, sequence-network connections for SLG, LL, and DLG faults, and X/R asymmetry.

3-φ fault (per-unit)
If=VpuXpu×IbaseI_f = \dfrac{V_{pu}}{X_{pu}} \times I_{base}
Positive-sequence network only.
Fault MVA
MVAsc=MVAratedZpu,If=MVAsc3VLLMVA_{sc} = \dfrac{MVA_{rated}}{Z_{pu}}, \quad I_f = \dfrac{MVA_{sc}}{\sqrt{3}V_{LL}}
Series elements: reciprocals add; parallel sources: add.
SLG fault
If=3I0=3VfZ1+Z2+Z0+3Zf+3ZnI_f = 3I_0 = \dfrac{3V_f}{Z_1+Z_2+Z_0+3Z_f+3Z_n}
All three networks in series.
LL fault
If=3 VfZ1+Z2≈0.866 I3ϕI_f = \dfrac{\sqrt{3}\,V_f}{Z_1+Z_2} \approx 0.866\,I_{3\phi}
Positive ∥ negative; no zero sequence. The 0.866 ratio assumes Z₁ = Z₂.
DLG positive-sequence current
Ia1=VfZ1+Z2 (Z0+3Zf+3Zn)/(Z2+Z0+3Zf+3Zn)I_{a1} = \dfrac{V_f}{Z_1 + Z_2\,(Z_0{+}3Z_f{+}3Z_n)/(Z_2{+}Z_0{+}3Z_f{+}3Z_n)}
Zero and negative branches in parallel.
Asymmetry
Iasym,1st cycle≈1.6 Isym,Ipeak≈2.6 IsymI_{asym,1st\,cycle} \approx 1.6\,I_{sym}, \quad I_{peak} \approx 2.6\,I_{sym}
Classic ANSI anchors (2.6 peak at X/R = 17; about 2.7 peak and 1.6 total rms at X/R ≈ 25), not upper bounds: both keep rising with X/R toward 2√2 ≈ 2.83 and √3 ≈ 1.73, so use the exact e−ωtR/X expression when X/R is given.
Utility source impedance
Xutility,pu=SbaseMVAsc,utilityX_{utility,pu} = \dfrac{S_{base}}{MVA_{sc,utility}}
On the study base, from the utility's available fault MVA.

Where it lives: NCEES PE Power Reference Handbook §5.1.5 — Fault Current Analysis · IEEE Std 141 / 399 (Red/Brown Books) · IEEE Std C37.010 · Blackburn & Domin, Protective Relaying

Read the Transmission & Distribution chapter

Protection
15% of the exam

Overcurrent, protective relaying, fuses/breakers/reclosers, coordination.

Overcurrent Protection: 50/51 Settings, Inverse Curves, and CT Selection

Set 50/51 pickup and time dial with the IEEE C37.112 curve family, choose CT ratios that will not saturate, and apply residual ground elements.

IEEE inverse-time operating equation
top=TD(AMp−1+B)t_{op} = TD\left(\dfrac{A}{M^{p}-1}+B\right)
Operating time in seconds. TDTD = time-dial setting, MM = multiple of pickup, AA, BB, pp = curve constants per IEEE C37.112.
Multiple of pickup
M=IfaultIpickup=IsecItapM = \dfrac{I_{fault}}{I_{pickup}} = \dfrac{I_{sec}}{I_{tap}}
Dimensionless; must exceed 1 for the element to operate. Use fault and pickup on the same side (both primary A or both secondary A).
Moderately inverse (IEEE)
t=TD(0.0515M0.02−1+0.114)t = TD\left(\dfrac{0.0515}{M^{0.02}-1}+0.114\right)
Flattest IEEE shape; time changes least with current. Constants from IEEE C37.112 Table 1.
Very inverse (IEEE)
t=TD(19.61M2−1+0.491)t = TD\left(\dfrac{19.61}{M^{2}-1}+0.491\right)
Standard choice where fault current falls off along the zone.
Extremely inverse (IEEE)
t=TD(28.2M2−1+0.1217)t = TD\left(\dfrac{28.2}{M^{2}-1}+0.1217\right)
Steepest IEEE shape; best match to fuse melting curves and inrush/cold-load tolerance.
IEC 60255 curve family
t=TMS⋅KMα−1t = TMS\cdot\dfrac{K}{M^{\alpha}-1}
Standard inverse: K=0.14K=0.14, α=0.02\alpha=0.02. Very inverse: K=13.5K=13.5, α=1\alpha=1. Extremely inverse: K=80K=80, α=2\alpha=2. TMSTMS = time multiplier setting (not interchangeable with IEEE TDTD).
Primary pickup from tap
Ipickup,pri=tap×CTRI_{pickup,pri} = \text{tap}\times CTR
Tap in secondary amps; CTRCTR = CT turns ratio (e.g., 600:5 gives CTR=120CTR = 120).
Reset (disk-emulation) time
treset=TD⋅tr1−M2(M<1)t_{reset} = TD\cdot\dfrac{t_r}{1-M^{2}}\quad (M<1)
C37.112 reset characteristic that emulates the electromechanical disk's reset (its return travel after current falls below pickup), so a numerical relay coordinates with disk relays on repeated or reclosed faults; trt_r is the curve-family reset constant (4.85 moderately, 21.6 very, 29.1 extremely inverse).
C-class terminal voltage
VC=20 Is,rated ZB,stdV_{C} = 20\,I_{s,rated}\,Z_{B,std}
A C400 CT delivers 20×5 A=100 A20\times5\,\text{A} = 100\,\text{A} into the 4 Ω4\,\Omega standard burden = 400 V400\,\text{V} with ≤10%\le 10\% error.
CT saturation voltage check
Vs=Isec(RCT+Rlead+Zrelay)V_s = I_{sec}\left(R_{CT}+R_{lead}+Z_{relay}\right)
Symmetrical requirement; multiply by (1+X/R)(1+X/R) to cover full DC offset. Compare against the C-class voltage.
Instantaneous setting rule
I50≥(1.25–1.3) Ifault, next busmaxI_{50} \ge (1.25\text{–}1.3)\,I_{fault,\,next\ bus}^{max}
Keeps the 50 from overreaching the next downstream bus; on transformer primaries also require I50>12 IFLAI_{50} > 12\,I_{FLA} (inrush).
Residual (ground) current
3I0=Ia+Ib+Ic3I_0 = \mathbf{I}_a+\mathbf{I}_b+\mathbf{I}_c
Phasor sum measured by a 51N in the wye-CT residual; zero for balanced conditions.

Where it lives: NCEES PE Power Reference Handbook §3.1.6 — Single-Line Diagrams · IEEE Std C37.112-2018 — Inverse-Time Characteristic Equations for Overcurrent Relays · IEEE Std C57.13 / C37.110 — Instrument Transformers and CT Application · IEEE Std 242 (Buff Book) — Protection and Coordination of Industrial and Commercial Power Systems · NEC 2020 Art. 240 — Overcurrent Protection

Read the Protection chapter

Power Electronics & Controls
7% of the exam

Converters, inverter-based resources, VFDs, relays & ladder logic.

Rectifiers and Power Semiconductor Devices

Convert AC to DC: diode and SCR behavior, half-wave, full-wave, and six-pulse outputs, firing-angle control, ripple, and the harmonics rectifiers inject.

Half-wave average output
Vdc=Vmπ=0.318 VmV_{dc} = \frac{V_m}{\pi} = 0.318\,V_m
VmV_m = peak (not RMS) input voltage in volts; single diode, resistive load.
Half-wave RMS output
Vrms=Vm2V_{rms} = \frac{V_m}{2}
RMS of the half-wave rectified sine; ripple at line frequency ff.
Full-wave average output
Vdc=2Vmπ=0.637 Vm=0.90 VrmsV_{dc} = \frac{2V_m}{\pi} = 0.637\,V_m = 0.90\,V_{rms}
Bridge or center-tap; VrmsV_{rms} here is the RMS input voltage.
Full-wave RMS output
Vrms,out=Vm2V_{rms,out} = \frac{V_m}{\sqrt{2}}
Same RMS as the input sine wave; ripple at 2f2f (120 Hz on a 60 Hz system).
Controlled half-wave (SCR)
Vdc=Vm2π(1+cos⁡α)V_{dc} = \frac{V_m}{2\pi}\left(1 + \cos\alpha\right)
α\alpha = firing (delay) angle in degrees or radians, measured from the natural conduction point; resistive load (the handbook's case — an inductive load keeps the SCR conducting past π\pi and lowers the average).
Controlled full-wave, continuous current
Vdc=2Vmπcos⁡αV_{dc} = \frac{2V_m}{\pi}\cos\alpha
Single-phase full converter with continuous load current.
Six-pulse bridge average output
Vdc=32π VLL=1.35 VLLV_{dc} = \frac{3\sqrt{2}}{\pi}\,V_{LL} = 1.35\,V_{LL}
VLLV_{LL} = RMS line-to-line voltage in volts; diode (uncontrolled) bridge.
Controlled six-pulse bridge
Vdc=1.35 VLLcos⁡αV_{dc} = 1.35\,V_{LL}\cos\alpha
Continuous current; goes negative for α>90∘\alpha > 90^\circ (line-commutated inverter region).
Three-pulse (3-phase half-wave) average
Vdc=332π Vm,LN=1.17 VLNV_{dc} = \frac{3\sqrt{3}}{2\pi}\,V_{m,LN} = 1.17\,V_{LN}
Vm,LNV_{m,LN} = peak line-to-neutral voltage; VLNV_{LN} = RMS line-to-neutral voltage.
Form factor and ripple factor
FF=VrmsVdc,RF=FF2−1FF = \frac{V_{rms}}{V_{dc}}, \qquad RF = \sqrt{FF^2 - 1}
RF benchmarks: half-wave 1.21, single-phase full-wave 0.483, six-pulse about 0.04.
Ripple frequency
fr=p ff_r = p\,f
pp = pulse number (1, 2, 6, 12), ff = line frequency in Hz.
Characteristic harmonics
h=np±1,Ih≈I1hh = np \pm 1, \qquad I_h \approx \frac{I_1}{h}
pp = pulse number, n=1,2,3,…n = 1, 2, 3, \ldots; six-pulse gives h=5,7,11,13,…h = 5, 7, 11, 13, \ldots

Where it lives: NCEES PE Power Reference Handbook §3.2.2 — Power Supplies and Converters · IEEE Std 519-2014 — Harmonic Control in Electric Power Systems · NCEES PE Power Reference Handbook §5.1.4 — Power Quality

Read the Power Electronics & Controls chapter

Measurement & Instrumentation
9% of the exam

Instrument transformers & metering, insulation & ground testing.

Three-Phase Power Measurement and the Two-Wattmeter Method

Blondel's theorem, the 1-, 2-, and 3-wattmeter methods, two-wattmeter math and sign rules below 0.5 PF, and kWh/demand metering with CT/VT multipliers.

Wattmeter response
W=∣Vcoil∣ ∣Icoil∣cos⁡ϕW = |V_{\text{coil}}|\,|I_{\text{coil}}|\cos\phi
ϕ\phi = angle between the potential-coil voltage and current-coil current. Polarity (±) terminals set the sign.
Blondel's theorem
Nelements=Nwires−1N_{\text{elements}} = N_{\text{wires}} - 1
Minimum wattmeter elements for exact total power, balanced or unbalanced. 4-wire → 3; 3-wire → 2.
Two-wattmeter readings (balanced, ABC sequence)
W1=VLL ILcos⁡(30∘+θ)W2=VLL ILcos⁡(30∘−θ)\begin{aligned} W_1 &= V_{LL}\, I_L \cos(30^\circ + \theta) \\ W_2 &= V_{LL}\, I_L \cos(30^\circ - \theta) \end{aligned}
θ\theta = load impedance angle (positive lagging). W2≥W1W_2 \ge W_1 for lagging PF with this labeling.
Total real power
P=W1+W2P = W_1 + W_2
Algebraic sum — include the sign of a negative reading. Exact even for unbalanced 3-wire loads.
Reactive power from the readings
Q=3 (W2−W1)Q = \sqrt{3}\,(W_2 - W_1)
Balanced loads only. Positive = lagging (absorbing vars) with the labeling above.
Power-factor angle
tan⁡θ=3 W2−W1W2+W1\tan\theta = \sqrt{3}\,\frac{W_2 - W_1}{W_2 + W_1}
Balanced loads only; then PF=cos⁡θ\text{PF} = \cos\theta.
Single wattmeter, balanced four-wire
P=3WP = 3W
One element reading line-to-neutral voltage and line current of the same phase.
Power from disk revolutions
P=3600 Kh NtP = \frac{3600\, K_h\, N}{t}
KhK_h = watthour constant (Wh per revolution), NN = revolutions, tt in seconds; PP in watts at the meter.
Transformer factor (meter multiplier)
TF=CTR×VTR\text{TF} = \text{CTR} \times \text{VTR}
Multiply every meter-side reading (kWh, kW, disk-derived power) by TF to get primary values.
Primary watthour constant
PKh=Kh×CTR×VTRPK_h = K_h \times \text{CTR} \times \text{VTR}
Watthours of primary energy per disk revolution on a transformer-rated installation.
Block interval demand
Pdem=Einterval×60ΔtminP_{\text{dem}} = \frac{E_{\text{interval}} \times 60}{\Delta t_{\text{min}}}
EE in kWh over a Δt\Delta t-minute demand interval gives demand in kW (15-min interval → multiply interval kWh by 4).

Where it lives: NCEES PE Power Reference Handbook §2.1 — Measurement and Instrumentation · NCEES PE Power Reference Handbook §3.1.1 — Three-Phase Circuits · ANSI C12.1 — Code for Electricity Metering

Read the Measurement & Instrumentation chapter

Electrical Safety
14% of the exam

Wiring methods, hazardous locations, special occupancies, shock & burns.

Conductor Sizing, Derating, and OCPD Selection

Size conductors from Table 310.16 with ambient and bundling derating, respect 110.14(C) terminal limits, apply the 125% rule, and pick the OCPD per 240.4/240.6.

Continuous-load minimum rating
Imin⁡=1.25 Icont+InoncontI_{\min}=1.25\,I_{cont}+I_{noncont}
Minimum conductor table ampacity (at the termination column) and minimum OCPD rating. NEC 210.19(A), 210.20(A), 215.2, 215.3.
Corrected (derated) ampacity
Icorr=Itable⋅Ftemp⋅FadjI_{corr}=I_{table}\cdot F_{temp}\cdot F_{adj}
ItableI_{table} from the conductor's insulation column of Table 310.16; must satisfy Icorr≥IloadI_{corr}\ge I_{load} (100% of actual load).
Ambient temperature correction factor
Ftemp=Tc−TaTc−30F_{temp}=\sqrt{\frac{T_c-T_a}{T_c-30}}
TcT_c = conductor insulation rating (°C), TaT_a = ambient (°C). Table 310.15(B)(1) (NEC 2020) is this formula evaluated at the upper temperature of each band.
Bundling adjustment factor
Fadj={0.804-6 CCC0.707-9 CCC0.5010-20 CCC0.4521-30 CCC0.4031-40 CCC0.3541+ CCCF_{adj}=\begin{cases}0.80 & \text{4-6 CCC}\\0.70 & \text{7-9 CCC}\\0.50 & \text{10-20 CCC}\\0.45 & \text{21-30 CCC}\\0.40 & \text{31-40 CCC}\\0.35 & \text{41+ CCC}\end{cases}
Table 310.15(C)(1). CCC = current-carrying conductors in one raceway or cable; EGC never counts.
Required table ampacity under derating
Itable≥IloadFtemp FadjI_{table}\ge\frac{I_{load}}{F_{temp}\,F_{adj}}
Work the derating backward to pick a conductor size directly.
10-ft tap minimum ampacity (leaving enclosure)
Itap≥IOCPD10I_{tap}\ge\frac{I_{OCPD}}{10}
NEC 240.21(B)(1); tap must also carry the calculated load and match the terminating device rating.
25-ft tap minimum ampacity
Itap≥IOCPD3I_{tap}\ge\frac{I_{OCPD}}{3}
NEC 240.21(B)(2); must terminate in a single OCPD rated no more than the tap ampacity.

Where it lives: NFPA 70-2020 (NEC) Table 310.16, 310.15(B)(1), 310.15(C)(1) · NFPA 70-2020 (NEC) 110.14(C) · NFPA 70-2020 (NEC) 210.19, 210.20, 215.2, 215.3 · NFPA 70-2020 (NEC) 240.4, 240.6, 240.21(B) · NCEES PE Electrical and Computer: Power exam specification (October 2025) — supplied codes and standards

Read the Electrical Safety chapter

General Applications
11% of the exam

Lightning & surge protection, illumination, demand, grounding.

Demand Calculations and Energy Management

Demand, load, diversity, and coincidence factors; billing demand and rate structures; load profiles and conservation economics.

Demand factor
DF=DmaxPconnected≤1DF = \dfrac{D_{max}}{P_{connected}} \le 1
Basis of NEC Art. 220 feeder/service demand tables.
Load factor
LF=kWhDmax×T≤1LF = \dfrac{kWh}{D_{max}\times T} \le 1
T = hours in the period (730/month, 8760/yr).
Diversity factor
Fdiv=∑Dmax,iDmax,sys≥1F_{div} = \dfrac{\sum D_{max,i}}{D_{max,sys}} \ge 1
Coincidence factor = 1/Fdiv ≤ 1.
System peak from parts
Dsys=∑Dmax,iFdiv=Fcoin∑Dmax,iD_{sys} = \dfrac{\sum D_{max,i}}{F_{div}} = F_{coin}\sum D_{max,i}
Transformer/feeder sizing with diversity.
Average demand
Davg=kWhTD_{avg} = \dfrac{kWh}{T}
The flat-line equivalent of the load profile.
Ratchet demand
Dbilled=max⁡(Dmonth, k×Dmax,11mo)D_{billed} = \max(D_{month},\ k \times D_{max,11mo})
k = ratchet percentage, e.g. 0.8.
Simple payback
PB=Cannual savingsPB = \dfrac{C}{\text{annual savings}}
Screening metric for demand-side efficiency comparisons.

Where it lives: NCEES PE Power Reference Handbook §2.2.4 — Demand Calculations / §2.2.5 — Energy Management · NEC 2020 Article 220 · IEEE Std 141 (Red Book)

Read the General Applications chapter

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Ten free PE Electrical and Computer: Power questions with full worked solutions and figures where needed, no account required. Then the full bank, timed mock exams and the complete study handbook when you are ready.

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Many equations follow the PE Electrical and Computer: Power Reference Handbook as NCEES prints it; the rest come from the code, design standard or textbook named in their “Where it lives” line or note. The on-screen reference NCEES supplies on exam day is the copy that counts, so check the edition for your sitting. Independent study resource; not affiliated with, endorsed by, or sponsored by NCEES.