Skip to content
Engineering reference // electrical engineering

Drawing No. EH–EE–008 // Electrical Engineering

Electrical Engineering Formula Sheet

A practical circuit and power reference from DC fundamentals through AC impedance, three-phase systems and ideal transformer relationships. The equations are suitable for engineering screening; conductor sizing, protection and installation still require the applicable electrical code and equipment standards.

Fast reference, with engineering context

Use the equations directly for screening calculations, then open the linked EngineerHub tools for input handling and unit conversion. Formula applicability and major limitations are stated beside each relation.

Reference conventions

V = IROhm’s law
P = VIElectrical power
ω = 2πfAngular frequency
√3Line-to-line three-phase factor

The power triangle

THE POWER TRIANGLE φ P — real power (W) P = S cos φ, does the work Q — reactive (var) Q = S sin φ S — apparent (VA) S = √(P² + Q²) POWER FACTOR pf = cos φ = P / S Lagging — inductive load, current behind voltage Leading — capacitive load, current ahead of voltage Correcting pf lowers S for the same P, so cables carry less current.
Real power P performs net work; reactive power Q represents alternating energy exchange; apparent power S is the RMS volt-ampere product.

01 // Ohm’s law

For an ohmic element at a stated operating condition, voltage, current and resistance are related linearly.

Core relation
V = IR
I = V/R   ;   R = V/I
SymbolMeaningSI unitsUS customary
VVoltageVV
ICurrentAA
RResistanceΩΩ
Worked example — 24 V across an 8 Ω load

Given: V = 24 V, R = 8 Ω.

I = V/R = 24/8 = 3 A
Answer: I = 3.00 A.

Resistance can change with temperature; nonlinear devices are not described by one constant R.

For AC circuits, Ohm’s law generalizes to phasors: V̲ = I̲Z̲, where impedance can contain resistance and reactance.

02 // Series and parallel resistance

Series resistors carry the same current; parallel resistors have the same voltage.

Core relation
Rseries = ΣRᵢ
1/Rparallel = Σ(1/Rᵢ)
SymbolMeaningSI unitsUS customary
RᵢIndividual resistanceΩΩ
ReqEquivalent resistanceΩΩ
Worked example — 4 Ω in series with 6 Ω ∥ 3 Ω

Given: R₁ = 4 Ω; R₂ = 6 Ω; R₃ = 3 Ω.

Rparallel = (1/6 + 1/3)⁻¹ = 2 Ω
Req = 4 + 2 = 6 Ω
Answer: Req = 6 Ω.

These relations apply directly to ideal lumped resistors.

Capacitors and inductors combine differently because their impedance depends on frequency. Keep complex phase information for AC networks.

03 // DC / resistive power

Electrical power is the rate of energy transfer. For a resistor, Ohm’s law gives three equivalent forms.

Core relation
P = VI
P = I²R = V²/R   ;   E = Pt
SymbolMeaningSI unitsUS customary
PReal powerWW or hp equivalent
VVoltageVV
ICurrentAA
RResistanceΩΩ
EEnergyJ, Wh, kWhWh, kWh, Btu
Worked example — 230 V resistive load drawing 10 A

Given: V = 230 V, I = 10 A.

P = 230×10 = 2300 W
Energy for 8 h = 2.3×8 = 18.4 kWh
Answer: P = 2.30 kW; 8 h consumes 18.4 kWh.

Power and energy are different quantities: kW is a rate; kWh is accumulated energy.

For non-unity-power-factor AC loads, VI is apparent power rather than real power. Use the P–Q–S relations below.

04 // Series RLC impedance

In sinusoidal steady state, a series RLC circuit has resistance plus inductive and capacitive reactance.

Core relation
Z̲ = R + j(XL − XC)
XL = 2πfL   ;   XC = 1/(2πfC)
SymbolMeaningSI unitsUS customary
ZComplex impedanceΩΩ
RResistanceΩΩ
XLInductive reactanceΩΩ
XCCapacitive reactanceΩΩ
fFrequencyHzHz
LInductanceHH
CCapacitanceFF
Worked example — series RLC at 50 Hz

Given: R = 20 Ω, L = 50 mH, C = 100 µF, f = 50 Hz, V = 230 V RMS.

XL = 2π50(0.05) = 15.708 Ω
XC = 1/[2π50(100µF)] = 31.831 Ω
Z = 20 − j16.123 Ω; |Z| = 25.690 Ω
I = 230/25.690 = 8.95 A
Answer: |Z| ≈ 25.69 Ω, phase ≈ −38.9°, current ≈ 8.95 A.

A negative phase angle here means the net series load is capacitive.

RMS values are normally used for AC power calculations. Harmonic/non-sinusoidal systems require more than a single-frequency phasor model.

05 // Single-phase AC power triangle

For sinusoidal single-phase systems, real, reactive and apparent power form a right triangle.

Core relation
S = VI
P = VI cosφ   ;   Q = VI sinφ   ;   S² = P² + Q²
SymbolMeaningSI unitsUS customary
SApparent powerVAVA
PReal powerWW
QReactive powervarvar
φVoltage-current phase angledeg or raddeg or rad
PFPower factor = cosφdimensionlessdimensionless
Worked example — single-phase motor load

Given: V = 230 V, I = 10 A, PF = 0.80 lagging.

S = 230×10 = 2300 VA
P = 2300×0.80 = 1840 W
Q = 2300×sin(cos⁻¹0.8) = 1380 var
Answer: P = 1.84 kW, Q = 1.38 kvar, S = 2.30 kVA.

Displacement power factor is not the whole story for distorted currents.

For nonlinear loads, true power factor also includes waveform distortion. The simple triangle is exact for sinusoidal steady state.

06 // Balanced three-phase power

For a balanced three-phase load using line-to-line voltage and line current, the √3 relation gives total power.

Core relation
P = √3 VLIL cosφ
S = √3 VLIL   ;   Q = √3 VLIL sinφ
SymbolMeaningSI unitsUS customary
VLLine-to-line RMS voltageVV
ILLine RMS currentAA
PThree-phase real powerWW
QThree-phase reactive powervarvar
SThree-phase apparent powerVAVA
Worked example — 400 V three-phase load

Given: VL = 400 V, IL = 32 A, PF = 0.90.

P = √3×400×32×0.90 = 19.95 kW
Answer: P ≈ 20.0 kW.

The formula assumes a balanced three-phase system.

For wye and delta circuits, line and phase quantities differ. Use the correct line/phase relationships before applying per-phase equations.

07 // Approximate feeder voltage drop

A common balanced three-phase approximation uses the conductor resistance and reactance at operating temperature and frequency.

Core relation
ΔV ≈ √3 I L(R cosφ + X sinφ)
%ΔV = 100 ΔV / VL
SymbolMeaningSI unitsUS customary
ILine currentAA
LOne-way route lengthmft
RAC resistance per length, per phaseΩ/mΩ/ft
XReactance per length, per phaseΩ/mΩ/ft
φLoad angledeg or raddeg or rad
VLNominal line voltageVV
Worked example — balanced 400 V feeder

Given: I = 50 A, L = 50 m, R = 0.00050 Ω/m, X = 0.00008 Ω/m, PF = 0.90.

sinφ = √(1−0.9²) = 0.4359
ΔV = √3×50×50×(0.00050×0.9 + 0.00008×0.4359) = 2.10 V
%ΔV = 100×2.10/400 = 0.525%
Answer: ΔV ≈ 2.10 V or 0.525%.

Conductor resistance rises with temperature; use the correct route/loop convention for the chosen formula.

Actual voltage-drop design depends on conductor material, temperature, harmonic content, cable arrangement and applicable electrical-code criteria.

08 // Ideal transformer ratios

For an ideal transformer, voltage ratio equals turns ratio and current ratio is inverse so apparent power is conserved.

Core relation
Vₚ/Vₛ = Nₚ/Nₛ
Iₚ/Iₛ = Nₛ/Nₚ   ;   VₚIₚ ≈ VₛIₛ
SymbolMeaningSI unitsUS customary
Vp, VsPrimary / secondary RMS voltageVV
Np, NsPrimary / secondary turnsturnsturns
Ip, IsPrimary / secondary RMS currentAA
Worked example — 400 V to 230 V ideal transformer

Given: Vp = 400 V, Vs = 230 V, Np = 1000 turns, Is = 10 A.

Ns = 1000×230/400 = 575 turns
Ip = 10×230/400 = 5.75 A
Answer: Ns = 575 turns and ideal Ip = 5.75 A.

Real transformers have winding resistance, leakage reactance, magnetizing current and core losses.

Use rated kVA, temperature rise, insulation class, impedance and protection requirements for actual transformer selection.

09 // Quick formula summary

Compact print reference. Use the detailed sections above for definitions and limitations.

TopicEquationPurposeTool
OhmV=IRDC/resistive relationOhm’s law
ResistorsRs=ΣR; 1/Rp=Σ1/REquivalent resistanceSeries/parallel
PowerP=VIPower and energyElectrical power
RLCZ=R+j(XL−XC)AC current and phaseCircuit load
AC P-Q-SP=VIcosφPower factorElectrical power
Three-phaseP=√3VLILcosφBalanced 3φ powerElectrical power
Voltage dropΔV≈√3IL(Rcosφ+Xsinφ)Feeder screeningVoltage drop
TransformerVp/Vs=Np/NsIdeal ratioTurns ratio

10 // Assumptions & limitations

Fundamental equations are only useful when their assumptions match the actual problem.

11 // Technical references

The DOE handbooks are archived fundamentals/training references and are not current installation codes. Use current local electrical standards for design compliance.

12 // Related EngineerHub tools