electric power formulas

Ampere - A

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2 x 10-7 Newton per meter of length.

Coulomb - C

The standard unit of quantity in electrical measurements. It is the quantity of electricity conveyed in one second by the current produced by an electro-motive force of one volt acting in a circuit having a resistance of one ohm, or the quantity transferred by one ampere in one second.

Farad - F

The farad is the standard unit of capacitance. Reduced to base SI units one farad is the equivalent of one second to the fourth power ampere squared per kilogram per meter squared (s4 A2/kg m2).

When the voltage across a 1 F capacitor changes at a rate of one volt per second (1 V/s) a current flow of 1 A results. A capacitance of 1 F produces 1 V of potential difference for an electric charge of one coulomb (1 C).

In common electrical and electronic circuits units of microfarads μF (1 μF = 10-6 F) and picofarads pF (1 pF = 10-12 F) are used.

Ohm - Ω

The derived SI unit of electrical resistance - the resistance between two points on a conductor when a constant potential difference of 1 volt between them produces a current of 1 ampere.

Henry - H

The Henry is the unit of inductance. Reduced to base SI units one henry is the equivalent of one kilogram meter squared per second squared per ampere squared (kg m2 s-2 A-2).

Inductance

An inductor is a passive electronic component that stores energy in the form of a magnetic field.

The standard unit of inductance is the henry abbreviated H. This is a large unit and more commonly used units are the microhenry abbreviated μH (1 μH =10-6H) and the millihenry abbreviated mH (1 mH =10-3 H). Occasionally, the nanohenry abbreviated nH (1 nH = 10-9 H) is used.

Joule - J

The unit of energy work or quantity of heat done when a force of one Newton is applied over a displacement of one meter. One joule is the equivalent of one watt of power radiated or dissipated for one second.

In imperial units the British Thermal Unit (Btu) is used to express energy. One Btu is equivalent to approximately 1,055 joules.

Siemens - S

The unit of electrical conductance S = A / V

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Watt

The watt is used to specify the rate at which electrical energy is dissipated, or the rate at which electromagnetic energy is radiated, absorbed, or dissipated.

The unit of power W or Joule/second

Weber - Wb

The unit of magnetic flux.

The flux that when linking a circuit of one turn, produces an Electro Motive Force - EMF - of 1 volt as it is reduced to zero at a uniform rate in one second.

1 Weber is equivalent to 108 Maxwells

Tesla - T

The unit of magnetic flux density the Tesla is equal to 1 Weber per square meter of circuit area.

Volt

The Volt - V - is the Standard International (SI) unit of electric potential or electromotive force. A potential of one volt appears across a resistance of one ohm when a current of one ampere flows through that resistance.

Reduced to SI base units,

1 (V) = 1 (kg m2 / s3 A)

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Real Power

Wapplied = 31/2 U I cos Φ

= 31/2 U I PF (1)

where

Wapplied = real power (watts)

U = voltage (volts)

I = current (Amps)

PF = cos Φ = power factor (0.7 - 0.95)

For purely resistive load: PF = cos Φ = 1

Total Power

W = 31/2 U I (2)

Brake Horsepower

WBHP = 31/2 U I PF μ / 746 (3)

where

WBHP = brake horse power (hp)

μ = device efficiency

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Electric Power Formulas

W = E I (1a)

W = R I2 (1b)

W = E2/ R (1c)

where

W = power (Watts)

E = voltage (Volts)

I = current (Amperes)

R = resistance (Ohms)

Electric Current Formulas

I = E / R (2a)

I = W / E (2b)

I = (W / R)1/2 (2c)

Electric Resistance Formulas

R = E / I (3a)

R = E2/ W (3b)

R = W / I2 (3c)

Electrical Potential Formulas - Ohms Law

Ohms law can be expressed as:

E = R I (4a)

E = W / I (4b)

E = (W R)1/2 (4c)

Example - Ohm's law

A 12 volt battery supplies power to a resistance of 18 ohms.

I = (12 Volts) / (18 ohms)

= 0.67 Ampere

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Electrical Motor Formulas

Electrical Motor Efficiency

μ = 746 Php / Winput (5)

where

μ = efficiency

Php = output horsepower (hp)

Winput = input electrical power (Watts)

or alternatively

μ = 746 Php / (1.732 E I PF) (5b)

Electrical Motor - Power

W3-phase = (E I PF 1.732) / 1,000 (6)

where

W3-phase = electrical power 3-phase motor (kW)

PF = power factor electrical motor

Electrical Motor - Amps

I3-phase = (746 Php) / (1.732 E μ PF) (7)

where

I3-phase = electrical current 3-phase motor (Amps)

PF = power factor electrical motor

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E = Voltage / I = Amps /W = Watts / PF = Power Factor / Eff = Efficiency / HP = Horsepower

AC/DC FormulasTo Find Direct Current AC / 1phase

115v or 120vAC / 1phase

208,230, or 240vAC 3 phaseAll Voltages

Amps whenHorsepower is Known

HP x 746E x Eff

HP x 746E x Eff X PF

HP x 746E x Eff x PF

HP x 7461.73 x E x Eff x PF

Amps whenKilowatts is known

kW x 1000E

kW x 1000E x PF

kW x 1000E x PF

kW x 10001.73 x E x PF

Amps whenkVA is known

kVA x 1000E

kVA x 1000E

kVA x 10001.73 x E

Kilowatts I x E1000

I x E x PF1000

I x E x PF1000

I x E x 1.73 PF1000

Kilovolt-Amps I x E1000

I x E1000

I x E x 1.731000

Horsepower(output)

I x E x Eff746

I x E x Eff x PF746

I x E x Eff x PF746

I x E x Eff x 1.73 x PF746

Three Phase ValuesFor 208 volts x 1.732, use 360For 230 volts x 1.732, use 398For 240 volts x 1.732, use 416For 440 volts x 1.732, use 762For 460 volts x 1.732, use 797For 480 Volts x 1.732, use 831

E = Voltage / I = Amps /W = Watts / PF = Power Factor / Eff = Efficiency / HP = Horsepower

AC Efficiency and Power Factor Formulas

To FindSingle Phase

Three Phase

Efficiency

746 x HP

E x I x PF

746 x HPE x I x PF

x 1.732

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Power Factor

Input WattsV x A

Input WattsE x I x 1.732

Power - DC Circuits

Watts = E xI

Amps = W / E

Ohm's Law / Power Formulas

P = watts

I = amps

R = ohms

E = Volts

Voltage Drop Formulas

Single Phase(2 or 3 wire)

VD = 2 x K x I x LCM

K = ohms per mil foot

(Copper = 12.9 at 75°)

(Alum = 21.2 at 75°) Note: K value changes with temperature. See Code chapter 9, Table 8

L = Length of conductor in feet

I = Current in conductor (amperes)

CM = Circular mil area of conductor

CM= 2K x L x IVD

Three Phase

VD= 1.73 x K x I x LCM

CM= 1.73 x K x L x IVD

Check out these Online Calculators!

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To better understand the following formulas review the rule of transposition in equations.A multiplier may be removed from one side of an equation by making it a division on the other side, or a division may be removed from one side of an equation by making it a multiplier on the other side.

1. Voltage and Current: Primary (p) secondary (s)Power(p) = power (s) or Ep x Ip = Es x Is

A. Ep =Es x Is

IpB. Ip =

Es x IsEp

C. Is =Ep x Ip

EsD. Es =

Ep x IpIs

2. Voltage and Turns in Coil:Voltage (p) x Turns (s) = Voltage (s) x Turns (p)

or Ep x Ts = Es x Ip

A.Ep =Es x Ip

TsB.Ts =

Es x TpEp

C.Tp =Ep x Ts

EsD.Es =

Ep x TsTp

3. Amperes and Turns in Coil:Amperes (p) x Turns (p) = Amperes (s) x Turns (s)

or Ip x Tp = Is x Ts

A.

Ip =

Is x TsTp

B.Tp =

Is x TsIp

C.

Ts =

Ip x Tp

D.

Is = Ip x Tp

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Is Ts

OHM’S LAW CALCULATOR

VOLTS=Voltage, AMPS=Current, OHMS=Resistance, WATTS=Power

Give me any TWO numeric values and I'll give you all FOUR. Press the Ohm's Law button after you have made your entries:

VOLTS AMPS OHMS WATTS

Fundamentals: Electric Laws − Formulary − Equations

Formula wheel ▼ Important formulasElectrical engineering laws Electronic engineering laws

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Formula 1 − Electrical (electric) power equation: Power P = I × V = R × I2 = V2 ⁄ R

where power P is in watts, voltage V is in volts and current I is in amperes (DC). If there is AC, look also at the power factor PF = cos φ and φ = power factor

angle (phase angle) between voltage and amperage.

Formula 2 − Mechanical (mechanic) power equation: Power P = E ⁄ t = W ⁄ t where power P is in watts, Energy E is in joules, and time t is in seconds. 1 W =

1 J/s. Power = force times displacement divided by time P = F · s / t or:

Power = force times speed (velocity) P = F · v. Electric (electrical) Energy is E = P × t − measured in watthours, or also in kWh.

Tip: The electrical power triangle (power formula)

The magic triangle can be used to calculate all formulas of the "electric power law". You hide with

a finger the value to be calculated. The other two values show then how to do the calculation.

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CURRENT(3 phase):

I = P/√3(Vl cos(f)) or I = P/3(Vp cos(f))

POWER(single phase):

P = (VpIp cos(f))

POWER(3 phase):

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ENGINEERING FORMULA

ELECTRICAL

DIRECT CURRENT(DC)

I = current(amps.), V = voltage(volts), R = resistance(ohms), P = power(watts) CURRENT:

I = V/R or I = P/V

VOLTAGE:

V= P/I or V = IR

POWER:

I2R or VI

RESISTANCE:

R = V/I

ALTERNATING CURRENT(AC):

Il = line current(amps.), Ip = phase current(amps.), Vp = phase voltage(volts), Vl = line voltage(volts), Z = impedance(ohms), P = power(watts), f = power factor(angle), VA = volt ampers

CURRENT(single phase): I = P/(Vp cos(f)

P = √3(VlIl cos(f)) or P = 3(VpIp cos(f))

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Electrical Formula

Legend

R = Resistance, in ohmsP = Power, in wattsV = Voltage, in voltsI = Current, in ampsHP = HorsepowerL = LengthEff = Efficiency e.g. 85% = 0.85pF = Power FactorOutput kW = Net effective mechanical power of a motor

Ohms Law

Power = V2 or

R x I2 or

V x I

R

Voltage =

P or

R x I or

I

Resistance =

V or

V2 or

P

I P I2

Current =

P or

V or

V R

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Three Phase, Single Phase & DC FormulaThree Phase Single Phase Direct Current

kW =

I x V x 1.732 x pF kW =

I x V x pFkW =

I x V

1000 1000 1000

kW = KVA x pF kW = KVA x pF

kW =HP x 746

1000 x Eff

OutputkW =

I x V x 1.732 x pF x Eff Output

kW =

I x V x pF x Eff Output

kW =

I x V x Eff

1000 1000 1000

kVA =I x V x 1.732

kVA =I x V

1000 1000

kVA =kW

pF

I =

kW x 1000

I =

kW x 1000

I =

kW x 1000

V x 1.732 x pF x Eff

V x pF x Eff V x Eff

I =kVA X 1000

I =kVA X 1000

V x 1.732 V

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I =

HP x 746

I =

HP x 746

I =

HP x 746

V x 1.732 x pF x Eff

V x pF x Eff V x Eff

HP =

I x V x 1.732 x pF x Eff HP =

I x V x pF x Eff HP =

I x V x Eff

746 746 746

HP =kW x Eff x 1000

746

HP =

kVA x Eff x 1000 x pF

746

pF =kW x 1000

pF =kW x 1000

I x V x 1.732 I x V

Eff =

HP x 746

Eff =

HP x 746

Eff =

HP x 746

I x V x 1.732 x pF

I x V x pF I x V

Voltage Drop

Vc =1000 x Vd

or Vd =

Vc x L x I

where: Vc = mV/AmL = LengthI = CurrentVd = Voltage Drop L x I 1000

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Australian Standards specify a maximum of 5% voltage drop across the entire circuit (including mains, submains and final subcircuit).

Maximum Circuit Length Calculation

Lmax =

where: Tha = System Impedance FactorUo = Phase VoltageSph = Cable CSASpe = Cable Earth CSAResistivity = Resistivity for CopperIa = Magnetic release operating current, or fuse interupting capacity

Tha x Uo x Sph x Spe

Ia x Resistivity x (Sph+Spe)

DC circuits

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Library - Electrical Circuit Formulae

ResistanceThe resistance R of a circuit is equal to the applied direct voltage E divided by the resulting steady current I:R = E / I

Resistances in SeriesWhen resistances R1, R2, R3, ... are connected in series, the total resistance RS is:RS = R1 + R2 + R3 + ...

Voltage Division by Series ResistancesWhen a total voltage ES is applied across series connected resistances R1 and R2, the current IS which flows through the series circuit is:IS = ES / RS = ES / (R1 + R2)

The voltages V1 and V2 which appear across the respective resistances R1 and R2 are:V1 = ISR1 = ESR1 / RS = ESR1 / (R1 + R2)V2 = ISR2 = ESR2 / RS = ESR2 / (R1 + R2)

In general terms, for resistances R1, R2, R3, ... connected in series:IS = ES / RS = ES / (R1 + R2 + R3 + ...)Vn = ISRn = ESRn / RS = ESRn / (R1 + R2 + R3 + ...)Note that the highest voltage drop appears across the highest resistance.

Resistances in ParallelWhen resistances R1, R2, R3, ... are connected in parallel, the total resistance RP is:1 / RP = 1 / R1 + 1 / R2 + 1 / R3 + ...

Alternatively, when conductances G1, G2, G3, ... are connected in parallel, the total conductance GP is:GP = G1 + G2 + G3 + ...where Gn = 1 / Rn

For two resistances R1 and R2 connected in parallel, the total resistance RP is:RP = R1R2 / (R1 + R2)RP = product / sum

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The resistance R2 to be connected in parallel with resistance R1 to give a total resistance RP is:R2 = R1RP / (R1 - RP)R2 = product / difference

Current Division by Parallel ResistancesWhen a total current IP is passed through parallel connected resistances R1 and R2, the voltage VP which appears across the parallel circuit is:VP = IPRP = IPR1R2 / (R1 + R2)

The currents I1 and I2 which pass through the respective resistances R1 and R2 are:I1 = VP / R1 = IPRP / R1 = IPR2 / (R1 + R2)I2 = VP / R2 = IPRP / R2 = IPR1 / (R1 + R2)

In general terms, for resistances R1, R2, R3, ... (with conductances G1, G2, G3, ...) connected in parallel:VP = IPRP = IP / GP = IP / (G1 + G2 + G3 + ...)In = VP / Rn = VPGn = IPGn / GP = IPGn / (G1 + G2 + G3 + ...)where Gn = 1 / Rn

Note that the highest current passes through the highest conductance (with the lowest resistance).

CapacitanceWhen a voltage is applied to a circuit containing capacitance, current flows to accumulate charge in the capacitance:

Q = idt = CV

Alternatively, by differentiation with respect to time:dq/dt = i = C dv/dtNote that the rate of change of voltage has a polarity which opposes the flow of current.

The capacitance C of a circuit is equal to the charge divided by the voltage:

C = Q / V = idt / V

Alternatively, the capacitance C of a circuit is equal to the charging current divided by the rate of change of voltage:C = i / dv/dt = dq/dt / dv/dt = dq/dv

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Capacitances in SeriesWhen capacitances C1, C2, C3, ... are connected in series, the total capacitance CS is:1 / CS = 1 / C1 + 1 / C2 + 1 / C3 + ...

For two capacitances C1 and C2 connected in series, the total capacitance CS is:CS = C1C2 / (C1 + C2)CS = product / sum

Voltage Division by Series CapacitancesWhen a total voltage ES is applied to series connected capacitances C1 and C2, the charge QS which accumulates in the series circuit is:

QS = iSdt = ESCS = ESC1C2 / (C1 + C2)

The voltages V1 and V2 which appear across the respective capacitances C1 and C2 are:

V1 = iSdt / C1 = ESCS / C1 = ESC2 / (C1 + C2)

V2 = iSdt / C2 = ESCS / C2 = ESC1 / (C1 + C2)

In general terms, for capacitances C1, C2, C3, ... connected in series:

QS = iSdt = ESCS = ES / (1 / CS) = ES / (1 / C1 + 1 / C2 + 1 / C3 + ...)

Vn = iSdt / Cn = ESCS / Cn = ES / Cn(1 / CS) = ES / Cn(1 / C1 + 1 / C2 + 1 / C3 + ...)Note that the highest voltage appears across the lowest capacitance.

Capacitances in ParallelWhen capacitances C1, C2, C3, ... are connected in parallel, the total capacitance CP is:CP = C1 + C2 + C3 + ...

Charge Division by Parallel CapacitancesWhen a voltage EP is applied to parallel connected capacitances C1 and C2, the charge QP which accumulates in the parallel circuit is:

QP = iPdt = EPCP = EP(C1 + C2)

The charges Q1 and Q2 which accumulate in the respective capacitances C1 and C2 are:

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Q1 = i1dt = EPC1 = QPC1 / CP = QPC1 / (C1 + C2)

Q2 = i2dt = EPC2 = QPC2 / CP = QPC2 / (C1 + C2)

In general terms, for capacitances C1, C2, C3, ... connected in parallel:

QP = iPdt = EPCP = EP(C1 + C2 + C3 + ...)

Qn = indt = EPCn = QPCn / CP = QPCn / (C1 + C2 + C3 + ...)Note that the highest charge accumulates in the highest capacitance.

InductanceWhen the current changes in a circuit containing inductance, the magnetic linkage changes and induces a voltage in the inductance:d/dt = e = L di/dtNote that the induced voltage has a polarity which opposes the rate of change of current.

Alternatively, by integration with respect to time:

= edt = LI

The inductance L of a circuit is equal to the induced voltage divided by the rate of change of current:L = e / di/dt = d/dt / di/dt = d/di

Alternatively, the inductance L of a circuit is equal to the magnetic linkage divided by the current:L = / I

Note that the magnetic linkage is equal to the product of the number of turns N and the magnetic flux : = N = LI

Mutual InductanceThe mutual inductance M of two coupled inductances L1 and L2 is equal to the mutually induced voltage in one inductance divided by the rate of change of current in the other inductance:M = E2m / (di1/dt)M = E1m / (di2/dt)

If the self induced voltages of the inductances L1 and L2 are respectively E1s and E2s for the same rates of change of the current that produced the mutually induced voltages E1m and E2m, then:

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M = (E2m / E1s)L1

M = (E1m / E2s)L2

Combining these two equations:M = (E1mE2m / E1sE2s)½ (L1L2)½ = kM(L1L2)½

where kM is the mutual coupling coefficient of the two inductances L1 and L2.

If the coupling between the two inductances L1 and L2 is perfect, then the mutual inductance M is:M = (L1L2)½

Inductances in SeriesWhen uncoupled inductances L1, L2, L3, ... are connected in series, the total inductance LS is:LS = L1 + L2 + L3 + ...

When two coupled inductances L1 and L2 with mutual inductance M are connected in series, the total inductance LS is:LS = L1 + L2 ± 2MThe plus or minus sign indicates that the coupling is either additive or subtractive, depending on the connection polarity.

Inductances in ParallelWhen uncoupled inductances L1, L2, L3, ... are connected in parallel, the total inductance LP is:1 / LP = 1 / L1 + 1 / L2 + 1 / L3 + ...

Time ConstantsCapacitance and resistanceThe time constant of a capacitance C and a resistance R is equal to CR, and represents the time to change the voltage on the capacitance from zero to E at a constant charging current E / R (which produces a rate of change of voltage E / CR across the capacitance).

Similarly, the time constant CR represents the time to change the charge on the capacitance from zero to CE at a constant charging current E / R (which produces a rate of change of voltage E / CR across the capacitance).

If a voltage E is applied to a series circuit comprising a discharged capacitance C and a resistance R, then after time t the current i, the voltage vR across the resistance, the voltage vC across the capacitance and the charge qC on the capacitance are:

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i = (E / R)e - t / CR

vR = iR = Ee - t / CR

vC = E - vR = E(1 - e - t / CR)qC = CvC = CE(1 - e - t / CR)

If a capacitance C charged to voltage V is discharged through a resistance R, then after time t the current i, the voltage vR across the resistance, the voltage vC

across the capacitance and the charge qC on the capacitance are:i = (V / R)e - t / CR

vR = iR = Ve - t / CR

vC = vR = Ve - t / CR

qC = CvC = CVe - t / CR

Inductance and resistanceThe time constant of an inductance L and a resistance R is equal to L / R, and represents the time to change the current in the inductance from zero to E / R at a constant rate of change of current E / L (which produces an induced voltage E across the inductance).

If a voltage E is applied to a series circuit comprising an inductance L and a resistance R, then after time t the current i, the voltage vR across the resistance, the voltage vL across the inductance and the magnetic linkage L in the inductance are:i = (E / R)(1 - e - tR / L)vR = iR = E(1 - e - tR / L)vL = E - vR = Ee - tR / L

L = Li = (LE / R)(1 - e - tR / L)

If an inductance L carrying a current I is discharged through a resistance R, then after time t the current i, the voltage vR across the resistance, the voltage vL across the inductance and the magnetic linkage L in the inductance are:i = Ie - tR / L

vR = iR = IRe - tR / L

vL = vR = IRe - tR / L

L = Li = LIe - tR / L

Rise Time and Fall TimeThe rise time (or fall time) of a change is defined as the transition time between the 10% and 90% levels of the total change, so for an exponential rise (or fall) of time constant T, the rise time (or fall time) t10-90 is:t10-90 = (ln0.9 - ln0.1)T 2.2T

The half time of a change is defined as the transition time between the initial and 50% levels of the total change, so for an exponential change of time constant T, the half time t50 is :t50 = (ln1.0 - ln0.5)T 0.69T

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Note that for an exponential change of time constant T:- over time interval T, a rise changes by a factor 1 - e -1 ( 0.63) of the remaining change,- over time interval T, a fall changes by a factor e -1 ( 0.37) of the remaining change,- after time interval 3T, less than 5% of the total change remains,- after time interval 5T, less than 1% of the total change remains.

PowerThe power P dissipated by a resistance R carrying a current I with a voltage drop V is:P = V2 / R = VI = I2R

Similarly, the power P dissipated by a conductance G carrying a current I with a voltage drop V is:P = V2G = VI = I2 / G

The power P transferred by a capacitance C holding a changing voltage V with charge Q is:P = VI = CV(dv/dt) = Q(dv/dt) = Q(dq/dt) / C

The power P transferred by an inductance L carrying a changing current I with magnetic linkage is:P = VI = LI(di/dt) = (di/dt) = (d/dt) / L

EnergyThe energy W consumed over time t due to power P dissipated in a resistance R carrying a current I with a voltage drop V is:W = Pt = V2t / R = VIt = I2tR

Similarly, the energy W consumed over time t due to power P dissipated in a conductance G carrying a current I with a voltage drop V is:W = Pt = V2tG = VIt = I2t / G

The energy W stored in a capacitance C holding voltage V with charge Q is:W = CV2 / 2 = QV / 2 = Q2 / 2C

The energy W stored in an inductance L carrying current I with magnetic linkage is:W = LI2 / 2 = I / 2 = 2 / 2L

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BatteriesIf a battery of open-circuit voltage EB has a loaded voltage VL when supplying load current IL, the battery internal resistance RB is:RB = (EB - VL) / IL

The load voltage VL and load current IL for a load resistance RL are:VL = ILRL = EB - ILRB = EBRL / (RB + RL)IL = VL / RL = (EB - VL) / RB = EB / (RB + RL)

The battery short-circuit current Isc is:Isc = EB / RB = EBIL / (EB - VL)

Voltmeter MultiplierThe resistance RS to be connected in series with a voltmeter of full scale voltage VV and full scale current drain IV to increase the full scale voltage to V is:RS = (V - VV) / IV

The power P dissipated by the resistance RS with voltage drop (V - VV) carrying current IV is:P = (V - VV)2 / RS = (V - VV)IV = IV

2RS

Ammeter ShuntThe resistance RP to be connected in parallel with an ammeter of full scale current IA and full scale voltage drop VA to increase the full scale current to I is:RP = VA / (I - IA)

The power P dissipated by the resistance RP with voltage drop VA carrying current (I - IA) is:P = VA

2 / RP = VA(I - IA) = (I - IA)2RP

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Wheatstone BridgeThe Wheatstone Bridge consists of two resistive potential dividers connected to a common voltage source. If one potential divider has resistances R1 and R2 in series and the other potential divider has resistances R3 and R4 in series, with R1 and R3 connected to one side of the voltage source and R2 and R4 connected to the other side of the voltage source, then at the balance point where the two resistively divided voltages are equal:R1 / R2 = R3 / R4

If the value of resistance R4 is unknown and the values of resistances R3, R2 and R1 at the balance point are known, then:R4 = R3R2 / R1

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Library - Electrical Circuit Theorems

Ohm's LawWhen an applied voltage E causes a current I to flow through an impedance Z, the value of the impedance Z is equal to the voltage E divided by the current I.Impedance = Voltage / Current Z = E / I

Similarly, when a voltage E is applied across an impedance Z, the resulting current I through the impedance is equal to the voltage E divided by the impedance Z.

Current = Voltage / Impedance I = E / Z

Similarly, when a current I is passed through an impedance Z, the resulting voltage drop V across the impedance is equal to the current I multiplied by the impedance Z.

Voltage = Current * Impedance V = IZ

Alternatively, using admittance Y which is the reciprocal of impedance Z:

Voltage = Current / Admittance V = I / Y

Kirchhoff's LawsKirchhoff's Current LawAt any instant the sum of all the currents flowing into any circuit node is equal to the sum of all the currents flowing out of that node:Iin = Iout

Similarly, at any instant the algebraic sum of all the currents at any circuit node is zero:I = 0

Kirchhoff's Voltage LawAt any instant the sum of all the voltage sources in any closed circuit is equal to the sum of all the voltage drops in that circuit:E = IZ

Similarly, at any instant the algebraic sum of all the voltages around any closed circuit is zero:E - IZ = 0

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Thévenin's Theorem

Any linear voltage network which may be viewed from two terminals can be replaced by a voltage-source equivalent circuit comprising a single voltage source E and a single series impedance Z. The voltage E is the open-circuit voltage between the two terminals and the impedance Z is the impedance of the network viewed from the terminals with all voltage sources replaced by their internal impedances.

Norton's Theorem

Any linear current network which may be viewed from two terminals can be replaced by a current-source equivalent circuit comprising a single current source I and a single shunt admittance Y. The current I is the short-circuit current between the two terminals and the admittance Y is the admittance of the network viewed from the terminals with all current sources replaced by their internal admittances.

Thévenin and Norton EquivalenceThe open circuit, short circuit and load conditions of the Thévenin model are:Voc = EIsc = E / ZVload = E - IloadZIload = E / (Z + Zload)

The open circuit, short circuit and load conditions of the Norton model are:Voc = I / YIsc = IVload = I / (Y + Yload)Iload = I - VloadY

Thévenin model from Norton model

Voltage = Current / AdmittanceImpedance = 1 / Admittance

E = I / YZ = Y -1

Norton model from Thévenin model

Current = Voltage / ImpedanceAdmittance = 1 / Impedance

I = E / ZY = Z -1

When performing network reduction for a Thévenin or Norton model, note that:- nodes with zero voltage difference may be short-circuited with no effect on the

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network current distribution,- branches carrying zero current may be open-circuited with no effect on the network voltage distribution.

Superposition Theorem

In a linear network with multiple voltage sources, the current in any branch is the sum of the currents which would flow in that branch due to each voltage source acting alone with all other voltage sources replaced by their internal impedances.

Reciprocity Theorem

If a voltage source E acting in one branch of a network causes a current I to flow in another branch of the network, then the same voltage source E acting in the second branch would cause an identical current I to flow in the first branch.

Compensation Theorem

If the impedance Z of a branch in a network in which a current I flows is changed by a finite amount Z, then the change in the currents in all other branches of the network may be calculated by inserting a voltage source of -IZ into that branch with all other voltage sources replaced by their internal impedances.

Millman's Theorem (Parallel Generator Theorem)

If any number of admittances Y1, Y2, Y3, ... meet at a common point P, and the voltages from another point N to the free ends of these admittances are E1, E2, E3, ... then the voltage between points P and N is:VPN = (E1Y1 + E2Y2 + E3Y3 + ...) / (Y1 + Y2 + Y3 + ...)VPN = EY / Y

The short-circuit currents available between points P and N due to each of the voltages E1, E2, E3, ... acting through the respective admitances Y1, Y2, Y3, ... are E1Y1, E2Y2, E3Y3, ... so the voltage between points P and N may be expressed as:VPN = Isc / Y

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Joule's Law

When a current I is passed through a resistance R, the resulting power P dissipated in the resistance is equal to the square of the current I multiplied by the resistance R:P = I2R

By substitution using Ohm's Law for the corresponding voltage drop V (= IR) across the resistance:P = V2 / R = VI = I2R

Maximum Power Transfer TheoremWhen the impedance of a load connected to a power source is varied from open-circuit to short-circuit, the power absorbed by the load has a maximum value at a load impedance which is dependent on the impedance of the power source.

Note that power is zero for an open-circuit (zero current) and for a short-circuit (zero voltage).

Voltage SourceWhen a load resistance RT is connected to a voltage source ES with series resistance RS, maximum power transfer to the load occurs when RT is equal to RS.

Under maximum power transfer conditions, the load resistance RT, load voltage VT, load current IT and load power PT are:RT = RS

VT = ES / 2IT = VT / RT = ES / 2RS

PT = VT2 / RT = ES

2 / 4RS

Current SourceWhen a load conductance GT is connected to a current source IS with shunt conductance GS, maximum power transfer to the load occurs when GT is equal to GS.

Under maximum power transfer conditions, the load conductance GT, load current IT, load voltage VT and load power PT are:GT = GS

IT = IS / 2VT = IT / GT = IS / 2GS

PT = IT2 / GT = IS

2 / 4GS

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Complex ImpedancesWhen a load impedance ZT (comprising variable resistance RT and variable reactance XT) is connected to an alternating voltage source ES with series impedance ZS (comprising resistance RS and reactance XS), maximum power transfer to the load occurs when ZT is equal to ZS

* (the complex conjugate of ZS) such that RT and RS are equal and XT and XS are equal in magnitude but of opposite sign (one inductive and the other capacitive).

When a load impedance ZT (comprising variable resistance RT and constant reactance XT) is connected to an alternating voltage source ES with series impedance ZS (comprising resistance RS and reactance XS), maximum power transfer to the load occurs when RT is equal to the magnitude of the impedance comprising ZS in series with XT:RT = |ZS + XT| = (RS

2 + (XS + XT)2)½

Note that if XT is zero, maximum power transfer occurs when RT is equal to the magnitude of ZS:RT = |ZS| = (RS

2 + XS2)½

When a load impedance ZT with variable magnitude and constant phase angle (constant power factor) is connected to an alternating voltage source ES with series impedance ZS, maximum power transfer to the load occurs when the magnitude of ZT is equal to the magnitude of ZS:(RT

2 + XT2)½ = |ZT| = |ZS| = (RS

2 + XS2)½

Kennelly's Star-Delta TransformationA star network of three impedances ZAN, ZBN and ZCN connected together at common node N can be transformed into a delta network of three impedances ZAB, ZBC and ZCA by the following equations:ZAB = ZAN + ZBN + (ZANZBN / ZCN) = (ZANZBN + ZBNZCN + ZCNZAN) / ZCN

ZBC = ZBN + ZCN + (ZBNZCN / ZAN) = (ZANZBN + ZBNZCN + ZCNZAN) / ZAN

ZCA = ZCN + ZAN + (ZCNZAN / ZBN) = (ZANZBN + ZBNZCN + ZCNZAN) / ZBN

Similarly, using admittances:YAB = YANYBN / (YAN + YBN + YCN)YBC = YBNYCN / (YAN + YBN + YCN)YCA = YCNYAN / (YAN + YBN + YCN)

In general terms:Zdelta = (sum of Zstar pair products) / (opposite Zstar)Ydelta = (adjacent Ystar pair product) / (sum of Ystar)

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Kennelly's Delta-Star TransformationA delta network of three impedances ZAB, ZBC and ZCA can be transformed into a star network of three impedances ZAN, ZBN and ZCN connected together at common node N by the following equations:ZAN = ZCAZAB / (ZAB + ZBC + ZCA)ZBN = ZABZBC / (ZAB + ZBC + ZCA)ZCN = ZBCZCA / (ZAB + ZBC + ZCA)

Similarly, using admittances:YAN = YCA + YAB + (YCAYAB / YBC) = (YABYBC + YBCYCA + YCAYAB) / YBC

YBN = YAB + YBC + (YABYBC / YCA) = (YABYBC + YBCYCA + YCAYAB) / YCA

YCN = YBC + YCA + (YBCYCA / YAB) = (YABYBC + YBCYCA + YCAYAB) / YAB

In general terms:Zstar = (adjacent Zdelta pair product) / (sum of Zdelta)Ystar = (sum of Ydelta pair products) / (opposite Ydelta)

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Library - Electrical System Formulae

ImpedanceThe impedance Z of a resistance R in series with a reactance X is:Z = R + jX

Rectangular and polar forms of impedance Z:Z = R + jX = (R2 + X2)½tan-1(X / R) = |Z|f = |Z|cosf + j|Z|sinf

Addition of impedances Z1 and Z2:Z1 + Z2 = (R1 + jX1) + (R2 + jX2) = (R1 + R2) + j(X1 + X2)

Subtraction of impedances Z1 and Z2:Z1 - Z2 = (R1 + jX1) - (R2 + jX2) = (R1 - R2) + j(X1 - X2)

Multiplication of impedances Z1 and Z2:Z1 * Z2 = |Z1|f * |Z2|f = ( |Z1| * |Z2| )(ff)

Division of impedances Z1 and Z2:Z1 / Z2 = |Z1|f / |Z2|f = ( |Z1| / |Z2| )(ff)

In summary:- use the rectangular form for addition and subtraction,- use the polar form for multiplication and division.

AdmittanceAn impedance Z comprising a resistance R in series with a reactance X can be converted to an admittance Y comprising a conductance G in parallel with a susceptance B:Y = Z -1 = 1 / (R + jX) = (R - jX) / (R2 + X2) = R / (R2 + X2) - jX / (R2 + X2) = G - jBG = R / (R2 + X2) = R / |Z|2

B = X / (R2 + X2) = X / |Z|2

Using the polar form of impedance Z:Y = 1 / |Z|f = |Z| -1f = |Y|f = |Y|cosf - j|Y|sinf

Conversely, an admittance Y comprising a conductance G in parallel with a susceptance B can be converted to an impedance Z comprising a resistance R in series with a reactance X:Z = Y -1 = 1 / (G - jB) = (G + jB) / (G2 + B2) = G / (G2 + B2) + jB / (G2 + B2) = R + jXR = G / (G2 + B2) = G / |Y|2

X = B / (G2 + B2) = B / |Y|2

Using the polar form of admittance Y:Z = 1 / |Y|f = |Y| -1f = |Z|f = |Z|cosf + j|Z|sinf

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The total impedance ZS of impedances Z1, Z2, Z3,... connected in series is:ZS = Z1 + Z1 + Z1 +...The total admittance YP of admittances Y1, Y2, Y3,... connected in parallel is:YP = Y1 + Y1 + Y1 +...

In summary:- use impedances when operating on series circuits,- use admittances when operating on parallel circuits.

ReactanceInductive ReactanceThe inductive reactance XL of an inductance L at angular frequency and frequency f is:XL = L = 2fL

For a sinusoidal current i of amplitude I and angular frequency :i = I sintIf sinusoidal current i is passed through an inductance L, the voltage e across the inductance is:e = L di/dt = LI cost = XLI cost

The current through an inductance lags the voltage across it by 90°.

Capacitive ReactanceThe capacitive reactance XC of a capacitance C at angular frequency and frequency f is:XC = 1 / C = 1 / 2fC

For a sinusoidal voltage v of amplitude V and angular frequency :v = V sintIf sinusoidal voltage v is applied across a capacitance C, the current i through the capacitance is:i = C dv/dt = CV cost = V cost / XC

The current through a capacitance leads the voltage across it by 90°.

ResonanceSeries ResonanceA series circuit comprising an inductance L, a resistance R and a capacitance C has an impedance ZS of:ZS = R + j(XL - XC)where XL = L and XC = 1 / C

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At resonance, the imaginary part of ZS is zero:XC = XL

ZSr = Rr = (1 / LC)½ = 2fr

The quality factor at resonance Qr is:Qr = rL / R = (L / CR2)½ = (1 / R )(L / C)½ = 1 / rCR

Parallel resonanceA parallel circuit comprising an inductance L with a series resistance R, connected in parallel with a capacitance C, has an admittance YP of:YP = 1 / (R + jXL) + 1 / (- jXC) = (R / (R2 + XL

2)) - j(XL / (R2 + XL2) - 1 / XC)

where XL = L and XC = 1 / C

At resonance, the imaginary part of YP is zero:XC = (R2 + XL

2) / XL = XL + R2 / XL = XL(1 + R2 / XL2)

ZPr = YPr-1 = (R2 + XL

2) / R = XLXC / R = L / CRr = (1 / LC - R2 / L2)½ = 2fr

The quality factor at resonance Qr is:Qr = rL / R = (L / CR2 - 1)½ = (1 / R )(L / C - R2)½

Note that for the same values of L, R and C, the parallel resonance frequency is lower than the series resonance frequency, but if the ratio R / L is small then the parallel resonance frequency is close to the series resonance frequency.

Reactive Loads and Power FactorResistance and Series ReactanceThe impedance Z of a reactive load comprising resistance R and series reactance X is:Z = R + jX = |Z|fConverting to the equivalent admittance Y:Y = 1 / Z = 1 / (R + jX) = (R - jX) / (R2 + X2) = R / |Z|2 - jX / |Z|2

When a voltage V (taken as reference) is applied across the reactive load Z, the current I is:I = VY = V(R / |Z|2 - jX / |Z|2) = VR / |Z|2 - jVX / |Z|2 = IP - jIQ

The active current IP and the reactive current IQ are:IP = VR / |Z|2 = |I|cosfIQ = VX / |Z|2 = |I|sinf

The apparent power S, active power P and reactive power Q are:S = V|I| = V2 / |Z| = |I|2|Z|P = VIP = IP

2|Z|2 / R = V2R / |Z|2 = |I|2RQ = VIQ = IQ

2|Z|2 / X = V2X / |Z|2 = |I|2X

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The power factor cosf and reactive factor sinf are:cosf = IP / |I| = P / S = R / |Z|sinf = IQ / |I| = Q / S = X / |Z|

Resistance and Shunt ReactanceThe impedance Z of a reactive load comprising resistance R and shunt reactance X is found from:1 / Z = 1 / R + 1 / jXConverting to the equivalent admittance Y comprising conductance G and shunt susceptance B:Y = 1 / Z = 1 / R - j / X = G - jB = |Y|f

When a voltage V (taken as reference) is applied across the reactive load Y, the current I is:I = VY = V(G - jB) = VG - jVB = IP - jIQ

The active current IP and the reactive current IQ are:IP = VG = V / R = |I|cosfIQ = VB = V / X = |I|sinf

The apparent power S, active power P and reactive power Q are:S = V|I| = |I|2 / |Y| = V2|Y|P = VIP = IP

2 / G = |I|2G / |Y|2 = V2GQ = VIQ = IQ

2 / B = |I|2B / |Y|2 = V2B

The power factor cosf and reactive factor sinf are:cosf = IP / |I| = P / S = G / |Y|sinf = IQ / |I| = Q / S = B / |Y|

Complex PowerWhen a voltage V causes a current I to flow through a reactive load Z, the complex power S is:S = VI* where I* is the conjugate of the complex current I.

Inductive LoadZ = R + jXL

I = IP - jIQ

cosf = R / |Z| (lagging)I* = IP + jIQ

S = P + jQAn inductive load is a sink of lagging VArs (a source of leading VArs).

Capacitive LoadZ = R - jXC

I = IP + jIQ

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cosf = R / |Z| (leading)I* = IP - jIQ

S = P - jQA capacitive load is a source of lagging VArs (a sink of leading VArs).

Three Phase PowerFor a balanced star connected load with line voltage Vline and line current Iline:Vstar = Vline / 3Istar = Iline

Zstar = Vstar / Istar = Vline / 3Iline

Sstar = 3VstarIstar = 3VlineIline = Vline2 / Zstar = 3Iline

2Zstar

For a balanced delta connected load with line voltage Vline and line current Iline:Vdelta = Vline

Idelta = Iline / 3Zdelta = Vdelta / Idelta = 3Vline / Iline

Sdelta = 3VdeltaIdelta = 3VlineIline = 3Vline2 / Zdelta = Iline

2Zdelta

The apparent power S, active power P and reactive power Q are related by:S2 = P2 + Q2

P = ScosfQ = Ssinfwhere cosf is the power factor and sinf is the reactive factor

Note that for equivalence between balanced star and delta connected loads:Zdelta = 3Zstar

Per-unit SystemFor each system parameter, per-unit value is equal to the actual value divided by a base value:Epu = E / Ebase

Ipu = I / Ibase

Zpu = Z / Zbase

Select rated values as base values, usually rated power in MVA and rated phase voltage in kV:Sbase = Srated = 3ElineIline

Ebase = Ephase = Eline/ 3

The base values for line current in kA and per-phase star impedance in ohms/phase are:

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Ibase = Sbase / 3Ebase ( = Sbase / 3Eline)Zbase = Ebase / Ibase = 3Ebase

2 / Sbase ( = Eline2 / Sbase)

Note that selecting the base values for any two of Sbase, Ebase, Ibase or Zbase fixes the base values of all four. Note also that Ohm's Law is satisfied by each of the sets of actual, base and per-unit values for voltage, current and impedance.

TransformersThe primary and secondary MVA ratings of a transformer are equal, but the voltages and currents in the primary (subscript 1) and the secondary (subscript 2) are usually different:3E1lineI1line = S = 3E2lineI2line

Converting to base (per-phase star) values:3E1baseI1base = Sbase = 3E2baseI2base

E1base / E2base = I2base / I1base

Z1base / Z2base = (E1base / E2base)2

The impedance Z21pu referred to the primary side, equivalent to an impedance Z2pu on the secondary side, is:Z21pu = Z2pu(E1base / E2base)2

The impedance Z12pu referred to the secondary side, equivalent to an impedance Z1pu on the primary side, is:Z12pu = Z1pu(E2base / E1base)2

Note that per-unit and percentage values are related by:Zpu = Z% / 100

Symmetrical ComponentsIn any three phase system, the line currents Ia, Ib and Ic may be expressed as the phasor sum of:- a set of balanced positive phase sequence currents Ia1, Ib1 and Ic1 (phase sequence a-b-c),- a set of balanced negative phase sequence currents Ia2, Ib2 and Ic2 (phase sequence a-c-b),- a set of identical zero phase sequence currents Ia0, Ib0 and Ic0 (cophasal, no phase sequence).

The positive, negative and zero sequence currents are calculated from the line currents using:Ia1 = (Ia + hIb + h2Ic) / 3Ia2 = (Ia + h2Ib + hIc) / 3Ia0 = (Ia + Ib + Ic) / 3

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The positive, negative and zero sequence currents are combined to give the line currents using:Ia = Ia1 + Ia2 + Ia0

Ib = Ib1 + Ib2 + Ib0 = h2Ia1 + hIa2 + Ia0

Ic = Ic1 + Ic2 + Ic0 = hIa1 + h2Ia2 + Ia0

The residual current Ir is equal to the total zero sequence current:Ir = Ia0 + Ib0 + Ic0 = 3Ia0 = Ia + Ib + Ic = Ie

which is measured using three current transformers with parallel connected secondaries.Ie is the earth fault current of the system.

Similarly, for phase-to-earth voltages Vae, Vbe and Vce, the residual voltage Vr is equal to the total zero sequence voltage:Vr = Va0 + Vb0 + Vc0 = 3Va0 = Vae + Vbe + Vce = 3Vne

which is measured using an earthed-star / open-delta connected voltage transformer.Vne is the neutral displacement voltage of the system.

The h-operatorThe h-operator (1120°) is the complex cube root of unity:h = - 1 / 2 + j3 / 2 = 1120° = 1-240°h2 = - 1 / 2 - j3 / 2 = 1240° = 1-120°

Some useful properties of h are:1 + h + h2 = 0h + h2 = - 1 = 1180°h - h2 = j3 = 390°h2 - h = - j3 = 3-90°

Fault CalculationsThe different types of short-circuit fault which occur on a power system are:- single phase to earth,- double phase,- double phase to earth,- three phase,- three phase to earth.

For each type of short-circuit fault occurring on an unloaded system:- the first column states the phase voltage and line current conditions at the fault,- the second column states the phase 'a' sequence current and voltage conditions at the fault,- the third column provides formulae for the phase 'a' sequence currents at the fault,

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- the fourth column provides formulae for the fault current and the resulting line currents.By convention, the faulted phases are selected for fault symmetry with respect to reference phase 'a'.

I f = fault currentIe = earth fault currentEa = normal phase voltage at the fault locationZ1 = positive phase sequence network impedance to the faultZ2 = negative phase sequence network impedance to the faultZ0 = zero phase sequence network impedance to the fault

Single phase to earth - fault from phase 'a' to earth:

Va = 0Ib = Ic = 0I f = Ia = Ie

Ia1 = Ia2 = Ia0 = Ia / 3Va1 + Va2 + Va0 = 0

Ia1 = Ea / (Z1 + Z2 + Z0)Ia2 = Ia1

Ia0 = Ia1

I f = 3Ia0 = 3Ea / (Z1 + Z2 + Z0) = Ie

Ia = I f = 3Ea / (Z1 + Z2 + Z0)

Double phase - fault from phase 'b' to phase 'c':

Vb = Vc

Ia = 0I f = Ib = - Ic

Ia1 + Ia2 = 0Ia0 = 0Va1 = Va2

Ia1 = Ea / (Z1 + Z2)Ia2 = - Ia1

Ia0 = 0

I f = - j3Ia1 = - j3Ea / (Z1 + Z2)Ib = I f = - j3Ea / (Z1 + Z2)Ic = - I f = j3Ea / (Z1 + Z2)

Double phase to earth - fault from phase 'b' to phase 'c' to earth:

Vb = Vc = 0Ia = 0I f = Ib + Ic = Ie

Ia1 + Ia2 + Ia0 = 0Va1 = Va2 = Va0

Ia1 = Ea / Znet

Ia2 = - Ia1Z0 / (Z2 + Z0)Ia0 = - Ia1Z2 / (Z2 + Z0)

I f = 3Ia0 = - 3EaZ2 / zz = Ie

Ib = I f / 2 - j3Ea(Z2 / 2 + Z0) / zz

Ic = I f / 2 + j3Ea(Z2 / 2 + Z0) / zz

Znet = Z1 + Z2Z0 / (Z2 + Z0) and zz = Z1Z2 + Z2Z0 + Z0Z1 = (Z2 + Z0)Znet

Three phase (and three phase to earth) - fault from phase 'a' to phase 'b' to phase 'c' (to earth):

Va = Vb = Vc (= 0)Ia + Ib + Ic = 0 (= Ie)I f = Ia = hIb = h2Ic

Va0 = Va (= 0)Va1 = Va2 = 0

Ia1 = Ea / Z1

Ia2 = 0Ia0 = 0

I f = Ia1 = Ea / Z1 = Ia

Ib = Eb / Z1

Ic = Ec / Z1

The values of Z1, Z2 and Z0 are each determined from the respective positive, negative and zero sequence impedance networks by network reduction to a single impedance.

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Note that the single phase fault current is greater than the three phase fault current if Z0 is less than (2Z1 - Z2).

Note also that if the system is earthed through an impedance Zn (carrying current 3I0) then an impedance 3Zn (carrying current I0) must be included in the zero sequence impedance network.

Three Phase Fault LevelThe symmetrical three phase short-circuit current Isc of a power system with no-load line and phase voltages Eline and Ephase and source impedance ZS per-phase star is:Isc = Ephase / ZS = Eline / 3ZS

The three phase fault level Ssc of the power system is:Ssc = 3Isc

2ZS = 3EphaseIsc = 3Ephase2 / ZS = Eline

2 / ZS

Note that if the X / R ratio of the source impedance ZS (comprising resistance RS and reactance XS) is sufficiently large, then ZS XS.

TransformersIf a transformer of rating ST (taken as base) and per-unit impedance ZTpu is fed from a source with unlimited fault level (infinite busbars), then the per-unit secondary short-circuit current I2pu and fault level S2pu are:I2pu = E2pu / ZTpu = 1.0 / ZTpu

S2pu = I2pu = 1.0 / ZTpu

If the source fault level is limited to SS by per-unit source impedance ZSpu (to the same base as ZTpu), then the secondary short-circuit current I2pu and fault level S2pu are reduced to:I2pu = E2pu / (ZTpu + ZSpu) = 1.0 / (ZTpu + ZSpu)S2pu = I2pu = 1.0 / (ZTpu + ZSpu)where ZSpu = ST / SS

Thermal Short-time RatingIf a conductor which is rated to carry full load current Iload continuously is rated to carry a maximum fault current Ilimit for a time tlimit, then a lower fault current Ifault can be carried for a longer time tfault according to:( Ilimit - Iload )2 tlimit = ( Ifault - Iload )2 tfault

Rearranging for Ifault and tfault:Ifault = ( Ilimit - Iload ) ( tlimit / tfault )½ + Iload

tfault = tlimit ( Ilimit - Iload )2 / ( Ifault - Iload )2

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If Iload is small compared with Ilimit and Ifault, then:Ilimit

2 tlimit Ifault2 tfault

Ifault Ilimit ( tlimit / tfault )½

tfault tlimit ( Ilimit / Ifault )2

Note that if the current Ifault is reduced by a factor of two, then the time tfault is increased by a factor of four.

Instrument TransformersVoltage TransformerFor a voltage transformer of voltampere rating S, rated primary voltage VP and rated secondary voltage VS, the maximum secondary current ISmax, maximum secondary burden conductance GBmax and maximum primary current IPmax are:ISmax = S / VS

GBmax = ISmax / VS = S / VS2

IPmax = S / VP = ISmaxVS / VP

Current TransformerFor a measurement current transformer of voltampere rating S, rated primary current IP and rated secondary current IS, the maximum secondary voltage VSmax, maximum secondary burden resistance RBmax and maximum primary voltage VPmax are:VSmax = S / IS

RBmax = VSmax / IS = S / IS2

VPmax = S / IP = VSmaxIS / IP

For a protection current transformer of voltampere rating S, rated primary current IP, rated secondary current IS and rated accuracy limit factor F, the rated secondary reference voltage VSF, maximum secondary burden resistance RBmax and equivalent primary reference voltage VPF are:VSF = SF / IS

RBmax = VSF / ISF = S / IS2

VPF = SF / IP = VSFIS / IP

Impedance MeasurementIf the primary voltage Vpri and the primary current Ipri are measured at a point in a system, then the primary impedance Zpri at that point is:Zpri = Vpri / Ipri

If the measured voltage is the secondary voltage Vsec of a voltage transformer of primary/secondary ratio NV and the measured current is the secondary current Isec of a current transformer of primary/secondary ratio NI, then the primary impedance Zpri is related to the secondary impedance Zsec by:

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Zpri = Vpri / Ipri = VsecNV / IsecNI = ZsecNV / NI = ZsecNZ

where NZ = NV / NI

If the no-load (source) voltage Epri is also measured at the point, then the source impedance ZTpri to the point is:ZTpri = (Epri - Vpri) / Ipri = (Esec - Vsec)NV / IsecNI = ZTsecNV / NI = ZTsecNZ

Power Factor CorrectionIf an inductive load with an active power demand P has an uncorrected power factor of cosf1 lagging, and is required to have a corrected power factor of cosf2 lagging, the uncorrected and corrected reactive power demands, Q1 and Q2, are:Q1 = P tanf1

Q2 = P tanf2

where tanfn = (1 / cos2fn - 1)½

The leading (capacitive) reactive power demand QC which must be connected across the load is:QC = Q1 - Q2 = P (tanf1 - tanf2)

The uncorrected and corrected apparent power demands, S1 and S2, are related by:S1cosf1 = P = S2cosf2

Comparing corrected and uncorrected load currents and apparent power demands:I2 / I1 = S2 / S1 = cosf1 / cosf2

If the load is required to have a corrected power factor of unity, Q2 is zero and:QC = Q1 = P tanf1

I2 / I1 = S2 / S1 = cosf1 = P / S1

Shunt CapacitorsFor star-connected shunt capacitors each of capacitance Cstar on a three phase system of line voltage Vline and frequency f, the leading reactive power demand QCstar and the leading reactive line current Iline are:QCstar = Vline

2 / XCstar = 2fCstarVline2

Iline = QCstar / 3Vline = Vline / 3XCstar

Cstar = QCstar / 2fVline2

For delta-connected shunt capacitors each of capacitance Cdelta on a three phase system of line voltage Vline and frequency f, the leading reactive power demand QCdelta and the leading reactive line current Iline are:QCdelta = 3Vline

2 / XCdelta = 6fCdeltaVline2

Iline = QCdelta / 3Vline = 3Vline / XCdelta

Cdelta = QCdelta / 6fVline2

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Note that for the same leading reactive power QC:XCdelta = 3XCstar

Cdelta = Cstar / 3

Series CapacitorsFor series line capacitors each of capacitance Cseries carrying line current Iline on a three phase system of frequency f, the voltage drop Vdrop across each line capacitor and the total leading reactive power demand QCseries of the set of three line capacitors are:Vdrop = IlineXCseries = Iline / 2fCseries

QCseries = 3Vdrop2 / XCseries = 3VdropIline = 3Iline

2XCseries = 3Iline2 / 2fCseries

Cseries = 3Iline2 / 2fQCseries

Note that the apparent power rating Srating of the set of three series line capacitors is based on the line voltage Vline and not the voltage drop Vdrop:Srating = 3VlineIline

ReactorsShunt ReactorsFor star-connected shunt reactors each of inductance Lstar on a three phase system of line voltage Vline and frequency f, the lagging reactive power demand QLstar and the lagging reactive line current Iline are:QLstar = Vline

2 / XLstar = Vline2 / 2fLstar

Iline = QLstar / 3Vline = Vline / 3XLstar

Lstar = Vline2 / 2fQLstar

For delta-connected shunt reactors each of inductance Ldelta on a three phase system of line voltage Vline and frequency f, the lagging reactive power demand QLdelta and the lagging reactive line current Iline are:QLdelta = 3Vline

2 / XLdelta = 3Vline2 / 2fLdelta

Iline = QLdelta / 3Vline = 3Vline / XLdelta

Ldelta = 3Vline2 / 2fQLdelta

Note that for the same lagging reactive power QL:XLdelta = 3XLstar

Ldelta = 3Lstar

Series ReactorsFor series line reactors each of inductance Lseries carrying line current Iline on a three phase system of frequency f, the voltage drop Vdrop across each line reactor and the total lagging reactive power demand QLseries of the set of three line reactors are:Vdrop = IlineXLseries = 2fLseriesIline

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QLseries = 3Vdrop2 / XLseries = 3VdropIline = 3Iline

2XLseries = 6fLseriesIline2

Lseries = QLseries / 6fIline2

Note that the apparent power rating Srating of the set of three series line reactors is based on the line voltage Vline and not the voltage drop Vdrop:Srating = 3VlineIline

Harmonic ResonanceIf a node in a power system operating at frequency f has a inductive source reactance XL per phase and has power factor correction with a capacitive reactance XC per phase, the source inductance L and the correction capacitance C are:L = XL / C = 1 / XC

where f

The series resonance angular frequency r of an inductance L with a capacitance C is:r = (1 / LC)½ = (XC / XL)½

The three phase fault level Ssc at the node for no-load phase voltage E and source impedance Z per-phase star is:Ssc = 3E2 / |Z| = 3E2 / |R + jXL|If the ratio XL / R of the source impedance Z is sufficiently large, |Z| XL so that:Ssc 3E2 / XL

The reactive power rating QC of the power factor correction capacitors for a capacitive reactance XC per phase at phase voltage E is:QC = 3E2 / XC

The harmonic number fr / f of the series resonance of XL with XC is:fr / f = r / = (XC / XL)½ (Ssc / QC)½

Note that the ratio XL / XC which results in a harmonic number fr / f is:XL / XC = 1 / ( fr / f )2

so for fr / f to be equal to the geometric mean of the third and fifth harmonics:fr / f = 15 = 3.873XL / XC = 1 / 15 = 0.067

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Library - Electrical Machine Applications

TransformersFor an ideal two-winding transformer with primary voltage V1 applied across N1 primary turns and secondary voltage V2 appearing across N2 secondary turns:V1 / V2 = N1 / N2

The primary current I1 and secondary current I2 are related by:I1 / I2 = N2 / N1 = V2 / V1

For an ideal step-down auto-transformer with primary voltage V1 applied across (N1 + N2) primary turns and secondary voltage V2 appearing across N2 secondary turns:V1 / V2 = (N1 + N2) / N2

The primary (input) current I1 and secondary (output) current I2 are related by:I1 / I2 = N2 / (N1 + N2) = V2 / V1

Note that the winding current is I1 through the N1 section and (I2 - I1) through the N2 section.

For a single-phase transformer with rated primary voltage V1, rated primary current I1, rated secondary voltage V2 and rated secondary current I2, the voltampere rating S is:S = V1I1 = V2I2

For a balanced m-phase transformer with rated primary phase voltage V1, rated primary current I1, rated secondary phase voltage V2 and rated secondary current I2, the voltampere rating S is:S = mV1I1 = mV2I2

The primary circuit impedance Z1 referred to the secondary circuit for an ideal transformer with N1 primary turns and N2 secondary turns is:Z12 = Z1(N2 / N1)2

The secondary circuit impedance Z2 referred to the primary circuit for an ideal transformer with N1 primary turns and N2 secondary turns is:Z21 = Z2(N1 / N2)2

The voltage regulation V2 of a transformer is the rise in secondary voltage which occurs when rated load is disconnected from the secondary with rated voltage applied to the primary. For a transformer with a secondary voltage E2 unloaded and V2 at rated load, the per-unit voltage regulation V2pu is:V2pu = (E2 - V2) / V2

Note that the per-unit base voltage is usually V2 and not E2.

Open Circuit TestIf a transformer with its secondary open-circuited is energised at rated primary

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voltage, then the input power Poc represents the core loss (iron loss PFe) of the transformer:Poc = PFe

The per-phase star values of the shunt magnetising admittance Ym, conductance Gm and susceptance Bm of an m-phase transformer are calculated from the open-circuit test results for the per-phase primary voltage V1oc, per-phase primary current I1oc and input power Poc using:Ym = I1oc / V1oc

Gm = mV1oc2 / Poc

Bm = (Ym2 - Gm

2)½

Short Circuit TestIf a transformer with its secondary short-circuited is energised at a reduced primary voltage which causes rated secondary current to flow through the short-circuit, then the input power Psc represents the load loss (primary copper loss P1Cu, secondary copper loss P2Cu and stray loss Pstray) of the transformer:Psc = P1Cu + P2Cu + Pstray

Note that the temperature rise should be allowed to stabilise because conductor resistance varies with temperature.

If the resistance of each winding is determined by winding resistance tests immediately after the short circuit test, then the load loss of an m-phase transformer may be split into primary copper loss P1Cu, secondary copper loss P2Cu and stray loss Pstray:P1Cu = mI1sc

2R1star

P2Cu = mI2sc2R2star

Pstray = Psc - P1Cu - P2Cu

If the stray loss is neglected, the per-phase star values referred to the primary of the total series impedance Zs1, resistance Rs1 and reactance Xs1 of an m-phase transformer are calculated from the short-circuit test results for the per-phase primary voltage V1sc, per-phase primary current I1sc and input power Psc using:Zs1 = V1sc / I1sc = Z1 + Z2(N1

2 / N22)

Rs1 = Psc / mI1sc2 = R1 + R2(N1

2 / N22)

Xs1 = (Zs12 - Rs1

2)½ = X1 + X2(N12 / N2

2)where Z1, R1 and X1 are primary values and Z2, R2 and X2 are secondary values

Winding Resistance TestThe resistance of each winding is measured using a small direct current to avoid thermal and inductive effects. If a voltage Vdc causes current Idc to flow, then the resistance R is:R = Vdc / Idc

If the winding under test is a fully connected balanced star or delta and the resistance measured between any two phases is Rtest, then the equivalent

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winding resistances Rstar or Rdelta are:Rstar = Rtest / 2Rdelta = 3Rtest / 2

The per-phase star primary and secondary winding resistances R1star and R2star of an m-phase transformer may be used to calculate the separate primary and secondary copper losses P1Cu and P2Cu:P1Cu = mI1

2R1star

P2Cu = mI22R2star

Note that if the primary and secondary copper losses are equal, then the primary and secondary resistances R1star and R2star are related by:R1star / R2star = I2

2 / I12 = N1

2 / N22

The primary and secondary winding resistances R1 and R2 may also be used to check the effect of stray loss on the total series resistance referred to the primary, Rs1, calculated from the short circuit test results:Rs1 = R1 + R2(N1

2 / N22)

Induction MachinesThe synchronous rotational speed ns and synchronous angular speed s of a machine with p pole pairs running on a supply of frequency fs are:ns = 60fs / ps = 2fs / p = 2ns / 60

The per-unit slip s of an induction machine of synchronous rotational speed ns running at rotational speed nm is:s = (ns - nm) / ns

Rearranging for rotational speed nm:nm = (1 - s)ns

Using angular speed instead of rotational speed n:m = (1 - s)s

The rated load torque TM for a rated output power PM is:TM = PM / m = 60PM / 2nm

For an induction machine with Ns stator turns and Nr rotor turns running at slip s on a supply of voltage Es and frequency fs, the rotor induced voltage and frequency Er and fr are:Er = sEsNr / Ns

fr = sfs

For a rotor current Ir, the equivalent stator current Irs is:Irs = IrNr / Ns

Note that the rotor / stator ratios are Ns / Nr for current, sNr / Ns for voltage and s for frequency.

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For an induction machine with rotor resistance Rr and locked rotor leakage reactance Xr, the rotor impedance Zr at slip s is:Zr = Rr + jsXr

The stator circuit equivalent impedance Zrf for a rotor / stator frequency ratio s is:Zrf = Rrs / s + jXrs

For an induction motor with synchronous angular speed s running at angular speed m and slip s, the airgap transfer power Pt, rotor copper loss Pr and gross output power Pm for a gross output torque Tm are related by:Pt = sTm = Pr / s = Pm / (1 - s)Pr = sPt = sPm / (1 - s)Pm = mTm = (1 - s)Pt

The power ratios are:Pt : Pr : Pm = 1 : s : (1 - s)The gross motor efficiency m (neglecting stator and mechanical losses) is:m = Pm / Pt = 1 - s

An induction machine can be operated as a generator, a motor or a brake:- for negative slip (speed above synchronous) the machine is a generator,- for positive slip between 0 and 1 (speed below synchronous) the machine is a motor,- for positive slip greater than 1 (speed negative) the machine is a brake,In all cases the magnetizing current (at lagging power factor) is provided by the supply system.

No Load TestIf an induction machine with its rotor unloaded is energised at rated voltage, then the input power represents the sum of the iron loss and mechanical loss of the machine.

Locked Rotor TestIf an induction machine with its rotor locked is energised at a reduced voltage which causes rated current input, then the input power represents the sum of the full load copper loss and stray loss of the machine.

Stator Resistance TestThe resistance of the stator winding is measured using a small direct current.

Synchronous MachinesThe synchronous rotational speed ns and synchronous angular speed s of a machine with p pole pairs running on a supply of frequency fs are:ns = 60fs / ps = 2fs / p

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The output power Pm for a load torque Tm is:Pm = sTm

The rated load torque TM for a rated output power PM is:TM = PM / s = PMp / 2fs = 60PM / 2ns

Synchronous GeneratorFor a synchronous generator with stator induced voltage Es, stator current Is and synchronous impedance Zs, the terminal voltage V is:V = Es - IsZs = Es - Is(Rs + jXs)where Rs is the stator resistance and Xs is the synchronous reactance

Synchronous MotorFor a synchronous motor with stator induced voltage Es, stator current Is and synchronous impedance Zs, the terminal voltage V is:V = Es + IsZs = Es + Is(Rs + jXs)where Rs is the stator resistance and Xs is the synchronous reactance

Note that the field excitation of a parallelled synchronous machine determines its power factor:- an under-excited machine operates with a leading power factor,- an over-excited machine operates with a lagging power factor.The field excitation of an isolated synchronous generator determines its output voltage.

Direct Current MachinesShunt GeneratorFor a shunt generator with armature induced voltage Ea, armature current Ia and armature resistance Ra, the terminal voltage V is:V = Ea - IaRa

The field current I f for a field resistance R f is:I f = V / R f

The armature induced voltage Ea and torque T with magnetic flux at angular speed are:Ea = k f = kmT = k fIa = kmIa

where k f and km are design coefficients of the machine.

Note that for a shunt generator:- induced voltage is proportional to speed,- torque is proportional to armature current.

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The airgap power Pe for a shunt generator is:Pe = T = EaIa = km Ia

Shunt MotorFor a shunt motor with armature induced voltage Ea, armature current Ia and armature resistance Ra, the terminal voltage V is:V = Ea + IaRa

The field current I f for a field resistance R f is:I f = V / R f

The armature induced voltage Ea and torque T with magnetic flux at angular speed are:Ea = k f = kmT = k fIa = kmIa


Note that for a shunt motor:- induced voltage is proportional to speed,- torque is proportional to armature current.

The airgap power Pe for a shunt motor is:Pe = T = EaIa = km Ia

Series MotorFor a series motor with armature induced voltage Ea, armature current Ia, armature resistance Ra and field resistance R f, the terminal voltage V is:V = Ea + IaRa + IaR f = Ea + Ia(Ra + R f)The field current is equal to the armature current.

The armature induced voltage Ea and torque T with magnetic flux at angular speed are:Ea = k f Ia = km Ia

T = k fIa2 = kmIa

2


Note that for a series motor:- induced voltage is proportional to both speed and armature current,- torque is proportional to the square of armature current,- armature current is inversely proportional to speed for a constant induced voltage.

The airgap power Pe for a series motor is:Pe = T = EaIa = km Ia

2

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EfficiencyThe per-unit efficiency of an electrical machine with input power Pin, output power Pout and power loss Ploss is: = Pout / Pin = Pout / (Pout + Ploss) = (Pin - Ploss) / Pin

Rearranging the efficiency equations:Pin = Pout + Ploss = Pout / = Ploss / (1 - )Pout = Pin - Ploss = Pin = Ploss / (1 - )Ploss = Pin - Pout = (1 - )Pin = (1 - )Pout /

For an electrical machine with output power Pout (proportional to current) and power loss Ploss comprising a fixed loss Pfix (independent of current) plus a variable loss Pvar (proportional to square of current) the efficiency is a maximum when Pvar is equal to Pfix.

For a transformer, Pfix is the iron loss and Pvar is the copper loss plus the stray loss.

For an induction machine, Pfix is the iron loss plus the mechanical loss and Pvar is the copper loss plus the stray loss.

Energy ConversionComparing megawatt-hours and gigajoules, 1 MWh is equivalent to 3.6 GJ. For an energy conversion process with a per-unit efficiency , 1 MWh of energy output is obtained from (3.6 / ) GJ of energy input.

Temperature RiseThe resistance of copper and aluminium windings increases with temperature, and the relationship is quite linear over the normal range of operating temperatures. For a linear relationship, if the winding resistance is R1 at temperature 1 and R2 at temperature 2, then:R1 / (1 - 0) = R2 / (2 - 0) = (R2 - R1) / (2 - 1)where 0 is the extrapolated temperature for zero resistance.

The ratio of resistances R2 and R1 is:R2 / R1 = (2 - 0) / (1 - 0)

The average temperature rise of a winding under load may be estimated from measured values of the cold winding resistance R1 at temperature 1 (usually ambient temperature) and the hot winding resistance R2 at temperature 2, using: = 2 - 1 = (1 - 0) (R2 - R1) / R1

Rearranging for per-unit change in resistance Rpu relative to R1:Rpu = (R2 - R1) / R1 = (2 - 1) / (1 - 0) = / (1 - 0)

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Note that the resistance values are measured using a small direct current to avoid thermal and inductive effects.

Copper WindingsThe value of 0 for copper is - 234.5 °C, so that: = 2 - 1 = (1 + 234.5) (R2 - R1) / R1

If 1 is 20 °C and is 1 degC:Rpu = (R2 - R1) / R1 = / (1 - 0) = 1 / 254.5 = 0.00393The temperature coefficient of resistance of copper at 20 °C is 0.00393 per degC.

Aluminium WindingsThe value of 0 for aluminium is - 228 °C, so that: = 2 - 1 = (1 + 228) (R2 - R1) / R1

If 1 is 20 °C and is 1 degC:Rpu = (R2 - R1) / R1 = / (1 - 0) = 1 / 248 = 0.00403The temperature coefficient of resistance of aluminium at 20 °C is 0.00403 per degC.

Note that aluminium has 61% of the conductivity and 30% of the density of copper, therefore for the same conductance (and same resistance) an aluminium conductor has 164% of the cross-sectional area, 128% of the diameter and 49% of the mass of a copper conductor.

Dielectric Dissipation FactorIf an alternating voltage V of frequency f is applied across an insulation system comprising capacitance C and equivalent series loss resistance RS, then the voltage VR across RS and the voltage VC across C due to the resulting current I are:VR = IRS

VC = IXC

V = (VR2 + VC

2)½

The dielectric dissipation factor of the insulation system is the tangent of the dielectric loss angle between VC and V:tan = VR / VC = RS / XC = 2fCRS

RS = XCtan = tan / 2fCNote that an increase in the dielectric losses of a insulation system (from an increase in the series loss resistance RS) results in an increase in tan. Note also that tan increases with frequency.

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The dielectric power loss P is related to the capacitive reactive power QC by:P = I2RS = I2XCtan = QCtan

The power factor of the insulation system is the cosine of the phase angle f between VR and V:cosf = VR / Vso that and f are related by: + f = 90°

tan and cosf are related by:tan = 1 / tanf = cosf / sinf = cosf / (1 - cos2f)½

so that when cosf is close to zero, tan cosf

Note that the series loss resistance RS is not related to the shunt leakage resistance of the insulation system (which is measured using direct current).

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electric power formulas

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