هذه مجموعة من المعادلات الكهربية الاساسية وفى الغالب تضم جميع الاساسيات
Impedance
The
impedance Z of a resistance R in series with a reactance X
is:
Z = R + jX
Z = R + jX
Rectangular
and polar forms of impedance Z:
Z = R + jX = (R2 + X2)½Ðtan-1(X / R) = |Z|Ðf = |Z|cosf + j|Z|sinf
Z = R + jX = (R2 + X2)½Ðtan-1(X / R) = |Z|Ðf = |Z|cosf + j|Z|sinf
Addition
of impedances Z1 and Z2:
Z1 + Z2 = (R1 + jX1) + (R2 + jX2) = (R1 + R2) + j(X1 + X2)
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)
Z1 - Z2 = (R1 + jX1) - (R2 + jX2) = (R1 - R2) + j(X1 - X2)
Multiplication
of impedances Z1 and Z2:
Z1 * Z2 = |Z1|Ðf1 * |Z2|Ðf2 = ( |Z1| * |Z2| )Ð(f1 + f2)
Z1 * Z2 = |Z1|Ðf1 * |Z2|Ðf2 = ( |Z1| * |Z2| )Ð(f1 + f2)
Division
of impedances Z1 and Z2:
Z1 / Z2 = |Z1|Ðf1 / |Z2|Ðf2 = ( |Z1| / |Z2| )Ð(f1 - f2)
Z1 / Z2 = |Z1|Ðf1 / |Z2|Ðf2 = ( |Z1| / |Z2| )Ð(f1 - f2)
In
summary:
- use the rectangular form for addition and subtraction,
- use the polar form for multiplication and division.
- use the rectangular form for addition and subtraction,
- use the polar form for multiplication and division.
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Admittance
An
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 - jB
G = R / (R2 + X2) = R / |Z|2
B = X / (R2 + X2) = X / |Z|2
Using the polar form of impedance Z:
Y = 1 / |Z|Ðf = |Z| -1Ð-f = |Y|Ð-f = |Y|cosf - j|Y|sinf
Y = Z -1 = 1 / (R + jX) = (R - jX) / (R2 + X2) = R / (R2 + X2) - jX / (R2 + X2) = G - jB
G = R / (R2 + X2) = R / |Z|2
B = X / (R2 + X2) = X / |Z|2
Using the polar form of impedance Z:
Y = 1 / |Z|Ðf = |Z| -1Ð-f = |Y|Ð-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 + jX
R = G / (G2 + B2) = G / |Y|2
X = B / (G2 + B2) = B / |Y|2
Using the polar form of admittance Y:
Z = 1 / |Y|Ð-f = |Y| -1Ðf = |Z|Ðf = |Z|cosf + j|Z|sinf
Z = Y -1 = 1 / (G - jB) = (G + jB) / (G2 + B2) = G / (G2 + B2) + jB / (G2 + B2) = R + jX
R = G / (G2 + B2) = G / |Y|2
X = B / (G2 + B2) = B / |Y|2
Using the polar form of admittance Y:
Z = 1 / |Y|Ð-f = |Y| -1Ðf = |Z|Ðf = |Z|cosf + j|Z|sinf
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 +...
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.
- use impedances when operating on series circuits,
- use admittances when operating on parallel circuits.
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Reactance
Inductive Reactance
The inductive reactance XL of an inductance L at angular frequency w and frequency f is:
XL = wL = 2pfL
The inductive reactance XL of an inductance L at angular frequency w and frequency f is:
XL = wL = 2pfL
For
a sinusoidal current i of amplitude I and angular frequency w:
i = I sinwt
If sinusoidal current i is passed through an inductance L, the voltage e across the inductance is:
e = L di/dt = wLI coswt = XLI coswt
i = I sinwt
If sinusoidal current i is passed through an inductance L, the voltage e across the inductance is:
e = L di/dt = wLI coswt = XLI coswt
The
current through an inductance lags the voltage across it by 90°.
Capacitive Reactance
The capacitive reactance XC of a capacitance C at angular frequency w and frequency f is:
XC = 1 / wC = 1 / 2pfC
The capacitive reactance XC of a capacitance C at angular frequency w and frequency f is:
XC = 1 / wC = 1 / 2pfC
For
a sinusoidal voltage v of amplitude V and angular frequency w:
v = V sinwt
If sinusoidal voltage v is applied across a capacitance C, the current i through the capacitance is:
i = C dv/dt = wCV coswt = V coswt / XC
v = V sinwt
If sinusoidal voltage v is applied across a capacitance C, the current i through the capacitance is:
i = C dv/dt = wCV coswt = V coswt / XC
The
current through a capacitance leads the voltage across it by 90°.
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Resonance
Series Resonance
A 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 = wL and XC = 1 / wC
A 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 = wL and XC = 1 / wC
At
resonance, the imaginary part of ZS is zero:
XC = XL
ZSr = R
wr = (1 / LC)½ = 2pfr
XC = XL
ZSr = R
wr = (1 / LC)½ = 2pfr
Parallel resonance
A 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 + XL2)) - j(XL / (R2 + XL2) - 1 / XC)
where XL = wL and XC = 1 / wC
A 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 + XL2)) - j(XL / (R2 + XL2) - 1 / XC)
where XL = wL and XC = 1 / wC
At
resonance, the imaginary part of YP is zero:
XC = (R2 + XL2) / XL = XL + R2 / XL = XL(1 + R2 / XL2)
ZPr = YPr-1 = (R2 + XL2) / R = XLXC / R = L / CR
wr = (1 / LC - R2 / L2)½ = 2pfr
XC = (R2 + XL2) / XL = XL + R2 / XL = XL(1 + R2 / XL2)
ZPr = YPr-1 = (R2 + XL2) / R = XLXC / R = L / CR
wr = (1 / LC - R2 / L2)½ = 2pfr
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.
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Reactive Loads and Power Factor
Resistance and Series Reactance
The impedance Z of a reactive load comprising resistance R and series reactance X is:
Z = R + jX = |Z|Ðf
Converting to the equivalent admittance Y:
Y = 1 / Z = 1 / (R + jX) = (R - jX) / (R2 + X2) = R / |Z|2 - jX / |Z|2
The impedance Z of a reactive load comprising resistance R and series reactance X is:
Z = R + jX = |Z|Ðf
Converting 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|cosf
IQ = VX / |Z|2 = |I|sinf
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|cosf
IQ = 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 = IP2|Z|2 / R = V2R / |Z|2 = |I|2R
Q = VIQ = IQ2|Z|2 / X = V2X / |Z|2 = |I|2X
S = V|I| = V2 / |Z| = |I|2|Z|
P = VIP = IP2|Z|2 / R = V2R / |Z|2 = |I|2R
Q = VIQ = IQ2|Z|2 / X = V2X / |Z|2 = |I|2X
The
power factor cosf and reactive factor sinf are:
cosf = IP / |I| = P / S = R / |Z|
sinf = IQ / |I| = Q / S = X / |Z|
cosf = IP / |I| = P / S = R / |Z|
sinf = IQ / |I| = Q / S = X / |Z|
Resistance and Shunt Reactance
The impedance Z of a reactive load comprising resistance R and shunt reactance X is found from:
1 / Z = 1 / R + 1 / jX
Converting to the equivalent admittance Y comprising conductance G and shunt susceptance B:
Y = 1 / Z = 1 / R - j / X = G - jB = |Y|Ð-f
The impedance Z of a reactive load comprising resistance R and shunt reactance X is found from:
1 / Z = 1 / R + 1 / jX
Converting to the equivalent admittance Y comprising conductance G and shunt susceptance B:
Y = 1 / Z = 1 / R - j / X = G - jB = |Y|Ð-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|cosf
IQ = VB = V / X = |I|sinf
I = VY = V(G - jB) = VG - jVB = IP - jIQ
The active current IP and the reactive current IQ are:
IP = VG = V / R = |I|cosf
IQ = 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 = IP2 / G = |I|2G / |Y|2 = V2G
Q = VIQ = IQ2 / B = |I|2B / |Y|2 = V2B
S = V|I| = |I|2 / |Y| = V2|Y|
P = VIP = IP2 / G = |I|2G / |Y|2 = V2G
Q = VIQ = IQ2 / 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|
cosf = IP / |I| = P / S = G / |Y|
sinf = IQ / |I| = Q / S = B / |Y|
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Complex Power
When
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.
S = VI* where I* is the conjugate of the complex current I.
Inductive Load
Z = R + jXL
I = IP - jIQ
cosf = R / |Z| (lagging)
I* = IP + jIQ
S = P + jQ
An inductive load is a sink of lagging VArs (a source of leading VArs).
Z = R + jXL
I = IP - jIQ
cosf = R / |Z| (lagging)
I* = IP + jIQ
S = P + jQ
An inductive load is a sink of lagging VArs (a source of leading VArs).
Capacitive Load
Z = R - jXC
I = IP + jIQ
cosf = R / |Z| (leading)
I* = IP - jIQ
S = P - jQ
A capacitive load is a source of lagging VArs (a sink of leading VArs).
Z = R - jXC
I = IP + jIQ
cosf = R / |Z| (leading)
I* = IP - jIQ
S = P - jQ
A capacitive load is a source of lagging VArs (a sink of leading VArs).
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Three Phase Power
For
a balanced star connected load with line voltage Vline
and line current Iline:
Vstar = Vline / Ö3
Istar = Iline
Zstar = Vstar / Istar = Vline / Ö3Iline
Sstar = 3VstarIstar = Ö3VlineIline = Vline2 / Zstar = 3Iline2Zstar
Vstar = Vline / Ö3
Istar = Iline
Zstar = Vstar / Istar = Vline / Ö3Iline
Sstar = 3VstarIstar = Ö3VlineIline = Vline2 / Zstar = 3Iline2Zstar
For
a balanced delta connected load with line voltage Vline
and line current Iline:
Vdelta = Vline
Idelta = Iline / Ö3
Zdelta = Vdelta / Idelta = Ö3Vline / Iline
Sdelta = 3VdeltaIdelta = Ö3VlineIline = 3Vline2 / Zdelta = Iline2Zdelta
Vdelta = Vline
Idelta = Iline / Ö3
Zdelta = Vdelta / Idelta = Ö3Vline / Iline
Sdelta = 3VdeltaIdelta = Ö3VlineIline = 3Vline2 / Zdelta = Iline2Zdelta
The
apparent power S, active power P and reactive power Q are
related by:
S2 = P2 + Q2
P = Scosf
Q = Ssinf
where cosf is the power factor and sinf is the reactive factor
S2 = P2 + Q2
P = Scosf
Q = Ssinf
where cosf is the power factor and sinf is the reactive factor
Note
that for equivalence between balanced star and delta connected loads:
Zdelta = 3Zstar
Zdelta = 3Zstar
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Per-unit System
For
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
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 = Ö3ElineIline
Ebase = Ephase = Eline/ Ö3
Sbase = Srated = Ö3ElineIline
Ebase = Ephase = Eline/ Ö3
The
base values for line current in kA and per-phase star impedance in Ohms/phase
are:
Ibase = Sbase / 3Ebase ( = Sbase / Ö3Eline)
Zbase = Ebase / Ibase = 3Ebase2 / Sbase ( = Eline2 / Sbase)
Ibase = Sbase / 3Ebase ( = Sbase / Ö3Eline)
Zbase = Ebase / Ibase = 3Ebase2 / 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.
Transformers
The 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:
Ö3E1lineI1line = S = Ö3E2lineI2line
The 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:
Ö3E1lineI1line = S = Ö3E2lineI2line
Converting
to base (per-phase star) values:
3E1baseI1base = Sbase = 3E2baseI2base
E1base / E2base = I2base / I1base
Z1base / Z2base = (E1base / E2base)2
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
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
Z12pu = Z1pu(E2base / E1base)2
Note
that per-unit and percentage values are related by:
Zpu = Z% / 100
Zpu = Z% / 100
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Symmetrical Components
In
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).
- 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) / 3
Ia2 = (Ia + h2Ib + hIc) / 3
Ia0 = (Ia + Ib + Ic) / 3
Ia1 = (Ia + hIb + h2Ic) / 3
Ia2 = (Ia + h2Ib + hIc) / 3
Ia0 = (Ia + Ib + Ic) / 3
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
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.
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.
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-operator
The h-operator (1Ð120°) is the complex cube root of unity:
h = - 1 / 2 + jÖ3 / 2 = 1Ð120° = 1Ð-240°
h2 = - 1 / 2 - jÖ3 / 2 = 1Ð240° = 1Ð-120°
Some useful properties of h are:
1 + h + h2 = 0
h + h2 = - 1 = 1Ð180°
h - h2 = jÖ3 = Ö3Ð90°
h2 - h = - jÖ3 = Ö3Ð-90°
The h-operator (1Ð120°) is the complex cube root of unity:
h = - 1 / 2 + jÖ3 / 2 = 1Ð120° = 1Ð-240°
h2 = - 1 / 2 - jÖ3 / 2 = 1Ð240° = 1Ð-120°
Some useful properties of h are:
1 + h + h2 = 0
h + h2 = - 1 = 1Ð180°
h - h2 = jÖ3 = Ö3Ð90°
h2 - h = - jÖ3 = Ö3Ð-90°
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Fault Calculations
The
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.
- 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,
- 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'.
- 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,
- 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 current
Ie = earth fault current
Ea = normal phase voltage at the fault location
Z1 = positive phase sequence network impedance to the fault
Z2 = negative phase sequence network impedance to the fault
Z0 = zero phase sequence network impedance to the fault
Ie = earth fault current
Ea = normal phase voltage at the fault location
Z1 = positive phase sequence network impedance to the fault
Z2 = negative phase sequence network impedance to the fault
Z0 = zero phase sequence network impedance to the fault
Single phase to earth - fault from phase
'a' to earth:
Va
= 0
Ib = Ic = 0 I f = Ia = Ie |
Ia1 = Ia2 = Ia0 = Ia
/ 3
Va1 + 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 = 0 I f = Ib = - Ic |
Ia1 + Ia2 = 0
Ia0 = 0 Va1 = Va2 |
Ia1 = Ea / (Z1 + Z2)
Ia2 = - Ia1 Ia0 = 0 |
I f = - jÖ3Ia1 = - jÖ3Ea / (Z1 + Z2)
Ib = I f = - jÖ3Ea / (Z1 + Z2) Ic = - I f = jÖ3Ea / (Z1 + Z2) |
Double phase to earth - fault from phase
'b' to phase 'c' to earth:
Vb
= Vc = 0
Ia = 0 I f = Ib + Ic = Ie |
Ia1 + Ia2 + Ia0 = 0
Va1 = Va2 = Va0 |
Ia1 = Ea / Znet
Ia2 = - Ia1Z0 / (Z2 + Z0) Ia0 = - Ia1Z2 / (Z2 + Z0) |
I f = 3Ia0 = - 3EaZ2
/ Szz = Ie
Ib = I f / 2 - jÖ3Ea(Z2 / 2 + Z0) / Szz Ic = I f / 2 + jÖ3Ea(Z2 / 2 + Z0) / Szz |
Znet
= Z1 + Z2Z0 / (Z2 + Z0)
and Szz
= 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 = 0 Ia0 = 0 |
I f = Ia1 = Ea / Z1
= Ia
Ib = Eb / Z1 Ic = Ec / Z1 |
Note
that the single phase fault current is greater than the three phase fault
current if Z0 is less than (2Z1 - Z2).
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.
Note
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.
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Three Phase Fault Level
The
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 / Ö3|ZS|
Isc = Ephase / |ZS| = Eline / Ö3|ZS|
The
three phase fault level Ssc of the power system is:
Ssc = 3Isc2|ZS| = 3EphaseIsc = 3Ephase2 / |ZS| = Eline2 / |ZS|
Ssc = 3Isc2|ZS| = 3EphaseIsc = 3Ephase2 / |ZS| = Eline2 / |ZS|
Note
that if the X / R ratio of the source impedance ZS
(comprising resistance RS and reactance XS)
is sufficiently large, |ZS| »
XS.
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Power Factor Correction
If
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
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)
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
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
QC = Q1 = P tanf1
I2 / I1 = S2 / S1 = cosf1 = P / S1
Shunt Capacitors
For 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 = Vline2 / XCstar = 2pfCstarVline2
Iline = QCstar / Ö3Vline = Vline / Ö3XCstar
Cstar = QCstar / 2pfVline2
For 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 = Vline2 / XCstar = 2pfCstarVline2
Iline = QCstar / Ö3Vline = Vline / Ö3XCstar
Cstar = QCstar / 2pfVline2
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 = 3Vline2 / XCdelta = 6pfCdeltaVline2
Iline = QCdelta / Ö3Vline = Ö3Vline / XCdelta
Cdelta = QCdelta / 6pfVline2
QCdelta = 3Vline2 / XCdelta = 6pfCdeltaVline2
Iline = QCdelta / Ö3Vline = Ö3Vline / XCdelta
Cdelta = QCdelta / 6pfVline2
Note
that for the same leading reactive power QC:
XCdelta = 3XCstar
Cdelta = Cstar / 3
XCdelta = 3XCstar
Cdelta = Cstar / 3
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Reactors
Shunt Reactors
For 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 = Vline2 / XLstar = Vline2 / 2pfLstar
Iline = QLstar / Ö3Vline = Vline / Ö3XLstar
Lstar = Vline2 / 2pfQLstar
For 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 = Vline2 / XLstar = Vline2 / 2pfLstar
Iline = QLstar / Ö3Vline = Vline / Ö3XLstar
Lstar = Vline2 / 2pfQLstar
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 = 3Vline2 / XLdelta = 3Vline2 / 2pfLdelta
Iline = QLdelta / Ö3Vline = Ö3Vline / XLdelta
Ldelta = 3Vline2 / 2pfQLdelta
QLdelta = 3Vline2 / XLdelta = 3Vline2 / 2pfLdelta
Iline = QLdelta / Ö3Vline = Ö3Vline / XLdelta
Ldelta = 3Vline2 / 2pfQLdelta
Note
that for the same lagging reactive power QL:
XLdelta = 3XLstar
Ldelta = 3Lstar
XLdelta = 3XLstar
Ldelta = 3Lstar
Series Reactors
For 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 = 2pfLseriesIline
QLseries = 3Vdrop2 / XLseries = 3VdropIline = 3Iline2XLseries = 6pfLseriesIline2
Lseries = QLseries / 6pfIline2
For 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 = 2pfLseriesIline
QLseries = 3Vdrop2 / XLseries = 3VdropIline = 3Iline2XLseries = 6pfLseriesIline2
Lseries = QLseries / 6pfIline2
Note
that the apparent power rating Srating of the set of three
line reactors is based on the line voltage Vline and not the
voltage drop Vdrop:
Srating = Ö3VlineIline
Srating = Ö3VlineIline
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Harmonic Resonance
If
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 / w
C = 1 / wXC
where w = 2pf
L = XL / w
C = 1 / wXC
where w = 2pf
The
series resonance angular frequency wr of an inductance L
with a capacitance C is:
wr = (1 / LC)½ = w(XC / XL)½
wr = (1 / LC)½ = w(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
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
QC = 3E2 / XC
The
harmonic number fr / f of the series resonance of XL
with XC is:
fr / f = wr / w = (XC / XL)½ » (Ssc / QC)½
fr / f = wr / w = (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 = Ö15 = 3.873
XL / XC = 1 / 15 = 0.067
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 = Ö15 = 3.873
XL / XC = 1 / 15 = 0.067
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Dielectric Dissipation Factor
If
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 + VC2)½
VR = IRS
VC = IXC
V = (VR2 + VC2)½
The
dielectric dissipation factor of the insulation system is the tangent of the
dielectric loss angle d between VC and V:
tand = VR / VC = RS / XC = 2pfCRS
RS = XCtand = tand / 2pfC
Note 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 tand. Note also that tand increases with frequency.
tand = VR / VC = RS / XC = 2pfCRS
RS = XCtand = tand / 2pfC
Note 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 tand. Note also that tand increases with frequency.
The
dielectric power loss P is related to the capacitive reactive power QC
by:
P = I2RS = I2XCtand = QCtand
P = I2RS = I2XCtand = QCtand
The
power factor of the insulation system is the cosine of the phase angle f between VR
and V:
cosf = VR / V
so that d and f are related by:
d + f = 90°
cosf = VR / V
so that d and f are related by:
d + f = 90°
tand and cosf are related
by:
tand = 1 / tanf = cosf / sinf = cosf / (1 - cos2f)½
so that when cosf is close to zero, tand » cosf
tand = 1 / tanf = cosf / sinf = cosf / (1 - cos2f)½
so that when cosf is close to zero, tand » 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).
Notation
The
library uses the symbol
font for some of the notation and formulae. If the symbols for the letters
'alpha beta delta' do not appear here [a b d] then the symbol font needs to be
installed before all notation and formulae will be displayed correctly.
C
E e G I i k L M N P |
capacitance
voltage source instantaneous E conductance current instantaneous I coefficient inductance mutual inductance number of turns power |
[farads, F]
[volts, V] [volts, V] [siemens, S] [amps, A] [amps, A] [number] [henrys, H] [henrys, H] [number] [watts, W] |
Q
q R T t V v W F Y y |
charge
instantaneous Q resistance time constant instantaneous time voltage drop instantaneous V energy magnetic flux magnetic linkage instantaneous Y |
[coulombs, C]
[coulombs, C] [ohms, W] [seconds, s] [seconds, s] [volts, V] [volts, V] [joules, J] [webers, Wb] [webers, Wb] [webers, Wb] |

Resistance
The
resistance R of a circuit is equal to the applied direct voltage E
divided by the resulting steady current I:
R = E / I
R = E / I
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Resistances in Series
When
resistances R1, R2, R3,
... are connected in series, the total resistance RS is:
RS = R1 + R2 + R3 + ...
RS = R1 + R2 + R3 + ...
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Voltage Division by Series Resistances
When
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)
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)
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.
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.
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Resistances in Parallel
When
resistances R1, R2, R3,
... are connected in parallel, the total resistance RP is:
1 / RP = 1 / R1 + 1 / R2 + 1 / R3 + ...
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
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
RP = R1R2 / (R1 + R2)
RP = product / sum
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
R2 = R1RP / (R1 - RP)
R2 = product / difference
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Current Division by Parallel Resistances
When
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)
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)
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).
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).
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Capacitance
When
a voltage is applied to a circuit containing capacitance, current flows to
accumulate charge in the capacitance:
Q = òidt = CV
Q = òidt = CV
Alternatively,
by differentiation with respect to time:
dq/dt = i = C dv/dt
Note that the rate of change of voltage has a polarity which opposes the flow of current.
dq/dt = i = C dv/dt
Note 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 = òidt / V
C = Q / V = òidt / 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
C = i / dv/dt = dq/dt / dv/dt = dq/dv
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Capacitances in Series
When
capacitances C1, C2, C3,
... are connected in series, the total capacitance CS is:
1 / CS = 1 / C1 + 1 / C2 + 1 / C3 + ...
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
CS = C1C2 / (C1 + C2)
CS = product / sum
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Voltage Division by Series Capacitances
When
a total voltage ES is applied to series connected
capacitances C1 and C2, the charge QS
which accumulates in the series circuit is:
QS = òiSdt = ESCS = ESC1C2 / (C1 + C2)
QS = òiSdt = ESCS = ESC1C2 / (C1 + C2)
The
voltages V1 and V2 which appear across the
respective capacitances C1 and C2 are:
V1 = òiSdt / C1 = ESCS / C1 = ESC2 / (C1 + C2)
V2 = òiSdt / C2 = ESCS / C2 = ESC1 / (C1 + C2)
V1 = òiSdt / C1 = ESCS / C1 = ESC2 / (C1 + C2)
V2 = òiSdt / C2 = ESCS / C2 = ESC1 / (C1 + C2)
In
general terms, for capacitances C1, C2, C3,
... connected in series:
QS = òiSdt = ESCS = ES / (1 / CS) = ES / (1 / C1 + 1 / C2 + 1 / C3 + ...)
Vn = òiSdt / 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.
QS = òiSdt = ESCS = ES / (1 / CS) = ES / (1 / C1 + 1 / C2 + 1 / C3 + ...)
Vn = òiSdt / 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.
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Capacitances in Parallel
When
capacitances C1, C2, C3,
... are connected in parallel, the total capacitance CP is:
CP = C1 + C2 + C3 + ...
CP = C1 + C2 + C3 + ...
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Charge Division by Parallel Capacitances
When
a voltage EP is applied to parallel connected capacitances C1
and C2, the charge QP which accumulates in
the parallel circuit is:
QP = òiPdt = EPCP = EP(C1 + C2)
QP = òiPdt = EPCP = EP(C1 + C2)
The
charges Q1 and Q2 which accumulate in the
respective capacitances C1 and C2 are:
Q1 = òi1dt = EPC1 = QPC1 / CP = QPC1 / (C1 + C2)
Q2 = òi2dt = EPC2 = QPC2 / CP = QPC2 / (C1 + C2)
Q1 = òi1dt = EPC1 = QPC1 / CP = QPC1 / (C1 + C2)
Q2 = òi2dt = EPC2 = QPC2 / CP = QPC2 / (C1 + C2)
In
general terms, for capacitances C1, C2, C3,
... connected in parallel:
QP = òiPdt = EPCP = EP(C1 + C2 + C3 + ...)
Qn = òindt = EPCn = QPCn / CP = QPCn / (C1 + C2 + C3 + ...)
Note that the highest charge accumulates in the highest capacitance.
QP = òiPdt = EPCP = EP(C1 + C2 + C3 + ...)
Qn = òindt = EPCn = QPCn / CP = QPCn / (C1 + C2 + C3 + ...)
Note that the highest charge accumulates in the highest capacitance.
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Inductance
When
the current changes in a circuit containing inductance, the magnetic linkage
changes and induces a voltage in the inductance:
dy/dt = e = L di/dt
Note that the induced voltage has a polarity which opposes the rate of change of current.
dy/dt = e = L di/dt
Note that the induced voltage has a polarity which opposes the rate of change of current.
Alternatively,
by integration with respect to time:
Y = òedt = LI
Y = òedt = 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 = dy/dt / di/dt = dy/di
L = e / di/dt = dy/dt / di/dt = dy/di
Alternatively,
the inductance L of a circuit is equal to the magnetic linkage divided
by the current:
L = Y / I
L = Y / I
Note
that the magnetic linkage Y is equal to the product of the number of turns N
and the magnetic flux F:
Y = NF = LI
Y = NF = LI
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Mutual Inductance
The
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)
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:
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.
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
M = (L1L2)½
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Inductances in Series
When
uncoupled inductances L1, L2, L3,
... are connected in series, the total inductance LS is:
LS = L1 + L2 + L3 + ...
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 ± 2M
The plus or minus sign indicates that the coupling is either additive or subtractive, depending on the connection polarity.
LS = L1 + L2 ± 2M
The plus or minus sign indicates that the coupling is either additive or subtractive, depending on the connection polarity.
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Inductances in Parallel
When
uncoupled inductances L1, L2, L3,
... are connected in parallel, the total inductance LP is:
1 / LP = 1 / L1 + 1 / L2 + 1 / L3 + ...
1 / LP = 1 / L1 + 1 / L2 + 1 / L3 + ...
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Time Constants
Capacitance and resistance
The 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).
The 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:
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)
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
i = (V / R)e - t / CR
vR = iR = Ve - t / CR
vC = vR = Ve - t / CR
qC = CvC = CVe - t / CR
Inductance and resistance
The 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).
The 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 yL in the inductance
are:
i = (E / R)(1 - e - tR / L)
vR = iR = E(1 - e - tR / L)
vL = E - vR = Ee - tR / L
yL = Li = (LE / R)(1 - e - tR / L)
i = (E / R)(1 - e - tR / L)
vR = iR = E(1 - e - tR / L)
vL = E - vR = Ee - tR / L
yL = 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 yL in the inductance
are:
i = Ie - tR / L
vR = iR = IRe - tR / L
vL = vR = IRe - tR / L
yL = Li = LIe - tR / L
i = Ie - tR / L
vR = iR = IRe - tR / L
vL = vR = IRe - tR / L
yL = Li = LIe - tR / L
Rise Time and Fall Time
The 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 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
t50 = (ln1.0 - ln0.5)T » 0.69T
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.
- 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.
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Power
The
power P dissipated by a resistance R carrying a current I
with a voltage drop V is:
P = V2 / R = VI = I2R
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
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
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 Y is:
P = VI = LI(di/dt) = Y(di/dt) = Y(dy/dt) / L
P = VI = LI(di/dt) = Y(di/dt) = Y(dy/dt) / L
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Energy
The
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
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
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
W = CV2 / 2 = QV / 2 = Q2 / 2C
The
energy W stored in an inductance L carrying current I with
magnetic linkage Y
is:
W = LI2 / 2 = YI / 2 = Y2 / 2L
W = LI2 / 2 = YI / 2 = Y2 / 2L
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Batteries
If
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
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)
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)
Isc = EB / RB = EBIL / (EB - VL)
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Voltmeter Multiplier
The
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
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 = IV2RS
P = (V - VV)2 / RS = (V - VV)IV = IV2RS
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Ammeter Shunt
The
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)
RP = VA / (I - IA)
The
power P dissipated by the resistance RP with voltage
drop VA carrying current (I - IA) is:
P = VA2 / RP = VA(I - IA) = (I - IA)2RP
P = VA2 / RP = VA(I - IA) = (I - IA)2RP
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