Tuesday, December 21, 2010

Basic Electrical

ABBREVATIONS

RESISTANCE

OHM’s LAW

V=I*R where V=Voltage

I=Current

R=Resistance

DC Resistance:

R= ρ*l/A Where ρ = Resistivity of teh material

l =Length of the conductor

A = Area of the conductor

ρ= 1 / Conductivity=Resistivity

Temperature dependance of a resistor

R =R0[α *(T-T0)+1] where R­­0=Resistance at Temperature T0

α =% change in resistivity / unit temperature

T =Temperature

AC resistance = 1.6*Dc resistance
note : AC resistance would be always greater than DC resistance because of 'SKIN EFFECT'

Resistance in series

When R1,R2, R3 are connected in series…..

Req=R1+R2+R3

Ø Voltages divides & Current remains same in a series circuit.

Resistance in parallel

When R1,R2,R3 are connected in parallel

1/Req=(1/R1+1/R2+1/R3) =

Req=(R1*R2*R3)/(R1+R2+R3)

Ø Current divides & Voltage remains the same in a parallel circuit.

Ø The equivalent resistance of two or more resistance is always less than the least value of the individual resistor.

VOLTAGE DIVIDER

When two resistance are connected as shown

1) The voltage across Resistor R2 is given as

Vout = R2/(R1+R2)*Vin Where Vout = Output voltage

Vin = Input voltage

The power P dissipated by a resistance R carrying a current I with a voltage drop V is:

P= (I^2)*R = VI = (V^2)/R

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=(I^2)*R*t


CAPACITANCE

C=Q/V Where C= Capacitance

Q= Charge

V= Voltage

C=(Ebsilon0)*(Ebsilonr)*A/D Where C=Capacitance

A=Area

D=Distance b/w the plates

The Energy stored in a capacitor

W(stored)= C(V^2)/2 = (Q^2)/2C= QV/2

Current-voltage relationship…

I(t)=C dV/Dt Where (dV/dT)=Change in Voltage

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

Capacitance in series….

When capacitances C1,C2,C3 are connected in series Equivalent capacitance is

1/Ceq=(1/C1+1/C2+1/C3)

Ceq= (C1*C2*C3)+(C1+C2+C3)

Capacitance in parallel….

When capacitances C1,C2 ,C3 are connected in parallel , the Equivalent capacitance is

Ceq=C1+C2+C3

Ø Capacitor blocks DC but passes AC

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


INDUCTANCE

Voltage-Current relationship…

V=L dI/dT where (dI/dT)=Change in current

Inductance in series…

When Inductors L1, L2, L3 are connected in series the equivalent inductance

Leq=L1+L2+L3

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

Inductance in parallel...

When inductance L1,L2,L3 are connected in parallel ,their equivalent inductance

1/Leq = (1/L1+1/L2+1/L3)

Leq = (L1*L2*L3)/L1+L2+L3

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

The energy W stored in an inductance L carrying current I with magnetic linkage Y is:
W = LI2 / 2 = YI / 2 = Y2 / 2L


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

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


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)

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


Wheatstone Bridge

The 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


Kennelly's Star-Delta Transformation

A 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)


Kennelly's Delta-Star Transformation

A 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)


KIRCHOFF’S LAW

Kirchhoff's Current Law
At 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:
SIin = SIout

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

Kirchhoff's Voltage Law
At 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:
SE = SIZ

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

AC CIRCUITS

Impedance

The 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)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)

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|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)

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

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

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

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.


Reactance

Inductive Reactance
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

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

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

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


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

At resonance, the imaginary part of ZS is zero:
XC = XL
ZSr = R
wr = (1 / LC)½ = 2pfr
The quality factor at resonance Qr is:
Qr = wrL / R = (L / CR2)½ = (1 / R )(L / C)½ = 1 / wrCR

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

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
The quality factor at resonance Qr is:
Qr = wrL / 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.

Three Phase Power

For a balanced star connected load with line voltage Vline and line current Iline:
Vline = Ö3 Vphase
Iphase = Iline
Zphase = Vphase / Iphase = Vline / Ö3Iline
Sphase = 3VphaseIphase = Ö3VlineIline = Vline2 / Zphase = 3Iline2Zphase

For a balanced delta connected load with line voltage Vline and line current Iline:
Vphase = Vline
Iphase = Iline / Ö3
Zphase = Vphase / Iphase = Ö3Vline / Iline
Sphase = 3VphaseIphase = Ö3VlineIline = 3Vline2 / Zphase = Iline2Zphase

Power Triangle

There are Three kind of power i.e

True power is a function of a circuit's dissipative elements, usually resistances (R).

Reactive power is a function of a circuit's reactance (X).

Apparent power is a function of a circuit's total impedance (Z).

The relation between all the three of them are given as

P=True power , Measured in watts (W)

P = (I^2)*R

Q=Reactive power , Measured in Volt-Ampere-Reactive (VAR)

Q=(I^2)*X

S=Apparent power ,Measured in Volt-Amps (VA)

S=(I^2)*Z

TRANSFORMERS

Two types according to construction

1) Core type

2) Shell type

Induced EMF in the primary winding:

E1 = 4.44 * f * N1 * [Bm] * A

= 4.11 * f * N1 * [fi] Where N1=Number of turns in Primary winding

F=Frequency

[fi]=maximum flux in core

A=Area

[fi]=A*[Bm]

Form Factor = R.M.S Value/Average Value = 1.11

Transformation ratio :K=V2/V1

For 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 volt-ampere 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 DV2 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 DV2pu is:
DV2pu = (E2 - V2) / V2
Note that the per-unit base voltage is usually V2 and not E2.

Tests done in a transformer…

1)Open circuit test (OC): Determines the core loss / High voltage left open

2)Closed circuit test (CC):

Losses in a transformer :

1) Core loss (Iron Loss) : Constant loss

1.1) Hysteresis loss: Wh=Bmax^1.6 * f ^ 2

1.2) Eddy Current loss: We=Bamx^2 * f ^ 2

2) Copper loss: I^2 * R loss. It is a variable loss

Voltage regulation

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