3. CONTENTS
• INTRODUCTION
• GENERATION OF SINUSOIDAL VOLTAGE
• STANDARD TERMINOLOGIES
• PHASOR DIAGRAM IN R,L AND C CIRCUIT.
• ANALYSIS OF R-L, R-C, R-L-C SERIES CIRCUITS
• CONCEPT OF IMPEDANCE
• POWER & POWER FACTOR
Dept. of E
CE, MITEMoodbidri
4. INTRODUCTION TO ACFUNDAMENTALS
• THE RESISTANCE, INDUCTANCE AND CAPACITANCE ARE THREE
BASIC ELEMENTSOF ANY ELECTRICALNETWORK. IN ORDERTO
NECESSAR
Y TO
ANALYSE ANY ELECTRIC CIRCUIT, IT IS
UNDERSTANDTHEFOLLOWING THREECASES,
• A.C. THROUGH PURERESISTIVECIRCUIT
.
• A.C. THROUGH PUREINDUCTIVECIRCUIT
.
• A.C. THROUGH PURECAPACITIVECIRCUIT.
Dept. of E
CE, MITEMoodbidri
5. INTRODUCTION TO AC FUNDAMENTALS
• IN EACH CASE, IT IS ASSUMED THAT A PURELY
SINUSOIDAL ALTERNATING VOLTAGE IS APPLIED TO THE
CIRCUIT.
• THE EQUATION OF THE CURRENT, POWER AND PHASE
SHIFT IS DEVELOPED IN EACH CASE. THE VOLTAGE
APPLIED HAVING ZERO PHASE ANGLE IS ASSUMED
REFERENCE WHILE PLOTTING THE PHASOR DIAGRAM IN
E
ACHCASE.
Dept. of E
CE, MITEMoodbidri
6. GENER
ATION OF A.C. VOLTAGE
• THEMACHINES WHICH AREUSEDTOGENERATEELECTRICAL VOLTAGES ARE
CALLEDGENERATORS.THEGENERATORSWHICH GENERATEPURELYSINUSOIDAL
A.C. VOL
TAGE
SARECALLEDAL
T
E
R
NATOR
S.
Dept. of E
CE, MITEMoodbidri
7. GENERATION OF A.C. VOLTAGE
•E= B L V SINƟ = E
M SINƟ
E= E
M SIN𝜔T= E
M SIN( 2ΠF)T
• WHERE B L V = EM = MAXIMUM VALUE OF THE EMF
INDUCED. SIMILAR EQUATION CAN BE WRITTEN FOR
CURRENTALSO , I= IM SIN(2ΠF)TOR IM SINƟ.
Dept. of E
CE, MITEMoodbidri
10. STANDARDTERMINOLOGIESRELATEDTOALTERNATING
QUANTITY
TIME PERIOD (T)
• TIME TAKEN BY AN ALTERNATING QUANTITY TO COMPLETE ITS ONE CYCLE IS
KNOWN AS ITS TIME PERIOD DENOTED BY T SECONDS. AFTER EVERY T
SECONDS, THE CYCLE OF AN ALTERNATING QUANTITY REPEATS.
FREQUENCY (F)
• THE NUMBER OF CYCLES COMPLETED BY AN ALTERNATING QUANTITY PER
SECOND IS KNOWN AS ITS FREQUENCY. IT IS DENOTED BY F AND IS
MEASURED IN CYCLES/SECOND WHICH IS KNOWN AS HERTZ, DENOTED AS HZ.
Dept. of E
CE, MITEMoodbidri
11. STANDARD TERMINOLOGIES RELATEDTO ALTERNATING
QUANTITY
AMPLITUDE
• THE MAXIMUM VALUE ATTAINED BY AN ALTERNATING QUANTITY DURING
POSITIVE OR NEGATIVE HALF CYCLE IS CALLED ITS AMPLITUDE.
ANGULAR FREQUENCY (Ω)
• IT IS THE FREQUENCY EXPRESSED IN RADIANS PER SECOND. AS ONE CYCLE
OF AN ALTERNATING QUANTITY CORRESPONDS TO 2Π RADIANS, THE
ANGULAR FREQUENCY CAN BE EXPRESSED AS 2ΠF (CYCLES/SEC). ITS UNIT IS
RADIANS/SEC.
Ω= 2 Π FR
ADIANS/SEC
Dept. of E
CE, MITEMoodbidri
12. EFFECTIVEVALUEOR R.M.S
VALUE OF AC SIGNAL
• THE EFFECTIVE VALUE OR R.M.S. VALUE OF AN ALTERNATING CURRENT IS GIVEN
BY THAT STEADY CURRENT WHICH, WHEN FLOWING THROUGH A GIVEN CIRCUIT
FOR A GIVEN TIME PRODUCES THE SAME AMOUNT OF HEAT AS PRODUCED BY THE
ALTERNATING CURRENT, WHEN FLOWING THROUGH THE SAME CIRCUIT FOR THE
SAME TIME.
• IR.M.S= 0.707IM
Dept. of E
CE, MITEMoodbidri
13. AVERAGE VALUEOFAC SIGNAL
• THE AVERAGE VALUE OF AN ALTERNATING QUANTITY IS DEFINED
AS THAT VALUE WHICH IS OBTAINED BY AVERAGING ALL THE
INSTANTANEOUS VALUES OVER A PERIOD OF HALF CYCLE.
• IAVG = 0.637 IM
Dept. of E
CE, MITEMoodbidri
14. FORM F
ACTOR (KF)
• THEFORM FACTOR OF AN ALTERNATING QUANTITY IS DEFINEDAS THERATIO
OF R
.M.S. VALUETO THEAVE
R
AGEVALUE
.
• KF =1.11 FORSINUSOIDALL
YVAR
YING QUANTITY
Dept. of E
CE, MITEMoodbidri
15. CRESTOR PEAKF
ACTOR(KP)
• THEPEAKVALUEOFAN ALTERNATINGQUANTITYISDEFINEDASTHERATIOOF
MAXIMUM VALUETO THER
.M.S. VALUE
.
• KP =1.414 FORSINUSOIDALL
YVAR
YING QUANTITY
Dept. of E
CE, MITEMoodbidri
16. PHASOR R
E
P
R
ESENTATION OF AN
ALTERNATING QUANTITY
CASE1: IN PHASE
• THEEQUATIONSFORVOL
TAGEAND
CUR
R
ENTAR
E
E= E
M SIN𝜔T
I= IMSIN𝜔T
Dept. of E
CE, MITEMoodbidri
17. PHASOR R
E
P
R
E
SENTATION OF AN ALTE
RNATING
QUANTITY
CASE2: LAGGING
•EQUATIONSFORVOL
TAGEAND CUR
R
ENTARE
E= E
M SIN𝜔T
I= IMSIN(𝜔T-ɸ)
Dept. of E
CE, MITEMoodbidri
18. PHASOR R
E
P
R
E
SENTATION OF AN ALTE
RNATING
QUANTITY
CASE3: LEADING
• EQUATIONSFORCUR
R
ENTAND VOL
TAGE
AR
E, E= EMSIN𝜔T
I= IMSIN(𝜔T+ɸ)
Dept. of E
CE, MITEMoodbidri
19. A.C. through Pure Resistance
• Consider a simple circuit consisting of a pure
resistance ‘R’ohms connected across anAC voltage
• Let v=instantaneous value of input voltage
i= instantaneous value of resulting current
The following relation shows that voltage and
current are in phase
Dept. of E
CE, MITEMoodbidri
20. A.C. THROUGH PURERESISTANCE
• THEWAVEFORM OF VOLTAGE AND CURRENTAND THECORRESPONDING PHASOR DIAGRAM
Dept. of E
CE, MITEMoodbidri
21. A.C.THROUGH PURERESISTANCE
P
ower=instantaneousvoltage x instantaneouscurrent
If p=instantaneous power
P=power for whole cycle
Where V=rms value of applied voltage I=rms value of resulting current
Dept. of E
CE, MITEMoodbidri
22. Voltage, Current and P
ower for
purely resistive circuit
Dept. of E
CE, MITEMoodbidri
23. A.C. through Pure Inductance
• Consider a simple circuit consisting of a pure inductance
of L Henries, connected across anAC voltage
• Pure inductor has zero ohmic resistance. Its internal
resistance is zero. The coil has pure inductance of L
Henries.
• Aback E.M.F. Is produced due to the self-inductance of
the coil.
• The back E.M.F., at every step, opposes the rise or fall of
current through the coil.
• The applied voltage has to overcome this self-induced
E.M.F
Dept. of E
CE, MITEMoodbidri
24. A.C. through Pure Inductance
• Consider a simple circuit consisting of a pure
inductance of L Henries, connected across an AC
voltage
• Pure inductor has zero ohmic resistance. Its internal
resistance is zero.
• A back E.M.F. Is produced due to the self-inductance of
the coil.
• The back E.M.F., at every step, opposes the rise or fall
of current through the coil.
• The applied voltage has to overcome this self-induced
E.M.F
Dept. of E
CE, MITEMoodbidri
25. A.C. THROUGH PUREINDUCTANCE
Current lags behind the applied
voltage by a quarter cycle
The phase difference between
the two is ℼ/2 with voltage
leading.
Dept. of E
CE, MITEMoodbidri
26. A.C.THROUGH PUREINDUCTANCE
Current lags behind the applied voltage by a quarter cycle
The phase difference between the two is ℼ/2 with voltage leading.
Dept. of E
CE, MITEMoodbidri
27. A.C. THROUGH PUREINDUCTANCE
POWER
• THEINSTANTANEOUSPOWERIN A.C. CIRCUITSCANB
EOBTAINED BYTAKING PRODUCT OF
THEINSTANTANEOUS VALUESOF CURRENTAND VOLTAGE.
Dept. of E
CE, MITEMoodbidri
28. A.C. THROUGH PUREINDUCTANCE
• IT CAN OBSERVED FROM IT THATWHEN POWER CURVEIS POSITIVE, ENERGY GETS STOREDIN
THE MAGNETIC FIELD ESTABLISHED DUE TO INCREASING CURRENT WHILE DURING NEGATIVE
POWER CURVE,THISPOWER IS RETURNEDBACKTO THESUPPLY.
Dept. of E
CE, MITEMoodbidri
29. AC THROUGH PURECAP
ACITANCE
• Consider a simple circuit consisting of
a pure capacitor of C-farads, connected
across anAC voltage source
• When an alternating voltage is applied
to the plates of a capacitor, the
one
first in
the opposite
capacitor is charged
direction and then in
direction.
• The current i charges the capacitor c.
The instantaneous charge ‘q’ on the
plates of the capacitor is given by
Dept. of E
CE, MITEMoodbidri
31. AC THROUGH PUR
ECAP
ACITANCE
The current in a pure capacitor leads its voltage by a quarter cycle
Phase difference between its voltage and current is ℼ/2 with the current
leading
Dept. of E
CE, MITEMoodbidri
32. AC THROUGH PURECAP
ACITANCE
POWER
• THEINSTANTANEOUSPOWERIN A.C. CIRCUITSCANB
EOBTAINED BYTAKING PRODUCT OF
THEINSTANTANEOUS VALUESOF CURRENTAND VOLTAGE.
Dept. of E
CE, MITEMoodbidri
34. A.C. THROUGH SERIES R-L CIRCUIT
• Consider a circuit consisting of pure resistance R ohms connected in series with a pure
inductance of L henries.
• The series combination is connected acrossA.C. Supply
If Circuit current is I, then there are two
voltage drops:
a) drop across pure resistance, VR=IR
b) drop across pure inductance, VL=I XL
where XL =2πf L =inductive reactance
I = r.M.S. Value of current drawn
VR,VL= r.M.S. Values of voltage drops
Dept. of E
CE, MITEMoodbidri
35. A.C. THROUGH SERIES R-L CIRCUIT
• The Kirchhoff's law can be applied to the A.C. Circuit but only the point to remember is
the addition of voltages should be vector addition.
V = VR + VL
v = i R + i XL
Dept. of E
CE, MITEMoodbidri
36. A.C. THROUGH SERIES R-L CIRCUIT
• THE PHASOR DIAGRAM AND THE VOLTAGE TRIANGLE ARE
SHOWNAS FOLLOWS:
Dept. of E
CE, MITEMoodbidri
40. A.C. THROUGH SERIES R-C CIRCUIT
• Consider a circuit consisting of pure resistance R-ohms connected in series with a pure
capacitor of C-farads.
• The series combination is connected acrossA.C. Supply.
If circuit current is I then there are two voltage drops:
a) drop across pure resistance, VR=IR
b) drop across pure Capacitor VC=I XC
where XC =1/(2πf C)=Capacitive reactance
I = r.M.S. Value of current drawn
VR,VC= r.M.S. Values of voltage drops
Dept. of E
CE, MITEMoodbidri
41. A.C. THROUGH SERIES R-C CIRCUIT
• The Kirchhoff's law can be applied to theA.C. Circuit but only the point to remember
is the addition of voltages should be vector addition.
V = VR + VC
V = I R + I XC
Dept. of E
CE, MITEMoodbidri
42. A.C. THROUGH SERIES R-C CIRCUIT
THE PHASOR DIAGRAM AND THE VOLTAGE
TRIANGLE
Dept. of E
CE, MITEMoodbidri
44. A.C. THROUGH SERIES R-C CIRCUIT
Impedance
• The impedance is nothing but opposition to the flow of
alternating current.
• It is measured in ohms given by
Dept. of E
CE, MITEMoodbidri
45. A.C. THROUGH SERIES R-L-C
CIRCUIT
Consider a circuit
consisting of
• resistance R ohms
• pure inductance L Henries
• Pure capacitance C farads
connected in series with
across AC
each other
Supply.
Dept. of E
CE, MITEMoodbidri
47. A.C. THROUGH SERIES R-L-C CIRCUIT
• TheAC supply is given by, 𝑣= 𝑉𝑚 𝑠
𝑖
𝑛
𝜔𝑡
According to Kirchhoff's law we can write V = V 𝑅 + V 𝐿 + VC
The + ve sign is to be used
when
current leads i.e. XC > XL.
The - ve sign is to be used
when
current lags i.e. when XL > XC.
Dept. of E
CE, MITEMoodbidri
48. A.C. THROUGH SERIES R-L-C CIRCUIT
• PHASEANGLE IS GIVENAS FOLLOWS:
• POWER FACTOR IS GIVENAS FOLLOWS:
Dept. of E
CE, MITEMoodbidri
49. A.C. THROUGH SERIES R-L-C CIRCUIT
• CASE(I): 𝑿𝑳> 𝑿C
SO, IF 𝑣 = 𝑉𝑚𝑠𝑖𝑛𝜔𝑡, THEN 𝑖 = 𝐼𝑚𝑠𝑖𝑛(𝜔𝑡– 𝜙)AS CURRENT LAGS VOLTAGE BYANGLE 𝜙.
Dept. of E
CE, MITEMoodbidri
50. A.C. THROUGH SERIE
SR
-L-CCIRCUIT
• CASE(II): 𝐗𝐋 < XC
• THECURRENTIS SAID TO BECAPACITIVEIN NATURE.THEPHASOR SUMOF VRAND(VC − VL) GIVES
THERESULTANTSUPPLYVOLTAGEV. CURRENTI LEADSTHEVOLTAGEV BY 𝝓.
Dept. of E
CE, MITEMoodbidri
51. A.C. THROUGH S
ERIE
SR
-L-CCIRCUIT
• CASE(III): 𝐗𝐋=𝐗𝐂
• WHEN𝐗𝐋= 𝐗𝐂, OBVIOUSL
Y𝐕𝐋= 𝐕𝐂. SO VL
AND VCWILLCANCE
LEACH OTHE
RAND
THEIRR
E
S
UL
TANTISZER
O.
• SO, 𝐕𝐑 = V IN SUCH CASEAND OVE
R
ALL
CIRCUITIS PURELYRESISTIVEIN NATURE.
• THEPHASORDIAGR
AM FORTHEABOVECASE
ISSHOWN
Dept. of E
CE, MITEMoodbidri
53. CONTENTS
Dept. of E
CE, MITEMoodbidri
• GENERATION OFTHR
E
EPHASEAC QUANTITY
• ADVANTAGESAND LIMITATIONS
• STARAND DE
L
TA CONNECTION
• R
ELA
TIONSHIPBE
T
WEEN LINEAND PHASEQUANTITIES
54. GENERATION OF THREEPHASEVOLTAGE SYSTEM
•The ends of each coil are
brought out through the slip
• Ring
arrangement
and brush
are used to
collect the induced e.M.F.
Dept. of E
CE, MITEMoodbidri
55. GENERATION OFTHRE
EPHASEVOLTAGE SYSTEM
•Let eR , eY and eB be the three independent voltages in coil R1R2, Y1Y2 and
B1B2 respectively.
•All of them will be displaced by one another by 120°.
Dept. of E
CE, MITEMoodbidri
56. ADVANTAGESOF THR
E
EPHASESYSTEM
• THEOUTPUT OF T
H
R
E
EPHASEMACHINE IS ALWAYSGREATERTHAN SINGLE
PHASEMACHINE OF SAMESIZE,APPROXIMATELY1.5 TIMES.
• FORA GIVEN SIZEAND VOLTAGE A T
H
R
E
EPHASEALTERNATOROCCUPIESLESS
SPACEAND HAS LESSCOST TOO THAN SINGLE PHASEHAVING SAMERATING.
• ITISPOSSIBLETOPRODUCEROTATINGMAGNETICFIELDWITHSTATIONARY
COILSBYUSING THR
E
EPHASESYST
E
M.
• THR
E
EPHASEMOTOR
SAR
ESELFSTAR
TING.
Dept. of E
CE, MITEMoodbidri
57. STARCONNECTION
• Starting or terminating ends
of all the three windings are
connected together.
• The ends R
1 - Y1 - B1 are
connected or ends R
2 - Y2 -
B2 are connected together.
• Thiscommonpoint iscalled
neutral point.
Dept. of E
CE, MITEMoodbidri
58. DELTA CONNECTION
• The delta is formed by
connecting one end of
winding to starting end of
other and connectionsare
continued to forma closed
loop.
• The supply terminals are
taken out from the three
junction points.
Dept. of E
CE, MITEMoodbidri
59. R
E
LATION FORSTARCONNECTEDLOAD
LINE VOLTAGES
VL = VRY= VYB = VBR
LINE CURRENTS
IL = IR= IY = IB
PHASE VOLTAGES
VPH= VR= VY=VB
PHASE CURRENTS
IPH=IR=IY=IB
FROM THE CIRCUIT DIAGRAM
𝐈𝐩𝐡= 𝐈𝐋
Dept. of E
CE, MITEMoodbidri
69. ACTIVEAND REA
CTIVECOMPONENTOF CURRENTI
ACTIVE COMPONENT: I COS Φ
REACTIVE COMPONENT: I SIN Φ
ACTUAL POWER=VI COS Φ (IN
WATTS/KW)
REACTIVE POWER= VI SIN Φ
(NO UNIT)
Dept. of E
CE, MITEMoodbidri