Basic electrical engineering GTU (3110005)

199 Pages·062.85 MB·English

by  J.N.Swamy, N.V.Sinha

As latest Syllabus of
per
Gujarat Technological University
srom 2018 Onwards
For, First Year Engineering Students
Semester - I & II
B a s i c E l e c t r i c a l
Engineering
(SUBJECT CODE:3110005)
AUTHORS
Prof. J. N. Swamy
Prof. N. V. Sinha
Asso. Prof., Electrical Engineering,
Asso. Prof., Electrical Engineering,
Government Engineering College, L.D. College of Engineering,
Modasa.
Ahmedabad.
MAHAJAN Publishing Hou8®
AHMEDABAD
(vi)
(vii)
Syllabus.
Index
Teaching and Examination Scheme
Examination Marks Total CHAPTER-1 D.C. CIRCUITS AND RESISTANCE 1.1 TO 1.42
Teaching Scheme Credits
L C Theory Marks Practical Marks Marks 1.1 Resistance, 1.1
ESE (E) PA (M) ESE (V) PA () 1.2 Inductance, 1.1
3 0 70 30 30 20 150 1.3 Capacitance, 1.2
1.4 Voltage and Current Sources, 1.3
Content 1.5 Kirchoffs Laws, 1.4
Sr. No. Topics Teaching Weightage 1.6 Bridge Circuit, 1.5
1.7 Delta/Star Transformation, 1.6
DCC ircuits:
1.8 Mesh Analysis, 1.30
Electrical circuit elements (R, L and C), voltage and current sources, KirchoffY's current 20
and voltage laws, analysis of simple circuits with de excitation. Superposition, Thevenin 1.9 Node Voltage Analysis, 1.31
and Norton Theorems. Time-domain analysis of first-order RL and RC circuits.
CHAPTER-2 NETWORK THEOREMS 2.1 TO 2.62
2. AC Circuits
Representationof sinusoidal waveforms, peak and RMS values, Phasor representation of 25 2.1 D.C. Network Theorems, 2.1I
AC quantities, real power, reactive power, apparent power, power factor. Analysis of 2.2 Cramer's Rule, 2.2
single-phase ac circuits consisting of R, L, C, RL, RC, RLC combinations (series and
parallel), Series and parallelresonance. Three phase balancedcircuits, voltageand cur- 2.3 Super Position Theorem, 2.2
rent relations in star and delta connections, Power measurement in three phase circuits. 2.4 Thevenin's Theorem, 2.18
3. Transformers: 2.5 Norton's Theorem, 2.34
Magneticmaterials, BH characteristics. Construction and workingprinciple of single 15 2.6 Growth of Current in an Inductive Circuit, 2.46
phase and three phase transformers. Ideal and practical transformer. Auto-transformer 2.7 Decay of Current in an Inductive Circuit., 2.48
and its applications.
2.8 Charging and Discharging of a Capacitor, 2.55
4 Electrical Machines Generation of rotatingmagneticfields. 20
Construction and workingof following machines: CHAPTER-3 AC CIRCUITS 3.1 TO 3.171
Three-phase induction motor
3.1 Introduction, 3.1
Single-phase induction motor.
3.2 Generation of Alternating Voltage, 3.2
Separately excitedDC motor.
3.3 Definitions related to A.C., 3.3
Synchronous generators.
3.4 Root Mean Square Value, 3.5
5. Electrical Installations Components of LT Switchgear: 3.5 Average Value of an Alternating Quantity, 3.8
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3.9 A.C. Cireuits Containing Only Resistance, 3.38
3.10 A.C. Through a Purely Induetive Circuit, 3.40
3.11 A.C. through a Purely Capacitive circuit, 3.42
3.12 Phasor Algebra applied to A.C. Circuits, 3.50
3.13 R-L Series Circuit, 3.54
3.14 Active or Real Power, 3.57
3.15 R-C Series Circuit, 3.73
(viil) (ix)
CHAPTER-6 ELECTRICAL MACHINES 6.1 TO 6.14
33..1167 RRe-s-oCna nSceer Riienes p RrCe-isLrec-nuCtiat ,tS ioe3nri. 7eos5f CRierscounita,n ce3,. 903 .91 66..12 STihnrgelee Pphhaassee IInndduuccttiioonn MMoottoorr,, 66..17
3.18 Graphical
33..1290 R-feasocntaonrc eo fC au rSveer,i es3 .R93e sonant Circuit, 3.94 66..43 SSyepncahrarotenloyu se xGcietende raDt.oCr., M6o.1to2r , 6.10
3.21 Bandwidth, 3.95
3.22 Methods of solving parallel A.C. Circuits, 3.107 CHAPTER-7 ELECTRICAL INSTALLATION, WIRING AND SAFETY 7.1 TO 7.18
3.23 Equivalent Impedance Method, 3.115 7.1 Introduction, 7.1
3.24 Admittance, 3.116
7.2 Earthing, 7.1
3.25 Admittance Method, 3.120 7.3 Various circuits explaining requirement of earthing, 7.1
33..2267 RCeosmopnaarnicseo ni no fp saerarilelesl &c irPcauritasl,l el 3R.1e5s0o nance, 3.154 7.4 MethoDs of Reducing Earth Resistance, 7.4
7.5 Electric Shock, 7.6
4.1 TO 4.63
CHAPTER-4 THREE PHASE A.C. CIRCUITS 7.6 First Aid for Electric Shock, 7.7
7.7 Equipment Grounding for Safety, 7.8
4.1 Introduction, 4.1
7.8 Circuit Protection Devices, 7.10
4.2 Advantage of Polyphase over Single Phase, 4.2 7.9 Earth leakage cireuit breaker, 7.12
4.3 Generation of Three Phase EMF, 4.2 7.10 Miniature Circuit Breaker (MCB), 7.13
4.4 Interconnection of the Three Phases, 4.4 7.11 Molded Case Circuit Breaker, 7.15
4.5 Important Definitions, 4.5 7.12 Wiring, 7.16
4.6 Voltage and Current Relationships in Star Connection, 4.6 7.13 Factors to be Considered while Selecting A Wiring system, 7.16
4.7 Delta (D) or Mesh Connection, 4.9 7.14 Cables, 7.17
44..89 MMeeaassuurreemmeenntt ooff PPoowweerr ainn d3 -PPohwaesre FCaicrtcouri tsb,y T4w.1o1 Wattmeter Method- Balanced Load, 4.14 7.15 Types of Cables, 7.18
4.10 Effect of Load Power Factor on the Readings of the Wattmeter, 4.17 CHAPTER-8 BATTERIES AND FUEL CELLS 8.1 TO 8.18
4.11 Two Wattmeter Method with Leading Power Factor Load, 4.18
8.1 Introduction
CHAPTER-5 TRANSFORMERS 5.1 TO 5.14 8.2 Batteries, 8.1
8.3 Grouping of Cels, 8.1
5.1 Introduction, 5.1
8.4 Rating of Battery, 8.4
5.2 What is Permeability ?, 5.2
8.5 Efficiency of Battery, 8.4
5.3 Classification of Magnetic Materials, 5.1
8.6 Charging Methods, 8.5
5.4 Relation between B and H, 5.2
8.7 Types of Charging, 8.6
5.5 Magnetisation Curve, 5.2
8.8 Polarization, 8.7
5.6 Magnetic Hysteresis, 5.3
8.9 Gassing, 8.7
5.7 Transformer, 5.5
8.10 Topping, 8.7
5.8 Construction, 5.5
8.11 Sedimentation, 8.7
5.9 Working Principle, 5.7
8.12 Sulphation, 8.7
5.10 Regulation, 5.10
8.13 Buckling of Plates, 8.7
5.11 Losses in a Transformer, 5.10
8.14 Short Circuiting of Cells, 8.7
5.12 Ideal and practical transformer, 5.11
8.15 Efect of Temperature, 8.8
5.13 Three Phase Transformer, 5.12
8.16 Indications of a Fully Charged Battery, 8.8
5.14 Auto Transformer and its Applications,
5.13
(x)
8.17 Safe Disposal of Batteries, 8.9
CHAPTER
88..1198 EFuleeml eCnetlalsr,y C8a.9lc ulations for Energy Consumption, 8.11 D.C. Circuits
8.20 Basics of Power Factor Improvement, 8.13 and Resistance
APPENDIX
P.1 TO Pa
GTU QUESTION PAPER
1.1 RESISTANCE
Summer 2019, P.1 Resistance is a property of an element which offers opposition to flow of current
R
ww
Fig. 1.1 Symbol of resistance
Every circuit exhibits some opposition to flow of current which is termed as resistance. Nornmally the free
electrons within the atom of a conductor vibrates due to thermal conditions. This does not constitute current.
When a voltage source is applied across such a conductor an electric field is created and electrons move in
direction opposite to that of the electric field and thus building current.
By Ohm's law we have
V
R
Resistance is also said to be
proportional to length of the conductor
inversely proportional to the cross sectional area
dependent on temperature
dependent on nature of material -V
Fig. 1.2 Graphical representation of Ohm's law
Ra or R = p
a a
w(Rhheor)e rpe f(eRrsh ot)o ist hce onnsattaunrte ooff pmroaptoerritaiol.n ality and it's called resistivity or specific resistance of the material. p
RIfe spiostwaenrc ea bmsaoyr baelds o bbye at hroeusgishtto ro fi sa sd penroopteedrt ya so fp a nth eelne ment which converts electrical energy into heat enerEy
W vi = (iR)i = i?R Watts
This is Joule's law.
2If ctu rrencat n isb ea wtirmitete nv aarsia nt quantity i', the amount of energy converted into heat during time interval
W R dt J
When current is a constant quantity "I' heat energy is written as
W RI2t J where, t = t2 - seconds
All resistors produce heat while carrying current which is given by PR.
1.2 INDUCTANCE
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current through it is changing wrt time. Thus although an element may possess
inductance it will not be exhibited if the element is connected in a circuit
where current remains constant with time. Fig. 1.3 Symbol of inductor
Basic Electrical Engine
D.C. Circuits and Resistance
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of current flowing through the source.
v
Induced emf = L
The unit of inductance is Henry. It is denoted by capital letter "H. V
Symbol of Inductance is L.
di
The power absorbed by inductor is P = iL W ()
The energy stored by inductor is W = 2
The voltage across the inductor would be zero if the current passed through it remains constant. This meas
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The inductor can store finite amount of energy. A pure inductor does not dissipate energy, but only stores i Small letter indicates time varying source (ac) while capital letter indicates time invariant source (dc). Fig (c)
shows the ideal characteristics of the ideal voltage source. We can see that and ideal voltage source does not
1.3 CAPACITANCE
Capacitance is ability of a capacitor to store electric charge. A capacitor stores electric energy in the fom possess and internal resiAstVan ce. This can be proved by using Ohm's law.
of electric field. R AI
The capacitance of a capacitor is defined as. It is the ratio of amount of charge required per unit potential differene|
where, AV denotes change in voltage whereas denotes change in current. If change in voltage remains constant
Q Charge required then R = 0.
C:
Potential difference The ideal current source is one which supplies a fixed current regardless of the value and direction of voltage
The Unit of capacitance is farad and it is denoted by capital letter 'F'. appearing across it.
One farad is too large for practical purposes. Hence, much smallerunit like microfarad , nanofarad or picola
is generally employed.
The capacitance is said to be one Farad when one coulomb of
charge is stored with one volt across the two plates of the
capacitor.
A capacitor consists of two conducting surfaces, separated by
an insulating medium called dielectric. The conducting surfaces Fig. 1.4 Symbol of capaci
may be either circular or rectangular.
+QC -QC
Consider a parallel plate capacitor and connected to a dc supply
shown in figure. Left hand side plate connected to a positive (a) ldeal d.c. (b) ldeal a.c. (C)
terminal of the battery and right hand side plate connected to currentsource currentsource
negative terminalof the battery. The electrons flow from plate Fig 1.7
fBAro, mtiot bApel,ac toietm beBes. cnoAemsg eanstievgpeao. tsiivHtievelnyec eac,nh daa rgapesod tethneetliseaecl tredolienfcfset rroaernnesc ewc o(iptlhl. edcdrta. )w oinns The internal resistaRnc e ofA aVn ideal current source is infinite. We can prove this by using Ohm's law as earlier
established between plates A and B. Al
If change in current remains zero due to any corresponding change in voltage R will tend to intinity.
The power absorbed by a capacitor P = VC dv w
dt
The energy stored by a capacitor W = CV2 J
Fig. 1.5 Capacitance
D.C. Circuits and Resista
1.4 1.5
1.5 KInI RsiCmHplOe FciFrc'Sui tLs AthWe Se quivalent resistance can be found by Ohm's law. In complex Circuits calculations are Bfoerf oarnea lwyrsiitningg ththee cKirVcuLi te. quation for the circuit shown in Fig. 1.9, let us understand the sign converntions necessary
done with the help of Kirchoff's laws.
Fall Rise
1.5.1 "ItIK ni isr acnah lsoeolfe fc'kstnr ioFcwailrn s cta irsLc uaKiwti,r caht oafnf'ys cpuorinret not fl atiwm e(,K tChLe )a. lgIet birsa isct asteudm aos f ftohlelo cwusr rent in all the conductors meeettiinng Motion Motion
at any point is zero." Fig. 1.10 Sign convention for KVL
i.e. 2I = 0 From a to b -Vs
From b to a +V,
The polarity of a voltage source is fixed and is independent of the direction of the current. The positive (+)
terminal is at higher potential and negative () termminal is at lower potential. Thus in Fig. 1.10 if we move
from a to b i.e. from the '+ terminal to the - terminal there is a voltage drop denoted by -V,. If we move
from b to a i.e. from the - terminal to the + teminal there is a voltage rise denoted by +V
2
R Motion
Ao-
R
MotionR
Fig. 1.11 Sign convention for KVL
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If the currents entering into a junction are assigned a positive sign, then the currents leaving the junction will Ndiorewc tiloent ufso rwtrriatvee rthsien gK VthLe ecqlousaetdio np aftho.r the circuit shown in the Fig. 1.11. Mostly we choose the clockwise
be assigned a negative sign. Path abcfa : E1 - I\Ri - 13R3 = 0
Consider Fig. 18. Here I1, h and I are entering the junction so they are assigned positive sign. Currents lb Path cdefc: +I>R2 - E + I^R3 = 0
and l are leaving the junction so they are assigned negative sign. Applying KCL.
I + - I - I4 + Is = 0 oSfin ecnee vrgoylt age is energy (or work) per unit charge, KVL is an alternative method of stating the law of conservation
1.6 BRIDGE CIRCUIT
or sum of incoming currents= Sum of outgoing currents
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1.5.2 Kirchoff's Second Law ww.
It is also known as Kirchoff's Voltage Law (KVL). It is stated as follows:
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i.e. EIR + EV 0 ww-
/11*s
2 www www.5 ww
R R3
TE2
Fig. 1.9 Kirchoff's
voltage law
Fig. 1.12(a) Fig. 1.12(b)
Basic EleGtrical Engineering
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1,R1 - 1,Rgt hR =0. R +Ra R23 l (R31 +RI2)
..)
Rewriting this equation, hR, +1Rz hR4 =0 R (Rg+R)
Similarly, applying KVL to path BCDB R2+ Rs R +Rz2 +Ri )
-1) Ra - (lh +1) Rs 1R, = 0 .) Similarly between terminals 3 and 1,
R + Ri Rs1 ll (Ri2 + R)
Ifl = 0 equations (i) and (i) can be written as follows R (R+R,)
1R I2R4 = 0 ii) R3 + R R +R23 +R
IR= 12R4
Adding equations (), (1) and (11) we get,
and LR I2R = 0 ..iv) 2(R R t+Rg R +R R)
2 (R + R, + R3) = R +Rt R
Dividing (ii) by (iv),
(Ri + R2 + Ra) = R2 +R23 +R1
Now, subtracting. (i) from (iv) we get,
RR3 R;R4
R K
The equation (V) gives the condition for a balanced bridge. Ri R2 +Rzt R ..(V)
1.7 DELTA/STAR TRANSFORMATION Similarly subtracting (i) from (iv) we get,
R R
R2R t+ Rt Ru ...(vi)
and subtracting (1) from (iv) we get,
RR
RR+ Rat R ...(vii)
www Equations (v), (vi) and (vii) are the formulaefor conversion froma delta to a star connection. In Fig. 1.14 the
delta circuit is superimposed on a star circuit.
ww
ww
R23
R12
R12 Ra
Rz R
w
R
ww www
Fig. 1.13(a) Star network
Fig. 1.13(b) Delta network
ww
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RReessiissttaannccee bbeettwweeeenn tthhee tteerrmmiinnaallss II aanndd 22 ooff tthhee dstealtra nneetwtwoorkrk = dIte ilst as ereens itshtaatn cthees raensids tadnivcied oinfg a ntyh ebmra nbcyh tohfe t hseu meq uoivf atlheen tt shtraere c odnelnteac trieosnis itsa nocbetas,i nieed. by multiplying the adjacent
Basic Electrical Engineering
D.C.Circuits and Resistance 1.11
1.10
250
RAC (0.857 +0.693) || 1.0 Current I 1025 0.244 A
R
- 1.55 | 10
Voltage across 200 V bulb = IRA
1.55 x 1.01.55
= 0.607 2 = 0.244 x 400 97.6 V
1.55+1.0 2.55
Ryy 0.6078 + 0.5 = 1.107 22 Voltage across 250 V bulb = IRB
= 0.244 x 625 152.5 V
EXAAM PISL 0E. 21 .V3 :a nAd ctuhrarte natc roofs s2 0B A isf lo0w.3s Vt.h rFoiungdh atsw too ahmomwe ttehres sAa maned c uBr rienn ts ewriielsl. d1i vhied peo rbeentwtiaele nd iAtt earenndc eB awcrhoesns EXAbMraPnLchE B1O.5 :a nIdn tthhee cnuertrwenotr kf loshwoiwngn tihnr oFuigg.h Eit.1 w.5h(ae)n ctahlec uclautrer etnhte inv atlhuee obrfa tnhceh uOnCkn oisw zne rroe.s istance R in the
they are connected in parallel.
SOLUTION
I = 20 A VA = 0.2 V VB 0.3 V
RA = Va/I =0 .2 20 0.01 Q
Rg Va/I = 0.3 20 = 0.015 2 1.52
Now I remains the same but Ra and Rg are connected in parallel.
R
IA =I* R +R
RO
Www www-
0.015
20 x 0.01 +0.015
wwww
1 12 A 10v 22
Ra Fig. E.1.5(a)
and IB X R+Rg
sOLUTION
0.01 The circuit is a wheatstone bridge. The current in branch OC is zero. Thus the bridge is balanced. At balance,
20 0.01+0.015
the circuit can be redrawn as in the Fig. E.1.5(5).
= 8.A 1 1.5
or Is = I - Ia 20 12 = 8 A ww- -
4 R-6
EXAMPLE 1.4: Two bulbs A and B are connected in series. A is a 100 W, 200 V bulb and B is 100 W, 250
V bulb. They are connected across a 250 V supply. Calculate () Circuit current (ii) Voltage across each lamp
assuming the bulb resistance to remain unchanged. (G.U. July 2002) www.
SOLUTION 10v
Fig. E.1.5(b)
100% 100W
250 At balance Ix R = 4 x 1.5
L
R 62
250v-
Total resistance 2 +(1 +1.5) || (4 +6)
Fig. E.1.4
= 2 +(2.5) || (10)
Resistance of bulb A = RA : VA = 2- 2.5x 10
2.5+10
(200) 25
400 9 = 2* 12.5
100
= 2 + 2 = 4 2
Resistance of bulb B = Rg
Current supplied by the battery
(250)
625 I= 2.5 A
100
Total resistance = 400 + 625 = 1025 2 This current divides into two parallel paths at point A
Basic Electrical Engineerina
1.8 D.C. Circuitsand Resistance
1.9
Product of adjacent A resistances .(vii)
Y resistance A resistance = Sum ott hep roduct of Y resistances taken two at a time
sum of A resistances
opposite Y resistance
A balanced delta circuit where A balanced star circuit i.e. where
RI2 Ra3 = R3 = R R R Rs = R
when converted to itsequivalent star yields, when converted to its equivalent delta yields
R R- R, R +R, R, +R, R
RR+Rz +Ri R12
3
Similarly, R = R3 = BR
Similarly, R = R1 = 3R
Hence for a balanced delta circuit the equivalent star resistances will be x (delta branch resistance) Hence for a balanced star circuit, the equivalent delta branch resistance will be
3x (star branch resistance)
Delta connected resistances can be converted their equivalents tarc onnected resistances. Here multiplying equations
() and (vi), (vi) and (vii), (vii) and (V) and adding the three we get,
EXAMPLES:
R R, R Ry R Ry, tR, R, R EXAMPLE1 .1 : The resistance of two coils is 25 when connected in series and 6 Qwhen connected in parall.
RR2 + RzR3 + R3R
(R +R +R, Determine the individual resistance of the two coils.
Rp R Ra (R, +RtR,2 sOLUTION
(Ra +R+R,¥ Let Ri and R2 be the two resistances
R R2
R R R R+ R2 25S2 ana R, +R2 62
(R +R t R ...ix)
R, R
From eqn (vii), we get 25 = 6
RIR2 150
RR2 +RR + R^R = Ri2 |R +Rz +R1 Now R2 = (25 R1)
Ri2 R R (25 R,) = 150
R R, +R,R +R, R or (R- 25R + 150) 0
R2 R ...) Solving this we get
From eqn (v), we get (R- 10) (Ri - 15) = 0
R 10 or 15 ohms
R,R +R,R, + RR =R z R+R+R Corresponding R= 15 or 10 ohms
= Rz3 R EXAMPLE 1.2 : Determine the resistance between the points X and Y in the network shown in the Fig E.1.2.
TKR , +R, R, 1.02
.R23
R .xi) wwww
From eqn (vi), we get
1.5 2.22
w- W-
R,R2 + RR + RaR = R R+R +Ry 2.52 0.50
w wW y
- R R 2.02 1.7
www-
R, R2*R R, *R, R
R3 Fig. E.1.2
R2 xi) SOLUTION
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Basic Electrical Engineering 1.13
D.C. Circuits and Resistance
1.12 Substituting in (i),
2.5 140 124 Sl2
I1 2.5+10 h = 2A
Curent in 32 > 4A
2.5 2.125.5 82 2A
0.5 A 282 (I - I1) = 5 -4 1A
EAA' MPLCEu r1re.6nt: iCn ablcruanlacthe cBuOrr e= nt2 .i5n -eac0h.5 re= si25s AtoA r. fworw thwe- network shown in the Fig. E.I.6(a). EXAMPLE 1.7: Calcu41l24at2 e the((IIv o l--taIg1Ie2 )+ alc=2 ro)4 s s= A52 B -i4n 2 At+h e2 circ3uAit shown in the Fig. E.I.7(a).
ww 6
www
ww2w9 - ww
10V
40v wwwwwww 22 359
20
www.-
www
62
Fig. E.1.7(a)
Fig. E.1.6(a)
sOLUTION
SOLUTION: Redrawing the circuit.
Redrawing the given circuit and assuming the currents as shown, 62
1 5A ww
10v
M.
www-
w. 2 52
wv 20
82
40v D ww-
( 2 ww( 1-1t1) ww
62
ww
22
Fig. E.1.6(b)
Fig. E.1.7b)
Applying KVL to loop ADCA Applying KVL to meshes I, II and I1, we get
40 31 + 14 (1 h) 10 (2 +6 + 2) I - 21 - 6l
-17 141 101 21- 61
Applying KVL to loop ABDA 5 SI-h -3l3 ...(1)
0 (6 + 5+ 2) 2 - 211 Sl3
0 28 (1 - 1) - 81 -3
28I 311 -82 0-21 + 131 5l ...(1)
But, I =:5 A 0= (6 + 5 + 2) I - 6l1 5l2
0 61- 5l2 + 13l3 ii)
140 31l, + 812
Multiplying eqn (i) by 2 and (i) by 5 and adding them we get
Multiplying (i) by 4 and (i) by 7 and adding them we get, 1011 212 - 6l3 = 10
160 = 681 - 56l2
-1011 + 6512 - 251, = 0
980 2171, + 56l2
631-3113 = 10
1140 285
I 4A I,= 31,-1 0 ..(iv)
31