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Vijay Balu Raskar E.T. Notes
 

Vijay Balu Raskar E.T. Notes

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Electrical Technology Notes for preparation.

Electrical Technology Notes for preparation.

Kindly note :- Don't forget to use class note book while studying.

Use my theory questions to clear ET and solved all the problems.

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    Vijay Balu Raskar E.T. Notes Vijay Balu Raskar E.T. Notes Document Transcript

    • Electrical Notes….. EELDAD“Vijay B. Raskar  What qualities do successful engineers have? Strong in mathematics and science. Highly analytical and detail oriented. Imaginative and creative. Good communication skills. Enjoy working in teams. Enjoy building or improving the way things work. TECHNIQUE TO CLEAR ELECTRICAL TECHNOLOGY SUBJECT IMPORTANT CHAPTERS IN ELECTRICAL TECHNOLOGY WITH HIGHER WAITAGE: ( HIGHER TO LOWER PRIORITY TO STUDY) 1) SINGLE PHASE A.C.CIRCUITS 2) SINGLE PHASE TRANSFORMERS 3) D.C.MOTORS 4) D.C.CIRCUITS 5) THREE PHASE A.C. CIRCUITS 6) A.C. FUNDAMENTALS 7) INSTALLATION, EARTHING AND TROUBLESHOOTING 8) FRACTIONAL HORSE POWER MOTORS Vijay Raskar 1
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ INDEX S. CHAPTER NUMBER PAGE WEIGHTAGE NO. NUMBER 1 IMP FORMULAES 2 D.C. MOTORS 3 INSTALLATIONS, EARTHING AND TROUBLESHOOTING 4 SINGLE PHASE TRANSFORMER 5 A.C. FUNDAMENTALS 6 D.C. CIRCUITS 7 SINGLE PHASE A.C. CIRCUITS 8 THREE PHASE A.C. CIRCUITS 9 FRACTIONAL HORSE POWER MOTORS 10 Theorems (Problems) NOTE: Problems are very important with theory. Subject is very easy if practice. Use technique to solve the problems. More practice = More marks. Use calculator and it is very essential to clear “ELECTRICAL SUBJECT”. 2
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ Derivation is also important. Solve previous question papers. Any doubts or query ask any time. 1. IMPORTANT FORMULAES D.C. CIRCUITS V = I*R R =  (/ A ) G = 1/ R SERIES CIRCUIT: R = R1 + R2 + …..Rn PARALLEL CIRCULT: R = (R1 + R2) / (R1*R2) KCL = I 0 KVL = V  0 LOOP METHOD = MESH ANALYSIS = IR METHOD USE CLOCK WISE DIRECTION R = + to - Take –ve R = - to + Take +ve V = OPPOSITE SYMBOLE I = SAME SYMBOLE NODAL ANALYSIS = V/R METHOD USE OUTGOING EVERYTIME I = INCOMING = TAKE +VE I = OUTGOING = TAKE –VE V = OPPOSITE SYMBOLE STAR –DELTA: R12 = R1R2 + R2R3 + R3R1 R3 R23 = R1R2 + R2R3 + R3R1 R1 R31 = R1R2 + R2R3 + R3R1 R2 DELTA-STAR: R1 = R12 * R31 R12 + R23 + R31 R2 = R12 * R23 R12 + R23 + R31 R3 = R23 * R31 3
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ R12 + R23 + R31 A.C. FUNDAMENTALS f=1/T   t   2f Instantaneous value of e.m.f. or standard equation of e.m.f. = e = Em Sin  e = Em Sin t e = Em Sin 2f (NOTE: e = V) Instantaneous value of current or standard equation of current = i = Im Sin  i = Im Sin t i = Im Sin 2f Heat produced in the nth interval = i₂²R(t / n) UNIT: Joules Irms = 0.707 * Imax Imax = 2 * Irms Vrms = 0.707 * Vmax Vmax = 2 * Vrms Maximulvalue Peak factor = = 1.414 R.M .S .value R.M .S .value Form factor = = 1.11 Averagevalue V 1  V 2  V 3  ...  Vn Vavg = n SINGLE-PHASE A.C. CIRCUITS PURELY RESISTIVE A.C.CIRCUIT:  i = ImSin t  v = VmSin t  P = V*I = V²/R = I²*R PURELY INDUCTIVE A.C.CIRCUIT:   i = ImSin( t - ) 2 4
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  v = VmSin t PURELY CAPACITIVE A.C.CIRCUIT:   i = ImSin( t + ) 2  v = VmSin t V = I*R VL = I*XL Vc = I*XC XL = L = 2fL 1 1 XC = = C 2fC FOR R-L CIRCUIT  VR = I*R  VL = I*XL  V = I*Z  Z² = R² + XL² R  POWER FACTOR = COSᴓ = Z  P = V*I*COSᴓ FOR R-C CIRCUIT  VR = I*R  Vc = I*Xc  V = I*Z  Z² = R² + Xc² R  POWER FACTOR = COSᴓ = Z  P = V*I*COSᴓ FOR R-L-C CIRCUIT  VR = I*R  VL = I*XL  Vc = I*Xc  V = I*Z  Z² = R² + (XL-Xc)² If XL < Xc  Z² = R² + X² If XL = Xc R  POWER FACTOR = COSᴓ = Z  P = V*I*COSᴓ THREE PHASE A.C. CIRCUITS 5
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ STAR CONNECTION:  LINE CURRENT = PHASE CURRENT  LINE E.M.F. = 3 * PHASE E.M.F. (VOLTAGE = E.M.F.) DELTA CONNECTION:  LINE CURRENT = 3 * PHASE CURRENT  LINE E.M.F. = PHASE E.M.F. (VOLTAGE = E.M.F.) POWER FOR STAR & DELTA: P = 3 * LINE VOLTAGE * LINE CURRENT * COSᴓ P = 3 * VL * IL * COSᴓ R = Z * COSᴓ X = Z * Sinᴓ Output Efficiency = *100 Input Reactive power = Q = 3 * VL * IL * Sinᴓ SINGLE PHASE TRANSFORMER E.M.F. EQUATION OF TRANSFORMER:  e = 4.44 * ᴓm * f * N  For primary winding: e1 = 4.44 * ᴓm * f * N₁  For secondary winding: e2 = 4.44 * ᴓm * f * N₂ m  Bm = Maximum flux density = Area V 2 V1 Voltage regulation: V2 EFFICIENCY OF TRANSFORMER:  OUTPUT POWER = INPUT POWER – LOSSES  INPUT POWER = OUTPUT POWER + LOSSES  INPUT POWER = V2I2 * COSᴓ +Pi + Pcu Output  Efficiency = *100 Input  EFFICIENCY AT FULL LOAD = X * FULLLOADRATING * p. f . ( X * FULLLOADRATING * p. f .)  Pi  ( X * X * Pcu ) TOTAL COPPER LOSS = I₁²R₁ + I₂²R₂ HYSTERESIS LOSS = Ph = Kh * Bm¹·⁶ * f * v 6
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ EDDY CURRENT LOSS = Pe = Ke * Bm² * f² * t² * v OUTPUT POWER = kVA * p.f. 2. D.C. MOTORS  Voltage equation of the motor:  V = Eb  IaRa  Eb = ZNP FOR LAP (A = P) & WAVE WINDING (A=2) 60 A  T = 0.159 *  * Z * Ia * P A  SHUNT MOTOR:  Ish = Vsh Rsh  Ia = I  Ish  Eb  V  IaRa  SERIES MOTOR:  Ia  I  Eb  V  Ia( Ra  Rse)  SHORT-SHUNT COMPOUND MOTOR:  V’ = V  I * Rse  Ish  V Rsh  Ia  I  Ish  Eb  V  IaRa  LONG-SHUNT COMPOUND MOTOR:  Ish  V Rsh  Ia  I  Ish  Eb  V  Ia( Ra  Rse)  If  Ia Then, N1  Eb1* 2 = Eb1 * Ia 2 N2 Eb 2 * 1 Eb 2 * Ia1 7
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ (B.E.ELECTRICAL) 2. SINGLE PHASE A.C. CIRCUITS Q.1 Derive the relationship between voltage and current in pure resistive circuits. ANS: 8
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Consider a purely resistive circuit.  It contains pure resistance.  Alternating voltage is applied across resistance.  v = VmSin     t ………………………………………………………(1)  Current is set up in the circuit.  By ohm’s law, V = IR VmSin  = IR VmSin :. I = R  The current is maximum when Sin  1  :. Im = Maximum current Vm  Im = R  i = Im Sin  i  Im Sint  From phasor diagram, Current is in phase with voltage.  Angle between current and voltage is zero.    0 CONCLUSION:  In purely a.c. resistive circuit, the current is in phase with voltage.  Ohm’s law is applicable.POWER:  Power is a product of current and voltage.  P = V*I  P = VmSin * Im Sin = Vm Im sin ²  = Vm Im sin ² t Vm Im 1  cos t = (1  cos 2t ) …………………..( sin ² t  ) 2 2 9
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ Vm Im Vm Im =  cos 2t 2 2  Fluctuating part is a cosine wave.  Cosine curve of frequency is double that of voltage and current waves.  Therefore, average power is zero. Vm Im  P= 0 2 Vm Im P= 2 Vm Im Maximumvalue P= (R.M.S. Value = ) 2 2 2 P = V*I P = VI = I²R = V² / R WATTSQ.2 Derive the relationship between voltage and current in pure inductive circuits.ANS:  Consider a purely inductive coil* with Inductance L henries.  Inductance having negligible ohmic resistance.  A.c. supply is connected across inductance. 10
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  V  VmSin  V  VmSint ………………………(1)  The resulting a.c is set up in the coil.  E.m.f. is induced in it.  E.m.f. of self induction is, di e  L dt di v = e  L …………………….(2) dt  Ohmic resistance is negligible.  Substitute (1) in (2) di VmSint  L dt Vm di  Sint.dt L Integratin g ,  cos t NOTE: Integration of  sin t   Vm  di  L Sint.dt Vm (  cos t ) i= L    VmSin (  t ) i 2 L  VmSin (t  ) i 2 ………………………(3) L  Im is Maximum when Sin (t  ) 1 2 Vm Im  L Substitute in (3)  i  Im Sin (t  ) 2 This equation is shows,  Current will be sinusoidal.  Current lag behind the voltage by 90 11
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ VmIm  L I V 0.707 0.707L VI LXL = LPOWER:P = VI P  VmSint * Im Sin (t  ) 2 P  Vm Im Sint * Sin (t  ) 2P  Vm Im Sint * ( Cost )P  Vm Im Sint * Cost Vm ImP Sin 2t 2 :. Sin curve of frequency double that of voltage and current waves. Therefore, Average power is zero.CONCLUSION:The average demand for power in a purely inductive circuit over the whole period is alwayszero.Q.3 Derive the relationship between voltage and current in pure capacitive circuits.ANS: 12
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Capacitor is connected across a.c. supply.  When applied voltage increases capacitor charging.  When applied voltage decreases capacitor discharging.  A.c. supply is connected across capacitor.  V  VmSin  V  VmSint ………………………(1)  C = Capacitor  We have, Charge = Capacitor * Voltage Q=C*V Q = C* VmSint  We have, Charge = Current * Time  Small charge = dQ on a capacitor plate in small time (dt)  Therefore, Charge = Current * Time dQ = i*dt dQ i dt d i  (CVmSint ) dt 13
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ d i  CVm ( Sint ) d dt ………….( Sint  Cost dt i  CVm (Cost )  i  CVmSin (t  ) 2  Current reaches maximum value, when Sin (t  ) 1 2 Im  CVm Substitute ,  i  Im Sin (t  ) 2 CONCLUSION: Current varies sinusoidally and leads the applied voltage by 90. POWER: P = VI  P  VmSint * Im Sin (t  ) 2  P  Vm Im Sint * Sin (t  ) 2 P  Vm Im Sint * Cost Sin 2t P  Vm Im 2 Power curve is purely sin wave of frequency double that of voltage and current. Average power is zero.Q.4 Explain power factor in detail.ANS: R  Power factor = Cos  Z TruePower VICos  Power factor =   Cos ApparantPo wer VI  Power factor is defined as the ratio of True power to Apparent Power.  The numerical value of the cosine of the phase angle between voltage and current.  Power factor is always less than one.  Nature of power factor is decided by lead or lag of the current with respect to voltage.  Capacitive circuit = Lead  Inductive circuit = Lag  Power factor is also expressed as percentage. 14
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________Q.5 What is a Choking coil?ANS:  It consists of a coil wound on a laminated iron core.  In practice, a.c. circuit have negligible small capacitance.  Such circuits come under R-L series circuit.  Its inductance is very high in comparison with its ohmic resistance.  Resistance and inductance are not physically separate.  Such a coil is always equal to R-L series circuit.  In many cases, we require low voltage. Hence, Reduction is obtained by choking coil.  Choking coil is connected in series with circuit.  Choking coil is used in fluorescent tube circuit.  Its having very little loss as compare with simple resistor used in phase.  Choking coil reduces the circuit current.  Choking coil is also used as current limiting device.  It also used as simple resistor due to small loss.Q.6 What is Active power, Reactive power and Power triangle?ANS:  Active power (P):  It is the product of applied voltage and active component of the current.  P  VICos  Unit: Watts or Kilo-watts  Reactive Powers (Q):  It is the product of applied voltage and reactive component of the current.  P  VISin   Unit: Volt-Ampere or Kili-Volt-Ampere  Power Triangle:  All three sides of impedance triangle are multiplied by I²  From power triangle, 15
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ S  (P2  Q 2 ) P  SCos Q  SSin  Inductive Reactive power is always negative.  Capacitive Reactive power is always positive. 3. SINGLE PHASE TRANSFORMERSQ.1 State and explain the principle of transformer?ANS: STATEMENT: The operation of a transformer is based on the principle of mutualinduction between two circuits linked by a common magnetic field.EXPLAINATION:  See the elementary transformer.  There are two windings PRIMARY and SECONDARY.  They are separated and wound on common laminated steel core.  Vertical portion of core where windings are placed is called as LIMB.  Top and bottom portions are called as YOKES.  Primary winding is connected across mains and output is receives from secondary windings. 16
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  “ When the supply is flows from primary windings, alternate voltage circulates an alternating current through it.  The current flowing through primary windings produces the alternating flux (ᴓ)  This varying flux is linked with secondary windings through iron core and induces an e.m.f. (Electro Magnetic Force).  E.m.f. is induced accordance to Faraday’s law of electromagnetic induction.  a.c. current in the primary winding produces an e.m.f. in the secondary winding. This phenomenon is called as mutual induction.  E.m.f. induced in the secondary winding is known as mutual induce e.m.f. or e.m.f. of mutual induction.  Frequency remains same at both sides.  Transformer operation is possible only in alternating currents and not for direct current.”Q.2 Explain types of transformer.ANS:There are three types of transformers depending upon the core and windings. a) Core Type Transformer b) Shell type transformer c) Berry type transformerEXPLAINATION: 1) Core Type Transformer: 17
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ Figure: Core type transformer  This core transformer is having rectangular frame and provides a single magnetic circuit.  The windings are normally cylindrical in form and concentric, low-voltage winding being placed near the core.  Both windings are uniformly distributed over two limbs of the core.  Therefore, Neutral core is very effective.  The coils can be withdrawn for repairs by dismantling lamination of the top yoke. 2 ) Shell type transformer: 18
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Shell type transformer provides double magnetic circuit.  The windings are normally sandwitch type.  Windings are placed on central limb of the core.  H.V. and L.V. coils are wound in the form of pancakes and they are interleaved.  The windings are near to the yoke of the core is L.V. windings only.  The windings are placed on the central limb which gives mechanical protection to the winding.  Natural cooing is poor.  During repairing, large numbers of laminations are to be dismantled. 3) Berry Type Transformer:  Berry type transformer distributes the magnetic circuit.  Magnetic circuits are independent.  The core is constructed in such a fashion that its yokes radiate out from the centre just like the spokes of a wheel.  The windings are near to the yoke of the core is L.V. windings only. 19
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  The windings are placed on the central limb which gives mechanical protection to the winding.  Natural cooing is poor.  During repairing, large numbers of laminations are to be dismantled.Q.3 Derive the e.m.f. equation of the transformer?ANS:  We know that, primary winding is connected to a.c. supply and secondary winding is connected to the load.  When current is flowing through the primary winding flux is induced in the core.  Flux is linked with secondary winding and produced e.m.f. by mutual induction.  In the primary winding, E.m.f is induced due to self induction.MATHEMATICAL EXPRESSIONS: Where,  N1 = Number of turns on the primary winding. o N2 = Number of turns on the secondry winding.  ᴓm = Maximum flux in weber  f = Frequency in Hz  Figure shows one cycle of the sinusoidal flux.  :. Average value of change of flux = maximum flux / time = ᴓm / (1/4f)  Average value of change of flux = 4 X ᴓm X f weber per second.  Average e.m.f. = 4 X ᴓm X f Volts  We have, form factor = 1.11  :. R.M.S. value of e.m.f. induced in each turn : 1.11 X average value : 1.11 X 4 X ᴓm X f 20
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ R.M.S. value of e.m.f. induced in each turn : 4.44 X ᴓm X f …….(1)  R.M.S. value of induced e.m.f. for primary winding: E1  E1 = Induced e.m.f. each turn X number of turns of P Winding (from (1))  E1= 4.44 X ᴓm X f X N1 VOLTS  Similarly,  E2= 4.44 X ᴓm X f X N2 VOLTS This is the transformer equation. OR  ᴓ = ᴓm sinwt ………………………..(1)  according to Faraday’s electromagnetic induction,  e = -N dᴓ dt  N = 1 (For e.m.f. per turn)  e = - dᴓ dt  e = - d (ᴓm sinwt) (From 1) dt  e = -w ᴓm coswt (derivative of sinwt = wcoswt)  e = w ᴓm sin(wt – (π/2)) volts ………….(2)  maximum induced e.m.f. per turn = w ᴓm = 2πf ᴓm volts (w = 2πf) ….(3)  We have, R.M.S. value =0.707 X Maximum value  R.M.S. value =0.707 X 2πf ᴓm volts  E1= 4.44 X ᴓm X f X N1 VOLTS  Similarly,  E2= 4.44 X ᴓm X f X N2 VOLTS This is the transformer equatioQ.4 Explain properties of ideal transformer or define ideal transformer?ANS: PROPERITIES OF IDEAL TRANSFORMER:  It has no losses.  Its winding has no ohmic resistance.  There is no magnetic leakage.  The permeability f the core is so high.  Required very low current to produce flux in the core.Q.5 EXPLAIN Transformer losses or its type.ANS: 21
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Electrical energy is transfer from one circuit to another.  Some energy is always lost.  All the losses in the transformer are ultimately converted into heat.  Therefore, transformer is provided with cooling system to protect the insulation of the windings.  There are two types of losses in the transformer: 1) COPPER LOSS:  TOTAL COPPER LOSS = I₁² R₁ + I₂²R₂ Where, I₁ & R₁ : For primary I₂ & R₂ : For secondary  Copper conductor is used for winding.  Loss of power caused by resistance of the winding.  Power loss is proportional to the square of the current flowing through the windings.  It is ultimately utilized in heating the windings.  Copper loss is proportional to the current and square of the KVA output. 2) CORE OR IRON LOSS: a) Hysteresis loss:  Hysteresis loss = Ph = Kh X Bm¹·⁶ X f X v volts Where, Kh = constant Bm = Maximum flux density in tesla f = frequency v = volume of the magnetic material  This loss is takes place in the transformer core. 22
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  It is continuously subjected to rapid reversals of magnetization by the alternating flux. b) Eddy current loss:  Eddy current loss= Pe = Ke X Bm² X f² X v X t² watts Where, Ke = constant Bm = Maximum flux density in tesla f = frequency v = volume of the magnetic material t = thickness of the laminations  Core losses together are nearly constant and independent of the magnitude of the current.  They can be reduced by choosing silicon steel having small values of Kh and Ke.Q.6 Explain Auto-Transformers. Also, explain its applications.ANS:  Auto-transformer having only one winding on a laminated magnetic core.  Winding is common for both primary and secondary circuits.  Ordinary two winding transformer, it can be used as step up and step down transformer.  V1, N1 for primary side and V2, N2 For secondary side.  If neglecting the losses, leakage reactance and magnetizing current, We have, V2 = I1 = N2 = K V1 I2 N1APPLICATION:  For starting squirrel-cage induction motors & synchronous motors. 23
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  For interconnection of a.c. systems.  As booster to raise the voltage in a.c. feeders.  For furnace transformers for getting suitable supply voltage.  As variacs for getting continuously variable supply a.c. voltage.Q.7 What are the advantages and disadvantages of auto-transformers?ANS: ADVANTAGES OF AUTO-TRANSFORMER:  Its cheaper.  Copper require very less for windings.  Transformation ratio is approaches unity.  Its regulation is better than two winding transformer.  Efficiency of transformer is high.  I²R Losses are less.  The resistance and leakage reactance is less.DISADVANTAGES OF AUTO-TRANSFORMER:  There is always risk of electrical shock.  When it is used in high-voltage circuit, possibilities of electrical shocks.  Therefore, Low voltage and high voltage sides are not electrically separated.Q.8 Explain isolation transformers.ANS: Definition: The transformer which are specially designed to provide electrical isolationbetween the primary and secondary circuits without changing voltage and current levels.EXPLAINATION:  The isolation transformer is frequently used to isolate one portion of an electrical system from another.  These systems are physically separated.  They are magnetically couple to each others.  Electrical isolation provided between primary and secondary circuits.  Isolation transformer is a 1:1 transformer and is used only for the purpose of electrical isolation.FUNCTIONS OF ISOLATION TRANSFORMERS: 1) Isolation transformers isolate and protect the sensitive and expensive equipments from electrical system grounds. 2) Isolation transformers protect the delicate and expensive equipments against voltage spikes. 24
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ 4. D.C.MOTORSQ.1 Explain the working principle of operation of a d.c. generator.ANS: Generator: Which converts mechanical energy into electrical energy either ac or dc.Principle of Generator: “When a conductor is moved in magnetic field or magnetic field movedwith respect to the conductor. According to Faraday’s law of electromagnetic induction,electromotive force is set up in the conductor”  There is relative motion between a conductor and a magnetic field.  Voltage will always be generated in the conductor.  A.C. Generators are normally known as alternators.CONSTRUCTION:  It consists of a single turn rectangular conducting coil (AB) like copper or aluminum.  Coil / conductor is placed in between two poles (S & N POLE).  Coil is placed in such a way that to rotate at constant speed in uniform magnetic field.  The coil is also known as armature of the alternator.  The ends of the armature coil are connected to rings called slip rings. i.e. R1 & R2  Two carbon brushes pressed against the slip rings.  The slip rings collect the current and carry it to the external resistor (R).OPERATION: Current supplied by alternator is necessarily alternating.  Unidirectional or direct current is desired.  Before external circuit, alternating current must be rectified or commutated.  Rectification can be done by replacing the slip-rings R1 and R2 of the alternator.  Where, M is shown in figure “COMMUTATOR”.  Commutator is made by conducting material which is cuts in two equal halves or segments.  Segments are insulated by thin sheet of mica or some other insulating material.  The ends of the armature coil AB are joined to these segments.  The external load resistor R is connected to the split ring M through brushes B1 and B2. 25
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ACTION OF A COMMUTATOR:  When coil is undergoing for first half cycle, current flowing from load(R) (C to D).  For next half cycle, its reverses with position of a and b terminals.  Again for next cycle, current is flowing from C to D and so on.  Direction of current through the external circuit is same.  In the commercial d.c. generator, by using number of coils evenly distributed around the armature core and connecting these coils to commutator having a corresponding number of segments, a more steady direct current is obtained in the external circuit.Q.2 Explain the working principle of operation of a d.c. Motor.ANS: CONSTRUCTION:  It consists of a single turn rectangular conducting coil (AB) like copper or aluminum.  Coil / conductor is placed in between two poles (S & N POLE).  Coil is placed in such a way that to rotate at constant speed in uniform magnetic field.  Two carbon brushes pressed against the slip rings.  Armature coil is supplied with direct current through the brushes B1 and B2.WORKING / OPERATION:  Unidirectional torque is produced in the d.c. motor.  There are three positions to understand it.  Torque developed at different position of the coil.  By Flemming’s left hand rule, conductor moves upwards as well as downwards.  Conductor is placed between the two poles.  The computer plays very important role in the operation of the d.c.motor. 26
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________POSITION-1  Plane of the coil lies parallel to a magnetic field.  The direction of current in the conductor is plane of paper from North to South.  Therefore, conductor is moves downwards from N side and upwards from S side.  They are moves as per the Fleming’s left hand rule.  These two upward and downward forces are produces the torque.  Torque tending to rotate the coil about its axis in a counterclockwise direction.  Perpendicular distances between the two forces are maximum. Hence, torque is also maximum.POSITION-2  In this position, the two forces being opposite to each others.  Therefore, the perpendicular distances between two forces are zero.  Therefore, torque is zero.  Hence, coil tends to remain stationary in this position. Because of its inertia.POSITION-3  After crosses position-2, at same time. The direction of the current in the coil is reversed by the commutator.  Therefore, coil continues to experience the torque in the counterclockwise direction only.  Commutator reverse the current in each conductor of the armature as it passes from one pole to another.Q.3 Explain back e.m.f. or counter e.m.f. 27
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ANS: e.m.f. = Electro Motive Force.  Current carrying conductor of the armature winding is placed in the magnetic field.  Magnetic field is produced by field winding.  When armature conductor starts rotating, flux cuts and emf is induced.  Emf is induced according to Faraday’s law of electromagnetic induction.  According to Lenz’s law, emf is opposes the applied voltages.  Therefore, it is known as back emf .  Counter emf is opposes the applied voltage across the armature.  The net or total voltage in the armature circuit is difference between the two. i.e. V-Eb  By ohm’s law, V=IR  Eb = V-IaRa i.e. V=Eb+IaRa ……………..Voltage equation of motor Where, Ia = Armature current. Eb= back emf Ra=Armature resistance V=Applied voltage  Voltage equation of the transformer is shows, Voltage is not depends on resistance or Armature current but also, depends on back emf.  Back emf of motor = Eb = ᴓZNP 60AQ.4 Explain the types of d.c. motors. ANS: 28
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  There are three types of D.C.Motors. 1. A d.c. shunt motors 2. A d.c. series motors 3. Compound motors a) A short-shunt compound motor b) A long-shunt compound motorEXPLANATION: A) SHUNT MOTORS:  Z-ZZ is a field winding.  A-AA is a armature winding.  When field winding is parallel to armature winding. It is said to be “Shunt motor”. 29
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Shunt means parallel.  In shunt motor Z-ZZ (field) winding has large number of turns of the fine gauge copper wire.  The field winding has high resistance to limit field current to a small value.  This is helpful to reduce the power loss in the field circuit.  Ish = V / Rsh  Ia = I – Ish  Eb = V – IaRa B) SERIES MOTORS:  Y-YY is a field winding.  A-AA is a armature winding.  When field winding is series with armature winding. It is said to be “Series motor”.  In series motor Y-YY (field) winding has few turns of copper wire for a low resistance.  Field current being large which is equal to motor current.  The required flux can be produced by small number of turns.  Ia = I  Eb = V – Ia(Ra+Rse) C) COMPOUND MOTORS:  A compound motor is a combination of shunt and series motor.  Both shunt and series windings are employed.  Both windings are wounded on same poles.  Y-YY is a series field winding.  A-AA is a armature winding.  Z-ZZ is a shunt field winding.  Where, Y-YY is in series with A-AA and in parallel with Z-ZZ. It is said to be short shunt compound motor.  Short shunt compound: V’ = V-IRse Ish = V’ / Rsh Ia = I – Ish Eb = V’ – IaRa = V – Irse -IaRa  Where, Z-ZZ is connected across A-AA and in series with YY. It is said to be long- shunt compound motor  long-shunt compound motor: Ish = V / Rsh Ia = I – Ish Eb = V’ – Ia(Ra + Rse) Where, Eb = Back emf. Ia = Armature Current. 30
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ Rse = Series resistance. Ish = shunt field current.Q.5 Explain characteristics of D.C.Motors. (for 4-5 Marks)ANS: There are three characteristics of d.c. motors. 1) T vs Ia 2) N vs Ia 3) T vs NEXPLAINATION: 1) T vs Ia : T  Ia Where, T = Torque Ia = Armature current  In dc motor, torque is directly proportional to armature current.  The curve which shows the relation between the Torque (T) and the armature current (Ia). 2) N vs Ia : N  Ia Where, N = Speed Ia = Armature current  In dc motor, speed is directly proportional to armature current.  The curve which shows the relation between the Speed (N) and the armature current (Ia). 3) T vs N: T N Where, T = Torque N = Speed  In dc motor, torque is directly proportional to speed.  The curve which shows the relation between the Torque (T) and the Speed (N).Q.6 Explain the characteristics of D.C.Motor with one type in details.ANS: There are three characteristics of d.c. motors. 1) T vs Ia 2) N vs Ia 3) T vs NEXPLAINATION: 1) T vs Ia: T  Ia 31
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ Where, T = Torque Ia = Armature current  In dc motor, torque is directly proportional to armature current.  The curve which shows the relation between the Torque (T) and the armature current (Ia). 2) N vs Ia N  Ia Where, N = Speed Ia = Armature current  In dc motor, speed is directly proportional to armature current.  The curve which shows the relation between the Speed (N) and the armature current (Ia). 3) T vs N: T N Where, T = Torque N = Speed  In dc motor, torque is directly proportional to speed.  The curve which shows the relation between the Torque (T) and the Speed (N).There are three characteristics of d.c. shunt motors. 1) T vs Ia 2) N vs Ia 3) T vs NEXPLAINATION:Let us see D.C.SHUNT MOTOR CHARACTERISTICS 1) T vs Ia :  T  Ia Where, T = Torque Ia = Armature current  In dc motor, torque is directly proportional to armature current.  The curve which shows the relation between the Torque (T) and the armature current (Ia). 32
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Whenever, Torque increases as Armature current increases. 2) N vs Ia: N  Ia Where, N = Speed Ia = Armature current  In dc motor, Speed is directly proportional to armature current.  The curve which shows the relation between the speed (N) and the armature current (Ia).  Whenever, Armature current increases speed decreases.  N  ( Eb /  )  N  Eb  V - IaRa  Ia increases , back emf is reduces. Therefore, speed downs.  Hence, the curve is slightly drooping.  Drop in speed from no load to full load being very small. 3) T vs N: T N Where, N = Speed T = Torque  In dc motor, Torque (T) is directly proportional to speed (N).  The curve which shows the relation between the Torque (T) and the speed (N). 33
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Whenever, Torque increases speed decreases.  N  ( Eb /  )  N  Eb  V - IaRa  Ia increases , back emf is reduces. Therefore, speed downs.  Hence, the curve is slightly drooping.  Drop in speed from no load to full load being very small.  This type of speed characteristic is generally known as shunt characteristics.Q.7 Explain D.C. Series motor characteristics.ANS: There are three characteristics. 1) T vs Ia 2) N vs Ia 3) T vs N EXPLAINATION: Let us see D.C.SERIES MOTOR CHARACTERISTICS 1) T vs Ia:   Ia T   Ia  I²a  Before saturation of the magnetic circuit,   Ia  The torque is proportional to the square of armature current.  Characteristic of D.C.Series motor is like parabola.  After saturation of the magnetic circuit,  remains constant.  Hence, T    Characteristic is straight line after the saturation.  At low speed, series motors produced a high torque.  Where large torque is required, d.c. series motors are used. 2) N vs Ia: 34
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  N  ( 1 / Ia).  Speed is inversely proportional to the armature current.  N  (Eb /  )  (1 /  )  ( 1 / Ia)  Therefore, Speed is inversely proportional to the armature current.  Eb = V – Ia (Ra + Rse)  Ia (Ra + Rse) is very small.  Eb is nearly constant.  See the characteristics, Its rectangular.  The curve shows, the series motor is variable speed motor.  On load or no-load it can be rise the speed.  The centrifugal force developed at such a high speed can damage the motor.  Hence, the d.c. series motor is never started without some mechanical load on it or never run on light load. 3) T vs N:  Curve shows that, torque increases with speed deceases.  Speed of the motor falls as the torque increases.  This type of characteristics is known as series characteristics.Q.8 Explain speed control of d.c. motors or explain armature voltage control method.ANS: We have,  N  (Eb /  )  Eb = V – IaRa 35
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  N  ((V – IaRa )/  )  Armature Resistance is very small.  Therefore, IaRa is also small.  :. N  ( V /  )  Speed of d.c. motor is directly proportional to Applied voltage and inversely proportional to flux.There are two basic methods to vary the speed of d.c. motor. 1. Armature Voltage control 2. Flux controlLet us see in details. 1. Armature Voltage Control: 36
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________EXPLAINATION:  When Applied Voltage across the armature winding changes then speed changes.  Speed of d.c. motor is depends on applied voltage.  Applied voltage is constant.  Variable Resistance is connected across armature winding with voltage.  Variable resistance (R) is called as controller in this method.  As external resistance is increased , voltage applied across the armature decreases and speed falls.  i.e. R = Increases, V = Decreases N = Decreases or falls.  We can use this method for to use motor at different speeds.  This method having few merits and demerits.MERITS OF ARMATURE VOLTAGE CONTROL METHOD: 37
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  It is very simple method.  Any speed can obtained.  Below normal speed (to crawing)DEMERITS OF ARMATURE VOLTAGE CONTROL METHOD:  Efficiency of the motor reduces due to power wastage in the controller.  Controller is costly.  Proper heat dissipation is required.APPLICATION:  This method is used, where slow speed require.  Printing machine  Cranes  Trams, Hoist etc..Q.9 Explain speed control of d.c. motors or explain flux control method.ANS:  N  (Eb /  )  Eb = V – IaRa  N  ((V – IaRa )/  )  Armature Resistance is very small.  Therefore, IaRa is also small.  :. N  ( V /  )  Speed of d.c. motor is directly proportional to Applied voltage and inversely proportional to flux. There are two basic methods to vary the speed of d.c. motor. 3. Armature Voltage control 4. Flux control Let us see in details. 38
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ 2. FLUX CONTROL:  By changing flux speed can be vary.  The flux per pole is depends on field current.  To change the field current, variable resistor is used.  Variable resistor is called as field regulator in this method.  In shunt motor, Field resistor is connected in series with field winding of d.c. shunt motor.  When resistance increases, current and flux reduces. Then speed increases.  And vice versa for speed in d.c. shunt motor.  In series motor, Field resistor is connected in parallel with field winding of d.c. shunt motor.  Resistance is called as diverter in d.c. series motor.  When resistance increases, current and flux reduces. Then speed increases.MERITS OF ARMATURE FLUX CONTROL METHOD:  Speed above normal can be obtained with ease.  Method is more convenient.  This method is economical and efficient.  Low power loss due to low resistance value. 39
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________Q. 10 What are the application of D.C. Motors?ANS: APPLICATION OF D.C. MOTORS:  D.C. Shunt Motors: D.C. Shunt Motors having constant speed.  Drilling machines  Grinders  Planers  Metal cutting machines  Wood working machines  Fans  Blowers  Compressors  Conveyors  Vacuum cleaners  Centrifugal and reciprocal pumps etc.  D.C. Series Motors: Those machines require high starting torque.  Traction  Lifts  Cranes  Hoists  D.C. Series Motors are commonly used for conveyors and in rolling mills.  D.C. Compound Motors:  Driving of steel rolling mills  Elevators  Punches  Shears  Conveyors  Heavy planers  Presses etc.  Cumulative compound motors are suitable for high starting torque and for particular under fluctuating load conditions.  Differential Compound motors have low starting torque. They are in actual practice. 40
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ 5. D.C. CIRCUITSQ. 1 What is ohm’s law? Or what fundamental relation does Ohm’s law express? Give thethree forms in which the law may be expressed?ANS: OHM’S LAW: “The current flowing through a solid conductor is directly proportional tothe difference of potential across the conductor and inversely proportional to its resistance,temperature remains constant”.EXPRESSIONS: I  V/ R I = (V/ R)*CONSTANT : . V = IR …………..Ohm’s law equation Where, V= Voltage, Unit: Volt (V) I= Current, Unit: Ampere (A) R= Resistance, Unit: Ohm (  )Q.2 Define the terms resistance, specific resistance, conductance and specific conductance?ANS: (1) Definition of resistance: “It is defined as the property of a substance by virtue ofwhich it opposes the flow of current through it.Unit of resistance is Ohm (  )Pure metals means less resistance and vice versa.(2) Definition of specific resistance: “It is the resistance of a specimen piece of a materialhaving unit length and unit cross-sectional area”.:. It is also called as resistivity.:. R =  . L / AWhere, R = Resistance  = Specific resistance l = Length A = AreaUnit: Ohm-meter(3) Definition of conductance (G): “It is the reciprocal of resistance”Where resistance is measure of opposition offered by the conductor to the flow of current, theconductance is measure which current may flow through a conductor. Unit of conductance is Mho (  )¯¹(4) Definition of specific conductance: “It is the reciprocal of the specific resistance”G = 1 / R =  .A / lWhere, R = Resistance  = Specific resistance 41
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ l = Length A = AreaUnit of specific conductance or conductivity: siemens/meter.Q.3 what are the factors on which the resistance of a material depends?ANS: The resistance of a conductor depends upon the following factors: (i) Type and nature of material: The resistance varies from material to material. Based on capacity to carry free electrons. Example: silver is the best resistance and iron offers high resistance. (ii) Purity and hardness also affects the resistance. Pure metal less resistance and vice versa. (iii) Length of the conductor: The resistance of a conductor is directly proportional to the length. Resistance of the longer wire is greater. (iv) Cross sectional area: The resistance of a conductor is inversely proportional to the area. (v) Temperature: The resistance is directly proportional to the temperature. In most materials, resistance increases with temperature.  NOTE: CURRENT: The rate of transfer of electric charge per unit time.  CHARGE = CURRENT X TIME  VOLTAGE OR P.D.: The difference between the electric potentials or pressure at any two given points in the electric circuit.  e.m.f.(ELECTROMOTIVE FORCE): A p.d. generated by a source of electrical energy across its terminals which tends to produce an electric current in a circuit.  P.d = work done per charge. Q.4 Derive the expression for equivalent resistance of a circuit having several resistance connected (i) in series and (ii) parallel. ANS: (i) SERIES CIRCUITS: DERIVATION:  Consider the three resistances R1, R2 and R3 connected in series as shown in circuit diagram.  They are connected in series.  Current flowing through the circuit is “I” and voltage across each resistance is V1, V2 and V3 across R1, R2 and R3 respectively. 42
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ From circuit, V = V1 + V2 + V3 …………….(1) By Ohm’s law, V = IR (Substitute in equation 1) :. IR = IR1 + IR2 + IR3 IR = I ( R1 + R2 + R3 ) Hence, R = R1 + R2 +R3 In general, R = R1 + R2 + R3 +…..Rn SUMMARY  The same current flows through each resistance in turn.  The supply voltage is sum of the individual voltage drops across the resistances.  The total or equivalent resistance is sum of all the connected resistances. (ii) PARALLEL CIRCUITS:DERIVATION:  Consider the three resistances R1, R2 and R3 connected in series as shown in circuit diagram.  They are connected in series.  Current flowing through the circuit is “I” and voltage across each resistance is V1, V2 and V3 across R1, R2 and R3 respectively. I = I1 + I2 + I3 …….(1) By ohm’s law, V = IR I = V / R (Substitute in equation 1) 43
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ :. V/R = V/R1 + V/ R2 + V/R3 V/R = V1/ R1 + V2/ R2 + V 3/ R3 V/R = V/ R1 + V/ R2 + V / R3 …………………………. (V=V1=V2=V3) V/R = V(1/ R1 + 1 / R2 + 1 / R3) Hence, 1/R = 1 / R1 + 1 / R2 + 1 / R3 In general, 1/ R = 1 / R1 + 1 / R2 + 1 / R3 ……1 / Rn SUMMARY  The same potential difference is applied across all of the resistances.  The total current drawn from the supply is equal to the sum of the currents flowing through the individual resistances.  The total or equivalent resistance is sum of all individual reciprocals of resistances.Q.5 Classify the electrical networks. ANS: In an electrical networks or circuit, there are sources of energy and parameters like resistors, capacitors and inductors etc. The behavior of entire network depends on these elements. Accordingly, they can be classified as, (i) Linear Circuits (ii) Non-linear Circuits (iii) Bilateral Circuits (iv) Unilateral Circuits and (v) Active Networks (vi) Passive networks (i) Linear Circuits: Circuits whose parameters are always constant at any condition like variation in voltage and currents etc. Current is directly proportional to voltage. (ii) Non-linear Circuits: Circuits whose parameters changes with variation in current and voltages. (iii) Bilateral Circuits: Circuit whose characteristic is same as direction of operation like transmission line. It has same relationship between current and voltage for current flowing in either direction. (iv) Unilateral Circuits: Circuits whose characteristics are dependent on the direction of its operation is called unilateral circuits. E.g. vacuum diodes which allow the current at one direction. (v) Active Circuits: Having source of energy in the circuit. It may be either voltage or current. (vi) Passive networks: A networks contains no source of energy. 44
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ 6. THREE PHASE A.C. CIRCUITSQ. 1 Define symmetrical system and phase Sequence.ANS: Symmetrical system:  When e.m.f .s of the same frequency in different phases.  They are equal in magnitude.  They are displaced from one another by equal phase angles.Phase Sequence:  E.m.f. of three phases are reach at its maximum value.  It is essential for finding the direction of rotation of a.c. machines, parallel operation of alternator and transformer etc.Q.2 Write a short notes on “three phase supply system”.ANS:  THREE PHASE SUPPLY SYSTEM:  For single phase system, there are two wires. Phase and neutral.  For three phase alternator, there are three independent armature phases.  Total six wires are required for three phase alternator. See the figure.  The arrangement is more expensive.  In actual practice, three armature windings are interconnected to reduced the numbers of wires. (a) THREE PHASE AND THREE WIRE SYSTEM: 45
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Three phases are connected at a common point N.  Where, N = Neutral  R,Y & B are three phases.  Such an arrangement is called Ster or wye (Y) connection.  Fig (b) is star connected. (b) 3-phase,3-wire, Delta or mesh-connected System:  Each phase is connected to other phase end.  Its closed loop.  Fig (b) is known as delta or mesh connection. (C) 3-phase,4-wire system:  This system is used for low voltage distribution of electrical power.  It provides 3-phase medium voltage supply for heavy industrial loads.  It also provides a low voltage supply for lighting, heating and domestic loads.Q.3 Write a short notes on “balanced load”.ANS: 46
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  The load on a 3-phase system may be star or delta connected.  If all phases impedances of the three phases load are exactly identical in respect of magnitude and their nature. It is called as balanced load.  The lighting and heating loads are resistive in nature.  The lighting and heating loads are connected between line and neutral.  The electricity is distributed on three phase of equal loads.  In such condition, all single phase loads forms a balanced star-connected load.Q.3 Write a short notes on “balanced system”.ANS: A 3-phase system is balanced when it possesses following characteristics: 1) The three phase e.m.f.s are equal and displaced by 120 electrical degrees. 2) Impedance in any one phase should be identical as compare with other two phases. 3) The resulting currents in different phases with same phase angles. 4) Power and reactive power flow should be equal in each phase.Q.4 Derive the voltage, current and power relations in star connections in a star connection.ANS: EXPLAINATION:  Star connection is shown in figure (a). 47
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Em.f. generated in each phase on the armature of the alternator is known as phase e.m.f. and corresponding current as phase current.  PHASE e.m.f.s = Vph = Vr = Vy = Vb = coming out from two phases.  N = Neutral point.  PHASE CURRENT: Iph = Ir = Iy = Ib  LINE e.m.f.s = VL = VRY = VYB = VBR  LINE CURRENTS IL = IR = IY = IB  Balanced 3-phase inductive load is assumed.  :. Phasor diagram is shown in figure (b).  VRY = Vr – Vy  VYB = Vy – Vb  VBR = Vb – Vr  VOLTAGE: In star connection, From phasor diagram, VL = VRY = 2Vph*cos30 3 = 2*Vph* 2 :. VL = 3 Vph  CURRENT: Line current = Phase Current  IL = Iph  POWER: P = V*I P = 3*Power supplied by each phase P = 3* Vph*Iph* Cos VL P = 3* *IL* Cos 3 P = 3 * VL*IL* Cos Q.5 Derive the voltage, current and power relations in delta connections in a deltaconnection.ANS: 48
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________EXPLANATION:  Delta connection is shown in figure (a).  Em.f. generated in each phase on the armature of the alternator is known as phase e.m.f. and corresponding current as phase current.  PHASE e.m.f.s = Vph = Vr = Vy = Vb = coming out from two phases.  N = Neutral point.  PHASE CURRENT: Iph = Ir = Iy = Ib  LINE e.m.f.s = VL = VRY = VYB = VBR  LINE CURRENTS IL = IR = IY = IB  Balanced 3-phase inductive load is assumed.  :. Phasor diagram is shown in figure (b).  VOLTAGE: Line voltage = Phase voltage  CURRENT: In delta connection, From phasor diagram, IL = IR = 2Iph*cos30 3 = 2*Iph* 2 :. IL = 3 Iph  POWER: P = V*I P = 3*Power supplied by each phase P = 3* Vph*Iph* Cos VL P = 3* *IL* Cos 3 P = 3 * VL*IL* CosQ.6 Solve all problems based on star connection and delta connection. 49
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ 7. A.C. FUNDAMENTALQ.1 What is an alternating current? Compare it with a direct current.ANS: Alternating current is one which periodically passes through a definite cycle of changes inrespect of magnitude as well as direction.OrAlternating current is current which flows periodically through a definite cycle of changes inrespect of magnitude as well as direction.Direct current: d.c. is that current which flows continuously in one direction and has constantmagnitude with respect to time.Comparison between a.c. and d.c. or advantages of a.c. 1. The ac voltage can be raised or lowered very easily by using transformer. In dc, not so easy and economical. 2. Construction and operating costs per kilowatt are low. 3. We can build up high voltage, high speed a.c. generators with large capacities. It is not possible in dc. 4. Due to adoption of high voltages, a.c. transmission is always efficient and economical. Because, higher the voltages lesser the current through transmission. 5. ac motors are simple in construction, cheaper and more efficient than dc. 6. ac can be easily converted to dc for the application like traction, search lights, projections etc.Q.2 What is the principle of Generator? Also, Explain generation of alternating current andvoltage.ANS:Generator: Which converts mechanical energy into electrical energy either ac or dc.Principle of Generator:“When a conductor is moved in magnetic field or magnetic field moved with respect to theconductor. According to Faraday’s law of electromagnetic induction, electromotive force is setup in the conductor” 50
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  There is relative motion between a conductor and a magnetic field.  Voltage will always be generated in the conductor.  A.C. Generators are normally known as alternators.CONSTRUCTION:  It consists of a single turn rectangular conducting coil (AB) like copper or aluminum.  Coil / conductor is placed in between two poles (S & N POLE).  Coil is placed in such a way that to rotate at constant speed in uniform magnetic field.  The coil is also known as armature of the alternator.  The ends of the armature coil are connected to rings called slip rings. i.e. R1 & R2  Two carbon brushes pressed against the slip rings.  The slip rings collect the current and carry it to the external resistor (R).OPERATION:  Assume that the armature coil is rotating in an anticlockwise direction.  According to Faraday’s law of electromagnetic induction e.m.f. is produced in them. 51
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  The magnitude of this e.m.f. is depend on position of coil.  Let us consider few positions of the coil: POSITION 1:  At ϴ = 0  This is the initial position of the coil  The plane of the coil is perpendicular to the magnetic field .  The conductor of the coil move parallel to the magnetic field.  There is no flux cutting. So, no emf is generated in the conductor.  No current flows.  POSITION 2:  At ϴ= 0 to ϴ= 90  Coil is rotates from 0 to 90  Both the conductor of the coil move at right angles to the magnetic field.  There is maximum flux cutting. So, there is maximam emf generated in the conductor.  Emf is generated from 0 to maximum.  The emf’s in both conductor being in series. Therefore , total emf is sum of two emfs.  The voltages are equal and twice than one of the conductor.  The current is varies as the emf varies.  More and more lines of force are ct by the conductors.  According to fleming’s right hand rule, the direction of induced current is towards the observer (conductor A) and away from observer (conductor B).  Therefore,, current flowing through resistor from the terminal C to the terminal D. POSITION 3:  At ϴ= 90 to ϴ= 180  Coil is rotates from 90 to 180.  Conductor again moves parallel to the field.  The lines of force cut by the conductor gradually reduces.  Hence, emf is decreases.  Therefore, current through external resistor is zero.  In this, during first half cycle the conductor moves in the same direction through the magnetic field.  Therefore, polarity of generated e.m.f. is same and current flows through the external resistor from C to D.  POSITION 4:  At ϴ= 180 to ϴ=270  As the coil rotates beyond position 3.  The direction of the cutting lines of force through the magnetic field reverses. 52
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Conductor A cuts up and conductor B cuts down through the field.  Both the polarity of generated e.m.f. and current is reverse.  By fleming’s left hand rule, direction of current in conductor A is away and for B towards.  Current flows through D to C.  Similar to position 2, emf in the coil is maximum in position 4.  In general, the variations in the magnitude of emf of the alternator is exactly similar to those in first half revoltion.  Current supplied by alternator is necessarily alternating.  Unidirectional or direct current is desired.  Before external circuit, alternating current must be rectified or commutated.  Rectification can be done by replacing the slip-rings R1 and R2 of the alternator.  Where, M is shown in figure “COMMUTATOR”.  Commutator is made by conducting material which is cuts in two equal halves or segments.  Segments are insulated by thin sheet of mica or some other insulating material.  The ends of the armature coil AB are joined to these segments.The external load resistor R is connected to the split ring M through brushes B1 and B2. ACTION OF A COMMUTATOR:  When coil is undergoing for first half cycle, current flowing from load(R) (C to D).  For next half cycle, its reverses with position of a and b terminals.  Again for next cycle, current is flowing from C to D and so on.  Direction of current through the external circuit is same.  In the commercial d.c. generator, by using number of coils evenly distributed around the armature core and connecting these coils to commutator having a corresponding number of segments, a more steady direct current is obtained in the external circuit.GRAPHICAL REPRESENTATION OF E.M.F. OR CURRENT: 53
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________Q.3 Derive the equation of the alternating voltages and currents?ANS:Generator: Which converts mechanical energy into electrical energy either ac or dc.Principle of Generator:“When a conductor is moved in magnetic field or magnetic field moved with respect to theconductor. According to Faraday’s law of electromagnetic induction, electromotive force is setup in the conductor”  There is relative motion between a conductor and a magnetic field.  Voltage will always be generated in the conductor.  A.C. Generators are normally known as alternators.CONSTRUCTION:  It consists of a single turn rectangular conducting coil (AB) like copper or aluminum.  Coil / conductor is placed in between two poles (S & N POLE).  Coil is placed in such a way that to rotate at constant speed in uniform magnetic field.  The coil is also known as armature of the alternator.  The ends of the armature coil are connected to rings called slip rings. i.e. R1 & R2  Two carbon brushes pressed against the slip rings.  The slip rings collect the current and carry it to the external resistor (R).DERIVATION:Let,B= Flux density of the magnetic field (teslas)l = Length of the conductor (metres)r = Radius of circular path (metres)ῳ = Angular velocity of the coil (radians / second)v = Linear velocity of the coil sides ( metres / second)when the coil lies in YOY plane and zero e.m.f.then, angle ϴ = ῳt ……………………………………………………………..(1) 54
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________After some time coil rotates at angle ϴ. It resolved two components (1) horizontal (v.cosϴ) and(2) vertical (v.sinϴ).Therefore,E.M.F. generated in each coil side: B.l.v.sinϴ volts ……………….(2)Total e.m.f. generated in the coil = e:. e = 2 B.l.v.sinϴ volts ……………………………………….(3)When coil will be horizontal plane then ϴ = 90 and conductor cuts maximum flux (Em):. Em = 2 B.l.v.sinϴ = 2 B.l.v.sin90 = 2 B.l.v.1 ( sin90 = 1) = 2 B.l.v voltsTherefore, the standard equation of the voltage or e.m.f,e = Em.sinϴ e= Em.sinῳt e = Em.sin(2π.f.t) e = Em.sin(2π.t / T)Similarly, we can write for current,i = Im.sinϴi = Im.sinῳti = Im.sin(2π.f.t)i = Im.sin(2π.t / T)Q.4 Define peak factor, form factor, waveform, cycle, periodic time, frequency andamplitude.ANS:(i) PEAK FACTOR: “ It is the ratio of maximum value to the R.M.S. value of an alternating quantity”. It is also called as crest or amplitude factor.Peak factor = maximum value R.M.S. valueWhere, R.M.S. value = 0.707 X Maximum value Maximum value = √2 X R.M.S. valueSubstitute in peak factor,Peak factor = maximum value 0.707 X Maximum valuePeak factor = 1.4144(ii) Form factor: “ It is the ratio of R.M.S. value to the average value of an alternating quantity”.Form factor = R.M.S. value Average valueWhere, 55
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________Average value = 0.637 X maximum valueR.M.S. value = 0.707 X Maximum valueSubstitute in form factor,Form factor = 0.707 X Maximum value 0.637 X maximum value = 1.11(iii) waveform: “ It is the pictorial representation of an alternating quantity in a graphical form”.(iv) cycle:“Each repetition of a complete set of changes undergone by the alternating quantity is calledcycle”.(v) Periodic time:“ It is the time required by an alternating quantity to complete its one cycle”.It is denoted as “T” in seconds.(vi) Frequency:“ The number of cycle completed per second by an alternating quantity is known as frequency”.It is denoted as “f” in hurtz.f=1/T(vii) Amplitude: The maximum value attained by an alternating quantity during it’s positive or negative half cycle”. 56
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ 8. INSTALLATION, EARTHING AND TROUBLE SHOOTINGQ.1) Briefly describe the different types of wires used in the electrical wiring work. How thesewires normally specified?ANS: There are nearly (commonly) seven types of wires.They are as follows: 1) V.I.R. 2) C.T.S. / T.R.S. 3) Lead Sheathed Wires 4) P.V.C. Wires 5) Weather Proof Wires 6) Flexible Wires 7) CablesExplanation: 1) V.I.R. (Vulcanised India Rubber wires):  It consists of a tinned conductor coated with rubber insulation.  The thickness of rubber varies i.e. 250 or 660 V.  Insulation is not moisture & heat proof.  Conductor is covered with heat & moisture resistant bitumen compound.  It is finished with wax for cleanliness & smoothness.  These wires are single core.  These wires are rarely used.  Cheaper cost. 2) C.T.S. (Cab Tyre Sheathed ) or T.R.S.(Tough Rubber Sheathed):  It consists of rubber insulation conductor but additional tough rubber is used.  Tough rubber is provides protection against moisture ,chemical fumes, wear & tear.  These wires are single core, twin core & three core. 3) Lead Sheathed wires:  It consists of rubber or paper insulation conductor but additional tough rubber is used with an outer sheath of lead or lead alloy.  It gives protection against moisture, atmospheric corrosion, wear & tear.  These wires are single & multicore.  Quite heavy & costly. 4) P.V.C. (Poly Vinyl Chloride wires):  Conductors are insulated with P.V.C. 57
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  P.V.C is non-hygroscopic, tough,durable,resistant to corrosion & chemical inert.  These wires are sufficient to give mechanical protection.  P.V.C. being a thermoplastic, softens at high temperature.  It should not used for giving connection to heating appliance. 5) Weather Proof Wires:  It is used for outer service lines.  Wires are costly.  Its conductor consist of three braids of fibrous yarn with water proof compound are used.  They are cheap.  They are resistant to varying atmospheric conditions. 6) Flexible wires:  It consists of two separately insulated stranded conductors.  Stranded conductors are used for flexibility.  Rubber or P.V.C. are used or insulation.  They are available in the two parallel twin and twisted twin forms.  Easy to handles.  Good flexibility.  Used for holders, household portable appliances, like heater, irons, table lams, etc. 7) Cables: “It is defined as a length of a single insulated or two or more insulated conductors, each having its own insulation and all put together”.  It is used for distribution purpose.  They are laid along the wall poles (Aerial cables)  These cables are insulated with V.I.R. or P.V.C. or varnish cambric.  They are mechanically sound and weather proof. 58
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  V.I.R and P.V.C are used for outdoor works.  They are designed for underground laying (Underground cables).  They are costly.CONDUCTOR or CORE:  It has one central core or a number of cores i.e. 2,3,3⅟2 or 4.  It is made up of tinned stranded copper or aluminum conductor.  Stranding gives flexibility to the cable.INSULATION:  Insulating materials are impregnated paper or vulcanized bitumen or varnished cambric.  Impregnated paper is used for high-voltage cables.METALLIC SHEATH:  Lead alloy or aluminum sheath is provided over the insulation.  Insulation provides mechanical protection & preventing entry of moisture in the insulation.BEDDING:  Bedding consists of a layer of fibrous material.  It protects from corrosion & mechanical injury.ARMOURING:  It consist of one or two layers of galvanized steel wire or steel tape.  During laying, it provides protection from mechanical injury.SERVING:  A layer of fibrous material like jute cloth.  It protects from atmospheric condition.Q.2) What is fuse? Explain the principle of operation of a fuse and its type.ANS: FUSE:  Fuse is a current interrupting device or type of sacrificial over current protection device.  Fuse is an essential component which is a metal wire or strip that melts when too much current flows through the circuit.  It starts melts when fault is occurred, excess current flows, short circuit, overload, device failure etc. 59
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Short circuit means, where there is a direct connection between live wire to neutral wire.GENERAL CONSTRUCTION:  Small piece of wire connected between two terminals on the insulated base.FUNCTION OF THE FUSE:  When the fuse is inserted in a circuit to be protected.  Fuse carries normal current safely without heating.  Whenever excess current flows then it melts due to rapid overheating.  It disconnects the load during faults or over current or short circuit.  The fuse is in effect a safety value for the electrical circuit.FUSE ELEMENT:  It is metal like tin, lead, copper, aluminum, zinc, silver and antimony.  The most preferred lead-tin alloy is found more suitable for this purpose.  Tin-alloy = lead (37%) + alloy(63%)  Normally, tin-alloy wire is not used beyond 10 amps because of higher current.TYPES OF FUSES:There are two types. 1) Semi-enclosed or Rewirable type fuse. 2) H.R.C. Cartridge Fuse. EXPLAINATION: 1) Semi-enclosed or Rewirable type fuse: CONSTRUCTION:  Fuse element is semi-enclosed.  It is also known as kit-kat type fuse.  Where, E = fuse element 60
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ S = Screws C = contacts B’ = Porcelain fuse bridge B’’ = base.  E = Short length of fuse wire depending upon the ratings.  B’ = wire is threaded in the hole.  B’’ = Incoming and outgoing is connected to this bridge.  C = contactsOPERATION:  Fuse wire is in series with the circuit to be protected.  During fault, fuse element resist and heat is developed.  Current rises and fuse is starts melting.  After some time its breaks and circuits to be protects.APPLICATION:  Commonly used in domestic installation and the other circuits.  It is used to be protects low values of faults. 2) HIGH RUPTURING CAPACITY CARTRIDGE FUSES: Where, E = Fuse element C = cartridge M = Metal end caps P = Pure quartz B = Contact blades  E = Silver or copper made.  Fuse element is totally enclosed and hermetically sealed inside the container called Container.  C = Cearamic material or epoxy-resins made.  End of the enclosed fuse elements are connected to the metal end caps (M) normally of brass. 61
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  The fuse body is filled with powdered pure quartz.  Some form of the indication is provided on it.OPERATION:  When the fuse is inserted in the circuit to be protected.  It carries the normal working current safely without heating.  When a fault occurs, fuse element starts melting due to excess of current flow through it.  The metal released in the vapour form diffuses with the quartz powder.  The chemical reaction between the two substance of high electrical resistance like an insulator.  An indicator consists of resistance wire of fine gauge connected in parallel with the fuse element.APPLICATION:  Due to increasing loads and size of the network, H.R.C. replacing with rewirable fuse.  They are used in industrial installayion and low-voltage distribution.Q.3) Compare the different methods of wiring with respect to their application, cost and life?ANS: There are total five different types of wirings:PARTICULAR CLEAT WOODEN T.R.S./ C.T.S. LEAD CONDUIT WIRING CASING & WIRING SHEATHED WIRING CAPPING WIRINGDurability of Short life Fairly long life Long life Long life Very longlifeMechanical Poor Good Fair Good Very goodProtection protection protection protection protection protectionAppearance Shabby Neat Neat Neat GoodType of Semi-skilled Highly skilled Skilled Skilled Highly skilledlabourrequiredCost Low Medium Medium Costly than High capping or C.T.S. wiringPossibility of Nil Poor Good Very Good Poor or fairhazardsApplication Temporary General Most widely Damp Concealed installation Indoor work used in situation type wiring indoor wiring 62
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________Q.4) Explain conduit wiring? Or What are the various types of conduit used in conduit wiring?ANS: CONDUIT WIRING:V.I.R or P.V.C. wires are run through black enameled or galvansed metallic tubing calledconduit.There are two types of conduits: 1) Thin wall conduits 2) Rigid conduits 1) Thin wall conduits:  These are light gauge iron conduit with seam along its length.  The seam may be open type.  There is no mechanical adhesion between its two edges.  These edges are brazed together for damp proof. 2) Rigid conduit:  These are heavy gauge iron conduit.  Conduits are solid drawn or with welded seam.  Join fittings are always threaded.  Conduits are more costly.  Conduits are used for medium pressure circuits.  Also used in places where good mechanical protection & absolute protection from moisture is required.There are two types of Conduit wiring. They are as follows,  Surface conduit wiring  Concealed conduit wiring a) Surface conduit wiring:  Conduits are fixed on the walls.  By using saddles screwed to rawl plugs or wood plugs.  In damp situation, they are separated by small wooden blocks.  Conduit piping is kept at earth potential by proper earthing. 63
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ b) Concealed conduit wiring:  It employs heavy gauge rigid conduits buried under wall plaster.  This wiring is used where beauty is important.  Conduit piping is kept at earth potential by proper earthing.WIRING OF THE CONDUITS:  Drawing of wiring is most commonly adopted.  After planning the layout carefully, conduits are erected.  V.I.R or P.V.C. wires are used with steel draw wire.  Wires may be easily pulled.  They are rubbed with French chalk.APPLICATION:  It is used at indoor and outdoor.  It is permanent in nature for light and power like godowns, workshops and public buildings.ADVANTAGES:  Very long life.  It provides good protection, against mechanical injury, moisture and fire hazards.  Neat appearance.  It gives beauty.  Less maintenance is required.DISADVANTAGES:  Most costly.  Required more time and highly skilled labour.  Repairing required more time and difficult.  There are possibilities of an electrical shock due to absence of earth continuity.  If proper precaution are not taken then insulation will be damage.Q.5 What are the advantages and disadvantages of Semi-enclosed fuse?ANS: ADVANTAGES:  They are very cheap.  Maintenance is easy.  We can rewire after melting and clearance the fault.DISADVANTAGES: 64
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Not highly sensible.  Can’t be used for higher values of faults.  Since the wire is exposed to air, it is subjected to deterioration due to oxidation caused by heating. This decreases the diameter of wire.  Slow speed for the current interruption.  Risk of fire hazards due to external flash on blowing.Q.6 What are the advantages and disadvantages of H.R.C. Cartridge fuse?ANS: ADVANTAGES:  High speed operation.  Ability to clear high values of fault current.  Its operation is silent and without flame, gas or smoke.  It is safe from fire hazards.  Performance and protection is reliable.  Being totally closed, there is no deterioration of the fuse element. DISADVANTAGES:  Costly  We have to replace after its operation.  Overheating is possible at the contacts.Q.7 Explain, why H.R.C. Cartridge fuses are preferred over the rewirable type, particularlyhigh values of current.ANS: There are many advantages of H.R.C. Cartridge fuses over rewirable fuses.Advantages of H.R.C. Cartridge fuse over rewirable fuse:  High speed operation.  Ability to clear high values of fault current.  Its operation is silent and without flame, gas or smoke.  It is safe from fire hazards.  Performance and protection is reliable.  Being totally closed, there is no deterioration of the fuse element.Q.8 Why fuses should always be provided in the live (phase) wire?ANS: Because, electrons flows in the live or phase wire. Neutral is used for return path. If wehave to protects the circuit from excess current and install the fuse at the live part only. If weinstall it at neutral path then circuit will not be protected and can be damage.Hence, fuses should always be provided in the live (phase) wire.Q.9 Define current rating of fusing elements?ANS: It is the value of continuous current which the using element can carry safely undueheating, melting and deterioration.Q.10 Define minimum fusing current? 65
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ANS: it is defined as the minimum value of current which causes melting of the fuse.Q.11 Define fusing factor?ANS: It is the ratio of Minimum fusing current to the Current rating of the fusing element.  Its value is always more than one.Q.9 Explain MCCB.ANS: EXPLAINATION:  MCCB is a Moulded Case Circuit Breaker.  It is compact type air circuit breaker.  It is enclosed in a moulded case.  Moulded means insulating body.  It uses air for arc quenching medium.  The resistance of the arc is increased by cooling, lengthening and splitting the arc with the help of arc chute.  It protects against the overload and short circuits.  Overload protection is given by bimetal strips.  Short-circuit protection is given by magnetic attraction.ADVANTAGES OF MCCB:  Practically, no maintenance is required.  It avoids single phasing.  No recurring cost.  Gives indication in case it trips on fault.RATINGS / SPECIFICATION OF MCCB:  It is available at 16A to 1600.  Its breaking capacity as high as 85Ka.  Breaking capacity ratings are same for 3P as well as 4P. (P = POLES)APPLICATIONS OF MCCB:  It is suitable for both a.c. as well as d.c. application.  Distribution feeders.  Distribution transformer.  Diesel generating sets.  L.T. capacitors.  Rectifier panels.  U.P.S.  Motors, furnaces and other electronics equipments. 66
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________Q.10 Explain MCB.ANS: EXPLAINATION:  MCB is a Miniature Circuit Breaker.  Tripping mechanism and contacts are assembled in a moulded case.  Moulded out of thermosetting powders.  Thermosetting powders gives high dielectric strength, mechanical strength and virtually no aging.  Depending upon the rating, conductor is made up of copper or aluminum.  The arc chute has a special construction.  The length of arc is increased by magnetic field.  For single phase, 1P or 2P versions of MCB are used.  For three phase, 3P or 4P versions of MCB are used.  MCB is used for overload and short-circuit protection.  Overload protection gives by bimetal strips.  Short circuit protection is given by magnetic attraction.ADVANTAGES OF MCB:  It can be used by skilled or unskilled workmen.  It can be used as switch or an isolator.  It is a cost effective device.  It’s a fully enclosed unit and hence no ageing problem.RATINGS OF MCB:  It is a available from 0.5 A to 100 A.  Voltage ratings of 240V/415V.  Breaking capacity upto 3kA.APPLICATION OF MCB:  Commonly used for lighting circuits.  Distribution feeders.  Switching of motors.  Capacitors.  Control transformers etc.Q.10 Explain ELCB.ANS: EXPLAINATION:  ELCB is an Earth Leakage Circuit Breaker.  When fault current flows through the earth return path, the fault is called Earth fault. 67
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  ELCB gives protection against direct or indirect contact with live circuits under such a earth faults.  Figure shows the diagrammatically representation of 2-pole earth leakage circuit breaker suitable for single phase load. Figure: E.L.C.B.PRINCIPLE OF OPERATION:  During normal condition, current is flowing through phase to neutral wires.  Current flow is equal and opposite in phase and neutral wire.  Hence, transformer fluxes are equal and opposite.  So, they are neutralizing each other.  But, during the earth fault condition, small leakage current is returning back t the earth through earthing conductor or human body.  This disturbs the balance between earth and neutral wire.  So, certain resultant flux is set up in the current transformer core which induces e.m.f. in the secondary winding of the C.T.  Output signal gives tripping command to the circuit breaker.  Circuit to be protected.RATINGS OF E.L.C.B.:  30mA : Domestic application.  100mA : Industry.  300mA :Industry where high leakage currents.APPLICATION: 68
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  A combination of ELCB and MCB can be used to give overload, short circuit and earth leakage protection.  Domestic, industrial and commericial.Q. 11 Why earthing is necessary in a wiring installation? How it is achieved?8ANS: EXPLAINATION:  Due to safety reason for living things.  For better performance of the circuit or devices. Let us see the utility of such earthings, 1) Fig: Non-earthed appliance connected to supply. 2) Fig: Fault with non-earthed appliance. 3) Fig: Faulty non-earthed appliance giving an electric shock. 4) Fig: Fault with earthed appliance.EXPLAINATION:  See the two figures non-earthed appliance connected to supply.  If fault develops causing contact between the conductor and the casing. 69
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  The casing becomes live with respect to earth at equal potential.  See the Fig: Fault with non-earthed appliance.  When fault is occurred then also appliances are under working condition.  But, if any human or living things touch the appliances they will get a shock.  Very dangerous to living things.  During fault, human being or living things get shocks.  See the Fig: Fault with earthed appliance.  If the fault is occurred then there is no harmful effect due to earthing.  Fault current or leakage current is flowing through the earthing conductor at the earth.  Circuit and living things to be protected.  Appliances under working without any damages.Q.12 Discuss in brief, the commonly used methods of earthing.ANS: EXPLAINATION: 70
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  For earthing, Any metal plate, pipe, rod and other conductors are used.  Mostly, copper conductors are used.  Connection between appliances and mass of earth is made by earth electrode.  For small installation, one earth electrode is sufficient.  For large installation, more earth electrodes are required. There are two commonly used methods. 1) Plate earthing 2) Pipe earthing 1) PLATE EARTHING: 71
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  The earth wire is securely bolted to the earth plate.  Earth plate of copper (minimum size: 60cm X60cm X3.18mm) or of galvanized iron (minimum size: 60cm X 60cm X 6.35mm).  Depth of 3m.  The plate is kept in vertical position and is embedded in an alternate layer of coke and salt.  Thickness of this layer is about 15cm.  This layer is reduce the earth resistance.  A galvanized iron pipe fitted with funnel at the top.  It is provided pure salty water in the pit of earth plate.  From time to time in the summer season. 2) PIPE EARTHING:  Galvanized iron pipe is used.  It is not less than 38.1mm diameter and 2m in length for ordinary soil and 2.75m for dry.  The depth at which the pipe should be buried in ground depends upon the soil condition.  The earth wire is fastened at top of the pipe with nuts and bolts.  Top of the funnel is filled with pour salty water in the pit electrode.  The earth wire connection with galvanized iron pipe being above ground level.  The advantage of this earthing is, we can check its continuity by continuity test. 72
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________Q.13 what are the special requirements of wiring installation for computer?ANS:EXPLAINATION:  Computers are in use in almost every field.  The computer needs uninterrupted power supply.  To control the temperature and dust free atmosphere.  Therefore, wiring installation for computer play very vital role.In the installation of computer, we require following equipments. They are as follows: 1) Power supply 5)MCCB 2) Distribution box 6) UPS 3) Switch 7) ELCB 4) Air conditioner 8) EARTHINGLet us see one by one: 73
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ 1) Power supply:  Power supply is very essential for computers.  Single phase supply is required. 2) Distribution box:  It requires distributing the supply to the computer.  There is one phase and neutral connection. 3) Switch:  To on or off the power supply. 4) Air conditioner:  Dusty atmosphere and high temperature effects the performance of the computers.  Therefore, cooling system is essential like air conditioner. 5) MCCB:  It is a Moulded Case Circuit Breaker.  It is use to protects the circuit from short circuit or excess of current flows or overload. 6) UPS:  UPS means Uninterrupted Power Supply.  It gives continuous power supply to the computer.  It gives protection against abnormal supply conditions like voltage spikes, surges, sags, loss of data in the case of power supply, etc.  It is running up to several times depending upon the types of UPS.  It saves the data in the case of power supply.  It shut down the computer properly. There are two types of UPS: 1) Standby Power Systems (SPSs):  Power is directly obtained by power line until the power fails.  After power failure, battery power inverter turns on to continue supplying power.  It is more common for small business or home use. 2) On-line Uninterrupted Power supplies:  Convertor, battery and inverter are used. 74
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Convertor converts a.c. to d.c. and charges the battery.  Battery provides a.c. power to the computer with the help of invertor.  On-line UPS provides protection against loss of data in the case of power failure.  On-line UPS is costly.Q.14 What are the essential tests to be made on a wiring installation before it is put intoservice?ANS:There are five tests. They are as follows: 1) Insulation resistance to earth. 2) Insulation resistance between conductors. 3) Polarity test for Single-pole Switches. 4) Continuity test for earth wire. 5) Test for earth Resistance Measurement.EXPLAINATION: 1) Insulation resistance to earth:-  We can test the insulation of the conductor by using Megger.  We can measure the resistance between conductor and earth by 500V megger.  Rotating part is their which gives readings of the megger.  Reading is noted down which shows the insulation level.  Insulation level is more than 1 megaohm for the entire installation. 75
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Minimum insulation resistance required may be calculated from limiting value of leakage current. 2) Insulation resistance between conductors:  Insulation between two conductors can be measured by Megger.  Two conductors like Phase and Neutral.  Megger is connected between phase and neutral.  The resistance should not be less than 1 megaohm.  All fuse links in position.  All Switches should be ON.  All lamps and application out.  Main switch should be ON.  L= LIVE WIRE  PH = PHASE WIRE  N = NEUTRAL WIRE 3) Polarity test for Single-pole Switches:  To test the all phase wires are connected to single pole switches.  Phase wires are not connected to neutral wires.  This test is performed with the help of test lamp.  Switch is in the position and lamp removed from holder.  Test lamp is connected between anyone terminal of the switch and the earth point. 76
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Lamp will glow, when switch is connected to phase wire correctly.  This test is performed for all the switches.  This test is also performed by testers with neon bulb. 4) Continuity test for earth wire:  To test the connectivity of the earthing.  Where there is low resistance and path is continuous.  Continuity is checked by ohmmeter and to measure the resistance between the earth electrode & remote end of the earth continuity conductor.  Resistance can be measured by voltmeter-ammeter method.  The resistance of the earth continuity conductor between any point in the system and the earth connection must not exceed 1ohm. 5) Test for earth Resistance Measurement:  By using “megger earth tester”, the earth resistance of the consumer’s ground connection is measured.  The earth resistance of the installation should be less than 5 ohm.Q.15 What is trouble shooting?ANS:  TROUBL SHOOTING is the process of detecting the circuit faults.  Faults can be detected with the help of megger, multimeter, simple tester with neon bulb and test lamp. 77
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Some special equipment may be necessary in some cases.  If fault is detected then it can be solved by appropriate work.  Fault can be solved by expert workers only.Q.16 What are the problems encountered in wiring installation or wiring installation forcomputers?ANS:GENERAL WIRING PROBLEMS:  Open grounds  Phase-neutral reversals  Discontinuity in wires / supply cords and components.  Blown fuses  Improper plugging etc.  By using multimeter, we can calculates the value of Voltage, Current and resistance.  In the case of computer or any installation, we have to check the potential difference between the neutral wire and earth.  The permissible limit is 3volts.  The value of potential may not be large.  If value is large then earthing is not proper.  Earthing is very important for such a equipments to protect them.There are many problems with installation of computers.Like problems with the supply systems for air conditioners and UPS.Let us see one by o ne. 1) Problems with the supply systems for air conditioners:  Starting current is very large as compare with normal working condition  Hence, ratings of the accessories should be carefully selected like fuses, switches, MCCB, ELCB Etc.  At same time, many air conditioners should not not be started simultaneously to avoid initial inrush of the current. 2) Problems with UPS:  Uninterrupted Power Supply can be failed due to failure of battery.  Battery is not charging properly.  If there is no plug in, then battery is not charging properly. So, mains kept on.  Overloading of the UPS.  This can be avoided by unplugging unnecessary equipments like printers.  Battery is weak due to regular wear and tear during the service.  Battery has to replace.  Regular maintenance of the batteries is very essential for long life. 78
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ 9. FRACTIONAL HORSE POWER MOTORQ.1 Explain the principal of working of a single-phase induction motor with the help of doublerevolving field theory.ANS:EXPLANATION:  Construction of 1-phase motor is similar to the polyphase motor.  Only difference is stator has distributed 1-phase winding.OPERATING PRINCIPLE:When a.c. voltage is applied to the motor, current is flowing through 1-phase stator winding.Stator winding produces a resultant field which alternates along the axis of the winding.  Such a field varying sinusoidally with time along with fixed space axis.  Rotor is at rest for understanding.  When supply is connected, current is flowing through the winding. 79
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________  Winding produces alternating flux along the axis of the winding.  By transformer action, field induces an e.m.f. in the rotor conductor. Q.2 What are the special features of servomotors? State their types.ANS: REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011Q.3 Write a short note on characteristics and application of stepper motors.ANS: REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011Q.4 Compare different types of single-phase induction motors.ANS: REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011Q.5 Write a short notes on D.C. Servomotors.ANS: REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011Out of notes:All theorems which is taught in the class room.LikeThevenian’s , Norton, Maximum power, etc………………..REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011 80
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ THEROEMS  SUPERPOSITION THEROEM  Steps depends on number of Voltage as well as current source.  Step 1: Only one source acting  other Voltage = short circuit and current = open circuit  Step 2: 2nd source is acting, others short or open.  Step last: Addition of all steps for final answer.  THEVENIN’S THEOREM  There are three steps only.  Step 1: Find Vth across load.  Step 2: Find Rth across load.  I Step 3: Find L across load.  NORTON’S THEOREM  There are three steps only.  I Step 1: Find SC across load. Short circuit the load to get short circuit current.  Step 2: Find Rth OR RN across load.  Step 3: Find IL across load.  MAXIMUM POWER TRANSFER THEOREM  There are four steps only.  Step 1: Find Vth across load.  Step 2: Find Rth = RL across load.  I Step 3: Find L across load.  Step 4: P = I²L RL 81
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ IMPORTANT THEORY QUESTIONSQ.1 State the duality between d.c. series and parallel circuit.Q.2 State the kirchhoff’s law.Q.3 Define active and passive elements.Q.4 State the condition of transferring maximum power from the source to load in d.c. circuits.Q.5 Define form factor. What is its value for sinusoidal waveform?Q.6 For star connected load state the numerical relationship between V & I for phase and line.Q.7 For delta connected load state the numerical relationship between V & I for phase and line.Q.8 Define ideal and practical voltage source.Q.9 With the help of waveforms and phasor diagrams, show the phase relationship betweenvoltage and current in pure inductive and pure capacitive circuits.Q.10 Define: a) Impedance b) current c) Power factor d) Power e) Resistance f) Form factor g)Peak factor h)True power i) Apparent power j) Reactive power k) Ideal transformer l)Isolationtransformer m)Auto transformer n)Regulation of a transformer o)Symmetrical system p)Phasesequence q) Current rating r)Minimum fusing current s)Fusing factor t)Ohm’s lawQ.11 What is the impedance of a.c. circuit? What is its unit? State factors on which it depends.Q.12 State the various losses in transformer. How these losses are minimized?Q.13 Draw schematic diagram of single-phase a.c. servometer and state its principle of working.Q.14 What are the factors on which the resistance of a material depends?Q.15 Derivation for series and parallel.Q.16 What is an a.c.? Compare it with a d.c. or Advantage of a.c. over d.c.Q.17Define rms value of an alternating quantity. Prove that in an alternating quantity varyingsinusoidally, the maximum value is 1.414 times the rms value.Q.18 Explain choking coil. State any two properties.Q.19 Explain the difference between a single-phase and a polyphase system.Q.20 Explain with the reference to a three phase system, the terms “balanced load” and“balanced supply system”.Q.21 What is the function of a transformer in ac circuit?Q.22 Derive the emf equation of a transformer.Q.23 Explain the meaning of kVA rating of a transformer. OR Why transformer rating is in kVA?Q.24 What are the types of transformer? Explain in detail with diagrams.Q.25 Explain the principle of operation of a dc motor.Q.26 Explain types of dc motors.Q.27 Explain characteristics of dc motor (a) Shunt (b) Series (c) Compound dc motor.Q.28 Derive equation of Back e.m.f.Q.29 Derive torque equation of motor.Q.30 Explain a.c. servometer and d.c. servometer.Q.31 Different characteristics of stepper motors & its application.Q.32 Explain the principle of working of a single-phase induction motor with the help of doublerevolving field theory.Q.33 Explain different types of wires.Q.34 Explain MCCB, MCB AND ELCB. 82
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________Q.35 Why earthing is necessary in a wiring installation? How it is achieved?Q.36 Explain types of fuses.Q.37 Explain classification of electrical network.Q.38 Explain concept of open and short circuits.Q.39 Explain the generation of ac current and voltage. Also explain generator principle.Q.40 Derive the equation of ac voltage and current. OR Derive the instantaneous value ofgenerated emf. PROBLEMSLOOP METHOD:Q.1 Find current flowing through each resistance.Solution: By loop method,Assuming current by clockwise direction for each loop.Apply KVL equation to loop 1 2 I 1  6 I 1  6 I 2  6  4 I 1  10  0 ………………(1) 12 I 1  6 I 2  4Apply KVL equation to loop 2 3I 2  2  5 I 2  6  6 I 2  6 I 1  0 …………………..(2)6 I 1  14 I 2  4Solving equation (1) and (2),By cramm’s rule, 83
    • Electrical Notes….. “Vijay B. Raskar” 8898443014 , vijayraskar2003@yahoo.co.in___________________________________________________________________ D = 132 D1 = 80 D2 = 72 D1 I1   0.61Amp D D2 I2   0.545 Amp DCurrent flowing through resistanceI 2   I 4   I 1  0.61AmpI 3  I 5  I 2   0.545 AmpI 6  I 1  I 2  0.065 Amp()orI 2  I 1  0.065 Amp() ALL THE BEST 84