Electrical Machines Past Paper PDF

# UNIT:- 4 ## Electrical Machines ### DC Machine - Principle and construction, types, emf equation of generator and torque eqn of motor. - Application of DC Motor (simple numerical problem) ### Three phase DC Motor - Principle and construction, types, slipped torque characteristics. - Application (...

# UNIT:- 4 ## Electrical Machines ### DC Machine - Principle and construction, types, emf equation of generator and torque eqn of motor. - Application of DC Motor (simple numerical problem) ### Three phase DC Motor - Principle and construction, types, slipped torque characteristics. - Application (Numerical Problem related to slip only) ### Single phase Inductance Motor - Principle of operation and introduction to methods of starting, Application. ### 3-synchronous Machine - Principle of operation of alternator and synchronous motor and their application. ## Electrical Machine - **DC** - Generator - Motor - **AC** - Induction Motor - Synchronous Machine (3φ) - Generator (Alternator) (3φ) - Motor (3φ) - Generator (Alternator) (1φ) - Motor (1φ) ## Supply - Stator - Stationary field winding - DC - Rotor - Rotating Armature - Supply - Hux - Lenz Rule's effect always oppose its cause, emf = -dф/dt ## DC Machines ### PRINCIPLE DC Machines works on the principle of faraday's law. ACC. to this law, whenever a conductor cuts magnetic field, emf is induced in it. This emf causes the current to flow if the conductor is closed. #### Essential Parts - Field Winding - Armature winding - Lap - Wave ### Types - **Series** - when field winding is in series with armature winding. - **Shunt** - when field winding is in parallel with armature winding. ## Induction Motor **Slip** - Let NR the speed of Rotor in rpm - Ns = Synchronous speed in rpm OR Speed of rotation of flux. - The relative speed b/w rotor & stator is called Slip. It always represents in percentage. - % Slip = (NS-NR)/Ns X100 - Ns = 120f/P - Where, f = frequency of supply = 50Hz - P= No. of poles **Quer:** A three phase 144, 50Hz in motor runs at 1460 rpm, determine its % slip. - P=4, f=50Hz, NR= 1460 rpm - Ns = 120f/P = (120 x 50)/4 ⇒ Ns = 1500 rpm - % slip = (NS-NR)/Ns X100 = (1500-1460)/1500 x100 = 2.67% **Quer:** Define slip in three-phase Induction valve what is it value at starling and synchronous speed. A 60 Hz induction motor has 2 poles & runs at 3510spm. Calculate Synchronous speed, the % Blip. - f=60 H2, P = 2, NR = 3510rpm, Ns =?, %slip=? - Ns = 120f/P = (120 x 60)/2 ⇒ Ns = 3600 rpm - ⇒ % slip : (N8-NR)/Ns X100 = (3600-3510) / 3600 ×100 - % slip = 2.5%. - At starting : - NR=0 - Slip = (NS-NR)x100/Ns = 100% - At synchronous speed ! - NR=NS - % Blip = (N8-NS)x100/N8 = 0% **NOTE:** Why never NR=NS? - Because slip will be zero and motor will not work. **Quer:** A 12th Pole 3-ø alternator coupled to an engine running at 500 spm, this alternator supplies an induction motor running at 14500pm. Find the '% slip, no. of Poler end of induction motor. - Ns = 120f/P = (120 x 50)/2 ⇒ Ns = 3000 rpm - again, Ns = 120f/P = (120×50)/4 ⇒ Ns = 1500 rpm - % slip = (NS-NR)x100/Ns = (1500-1450)/1500 x100 - % slip: 3.33% **Ques:** A 6 poles Induction motor operates from a supply whose frequency is 50 Hz. Calculate synchronous speed of motor, the speed of rotor and the slip is 0.04. - P=6, f=50 Hz, % slip=0.04 × 100 = 4 - Ns = 120f/P = (120 x 50)/6 ⇒ Ns = 1000 rpm - % slip = 4 = (1000-NR)/1000 x 100 = 40 - 1000-NR = 4 x 1000 / 100 - NR=1000-40 - NR = 960 rpm **Ques:** A 3φ 4 poles induction motor runs at 1440 spm at full load and is feed by 50 Hz supply. Calculate synchronous speed and full load slip of given motor.. - P=4, NR = 1440 rpm, f=50Hz, Ns =? - Ns = 120f/P = (120 x 50)/4 ⇒ Ns = 1500 rpm - % slip = (NS-NR)/NS X100 = (1500-1440) / 1500 ×100 - % Slip = 4% **Ques:** A particular load is to be driven at 700 spm, what should be the no. of polen for 3φ induction motor : f=60 Hz, f= 50 Hz, Calculate the actual speed is each case if rated Slip is 4%. - Ns = 700 rpm, P=?, f=60Hz - Ns = 120f/P ⇒ P= (120 x 60)/700 = 10 - Ns = 120f/P - Where, f = 50 Hz - P= (120 x 50)/700 ⇒ P= 8 - for, f = 60Hz, P=10, Ns = 120×60/10 = 720 rpm - for, f= 50 Hz and P=8, Ns = 120x50/8 = 750 rpm - Heme, % slip = 4% - given - NR for Ns = 720 pm, Ns = 750 rpm - 720-NR = 4 × 720/100 ⇒ NR = 28.8 720-28.8 / 100 = 691.2 - NR for NS = 750 rpm - 750-NR = 4 x 750/100 ⇒ NR = 750-30 = 720 spm # 18-May ## DC SHUNT GENERATOR - CL - isn bia : is a current arrow coming from terminal to the left. - awZRA: is a circle with current arrow going clockwise. - Rsh 2Fw: is labelled Rsh and has a line below Rsh - VT: is a terminal at the top going to the right. - Brushes (Carbon): is labelled with a current arrow going upwards on the left. - ia = isn + iL: is an equation written to the right of the diagram. - Eg = Vt + ia Ra + Loss: is an equation written to the right of the diagram. Vt = Ish x Rsh: is an equation written to the right of the diagram. - Question: A 20KW, 200V Shunt generator has an armature Resistance of 0.05 ohm and shunt field Resistance of 200 ohm. Calculate the power developed in the armature when it delivered 3000 W output. - Rsh=200 Ω , Ra=0.05 Ω - Vt = 200V, 20KW - Pg =? - Pg = Eg x ia - Eg = Vt + ia Ra - ish+iL = ia - ish = Vt /Rsh = 200/200 = 1A - iL = (20x1000-1000)/100 = 100A - so, ia = ish+iL = 1 +100 ⇒ ia = 101 A - Eg = vt + ia Ra = 200 + (101×0.05) = 205.05 - Pg = Egxia = 205.05 × 101 = 20.710KW - ia Ra CL Vt Rsh Eg is a diagram with a current arrow going downwards on the left - CL is a terminal on the left going upwards. - VT is a terminal on the right going upwards. **Ques:** A 4 poles DC shunt generator having field and armature resistance of 80 Ω and 0.1 Ω Resp. Supplies power to 50 lamp rated for 100V, 60W each. Calculate the total armature current and generated emf along a contract drop (Brush drop) IV per brush. - Rsn=80Ω, Ra=0.1 Ω - Vt=100V, P= 60x50= 3000 W - Eg=?, la=? - ia = 100/80 + 3000/100= 31.25 A - Eg = Vt + ia Ra = 100+ (31.25×0.1) + 2 = 105.125V. # 23/May ## EMF Equation of DC Generator - P = Potal no. of poles - Φ = flux per pole (unit = weber) - Z = Total no. of conductor - N = Rotation of Armature (speed); (rpm) - A = No. of parallel path in armature winding - E= emf induced in armature - spm = rotation per minute - e = average emf per conductor = dф/dt - Flux cut per conductor in one revolution = d = PΦ - no. of revolutions per second = N/60 - Time for one revolution = 60/N = dt - Generated emf 'e' = dф/dt = PΦN/60 = ΦZN/60 - emf for Z conductors = ΦZN/60 - Generated emf for each path = ΦZN/60 A - for LAP Winding :- A=P - so, E = ΦZN/60A = Φ2N/60A - ⇒ E = Φ2N/60 - A=2 - for Wave Winding: : - so, E = ΦZN/120 **Ques:** A 8 pole DC Shunt generator with 778 wave connected armature conductor running at 500 spm supplies a load of 12.5 Ω resistance at terminal voltage of 50 V. Armature Resistance is 0.24 Ω, field resistance is 250 Ω. Find Armature current, induced emf, and flux per pole. - Ra=0.24 Ω , Rsm=250 Ω, RL=12.5 Ω, V = 50V, P=8, A=2, N= 500rpm, Z=778 - Ia, E, Φ = ? - ia = isn + iL = ( 50+ 50)/12.5 + 250/12.5 = 8 + 20 = 28.7 + 28 = 4.2 - E = Vt + ia Ra = 50 + (4.2 × 0.24) = 51.008 - Φ = EX120/ PZN = 51.008 × 120 / 8 x 778 x 500 = 1.97 x 10^-3 weber - ia Ra RL V Rsh Eg - CL: is a terminal on the left going upwards. - is a current arrow going downwards on the left. - is a current arrow going downwards on the right. - is a terminal on the right going upwards. - is labelled RL with a current arrow going to the right. - is labelled Rsh with a current arrow going clockwise. - is labelled Eg with a current arrow going counter-clockwise **Ques:** A 6 Pole two circuit wave connected Armature of a DC Machine has 300 conductors and runs at 1000 rpm. The Emf generated is 400V. Determine the tux per pole. - P=6, E=400, A=4, N=1000 rpm, Z = 300 - E=ΦQ2N/120 ⇒ Φ = 120X2 / ΦZN = (120 x 400 x2)/ (6 x 1000 x 360) = 0.053 weber - Question: A 4 Pole generated with 400 Armature conductor have a useful fun of 0.04 weber per pole, what is the emf produce, if the machine is wave wound & runs at 1200rpm. what must be the speed at which the machine should be driven to generate same emf, if machine is Lap wound. - P=4, Φ=0.04 weber, N= 1200 rpm, Z = 400, E=? if: A=P, N = ? if, A, E = Φ2N/60 - E= 400 x 1200 x 0.04 x 2 / 60 = 320V x 2 = 640V - N = 60 E / Φ2N = (60 x 640) / (0.04 x 400) = 2400 rpm # Numerical on Compound: (Series + shunt) **Ques:** A shunt generator delivers 480 Ampere at 250V and the resistance of shunt field and armature are 50 Ω and 0.03 Ω resp. Calculate the generated emf. - Rsh = 50 Ω , Ra = 0.03 Ω - Vt = 230 V, iL = 450A - Eg= vt + ia Ra = 230 + (454.6 x 0.03) = 243.638V. - ia = isn + iL = Vt/Rsh + 450 = 230/50 + 450 = 454.6 A - ia Ra CL Vt Rsh Eg - CL: is a terminal on the left going upwards. - is a current arrow going downwards on the left. - is a current arrow going downwards on the right. - is a terminal on the right going upwards. - is labelled RL with a current arrow going to the right. - is labelled Rsh with a current arrow going clockwise. - is labelled Eg with a current arrow going counter-clockwise. **Ques:** A long shunt compound generator delivers a load current of 50A at 500 V and has armature, series field, shunt field resistance of 0.05 ohm, 0.03 ohm, 250 ohm resp. Calculate the generated voltage and armature current allow 1V per for contact drop. - Rsh = 250 Ω, Rp=0.03 Ω, Ra= 0.05 Ω, 1088= 2V - ia = ?, Eg=? - ia = isn + iL = Vt/Rsn + 50 = 500 / 250 + 50 = 2 + 50 = 52A - Eg = Vt + ia Ra + ia Rs + 1088 = 500+ (52×0.05)+(52×0.03) +2 = 506.16V. - ia Ra Rsh RB Eg VT is a diagram with a current arrow going downwards on the left. - CL is a terminal on the left going upwards. - VT is a terminal on the right going upwards. **Ques:** A short shunt compound generator delivers a load current of 30 A at 220 V and has armature, series field and shunt field resistance of 0.05 ohm, 0.30 ohm and 200 ohm resp. Calculate the induced emf and armature current allow 1V per brush drop. - iL = 30A, Vt=220V, Ra= 0.05 Ω, Rs= 0.30 Ω, Loss= 2V, Rsn = 200 Ω - Eg=?, ia =? - ia = isn + iL = Vsh/Rsn + 30 - Where , Vsh = (iL RS) + Vt - Vsh = (30x0.30) + 220 = 229 V - ia = 229/200 + 30 = 31.45A - Eg = Vt + ia Ra + iL Rs + Loss = 220+ (31.45× 0.05)+(30×0.30) + 2 = 232.95 V - ia Ra Rsn RB Eg VT is a diagram with a current arrow going downwards on the left. - CL is a terminal on the left going upwards. - VT is a terminal on the right going upwards. - is labelled RL with a current arrow going to the right. - is labelled Rsh with a current arrow going clockwise. - is labelled Eg with a current arrow going counter-clockwise. # DC Motor - Three Types - Series - Shunt (paralles) - Compound (series + parallel) - Torque: Output power = amature torque x speed rotation - Ebxia = Txw, then Eb = Eg = Φ2ZN/60A, T = Φ2ZN x ia / 60A, T ≈ Φ ia - Speed Relation-(N): Eb = Φ2ZN/60A, N x Eb /Φ ≈ N2 x Eb2/Φ2 - N2 / N1 = Eb2 x Φ1 / Eb1 x Φ2 ## Efficiency of a Motor - η = OIp Power / alp Power = Eb x ia / V x ia = Eb / V **Ques:** A 250 V DC shunt motor having an armature resistor of 0.25 Ω carries an armature current 50 A and runs at 750 rpm, If the flux is reduced by 10%. find the speed assume the torque remains constant. - Ra= 0.25 Ω, laF 50A, V= 250V, NF 750 rpm, N2 =? , T₁ = T₂ = const, Φ1=Φ, Φ2=0.9Φ1, loss = 0, Eb1 = V-ia, Ra = 250-(50x0.25) = 237.5 V, Eb2 = V-ia2 Ra = 250-(55.5x0.25) = 236.12 V, N2 / N1 = Eb1 x Φ2 / Φ1 x Eh2 = 237.5 x 0.9 / Φ1 x 236.12 = 750 x 236.12 / 237.5 x 0.9 = 828.5 rpm. - is a diagram with a current arrow going downwards on the left. - is labelled RL with a current arrow going to the right. - is labelled Rsh with a current arrow going clockwise. - is labelled Eg with a current arrow going counter-clockwise. **Ques:** A DC series motor has an armature resistance of 0.1 Ω and field resistance of 0.05 Ω and brush contact drop is equal to 3 V. It is operated at 250 V. When the line current (armature current) is 80 A and the speed is 600 rpm. find the speed when the line current is 100 A. - Ra=0.1 Ω, Rs= 0.05 Ω, Loss=3V, V = 250V, ia1 = 80A, N= 600 rpm, ia2 = 100 A, N2=?, Eb₁=V-ia Ra-iay Rs = 250-(80x0.1)-(80x0.05) -3 = 235V, Eb₂=V-iagra-ca₂ Rs = 250- (100×0.1) - (100×0.05) -3 = 232V, N2 = Eb2x Ioy x N, / Ia₂xEb1 = 232 x 600 x 80 / 235 x 100 = 473.87 rpm - is a diagram with a current arrow going downwards on the left. - is labelled RL with a current arrow going to the right. - is labelled Rsh with a current arrow going clockwise. - is labelled Eg with a current arrow going counter-clockwise. # 27/May/23 ## 3-ø Induction Motor - **Stator** - **Rotor** - **Squirrel cage IM** - Slipring induction Motor (3(external resistance)) (Star connection) - Endrings - Carbonbars - **Working:** - Stator (Star connection) - 3ø supply- a circle with an arrow going clockwise - RMF (Rotating Magnetic field) - Rotor (stationary conductor) - Speed NB = 120 f/P- a circle with an arrow going clockwise, a circle with an arrow going downwards. - Relative Motion - Right hand Thumb Rule - EMF (Principle) OR Faraday's law - Rotor CKt closed - Magnatic Field- a circle with an arrow going clockwise, a circle with an arrow going downwards. - Current - **Working Principle:** - Works on the principle of Electromagnetic Induction. - 3ø AC supply is given to 3ø stator winding. - Rotating magnatic field RMF is produced. - Speed of RMF ⇒ Ns = 120f/P - Let's assume the direction of RMF is clockwise. - **Now:** Rotor is stationary, stator flux (RMF) is Rotating. So, there is relative motion blwc rotor conductor and stator Hux. Which induced emf in rotor conductors (Electromagnetic Induction). - **As:** Rotor forms closed circuits, emf circulates current in rotor circuits called rotor current. - **Any:** current carrying conductor produces its own fluix called rotor flux in this case.... - Direction of fluix is clockwise (Right Hand Thumb rule). - **Now:** RMF and Rotor flux interect with each other, and increase flux area. - **Rotor:** Conductor experiences a force and start in the direction of RMF. - **Thus:** 3-ø induction motor is self starting motor. ## Construction: - A 3-ø IM essentially consists of two parts: - Stator - Rotor - **Stator** - Built up of high grade alloy steel lamination to reduce eddy current losses. - Stator frame is made up of cast iron or fabricated steel plate. - The insulated stator conductors are connected to form a 3-ø winding. - The phase winding may be either star or delta connected. - **Rotor** - Built up of thin laminationa of cast iron or fabricated steel - laminated cylinderical core is mounted directly on the shaft ### Types: - **Squirrel cage rotor** or **simply, cage rotor**: - Rotor Base with a current arrow going upwards - End rings on both sides with a current arrow going clockwise - Consist of cylinderical laminated come with slots nearly paralled to the shaft axis. - Each slot contains an Anis insulated bar conductors of aluminium or copper - Rotor bar conductors are short circuited by heavy end rings. - **Advantages:** - Uniform torque is produced. - Noise is reduced. - locking tendency of the rotor is reduced. - **Phase wound or wound rotor**: - **Rotor windings are connected in star** - Rotor - Shaft - Brush - Start slipring - Run Starter - **Rotor windings are connected in star.** - The open ends of the star circuit are brought outside the rotor and connected to three insulated sliprings. - The slipring' mounted on the shaft with brushes resting on them. - Slipring + Brushes provide a means for connecting external resistances. - **Purpose of Resistors:** - To increase the starting torque and decrease the starting current from the supply. - To control the speed of motor. ## Torque slip Characteristic: - Torqu Tmox Tfull D Tst C N=NR=1405:0 B S-1- N=0 → Slip - **Where:** OA → Stable Region, AC → Unstable Region, Point A → Max. Torque, Point D → Full Load Torque, Point C → Starting Torque - **Tα = SR2 / R2²+(3x2)2** - Where, S = slip, R2= Resistance of Rotor, X2 = Induction of Rotor - **As:** the Induction motor is loaded from no load to full speed decreases. Hence, slip increases. - **Due to:** Increased load, motor has to increase torque to satisfy load demand. - **The curve:** obtained by plotting torque against slip from S=1 to S=0 is called Torque-Slip characteristic. - **For constant voltage supply:** emf in rotor E2 is also constant, Tα = SR2 / R2²+(3x2)2. - **To understand:** the graph let us devide the torque-slip characteristic into two parts : - Low slip Torque Region - High slip Region. ## Unit – Continue - **Low Slip Region:** - In low slip Region S is very small, due to which (3x2)² is neglegible compare to R2 So it can be neglected, Tα = SR2 / R2². - So, load increased, speed decreases and slip increases. This increase the torque which satisfied the load demand. - Torque increases linearly in this region. - **High slip Region:** - In this Region Slip is, high, It can be assumed that R2 is small compare to X2, Tα = S / X2. - Tα 1/ X2 is constant. - It's graph is rectangular hyperbola. - When load increases, speed decreases and slip increases. As T is inversely proportional to slip, Torque (T) decreases as slip increases. - But torque must be increases to satisfy the load demand. - As torque decreases due to extra loading effect speed further decreases. Eventually, motor comes to Stand still condition. - Hence, High slip Region called unstable region. - **NOTE:** The maximum torque which motor can produce is called break down torave or pull out torque. - At S=1, N= 0, the produced torque by motor is called starting torque. ## Single phase Induction Motor - **Double Revolving field theory:** - According to this theory, any alternating quantity can be resolved into two Rotating components which rotates in opposite direction and each having magnitude as half of the max. magnitude of alternating quantity. - In case of signal phase Induction Motor, the stator winding produces an alternating magnetic field having max magnitude of Φm. - Acc. to double Revolving Field theory, consider the two components of stator fluse each having magnitude half of max magnitude of stotor fluse that is Φm. Both these components are rotating in opposite direction. - Let, Φf → forward component rotating in anti-clock wise direction. - ΦB→Backward component rotating in Clockwise direction. - Both the Components are Rotating hence, get cut by the rotor conductor... due to cutting of flux, emf get induced in rotor which circulates rotor current. - The Rotor current produces rotor flux. - This fluix interect with forward component to produce a torque in one perticular direction. Byt Rotor fux interect with backward component to produe a torque in opposite direction. - Each torque tries to rotate the rotor in its own direction. So, net torque experience by rotor is zero at starting : - Hence, single-phase Induction motor are not self starting. ## Types of Induction Motor: - **Method of starting of Induction Motor :** - **Single phase induction motor classified as :-.** - **Split phase Induction motor** - **Capacitor start Induction motor** - **Capacitor start**. ** Capacitor, Run IM** - **Shaded Pole IM** - **To Produce Rotating Magnetic feux, it is necessary to have minimum to alternating fluxes having x-phase difference.** - **The intereraction of two fluxer produce a resultant torque which rotates the motor in one direction** - **More the phase difference x, wore is the starting torque, once the motor start, auxiliary winding may be replaced.** - **Split Phase Induction Motor** - Im→ Main winding current - Ist → starting winding / Auxilary winding Current - Stator winding is called main winding. - Additional winding to stator is Called auxilary / starting winding. - Main winding is inductive in nature and auxiliary winding is resistive nature. - There, exist a phare difference 'x' between the two fixes. - The resultant fux is rotating in nature due to This starting torque is produced in one direction. - When a motor gathers a speed upto 75% to 80% of Synchronous speed, centrifugal Switch can be open mechanically. - **Disadvantage:** - gives poor starting Torque. - **Application** - Blower, washing Machine, Table fan, oil burner etc. - **Capacitor Start Induction Motor** - Im→ Main winding current - Ist → starting winding / Auxilary winding Current - Stator winding is called main winding. - Additional winding to stator is Called auxilary / starting winding. - Main winding is inductive in nature and auxiliary winding is resistive nature. - There, exist a phare difference 'x' between the two fixes. - The resultant fux is rotating in nature due to This starting torque is produced in one direction. - When a motor gathers a speed upto 75% to 80% of Synchronous speed, centrifugal Switch can be open mechanically. - **Disadvantage:** - gives poor starting Torque. - **Application** - Blower, washing Machine, Table fan, oil burner etc. - Im→ Main winding current - Ist → starting winding / Auxilary winding Current - Stator winding is called main winding. - Additional winding to stator is Called auxilary / starting winding. - Main winding is inductive in nature and auxiliary winding is resistive nature. - There, exist a phare difference 'x' between the two fixes. - The resultant fux is rotating in nature due to This starting torque is produced in one direction. - When a motor gathers a speed upto 75% to 80% of Synchronous speed, centrifugal Switch can be open mechanically. - **Disadvantage:** - gives poor starting Torque. - **Application** - Blower, washing Machine, Table fan, oil burner etc. - **The construction of this type motor is similay to the split phare type, the difference is that •itir in series with auxilary winding, the capacitor is connected.** - **The capacitère circuit draws leading current. This feature is used to increase a blw Im and Ist.** - pendending up whether capacitor remains in the circuit permanently or is this connected from the circuit using centrifugal switch, these motors are classified as : - Capacitor start Motor. - Capacitor start -- capacitor run motor. - **The starting torque is proportional to alphala) and hence such motor produce very high torque.** - **Buchen speed aproaches to 75% to 80% of synchron- our speed the starting winding get disconnected due to operation of centrifugal switch-** - **Hence the capacitor remains in the circuit only at start, that's why it is called capacitor start induction Motor.** - **NOTE:** in cave of capacitor start - capacitor Run Motor there is no centrifugal switch and capacitor, remain permanently in the circuit, this improver the power factor. **Applications:** - Compressor, Refrigerator, Air conditionals. - Capacitor Start - capacitor Run motor are used in ceiling fans, air circulator and lower. ## SHADED POLE INDUCTION MOTOR - **Salient Pole** - Copper band: A rectangle with a current arrow going right. - **This type of motor consists of squirrel gate rotor and stator consist of salient pole.** - **The Poles are shaded Means each pole carries a copper band on one of the unequally divided parts Called shaded band.** - **When single - ø He supply is given to stotor winding, due to shading it provide Rotating magnetic field.** **Advantage:** - **Due to absence of centrifugal switch, construction is robust (solid / hard).** **Disadvantage:** - **Due to shading band copper loss increase.** - **Efficiency is low.** - **starting torque is very poor** **Application:** - Used for very small fan, tuice, pin projector, photo copy machine. ## Comparison b/w Squirrel Cage and Slipering (or wound) cage - **Slipering (or wound) cage** - Rotor consist of wire nd similar to stator winding. - Construction is complicated. - Resistance can be added externally. - Sliperèng and Brushes are padd to external Resistance. present - The construction is delicate and due to bmshes grequent maintanance is neccasary. - High starting torque can be obtain - use for lift, train, aviator, compressor - **Squirrel Cage** - Rotors consists of baus which are shorted at the end cuith the help of end rings. - Construction is simple - As permanently shorted, external Resistan can not be added. - supering and Brusher are absent. - The construction is Robust and maintainance free. - Moderate starting torque that can not be control. - Pused for drilling triachine, Pumping machine, water pump, tanete. ## Synchronous Motor - **PRINCIPLE Synchronous motor works on the principle of Magnetic locking.** - **Magnetic Locking:** - When two unlike poles are brought near each other, if the magnets are strong, there exist a strong force of attraction between twopoles. - In such conditions two mag