NUMERICALS  

Q.1  A, power supply is having the following loads:-    Type of load     Max. demand (kW)    Diversity of group Demand factor     Domestic             1500        1.2                               0.8 Commercial        2000       1.1   0.9 

Industrial             10,000                    1.25                               1 

   If the overall system diversity factor is 1.35, determine the maximum demand and connected load of each type.  (8)  

  Ans:   The sum of maximum demands of three types of loads is = 1500 + 10,000+ 2000 = 13,500kW.  As the system diversity factor is 1.35,       Therefore, max. demand on the supply = 13,500 / 1.35 = 10,000 kW.       Each type of load has its own diversity factor among its consumers.         ∴ connected domestic load = 1500 X 1.2 / 0.8 = 2250 kW.        ∴ connected commercial load = 2000 X 1.1 / 0.9 = 2444 kW.        ∴ connected domestic load = 10,000 X 1.25 / 1 = 12, 500 kW.  

Q.2  A two-pole alternator runs at 3000 rpm and supplies power to a 10 –pole single – phase induction motor, which has full load slip of 5 %. Find the full load speed of the induction motor and the frequency of its rotor emf due to forward field. (8)   

  Ans: 

  

p

f

NS

× = 120 where NS = Synchronous speed and f = frequency of the supply voltage 

generated by the alternator, then  

  

2 120

3000

f× =  = 50. 120 30002 fHz = × = 

     600. 10 12050 Nrpm S = × = 

S

S N NN

S −

=  so, 

600 600

0.05

N− = = N = 570 rpm. 

     If fr is the rotor emf frequency, then fr  = 50 X 0.05 =2.5 Hz.  

Q.3  The voltage applied to a dc shunt motor is 220V. The armature current is 20A. The armature resistance is 0.5 Ω. The speed is 80 radians per second. Determine the induced emf, the electromagnetic torque and speed in rpm. (8)  

  Ans:   Given V= 220V, Ia = 20A where Ia = the armature current. Ra = Armature resistance = 0.5 Ω and ω = 8o rad. /s.       The back emf of the motor Eb = V - Ia Ra  = 220 – 20 X 0.5 = 210 V. 

     The electromagnetic torque Te = Eb Ia / ω = 210 X 20 / 80 = 52.5 N- m.  If N is the speed in rpm, then total angular distance covered in one minute = 2 π N radians.     Or angular distance covered in one second = 2 π N / 60 rad. / s       Hence 80 = 2 π N / 60 or 3.142 8060 × × =N = 764 rpm.  

Q.4  For the circuit shown in Fig.1, find the value of L R for maximum power transfer.  What will be the value of maximum power?  

Q.5  A series circuit of resistance 250Ω and inductance 0.25 H is excited from a pulse voltage of strength 10 V of duration 1 ms. Find the value of the current at 0.5 ms and 2 ms.    (8)   Ans: e= Em sinω t  ;  Em = 250V and 2 π f = 2 x 3.14 x 50.  e = 250 sin314t  When e =125V then sin314t= 125/250 =0.5 ; 314t = sin-10.5= 100 π t =300 t = (30) /100 x 1800= 1/600 = 1.667ms. 

Q.7  The electric mains in a house is marked as 230 V, 50 Hz.  Write down the equation for instantaneous voltage in sinusoidal form. (4)  

  Ans: Vrms =230V; f = 50Hz Vmax =√2 x Vrms = 325.22 V0lts. ω =2 π f = 2x 3.14 x 50            Hence e = 325.22 sin 314 t  

Q.8  The secondary of a 750 KVA, 11000/ 400 V, 50 Hz transformer has 160 turns. Determine the primary number of turns, primary and secondary full load current neglecting losses. If the area of cross section of the core is  100 cm2, what will be flux density in the core?   (8)  

  Ans:  

1

2

1

2

V VN N

= where N2 and N1 are the number of turns on the secondary and primary windings. 

    4400 400 11000160 2 12 1 = × == V VN N turns. N1=4400 turns. 

   7501000 11 =× IV =  11000 7501000    ∴

1× =I = 68.182 A. 

    7501000 22 =× IV = 400 7501000∴

2 × =I = 1875 A.  

E1 = 4.44 ϕmax f N1 volts =  ϕmax =

5044004.44 11000 = 0.01126 Wb. 

   ∴ Bmax = 2 4 126/1. 10010 011260. Wbm= × −      ∴ Bmax  = 1.126 Wb/ m2.  

Q.9  A 3 – phase transformer consisting of three 1 – phase transformers each with turn ratio of 10:1 (primary : secondary) is used to supply a 3 – phase load of 120 kVA at 400 V on the secondary side.  Calculate the primary line current and voltage if the transformer is connected   (i)   Y ∆  (ii)  ∆ Y .  What is the line-line transformation ratio in each case?    

 Ans: ∆/Y connection – I = (120 X 1000) /√ 3 X 400 = 173.2 A  Primary line-to line voltage = a V /√3  = 10 X 400/√3 = 2309 V;  where a = turns ratio Primary line current = √3 X I / a  = 1.732 X 173.2 x 1/10 = 30 A;  Line to line transformation ratio = a V / √3  / V  = a / √3  = 10 /√√3         Y/∆ = Primary line-to line voltage = √3 a V = √3 X 10 X 400 = 6928 V;  Primary line current = 173.2 / √3  = = 10 X 1.732 = 10 A;  Line to line transformation ratio =10√3   

Q.10  A separately excited dc motor is operating at an armature voltage of 300 V.  It’s no-load speed is 1200 rpm.  When fully loaded it delivers a motor torque of 350 N-m and its speed drops to 1100 rpm.  What is the full load current and power?  What is the armature resistance of the motor?  The motor is now fed with an armature voltage of 600 V, while its excitation is held fixed as before.  If it is once again fully loaded, find the motor torque, power and speed.   (8)  

  Ans: Given  Ea  = V = 300V, 300 = (Ka x Φ x 2 π x 1200) / 60 or Ka x Φ =2.39   Ea = (300 x 1100)/1200 =275 V.  Ka = Z P/ 2 π x A is a constant.    

 Ia = 350/2.39 = 146.4 A where Ia = the armature current.  Mechanical power developed = Ea Ia  = 275 x 146.7 = 40.3 KW.  Ra = (300 – 275) / 146.4 = 0.171Ω ΩΩ Ω  Armature voltage = 600V Ia = 350/2.39 = 146.4 A; T = 350 N-m The back emf of the motor Eb = V - Ia Ra  = 600 – 146.4 X 0.171 = 575 V. 575 = 2.39 x 2 π n / 60; n = 2297 rpm. Power = Ea Ra  = 575 x 146.4 = 84.2 kW  

Q.11  A coil, which has 10Ω resistance and 50mH inductance is connected to 230V, 50Hz supply. Calculate the current in the coil.  (5)  

  Ans: XL = 2 X 3.14 X 50 X 50 X10-3 = 15.7 Ω      ∴ 22 .7)(15)(10 =+Z = .5) (246100)( + = 18.6Ω.      ∴ . .3712 .618 230 IA == I = 12.37A.  

Q.12  A 3-phase induction motor which is wound for 4-poles, when running on full load, develops a useful torque of 100 Nm; also the rotor emf is observed to make 120-cycles/ min.  It is known that the torque lost on account of friction and core loss is 7 Nm.  Calculate the shaft power output, rotor copper loss, motor input and motor efficiency. (8)   

  Ans: f2= sf ; 120 / 60 = 2 Hz  where f2  = rotor frequency s(slip) = 2.5/ 50 =0.04   n s (synchronous speed) = 1500 rpm n = (1 – 0.04) x 1500 = 1440 rpm ω = 2 π X1440 / 60= 150.7 rad/s  Shaft power output = 100 x 150.7 = 15.07 KW. Pm = (100+7) x 150.7 = 16.12 kW. Rotor copper loss = 3 I22R2 = Pm (s /1-s) = 16.12 X 0.04/ (1- 0.04) = 0.67 kW Motor input = 16.12 + 0.67 + 0.7 = 17.49 kW. η = 15.07 / 17.49 = 86.16 %  

Q.13  When a coil is connected to a 230V, 50Hz supply, it takes a current of 2A and the power consumption is 150W. Calculate the resistance and inductance of the coil. (5)  

  Ans:  

I V Z = Where Z is the impedance, V is the voltage and I the current. 

       ∴ 2 230=Z = 115Ω. And P = I2 R; ∴ 2 I P R = 22 150 = 37.5 Ω ΩΩ Ω.  

       ∴ 22 .5)(37115)( =−LX) .25(140613225)( =−LX = .75 11818=LX = 108.71Ω. 

       Or   503.142 .71108 XX L = = 0.346 H. 

Q.15  A power station has a load cycle as under: 260 MW for 6 hr: 200MW for 8 hr; 160 MW for 4hr; 100MW for 6hr. If the power station is equipped with 4 sets of 75 MW each. Calculate the load factor and capacity factor from the above data. (8)  

  Ans:  Daily load factor = Units actually supplied in a day 

       Max. Demand X 24 

   MWh supplied per day = (260 X 6) + (200 X 8) + (160 X 4) + (100 X 6) =4,400 ∴ Station daily load factor =   4,400          =  0.704 or 70.4 % 

            260 X 24  

     Capacity factor = Average demand on station 

      Installed capacity of the station       No. of MWh supplied per day = 4,400  ∴Average power / day = 4,400/24 MW. 

Total installed capacity of the station = 75 X 4 = 300 MW. 

     Capacity factor = 4,400/24 = 0.611 or 61.1 % 

                                      300 

Q.16  A generating station has a maximum demand of 25 MW, a load factor of 60%, a plant capacity factor of 50% and a plant use factor of 72%.  Find     (i)  the daily energy produced (ii)  reserve capacity of the plant,    (iii) the maximum energy that could be produced daily, if the plant, while running as per schedule, were fully loaded.  (8)  

  Ans:  Load factor = Average demand / Max. demand  = 0.6 = Average demand / 25 

Average demand = 15 MW. 

Plant capacity factor = average demand / installed capacity = 0.50 = 15 / installed capacity  

       Installed capacity = 15 / 0.5 = 30 MW 

      Therefore reserve capacity of the plant = installed capacity – Maximum demand 

  = 30 - 25 = 5 MW. 

Daily energy produced = average demand x 24  = 15 x 24= 360 MWh 

Energy corresponding to installed capacity per day = 24 x 30 = 720 MWh 

Maximum energy that could be produced = actual energy produced in a day/ plant use factor = 360 / 0.72 = 500 MWh / day.  

Q.17  A 3-phase induction motor is wound for 4-poles and is supplied from a 50 Hz system.  Calculate        (i)  synchronous speed.     (ii) actual speed of the motor when running at 4 % slip.     (iii) frequency of emf induced in rotor. (6)  

  Ans:  Synchronous speed, NS = 120 X f  = 120 X 50 = 1500 r.p.m. Where f is the supply frequency. 

                                                      P                  4                               and  P is the No. of poles. 

     Actual Speed of motor = N = NS (1-S) Where S is the slip = 0.04 ∴ N = 1500 (1 – 0.04) = 1440 r.p.m. 

     Frequency of the rotor emf = fr = S f = 0.04 X 50 = 2 Hz.   

Q.18  Calculate the maximum power by a solar cell at an intensity of 200 2Wm.  Given 

0.14V max = V and 6mA Imax =− .  Also calculate the cell efficiency if the area is 4 2 cm .   (8)   Ans: For solar cell maximum power Pmax = Imax V max Pmax(output)  = -- 6 x 10-3 x0.14 = - 0.84 mW   = - 0.84x 10-3 W Pinput = intensity X area = 200x 4 x 10-4  Cell ή = (0.84 x 10-3) / (200 x 4 x10-4 )   = 1.05 %  

Q.19  A 6- pole lap wound shunt motor has 500 conductors in the armature. The resistance of the armature path is 0.05. The resistance of shunt field is 25Ω. Find the speed of the motor when it takes 120 A from the dc mains of 100 V supply. Flux per pole is 2 X 10 -2 wb.   (8)   Ans: Ish  = Vsh / Rsh = 100 / 25  = 4 A. Where Ish is the shunt field current, Vsh is the shunt field voltage and Rsh is the resistance of the shunt field. 

Ia   = IL  - Ish = 120 – 4 = 116 A. Where IL the line current and Ia is the armature current. 

Eb = V - Ia Ra where V is the applied voltage and Eb is back emf developed. 

    = 100 – 116 X 0.05 = 94.2 V. Eb  = P ϕ Z N / 60 A = 94.2 =  6 X 2 X 10(-2) X 500 X N   where Z = No. of conductors                                                                        60 X 6                      P = No.of poles : ϕ = flux per pole.    ∴ N = Speed of the motor is 565.2 r.p.m. 

 Q.21  In a 25 kVA, 2000 / 200V transformer the iron and full load copper losses are 350W and 400W respectively. Calculate the efficiency at unity power factor at     (i) full load and  (ii) half load. 

  Ans:  ηx =   x kVA X 1000 X cos ϕ                                    x kVA X 1000 X cos ϕ + Pi + x2 Pc 

        Where cos ϕ  = 1 and Pi = iron loss = 350 W; Pc = copper loss = 400 W. 

(i) At full load x =1        ∴η  =        1 X  25 X 1000 X 1                          X 100 = 97.087 %         1 X 25X 1000 X 1+ 350 + 1 X 1 X 400 

(ii)  At half load x =0.5   

         ∴η  =    0.5 X  25 X 1000 X 1                                     X 100 = 96.525 %          0.5 X 25X 1000 X 1+ 350 + 0.5 X 0.5 X 400 

Q.22  An a.c circuit consists of a pure resistance of 10 ohms and is connected across an a.c supply of 230V, 50 Hz.     Determine   (i) current flowing through the circuit.      (ii) Power consumed by the circuit.       (iii) Write down the equation for voltage and current. (8)          

  Ans: i) Current in the circuit, I = V/R = 230/10 = 23A ii) Power consumed by the circuit P = VI     = 230x23 = 5290 W iii) Maximum value of applied voltage VV Vmrms .27 23032522 ×=== Maximum valued of current IA Imrms .53 233222 ×=== Angular velocity rads f 314/2 == πω Equation for applied voltage Sint SintvV m 314 .27325== ω As in pure resistive circuit, current and voltage are in phase with each other, therefore equation for current is             Sint SintiI m 314 .5332== ω 

Q.24  The armature of a 4-pole, d.c shunt motor has a lap-connected armature winding with 740 conductors. The no load flux per pole is 30 mwb. If the armature current is 40A, determine the torque developed?   (8)      

  Ans:    Torque developed in a DC motor is given by   Nm PAZIT aa ). (/1590. =Φ   Here   Z = 740     P = 4     IA a 40 =     A = 4(for lap connected winding A = P)     mwbwb 3301030 −=×Φ=   Nm Ta . (4/4)7404030101590. 3 ××××=× −        = 141.14 Nm 

Q.26  A power station has a maximum demand of 15000kW.  The annual load factor is 50% and capacity factor is 40%.  Determine the reserve capacity of the plant.    (6)  

  Ans:   Energy generated / annum = Max. Demand X Load factor X Hours in a year.         = 15000 X 0.5 X 8760  = 65.7 X 106 kWh.   Capacity factor = Units generated/ annum           =        65.7 X 106     =   18,750 kW.                                            Plant capacity/ Hours in a year         0.4 X 8760   Reserve capacity = Plant capacity – Max. Demand. = 18,750 – 15,000 = 3750 kW.     

Q.27  A 100 MW power station delivers 100MW for 2 hours, 50 MW for 6 hours and is shut down for rest of each day. It is also shut down for maintenance for 45 days each year. Calculate its annual load factor.   (8)  

  Ans:   Energy supplied for each working day = (100 x 2) + (50 x 6) = 500 MWh.   Station operates for 365 – 45 = 320 days in a year.   Energy supplied / year = 500 x 320 = 160,000 MWh. 

  Annual load factor = 100 workinghoursinMW.demandMax perannumsuppliedMWh × × 

                                 = % .810020 24320100 ,000160 ×= ××   

Q.28  A 3-phase induction motor has 6-poles and runs at 960 rpm on full load.  It is supplied from an alternator having 4 poles and running at 1500 rpm.  Calculate the full load slip of the motor.   (6)  

  Ans:   No. of poles of the alternator = 4 Speed of the alternator = 1500 rpm. Therefore frequency f = N X P = 1500 X 4  = 50 Hz        120          120 

Therefore frequency generated by the alternator = 50 Hz. Induction motor has 6 poles (P).  Speed N of the motor = 960 rpm. Supply frequency of the alternator is 50 Hz. Synchronous speed of the motor Ns = 120 X 50 =120 X 50 = 1000 rpm.                                                                      P               6 % Slip S =  Ns – N X 100 = 1000 – 960 X 100 = 4 %.                      Ns                                 1000  

Q.30  A 25 KVA transformer has 500 turns on the primary and 40 turns on the secondary. If the primary is connected to a 3000V, 50 Hz mains, calculate (i) the primary and secondary currents at full load, (ii) the secondary e.m.f. and (iii) maximum flux in the core.  (8)  

  Ans:  At full load the current in the primary winding I1 = 25 X 103    =  8.33 A.                                                                                             3000 I1 = E2 =   N2           I2     E1      N1 I2   = N2    X I2  = 500 X 8.33  = 104.15 A is the current in the secondary winding.                     N1                40 N1 and N2 are the number of turns in the primary and secondary windings. E1 and E2 are the emfs of the primary and secondary windings. 

E2  = N2    X E1  =   40 X 3000  = 240 V.                       N1                  500 Using the relation  E1 = 4.44N1 f ϕm = 3000 = 4.44 X 500 X 50 X ϕm; where ϕm is the maximum flux and f is the frequency.   ϕ ϕϕ ϕm = 0.027 wb.  

Q.31  If a generating station has a maximum load for the year of 18,000 kW and a load factor of 30.5% and the maximum loads on the substations were 7500 kW, 5000 kW, 3400 kW, 4600 kW and 2800 kW. Calculate the units generated for the year and diversity factor of the generating station.   (8)  

  Ans:   Load factor = Maximumdemand Averagepower  

           0.305 = 18000 Averagepower   Average power = 18000 x .305 = 5490 kW   No. of hr/year = 36 x 24 = 8760 hr   No. of units generated/yr = 5490 x 8760 = 48,092,400 kWh 

  Diversity factor = yr.thewholedemandofMax. .demandindividualmaxoftheSum Diversityfactor =   Sum of the individual max. demand = 7500 + 5000 + 5400 + 4600 + 2800                                            = 23,300 kW    Max. load = 18000 kW 

  Therefore, Diversity factor = 18000 ,30023 = 1.3 approx.  

Q.32  A 230 V, 1150RPM, 4-pole, DC shunt motor has a total of 620 conductors arranged in two parallel paths, and yielding an armature circuit resistance of 0.2 Ω.  When it delivers rated power at rated speed, it draws a line current of 74.8 A, and a field current of 3A.  Calculate the flux per pole, torque developed, armature and field copper losses.  (8)    

  Ans: Ia (armature current) = IL (load current) – Ish (shunt field current). Ia = 74.8 – 3 = 71.8 A. Eb (back emf) = V (supply voltage) – Ia Ra. Where Ra is the armature resistance. Eb = 230 – 71.8 X 0.2 = 215.64 V. Eb = ϕ Z N P   where Z is the number of conductors, N is the speed in rpm, P the no. of poles.            60 A     where A is the number of parallel paths.  

ϕ = 60 X 2 X 215.64 = 9.073 mwb.         620 x 1150 X 4 

Ta(torque) = 0.159 X P X Ia X ϕ X Z = 0.159 X 4 X 71.8 X 9.073 X 10-3 X 620 = 133.8 N-m.                                           A                                                           2  Armature copper losses  = Ia2 Ra = (71.8) 2 X 0.2 = 1031 W.     Field copper losses = Ish 2 Rsh = Ish Vsh = 3 X 230 = 690 W. 

Q.34  A series AC circuit connected to 230V, 50Hz mains consists of a non- inductive resistance of 100 Ω and inductance of 100mH and a capacitance of 20µF. Calculate – impedance, current, power factor and power.  (8)  

Ans:   Inductive reactance, =Ω ××π×π== − .4 103150100L22fX 3 L   Capacitive reactance, =Ω ××π×= − .24 1015950201/2X 6 C   Impendance, ( ) ( ) ( ) =Ω +−−==+ .31 162.24159.410031XXZR 222 LC 2   Current, I = V/Z = 230/162.31 = 1.42A   Power factor cosφ = R/Z = 100/162.31 = 0.6

Q.36  A series R-L-C circuit consists of a 100Ω resistor, an inductor of 0.318H and a capacitor of 

unknown value.  When the circuit is energised by  2300V o∠ , 50 Hz sinusoidal a.c. supply, 

the current is found to be  2.3 0 A o∠ .  Find     (i)   value of capacitor in microfarad.    (ii)  voltage across the inductor.          (iii) total power consumed.    (14)  

  Ans:     Supply voltage, V= 230∠00 volts. Current, I = 2.3∠00 amperes. 

Impedance = Z = V/I = 230/2.3 = 100 Ω For R-L-C circuit = Z= √ 1002+ (XL- XC) 2 100 = √ 1002+ (XL- XC) 2 XL= XC 1/2πfC = 2πfL 1/ 2π X 50 X C  = 99.9 or , C = 1/ 2π X 50 X 99.9 = 31.85 µ µµ µF  XL = 2 X π X 50 X 0.318 = 99.9 Ω ΩΩ Ω Voltage across the inductor = VL = I X XL  = 2.3 X 99.9 = 229.77 V Power consumed P = V I Cos φ = 230 X 2.3 X 1 = 529 W Or P = I2 R = (2.3) 2 X 100 = 529 W   

Q.37  The emf per turn of 3300 /395, 50Hz single- phase core type transformer is 7.5V, if the maximum flux density is 1 tesla, then find a suitable number of primary and secondary turns and the net cross- sectional area of the core.                                                                            (8)    Ans:   Given V1 (primary voltage) = 3300 volts V2 (secondary voltage) = 395 volts.      Voltage per turn = 7.5 V, Bmax (max. flux density) = 1 tesla,   N1 (number of primary turns) = 3300 / 7.5 = 440 turns   N2 (number of primary turns) = 395 / 7.5 = 52.66 turns, N2 = 53 turns. 

  Therefore, primary number of turns may be taken = 443turns 395 330053 N1 = × = .   max2 222 fN BA4.44fNmax4.44VE ϕ=== 

  2 lcm.7335 5350144.4 395 A = ××× = where A = area of cross section of the core, f=frequency,   . maxmax =ϕ flux.  

Q.38  A 6- pole lap wound series motor has 60 slots; each slot consists of 12 conductors. If the armature current is 50 A, calculate the total torque in Nw -m. Flux per pole is 20 X 10-3 wb.   (4)    Ans:                                   Where Z = No. of conductors = 60 x 12 = 720; P = No. of poles = A (parallel paths) = 6:   φ = flux per pole = 20 mWb, Ia (armature current) = 50 A. 

  Torque = Nwm A l ZP a =− ××××× × ×= − .65 114 6 50 72062010 14.32 1 2 1 3 ϕ π

.  

Q.39   Two coils when connected in series have a resistance of 18 Ω and when connected in parallel have a resistance of 4Ω.  Find the resistance of each coil.  (8)      Ans:Let the resistances of the coils be R1and R2.  Equivalent resistance when connected in series = R1+ R2 = 18Ω. ---------(1) Equivalent resistance when connected in parallel = 1/R1+ 1/ R2 = 1/4Ω Or 4 =   R1 R2       ---------------------------(2)   

              R1+ R2 

 Multiplying (1) and (2)  R1 R2   = 72 Ω  ----------------------------(3)  R1 - R2 = √ (R1+ R2)2 – 4 R1 R2   = √ (18)2 – 4 X 72    = ± 6 Ω. ---------------(4)  Adding (1) and (4),  2R1 = 24 or 12 Ω , or  R1 = 12 or 6 Ω. , and      R2 = 6Ω ΩΩ Ω or 12Ω ΩΩ Ω.  

Q.41  A 12 pole, 50 Hz induction motor is running at 450 rpm.  Calculate the % slip of the motor on   account of forward field.    

  Ans: 

  Synchronous speed, m. r.p.500 12 12050 P 120f Ns = × = × = Where f is the supply frequency.                                Where P is the no. of poles. 

  10% 100 500 500450 100 N NN %S s s ×= − ×= − =  Q.42  A 50 kVA , 5000/500V, 50Hz, 1-phase transformer has the high voltage winding with a resistance of 8 ohms and low voltage winding with a resistance of 0.06 ohms. The no load losses of the transformer amount to 1000W. Calculate the efficiency of the transformer, when delivering its full rated output at a power factor of 0.8?  (10)  

  Ans: The no load loss in transformers is practically equal to the iron loss.   Hence Iron loss = 1000W   Full load loss = 02 2 2 IR   Now, K = 500/5000 = 1/10   1 2 022 kR RR =+   = 0.06 + (1/10)2 x 8     = 0.14Ω   Full load current I2 = 50,000/500 = 100A   Full load  Cu loss = 1002 x 0.14 = 1400W   Total Loss = 1000 + 1400            = 2400W            = 2.4kW   Full load output at 0.8pf = 50 x 0.8 = 40kW   Efficiency η = 40/(40 + 2.4) = 0.9434 = 94.34%    Q.43  A squirrel-cage induction motor has a full-load slip of 4%.  Its starting current is 5 times its full load current.  Calculate the starting torque in pu of the full load torque.  Neglect the stator impedance and the magnetizing current.  Also give a suitable remarks for the answer obtained.   (8)                 Ans: Example 12.14 , p 467 of textbook         Q.47    A 4:1 transformer supplies a bridge rectifier that is driving a load of 200 ohms.  If the transformer input is 230 V/ 50 Hz supply, calculate the dc output voltage, PIV, and the output frequency.  Assume the rectifier diodes to be ideal. (4)       Ans: 

  4 V V N N 1 2 2 1 == and V1= 230 V  Therefore V2 = 57.5  RL=200Ω,  VSmax=57.5 

 Imax= 0.287A 200 57.5 R V 2RR V L Smax FL Smax == = + 

 Idc= 0.182A 2Imax = π  Vdc=Idc.RL = 0.182 X 200 = 36.4 Volts  PIV = VSmax = 57.5 

 Irms = 0.202 Imax = 2   Fundamental frequency of ripple = 2.f = 100 Hz   

   Q.53  In an N-type semi conductor, the Fermi-level lies 0.3 eV below the conduction band at 

27C o

.  If the temperature is increased to 55C o

, find the new position of the Fermi-level.       (8)   Ans:    At temperature T= 3000K = 273+ 27,  

  EC-EF =0.3eV  

  W.K.T    

  EC-EF = KT loge nc/ND 

  So, 0.3 = 300K loge nc/ND;     

  K loge nc/ND= 0.3/300=0.001   At temperature = ′T 328 K= (273+ 55),  

  Let the new position of the Fermi level be EF,  

  so EC-EF = KT loge nc/ND 

  EC-EF = 328 x 0.001 = 0.328V.  

Q.54  In a transistor circuit load resistance is Ω 5k and quiescent current is 1.2 mA.  Determine the operating point when the battery voltage 12V VCC = .  How will the Q-point change when the load resistance is changed from Ω 5k to Ω 7.5k ? (8)  

  Ans:   Zero signal collector current IC =1.2mA, load resistance in collector circuit, RL= 5KΩ collector supply voltage VCC = 12V   Zero signal collector- emitter voltage  

  VCE = VCC – ICRC    =12- (1.2x 10-3x5x103) = 6V 

  Hence the operating point is (6V, 1.2mA)   When load resistance is changed from Ω 5k to Ω 7.5k 

  Zero signal collector- emitter voltage, VCE = VCC – ICRC    =12 – (1.2x 10-3x7.5x103) = 3V   Here the operating point is (3V, 1.2mA)