Chapter 20: Transformers

Example 1, page no. 317

In [1]:
from __future__ import division
import math
#initializing  the  variables:
N1  =  500;#  primary  turns
N2  =  3000;#  secondary  turns
V1  =  240;#  in  Volts

#calculation:
 #For  an  ideal  transformer,  voltage  ratio  =  turns  ratio
V2  =  V1*N2/N1


#Results
print  "\n\n  Result  \n\n"
print  "\n  secondary  voltage  ",round(V2,2),"V"


  Result  



  secondary  voltage   1440.0 V

Example 2, page no. 317

In [2]:
from __future__ import division
import math
#initializing  the  variables:
tr  =  2/7;#  turns  ratio
V1  =  240;#  in  Volts

#calculation:
 #A  turns  ratio  of  2:7  means  that  the  transformer  has  2  turns  on  the  primary  
    #for  every  7  turns  on  the  secondary
V2  =  V1/tr


#Results
print  "\n\n  Result  \n\n"
print  "\n  output  voltage  ",round(V2,2),"  V"


  Result  



  output  voltage   840.0   V

Example 3, page no. 317

In [3]:
from __future__ import division
import math
#initializing  the  variables:
tr  =  8/1;#  turns  ratio
I1  =  3;#  in  Amperes
V1  =  240;#  in  Volts

#calculation:
 #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  
    #primary  for  every  1turns  on  the  secondary
V2  =  V1/tr
 #secondary  current
I2  =  I1*tr


#Results
print  "\n\n  Result  \n\n"
print  "\n  secondary  voltage  is  ",round(V2,2),"  V  and  secondary  current  is  ", round(I2,2),"  A"


  Result  



  secondary  voltage  is   30.0   V  and  secondary  current  is   24.0   A

Example 4, page no. 318

In [4]:
from __future__ import division
import math
#initializing  the  variables:
V1  =  240;#  in  Volts
V2  =  12;#  in  Volts
P  =  150;#  in  Watts

#calculation:
I2  =  P/V2
 #A  turns  ratio  =  Vp/Vs
tr  =  V1/V2#  turn  ratio
 #    V1/V2  =  I2/I1
 #current  taken  from  the  supply
I1  =  I2*V2/V1


#Results
print  "\n\n  Result  \n\n"
print  "\n  turn  ratio  is  ",round(tr,2),"  and  current  taken  from  the  supply  is  ",round(I1,2),"  A"


  Result  



  turn  ratio  is   20.0   and  current  taken  from  the  supply  is   0.63   A

Example 5, page no. 318

In [4]:
from __future__ import division
import math
#initializing  the  variables:
S  =  5000;#  in  VA
tr  =  10;#  turn  ratio
V1  =  2500;#  in  Volts

#calculation:
 #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  primary  for  every  1turns  on  the  secondary
V2  =  V1/tr
 #transformer  rating  in  volt-amperes  =  Vs*Is
I2  =  S/V2
 #Minimum  value  of  load  resistance
RL  =  V2/I2
 #    tr  =  I2/I1
I1  =  I2/tr


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)full-load  secondary  current  is  ",round(I2,2),"  A"
print  "\n  (b)minimum  load  resistance  is  ",round(RL,2),"  ohm"
print  "\n  (c)  primary  current  is  ",round(I1,2),"  A"


  Result  



  (a)full-load  secondary  current  is   20.0   A

  (b)minimum  load  resistance  is   12.5   ohm

  (c)  primary  current  is   2.0   A

Example 6, page no. 319

In [7]:
from __future__ import division
import math
#initializing  the  variables:
V1  =  2400;#  in  Volts
V2  =  400;#  in  Volts
I0  =  0.5;#  in  Amperes
Pc  =  400;#  in  Watts

#calculation:
 #Core  loss  (i.e.  iron  loss)  P  =  V1*I0*cos(phi0)
pf  =  Pc/(V1*I0)
phi0  =  math.acos(pf)
 #Magnetizing  component
Im  =  I0*math.sin(phi0)
 #Core  loss  component
Ic  =  I0*math.cos(phi0)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)magnetizing  component  is  ",round(Im,3),"  A  and  Core  loss  component  is  ",round(Ic,3),"  A"


  Result  



  (a)magnetizing  component  is   0.471   A  and  Core  loss  component  is   0.167   A

Example 7, page no. 320

In [8]:
from __future__ import division
import math
#initializing  the  variables:
V  =  240;#  in  Volts
I0  =  0.8;#  in  Amperes
P  =  72;#  in  Watts
f  =  50;#  in  Hz

#calculation:
 #Power  absorbed  =  total  core  loss,  P  =  V*I0*cos(phi0)
 #Ic  =  I0*cos(phi0)
Ic  =  P/V
pf  =  Ic/I0
 #From  the  right-angled  triangle  in  Figure  20.2(b)  and  using
 #Pythagoras’  theorem,  
Im  =  (I0*I0  -  Ic*Ic)**0.5


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)  Core  loss  component  is  ",round(  Ic,2),"  A"
print  "\n  (b)  power  factor  is  ",round(  pf,2),""
print  "\n  (c)magnetizing  component  is  ",round(Im,2),"  A"


  Result  



  (a)  Core  loss  component  is   0.3   A

  (b)  power  factor  is   0.37 

  (c)magnetizing  component  is   0.74   A

Example 8, page no. 321

In [2]:
from __future__ import division
import math
#initializing  the  variables:
S  =  100000;#  in  VA
V1  =  4000;#  in  Volts
V2  =  200;#  in  Volts
N2  =  100;#  sec  turns
f  =  50;#  in  Hz

#calculation:
 #Transformer  rating  =  V1*I1  =  V2*I2
 #primary  current
I1  =  S/V1
 #secondary  current
I2  =  S/V2
 #primary  turns
N1  =  N2*V1/V2
 #maximum  flux
 #assuming  E2  =  V2
Phim  =  V2/(4.44*f*N2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)primary  current  is  ",round(  I1,2),"  A  and  secondary  current  is  ",round(  I2,2),"  A"
print  "\n  (b)number  of  primary  turns  is  ",round(  N1,2),""
print  "\n  (c)maximum  value  of  the  flux  is  ",round(Phim*1000,2),"mWb"


  Result  



  (a)primary  current  is   25.0   A  and  secondary  current  is   500.0   A

  (b)number  of  primary  turns  is   2000.0 

  (c)maximum  value  of  the  flux  is   9.01 mWb

Example 9, page no. 322

In [10]:
from __future__ import division
import math
#initializing  the  variables:
V1  =  250;#  in  Volts
A  =  0.03;#  in  m2
N2  =  300;#  sec  turns
N1  =  25;#  prim  turns
f  =  50;#  in  Hz

#calculation:
 #e.m.f.  E1  =  4.44*f*Phim*N1
 #maximum  flux  density,
Phim  =  V1/(4.44*f*N1)
 #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core
 #maximum  core  flux  density
Bm  =  Phim/A
 #voltage  induced  in  the  secondary  winding,
V2  =  V1*N2/N1


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)maximum  core  flux  density  ",round(  Bm,2),"  T"
print  "\n  (b)voltage  induced  in  the  secondary  winding  is  ",round(  V2,2),"  V"


  Result  



  (a)maximum  core  flux  density   1.5   T

  (b)voltage  induced  in  the  secondary  winding  is   3000.0   V

Example 10, page no. 323

In [10]:
from __future__ import division
import math
#initializing  the  variables:
V1  =  500;#  in  Volts
V2  =  100;#  in  Volts
Bm  =  1.5;#  in  Tesla
A  =  0.005;#  in  m2
f  =  50;#  in  Hz

#calculation:
 #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core
 #maximum  core  flux  density
Phim  =  Bm*A
 #e.m.f.  E1  =  4.44*f*Phim*N1
 #primary  turns,
N1  =  V1/(4.44*f*Phim)
 #secondary  turns,
N2  =  V2*N1/V1


#Results
print  "\n\n  Result  \n\n"
print  "\n  no.  of  primary  and  secondary  turns  are  ",round(N1,2),"  turns,  and  ",round(N2,2),"  turns  respectively"


  Result  



  no.  of  primary  and  secondary  turns  are   300.3   turns,  and   60.06   turns  respectively

Example 11, page no. 323

In [11]:
from __future__ import division
import math
#initializing  the  variables:
emfpt  =  15;#  in  Volts
V1  =  4500;#  in  Volts
V2  =  225;#  in  Volts
Bm  =  1.4;#  in  Tesla
f  =  50;#  in  Hz

#calculation:
 #E.m.f.  per  turn,  V1/N1  =  V2/N2  =  emfpt
 #primary  turns,
N1  =  V1/emfpt
 #secondary  turns,
N2  =  V2/emfpt
 #e.m.f.  E1  =  4.44*f*Phim*N1
 #maximum  flux  density,
Phim  =  V1/(4.44*f*N1)
 #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core
 #cross-sectional  area
A  =  Phim/Bm


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)no.  of  primary  and  secondary  turns  are  ",  N1,"  turns,  and  ",  N2,"  turns  respectively"
print  "\n  (b)cross-sectional  area  is  ",  round(A,4),"m2"


  Result  



  (a)no.  of  primary  and  secondary  turns  are   300.0   turns,  and   15.0   turns  respectively

  (b)cross-sectional  area  is   0.0483 m2

Example 12, page no. 324

In [12]:
from __future__ import division
import math
#initializing  the  variables:
N1  =  2000;#  prim  turns
N2  =  800;#  sec  turns
I0  =  5;#  in  Amperes
pf0  =  0.20;#  power  factor
I2  =  100;#  in  Amperes
pf2  =  0.85;#  power  factor

#calculation:
 #Let  I01  be  the  component  of  the  primary  current  which  provides  the  restoring  mmf.  Then  I01*N1  =  I2*N2
I01  =  I2*N2/N1
 #If  the  power  factor  of  the  secondary  is  0.85
phi2  =  math.acos(pf2)
 #If  the  power  factor  on  no-load  is  0.20,
phi0  =  math.acos(pf0)
I1h  =  I0*math.cos(phi0)  +  I01*math.cos(phi2)
I1v  =  I0*math.sin(phi0)  +  I01*math.sin(phi2)
 #Hence  the  magnitude  of  I1
I1  =  (I1h*I1h  +  I1v*I1v)**0.5
pf1  =  math.cos(math.atan(I1v/I1h))


#Results
print  "\n\n  Result  \n\n"
print  "\n  Primary  current  is  ", round(I1,2),"  A,  and  Power  factor  is  ",round(pf1,2)


  Result  



  Primary  current  is   43.58   A,  and  Power  factor  is   0.8

Example 13, page no. 328

In [13]:
from __future__ import division
import math
#initializing  the  variables:
N1  =  600;#  prim  turns
N2  =  150;#  sec  turns
R1  =  0.25;#  in  ohms
R2  =  0.01;#  in  ohms
X1  =  1.0;#  in  ohms
X2  =  0.04;#  in  ohms

#calculation:
tr  =  N1/N2#  turn  ratio
vr  =  tr#  voltage  ratio  =  turn  raio,  vr  =  V1/V2
 #equivalent  resistance  Re
Re  =  R1  +  R2*(vr**2)
 #equivalent  reactance,  Xe
Xe  =  X1  +  X2*(vr**2)
 #equivalent  impedance,  Ze
Ze  =  (Re*Re  +  Xe*Xe)**0.5
 #cos(phie)  =  Re/Ze
pfe  =  Re/Ze
phie  =  math.acos(pfe)
phied  =  phie*180/math.pi#  in  °(deg)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  equivalent  resistance  referred  to  the  primary  winding  is  ",round(  Re,2),"  ohm"
print  "\n  (b)the  equivalent  reactance  referred  to  the  primary  winding  is  ",round(  Xe,2),"  ohm"
print  "\n  (c)the  equivalent  impedance  referred  to  the  primary  winding  is  ",round(  Ze,2),"  ohm"
print  "\n  (d)phase  angle  is  ",round(  phied,2),"deg"


  Result  



  (a)the  equivalent  resistance  referred  to  the  primary  winding  is   0.41   ohm

  (b)the  equivalent  reactance  referred  to  the  primary  winding  is   1.64   ohm

  (c)the  equivalent  impedance  referred  to  the  primary  winding  is   1.69   ohm

  (d)phase  angle  is   75.96 deg

Example 14, page no. 329

In [14]:
from __future__ import division
import math
#initializing  the  variables:
V1  =  200;#  in  Volts
V2  =  400;#  in  Volts
V2L  =  387.6;#  in  Volts
S  =  5000;#  in  VA

#calculation:
 #regulation  =(No-load  secondary  voltage  -  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %
reg  =  (V2  -  V2L)*100/V2


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  regulation  of  the  transformer  is  ",round(reg,2),"  percent  "


  Result  



  the  regulation  of  the  transformer  is   3.1   percent

Example 15, page no. 329

In [15]:
from __future__ import division
import math
#initializing  the  variables:
VnL  =  240;#  in  Volts
reg  =  2.5;#  in  percent

#calculation:
 #regulation  =(No-load  secondary  voltage  -  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %
VL  =  VnL  -  reg*VnL/100


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  load  voltage  at  which  the  mechanism  operates  is  ",round(VL,2),"  V  "


  Result  



  the  load  voltage  at  which  the  mechanism  operates  is   234.0   V

Example 16, page no. 331

In [16]:
from __future__ import division
import math
#initializing  the  variables:
S  =  200000;#  in  VA
Pc  =  1500;#  in  Watt
Pi  =  1000;#  in  Watt
pf  =  0.85;#  power  factor

#calculation:
 #Efficiency  =  output  power/input  power  =  (input  power—losses)/input  power
 #Efficiency  =  1  -  losses/input  power
 #Full-load  output  power  =  V*I*pf
Po  =  S*pf
 #Total  losses
Pl  =  Pc  +  Pi
 #Input  power  =  output  power  +  losses
PI  =  Po  +  Pl
 #efficiency
eff  =  1-(Pl/PI)


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  transformer  efficiency  at  full  load  is  ",round(eff,2)


  Result  



  the  transformer  efficiency  at  full  load  is   0.99

Example 17, page no. 331

In [17]:
from __future__ import division
import math
#initializing  the  variables:
S  =  200000;#  in  VA
Pc  =  1500;#  in  Watt
Pi  =  1000;#  in  Watt
pf  =  0.85;#  power  factor

#calculation:
 #Efficiency  =  output  power/input  power  =  (input  power—losses)/input  power
 #Efficiency  =  1  -  losses/input  power
 #Half  full-load  power  output  =  V*I*pf/2
Po  =  S*pf/2
 #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared
 #Hence  the  copper  loss  at  half  full-load  is
Pch  =  Pc/(2*2)
 #Iron  loss  =  1000  W  (constant)
 #Total  losses
Pl  =  Pch  +  Pi
 #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses
PI  =  Po  +  Pl
 #efficiency
eff  =  (1-(Pl/PI))*100


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  transformer  efficiency  at  half  full  load  is  ",round(eff,2),"  percent"


  Result  



  the  transformer  efficiency  at  half  full  load  is   98.41   percent

Example 18, page no. 332

In [18]:
from __future__ import division
import math
#initializing  the  variables:
S  =  400000;#  in  VA
R1  =  0.5;#  in  Ohm
R2  =  0.001;#  in  Ohm
V1  =  5000;#  in  Volts
V2  =  320;#  in  Volts
Pi  =  2500;#  in  Watt
pf  =  0.85;#  power  factor

#calculation:
 #Rating  =  400  kVA  =  V1*I1  =  V2*I2
 #Hence  primary  current
I1  =  S/V1
 #secondary  current
I2  =  S/V2
 #Total  copper  loss  =  I1*I1*R1  +  I2*I2*R2,
Pcf  =  I1*I1*R1  +  I2*I2*R2
 #On  full  load,  total  loss  =  copper  loss  +  iron  loss
Plf  =  Pcf  +  Pi
 #  full-load  power  output  =  V2*I2*pf
Pof  =  S*pf
 #Input  power  at  full-load  =  output  power  at  full-load  +  losses
PIf  =  Pof  +  Plf
 #Efficiency  =  output  power/input  power  =  (input  power—losses)/input  power
 #Efficiency  =  1  -  losses/input  power
efff  =  (1-(Plf/PIf))*100

 #Half  full-load  power  output  =  V*I*pf/2
Poh  =  S*pf/2
 #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared
 #Hence  the  copper  loss  at  half  full-load  is
Pch  =  Pcf/(2*2)
 #Iron  loss  =  2500  W  (constant)
 #Total  losses
Plh  =  Pch  +  Pi
 #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses
PIh  =  Poh  +  Plh
 #efficiency
effh  =  (1-(Plh/PIh))*100


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  transformer  efficiency  at  full  load  is  ", round(efff,2),"  percent"
print  "\n  (b)the  transformer  efficiency  at  half  full  load  is  ",  round(effh,2),"  percent"


  Result  



  (a)the  transformer  efficiency  at  full  load  is   97.91   percent

  (b)the  transformer  efficiency  at  half  full  load  is   97.88   percent

Example 19, page no. 333

In [1]:
from __future__ import division
import math
#initializing  the  variables:
S  =  500000;#  in  VA
Pcf  =  4000;#  in  Watt
Pi  =  2500;#  in  Watt
pf  =  0.75;#  power  factor

#calculation:
 #Let  x  be  the  fraction  of  full  load  kVA  at  which  the  efficiency  is  a  maximum.
 #The  corresponding  total  copper  loss  =  (4  kW)*(x**2)
 #At  maximum  efficiency,  copper  loss  =  iron  loss  Hence
x  =  (Pi/Pcf)**0.5
 #Hence  the  output  kVA  at  maximum  efficiency
So  =  x*S
 #Total  loss  at  maximum  efficiency
Pl  =  2*Pi
 #Output  power
Po  =  So*pf
 #Input  power  =  output  power  +  losses
PI  =  Po  +  Pl
 #Efficiency  =  output  power/input  power  =  (input  power—losses)/input  power
 #Efficiency  =  1  -  losses/input  power
 #Maximum  efficiency
effm  =  (1  -  Pl/PI)*100


#Results
print  "\n\n  Result  \n\n"
print  "\n the output kVA at maximum efficiency is  ",round(So/1000,2),"kVA"
print  "\n  max.  efficiency  is  ",round(effm,2),"  pecent"


  Result  



 the output kVA at maximum efficiency is   395.28 kVA

  max.  efficiency  is   98.34   pecent

Example 20, page no. 335

In [20]:
from __future__ import division
import math
#initializing  the  variables:
tr  =  4;#  turn  ratio
RL  =  100;#  in  Ohms

#calculation:
 #the  equivalent  input  resistance,
Ri  =  RL*(tr**2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  equivalent  input  resistance  is  ",round(Ri,2),"  ohm"


  Result  



  the  equivalent  input  resistance  is   1600.0   ohm

Example 21, page no. 335

In [3]:
from __future__ import division
import math
#initializing  the  variables:
R1  =  112;#  in  Ohms
RL  =  7;#  in  Ohms

#calculation:
 #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  112  ohm  for  maximum  power  transfer.
 #R1  =  RL*(tr**2)
 #  tr  =  N1/N2  turn  ratio
tr  =  (R1/RL)**0.5


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  optimum  turns  ratio  is  ",tr,": 1.0"


  Result  



  the  optimum  turns  ratio  is   4.0 : 1.0

Example 22, page no. 335

In [22]:
from __future__ import division
import math
#initializing  the  variables:
tr  =  5;#  turn  ratio
R1  =  150;#  in  Ohms

#calculation:
 #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  150  ohm  for  maximum  power  transfer.
 #R1  =  RL*(tr**2)
RL  =  R1/(tr**2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  optimum  value  of  load  resistance  is  ",round(RL,2),"  ohm"


  Result  



  the  optimum  value  of  load  resistance  is   6.0   ohm

Example 23, page no. 335

In [23]:
from __future__ import division
import math
#initializing  the  variables:
V1  =  220;#  in  Volts
V2  =  1760;#  in  Volts
V  =  220;#  in  Volts
RL  =  1280;#  in  Ohms
R  =  2;#  in  Ohms

#calculation:
 #Turns  ratio,  tr  =  N1/N2  =  V1/V2
tr  =  V1/V2
 #Equivalent  input  resistance  of  the  transformer,
 #R1  =  RL*(tr**2)
R1  =  RL*(tr**2)
 #Total  input  resistance
Rin  =  R  +  R1
 #  Primary  current
I1  =  V1/Rin
 #For  an  ideal  transformer  V1/V2  =  I2/I1,
I2  =  I1*tr
 #Power  dissipated  in  load  resistor  RL
P  =  I2*I2*RL


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)  primary  current  flowing  is  ",round(I1,2),"  A"
print  "\n  (b)    power  dissipated  in  the  load  resistor  is  ",round(P,2),"  W"


  Result  



  (a)  primary  current  flowing  is   10.0   A

  (b)    power  dissipated  in  the  load  resistor  is   2000.0   W

Example 24, page no. 336

In [4]:
from __future__ import division
import math
#initializing  the  variables:
tr  =  25;#  teurn  ratio
V  =  24;#  in  Volts
R1  =  15000;#  in  Ohms
Rin  =  15000;#  in  ohms

#calculation:
 #Turns  ratio,  tr  =  N1/N2  =  V1/V2
 #For  maximum  power  transfer  R1  needs  to  be  equal  to  15  kohm
RL  =  R1/(tr**2)
 #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is
Rt  =  Rin  +  R1
 #Primary  current,
I1  =  V/Rt
 #N1/N2  =  I2/I1
I2  =  I1*tr
 #Power  dissipated  in  load  resistor  RL
P  =  I2*I2*RL


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)  the  load  resistance  is  ",round(RL,2),"ohm"
print  "\n  (b)    power  dissipated  in  the  load  resistor  is  ",round(P*1000,2),"mW"


  Result  



  (a)  the  load  resistance  is   24.0 ohm

  (b)    power  dissipated  in  the  load  resistor  is   9.6 mW

Example 25, page no. 337

In [25]:
from __future__ import division
import math
#initializing  the  variables:
V1  =  320;#  in  Volts
V2  =  250;#  in  Volts
S  =  20000;#  in  VA

#calculation:
 #Rating  =  20  kVA  =  V1*I1  =  V2*I2
 #Hence  primary  current,  I1
I1  =  S/V1
 #secondary  current,  I2
I2  =  S/V2
 #Hence  current  in  common  part  of  the  winding
I  =  I2  -  I1


#Results
print  "\n\n  Result  \n\n"
print  "\n  current  in  common  part  of  the  winding  is  ", round(I,2),"  A"
print  "\n  primary  current  and  secondary  current  are  ",round(I1,2),"  A  and  ",round(I2,2),"  A  respectively"


  Result  



  current  in  common  part  of  the  winding  is   17.5   A

  primary  current  and  secondary  current  are   62.5   A  and   80.0   A  respectively

Example 26, page no. 339

In [26]:
from __future__ import division
import math
#initializing  the  variables:
V1a  =  200;#  in  Volts
V2a  =  150;#  in  Volts
V1b  =  500;#  in  Volts
V2b  =  100;#  in  Volts

#calculation:
 #For  a  200  V:150  V  transformer,  xa
xa  =  V2a/V1a
 #volume  of  copper  in  auto  transformer
vca  =  (1  -  xa)*100#  of  copper  in  a  double-wound  transformer
 #the  saving  is
vsa  =  100  -  vca
 #For  a  500  V:100  V  transformer,  xb
xb  =  V2b/V1b
 #volume  of  copper  in  auto  transformer
vcb  =  (1  -  xb)*100#  of  copper  in  a  double-wound  transformer
 #the  saving  is
vsb  =  100  -  vcb


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)For  a  200  V:150  V  transformer,  the  saving  is  ", round(vsa,2),"  percent"
print  "\n  (b)For  a  500  V:100  V  transformer,  the  saving  is  ", round(vsb,2),"  percent"


  Result  



  (a)For  a  200  V:150  V  transformer,  the  saving  is   75.0   percent

  (b)For  a  500  V:100  V  transformer,  the  saving  is   20.0   percent

Example 27, page no. 340

In [1]:
from __future__ import division
import math
#initializing  the  variables:
N1  =  500;#  prim  turns
N2  =  50;#  sec  turns
VL  =  2400;#  in  Volts

#calculation:
 #For  a  star-connection,  VL  =  Vp*(3**0.5)
VL1s  =  VL
 #Primary  phase  voltage
Vp1s  =  VL1s/(3**0.5)
 #For  a  delta-connection,  VL  =  Vp
 #N1/N2  =  V1/V2,  from  which,
 #secondary  phase  voltage,  Vp2s
Vp2s  =  Vp1s*N2/N1
VL2d  =  Vp2s

 #For  a  delta-connection,  VL  =  Vp
VL1d  =  VL
 #primary  phase  voltage  Vp1d
Vp1d  =  VL1d
 #Secondary  phase  voltage,  Vp2d
Vp2d  =  Vp1d*N2/N1
 #For  a  star-connection,  VL  =  Vp*(3**0.5)
VL2s  =  Vp2d*(3**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  secondary  line  voltage  for  star  and  delta  connection  are  ",round(Vp2s,1),"  V "
print   " and  ",round(VL2s,0),"  V  respectively"


  Result  



  the  secondary  line  voltage  for  star  and  delta  connection  are   138.6   V 
 and   416.0   V  respectively

Example 28, page no. 343

In [28]:
from __future__ import division
import math
#initializing  the  variables:
N1  =  1;#  prim  turns
N2  =  60;#  sec  turns
I1  =  300;#  in  amperes
Ra  =  0.15;#  in  ohms
R2  =  0.25;#  in  ohms

#calculation:
 #Reading  on  the  ammeter,
I2  =  I1*(N1/N2)
 #P.d.  across  the  ammeter  =  I2*RA,  where  RA  is  the  ammeter  resistance
pd  =  I2*Ra
 #Total  resistance  of  secondary  circuit
Rt  =  Ra  +  R2
 #Induced  e.m.f.  in  secondary
V2  =  I2*Rt
 #Total  load  on  secondary
S  =  V2*I2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  reading  on  the  ammeter  is  ",round(I2,2),"  A  "
print  "\n  (b)potential  difference  across  the  ammeter  is  ",round(pd,2),"  V  "
print  "\n  (c)total  load  (in  VA)  on  the  secondary  is  ",round(S,2),"  VA  "


  Result  



  (a)the  reading  on  the  ammeter  is   5.0   A  

  (b)potential  difference  across  the  ammeter  is   0.75   V  

  (c)total  load  (in  VA)  on  the  secondary  is   10.0   VA