In [1]:

```
# Exa 9.1
#pg 305
#calculate the number of turns, load current
# Given data
V1 = 3000.;# in V
V2 = 300.;# in V
N2 = 86.;# in Turns
Rating = 60.*10**3;# in VA
#calculations
K = V2/V1;
#Transformer ratio, N2/N1 = K;
N1 = N2/K;# in turns
I2 = Rating/V2;# in A
I1 = Rating/V1;# in A
#results
print "The numbers of primary turns is",N1
print "The secondary full load current in A is",I2
print "The primary full load current in A is",I1
```

In [2]:

```
# Exa 9.2
#pg 306
#calculate the max flux density
# Given data
E1 = 3000.;# in V
E2 = 200.;# in V
f = 50.;# in Hz
a = 150.;# in cm**2
N2 = 80.;# turns
#calculations
#Formula E2 = 4.44*phi_m*f*N2;
phi_m = E2/(4.44*f*N2);# in Wb
Bm = phi_m/(a*10**-4);# in Wb/m**2
#results
print "The maximum flux density in Wb/m^2 is",round(Bm,2)
```

In [3]:

```
# Exa 9.3
#pg 306
#calculate the load current, emf and core flux
# Given data
N1 = 500.;
N2 = 40.;
E1 = 3000.;# in V
f = 50.;# in Hz
Rating = 25*10**3;# in VA
#calculations
K = N2/N1;
I1 = Rating/E1;# in A
print "The primary full load current in A is",round(I1,2)
I2 = I1/K;# in A
print "The secondary full load current in A is",round(I2,1)
# K = E2/E1;
E2 = K*E1;# in V
print "The secondary emf in V is",round(E2)
# e.m.f equation of the transformer, E1 = 4.44*phi_m*f*N1;
phi_m = E1/(4.44*f*N1);# in Wb
phi_m = phi_m*10**3;# in mWb
print "The maximum core flux in mWb is",round(phi_m)
```

In [4]:

```
# Exa 9.4
#pg 307
#calculate the current
# Given data
from math import acos,cos,sin
Rating = 25.;# in KVA
f = 50.;# in Hz
Io = 15.;# in A
Wo = 350.;# in W
Vo = 230.;# in V
#calculations
# No load power factor
phi_o = acos(Wo/(Vo*Io));
# active component of current
Ic = Io*cos(phi_o);# in A
print "The active component of current in A is",round(Ic,3)
# magnetizing component of current
Im = Io*sin(phi_o);# in A
print "The magnetizing component of current in A is",round(Im,3)
```

In [5]:

```
# Exa 9.5
#pg 307
#calculate the resistance and reactance
# Given data
V1 = 2200.;# in V
V2 = 110.;# in V
R1 = 1.75;# in ohm
R2 = 0.0045;# in ohm
X1 = 2.6;# in ohm
X2 = 0.0075;# in ohm
#calculations
K = V2/V1;
#R1e = R1+R_2 = R1 + (R2/(K**2));
R1e = R1 + (R2/(K**2));# in ohm
print "Equivalent resistance reffered to primary in ohm is",R1e
# R2e = R2+R_1 = R2+((K**2)*R1);
R2e = R2+((K**2)*R1);# in ohm
print "Equivalent resistance reffered to secondary in ohm is",R2e
#X1e = X1+X_2 = X1+(X2/(K**2));
X1e = X1+(X2/(K**2));# in ohm
print "Equivalent reactance reffered to primary in ohm is",X1e
# X2e = X2+X_1 = X2 + ((K**2)*X1);
X2e = X2 + ((K**2)*X1);# in ohm
print "Equivalent reactance reffered to secondary in ohm is",X2e
Z1e= R1e+1j*X1e;# in ohm
Z2e= R2e+1j*X2e;# in ohm
print "Equivalent impedance reffered to primary in ohm is : ",round(abs(Z1e),4)
print "Equivalent impedance reffered to secondary in ohm is : ",round(abs(Z2e),5)
```

In [6]:

```
# Exa 9.6
#pg 308
#calculate the impedance and copper loss
from math import sqrt
# Given data
V1 = 2200.;# in V
V2 = 440.;# in V
R1 = 0.3;# in ohm
R2 = 0.01;# in ohm
X1 = 1.1;# in ohm
X2 = 0.035;# in ohm
#calculations
K = V2/V1;
Rating = 100;# in KVA
I1 = (Rating*10**3)/V1;# in A
I2 = (Rating*10**3)/V2;# in A
R1e = R1 + (R2/(K**2));# in ohm
X1e = X1+(X2/(K**2));# in ohm
Z1e = sqrt( (R1e**2) + (X1e**2) );# in ohm
# Total copper loss
totalcopperloss = (I1**2)*R1e;# in W
#results
print "The equivalent impedance of the transformer reffered to primary in ohm is",round(Z1e,2)
print "The total copper loss in W is",round(totalcopperloss,2)
```

In [7]:

```
# Exa 9.7
#pg 309
#calculate the efficiency
from math import cos,acos
# Given data
Rating = 150000.;# in VA
phi= acos(0.8);# in radians
Pcu = 1600.;# in W
Pi = 1400.;# in W
n = 1/4.;
#calculations
# Total loss of 25% load
totalloss = Pi + (n**2)*Pcu;# in W
# efficiency of transformer of 25% load
Eta = n*Rating*cos(phi)/(n*Rating*cos(phi)+Pi+n**2*Pcu)*100;# in %
#results
print "The efficiency in percent is",round(Eta,2)
```

In [8]:

```
# Exa 9.8
#pg 309
#calculate the efficiency
# Given data
from math import acos
Rating = 25.;# in KVA
V1 = 2000.;# in V
V2 = 200.;# in V
Pi = 350.;# in W
Pi = Pi * 10**-3;# in kW
Pcu = 400.;# in W
Pcu = Pcu * 10**-3;# in kW
#calculations
phi= acos(1);# in radians
output = Rating;
losses = Pi+Pcu;
Eta = (output/(output + losses))*100;# %Eta in %
print "The efficiency of full load power in percent is",round(Eta,2)
# For half load
output = Rating/2;# in kW
h = 1.;
Pcu = Pcu*((h/2)**2);# in kW
losses = Pi+Pcu;
# efficiency of half load power
Eta = (output/(output+losses))*100;# in %
print "The efficiency of half load power in percent is",round(Eta,2)
```

In [9]:

```
# Exa 9.9
#pg 310
#calculate the efficiency
from math import acos,cos,sqrt
# Given data
Rating = 250*10**3;# in VA
Pi = 1.8;# in kW
Pi = Pi * 10**3;# in W
Pcu_f1 = 2000;# in W
phi= acos(0.8);# in radians
Eta = ((Rating*cos(phi))/((Rating*cos(phi))+Pi+Pcu_f1))*100;# %Eta in %
print "The efficiency at full load in percent is",round(Eta,2)
# The maximum efficiency
Eta_max = Rating * sqrt(Pi/Pcu_f1 );# in VA
Eta_max = Eta_max *10**-3;# in kVA
print "The maximum efficiency in kVA is",round(Eta_max,2)
Eta_max = Eta_max *10**3;# in VA
Pcu = Pi;# in W
Eta_max1 = ((Eta_max*cos(phi))/((Eta_max*cos(phi)) + Pi+Pcu ))*100;# in %
print "The maximum efficiency in percent is",round(Eta_max1,3)
```

In [10]:

```
# Exa 9.10
#pg 311
#calculate the iron loss, full load copper loss
# Given data
from math import acos,cos
phi= acos(1);# in radians
Pout = 500.;# in kW
Pout = Pout*10**3;# in W
Eta = 90.;# in %
n=1/2.;
#calculations
# For full load, Eta= Pout*100/(Pout+Pi+Pcu_f1) or Pi+Pcu_f1= (Pout*100-Eta*Pout)/Eta (i)
# For half load, Eta= n*Pout*100/(n*Pout+Pi+n**2*Pcu_f1) or Pi+n**2*Pcu_f1= (n*Pout*100-n*Eta*Pout)/Eta (ii)
# From eq(i) and (ii)
Pcu_fl= ((n*Pout*100-n*Eta*Pout)/Eta-(Pout*100-Eta*Pout)/Eta)/(n**2-1)
Pi=(Pout*100-Eta*Pout)/Eta-Pcu_fl
#results
print "The iron loss in W is : ",round(Pi,3)
print "The full load copper loss in watt",round(Pcu_fl,2)
```

In [11]:

```
# Exa 9.11
#pg 311
#calculate the flux, iron loass and load current
# Given data
from math import acos,sin,cos
Io = 10;# in A
phi_o= acos(0.25);# in radians
V1 = 400.;# in V
f = 50.;# in Hz
N1 =500.;
Im = Io*sin(phi_o);# in A
print "The magnetizing component of no load current in A is",round(Im,2)
Pi = V1*Io*cos(phi_o);# in W
print "The iron loss in W is",Pi
E1 = V1;# in V
#E1 v= 4.44*f*phi_m*N1;
phi_m = E1/(4.44*f*N1);# in Wb
phi_m=phi_m*10**3;# in mWb
print "The maximum value of flux in mWb is",round(phi_m,2)
```

In [12]:

```
# Exa 9.12
#pg 312
#calculate the equivalent resistance
# Given data
from math import sqrt
Rating = 30.*10**3;# in VA
V1 = 2000.;# in V
V2 = 200.;# in V
f = 50.;# in Hz
R1 = 3.5;# in ohm
X1 = 4.5;# in ohm
R2 = 0.015;# in ohm
X2 = 0.02;# in ohm
#calculations
K = V2/V1;
R1e = R1 + (R2/(K**2));# in ohm
print "The equivalent resistance to primary side in ohm is",R1e
X1e = X1 + (X2/(K**2));# in ohm
print "The equivalent reactance to primary side in ohm is",X1e
Z1e = sqrt( (R1e**2) + (X1e**2) );# in ohm
print "The equivalent impedance to primary side in ohm is",round(Z1e,1)
I1 = Rating/V1;# in A
# Total copper loss in transformer
Pcu_total = (I1**2)*R1e;# in W
print "Total copper loss in W is",Pcu_total
```

In [13]:

```
# Exa 9.13
#pg 313
#calculate the full load secondary voltage
from math import cos,sin,acos
# Given data
Rating = 10;# in KVA
phi= acos(0.8)
V1 = 2000.;# in V
V2 = 400.;# in V
R1 = 5.5;# in ohm
X1 = 12;# in ohm
R2 = 0.2;# in ohm
X2 = 0.45;# in ohm
K = V2/V1;
#R1e = R1 + R_2 = R1 + (R2/(K**2));
R1e = R1 + (R2/(K**2));# in ohm
#X1e = X1 + X_ = X1 + (X2/(K**2));
X1e = X1 + (X2/(K**2));# in ohm
I2 = (Rating*10**3)/V2;# in A
R2e = (K**2)*R1e;# in ohm
X2e = (K**2)*X1e;# in ohm
Vdrop = I2 * ( (R2e*cos(phi)) + (X2e*sin(phi)) );# voltage drop in V
#E2 = V2 +Vd;
E2 = V2;# in V
# The full load secondary voltage
V2 = E2-Vdrop;# in V
#results
print "The full load secondary voltage in V is",V2
```

In [14]:

```
# Exa 9.14
#pg 313
#calculate the efficiency
from math import cos,acos,sqrt
# Given data
Rating = 40*10**3;# in VA
Pi = 400.;# in W
Pcu_f1 = 800.;# in W
#calculations
phi= acos(0.9);# in radians
Eta_f1 = ((Rating*cos(phi))/( (Rating*cos(phi)) + Pi + Pcu_f1 ))*100;# in %
print "Full load efficiency in percent is",round(Eta_f1,2)
# percentage of the full load
Eta_max = Rating*sqrt( Pi/Pcu_f1);# in KVA
Eta_max = Eta_max/Rating*100;# in %
print "The percentage of the full load in percent is",round(Eta_max,2)
```

In [15]:

```
#pg 314
# Exa 9.15
#calculate the full load efficiency
from math import cos,acos
# Given data
Rating = 8*10**3;# in VA
phi= acos(0.8);# in radians
V1 = 400.;# in V
V2 = 100.;# in V
f = 50.;# in Hz
Pi = 60.;# in W
Wo = Pi;# in W
Pcu = 100.;# in W
#results
# The full load efficiency
Eta_f1 = ((Rating*cos(phi))/((Rating*cos(phi)) + Pi + Pcu))*100;# in %
#results
print "The full load efficiency in percent is",round(Eta_f1,2)
```

In [16]:

```
#pg 314
# Exa 9.16
# Given data
from math import acos,cos
Rating = 10*10**3;# in VA
phi= acos(0.8);# in radians
V1 = 500.;# in V
V2 = 250.;# in V
Pi = 200.;# in W
Pcu = 300.;# in W
Isc = 30.;# in A
#calculations
I1 = Rating/V1;# in A
# Pcu/(Pcu(f1)) = (Isc**2)/(I1**2);
Pcu_f1 = Pcu * ((I1**2)/(Isc**2));# in W
# The efficiency at full load
Eta_f1 = Rating*cos(phi)/(Rating*cos(phi) + Pi + Pcu_f1)*100;# in %
#results
print "The full load efficiency in percent is",Eta_f1
```

In [17]:

```
#pg 315
# Exa 9.17
#calculate the max efficiency
# Given data
from math import acos,cos,sqrt
Rating = 20*10**3;# in VA
phi= acos(0.8);# in radians
V1 = 2000.;# in V
V2 = 200.;# in V
Pi = 120.;# in W
Pcu = 300.;# in W
#calculations
Eta_max = Rating*(sqrt( Pi/Pcu ));# in VA
Pcu = Pi;# in W
# The maximum efficiency of transformer
Eta_max = ((Eta_max*cos(phi))/( Eta_max*cos(phi) + (2*Pi) ))*100;# in %
#results
print "The maximum efficiency of transformer in percent is",round(Eta_max,3)
```

In [18]:

```
#pg 315
# Exa 9.18
#calculate the resistances
# Given data
Turnratio = 5;
R1 = 0.5;# in ohm
R2 = 0.021;# in ohm
X1 = 3.2;# in ohm
X2 = 0.12;# in ohm
Rc = 350.;# in ohm
Xm = 98.;# in ohm
N1 = 5.;
N2 = 1.;
#calculations
K = N2/N1;
# Evaluation of the equivalent parameters referred to secondary side
R2e = R2 + ((K**2)*R1);# in ohm
print "The equivalent parameters referred to secondary side are : "
print "The value of R_2e is : ",R2e," ohm"
X2e = X2 + ((K**2)*X1);# in ohm
print "The value of X_2e is : ",X2e," ohm"
R_c = (K**2)*Rc;# in ohm
print "The value of R''c is : ",R_c," ohm"
X_m = (K**2)*Xm;# in ohm
print "The value of X''m is : ",X_m," ohm"
```

In [19]:

```
#pg 315
# Exa 9.19
#calculate the resistances
# Given data
from math import acos,cos,sin,sqrt
Rating = 100.*10**3;# in VA
V1 = 11000.;# in V
V2 = 220.;# in V
Wo = 2*10**3;# in W
Vo = 220.;# in V
Io = 45.;# in A
#calculations
phi_o = acos(Wo/(Vo*Io));
I_c = Io*cos(phi_o);# in A
I_m = Io*sin(phi_o);# in A
Ro= V2/I_c;# in ohm
Xo= V2/I_m;# in ohm
Wsc= 3*10**3;# in W
Vsc= 500.;# in V
Isc= 9.09;# in A
R1e= Wsc/Isc**2;# in ohm
Z1e= Vsc/Isc;# in ohm
X1e= sqrt(Z1e**2-R1e**2);# in ohm
K= V2/V1;
R2e= K**2*R1e;# in ohm
X2e= K**2*X1e;# in ohm
Z2e= K**2*Z1e;# in ohm
#results
print "The value of R''o is : ",Ro," ohm"
print "The value of X''o is : ",round(Xo)," ohm"
print "The value of R1e is : ",round(R1e,4)," ohm"
print "The value of Z1e is : ",round(Z1e,4)," ohm"
print "The value of X1e is : ",round(X1e,4)," ohm"
print "The value of R2e is : ",round(R2e,4)," ohm"
print "The value of X2e is : ",round(X2e,4)," ohm"
print "The value of Z2e is : ",round(Z2e,4)," ohm"
```

In [20]:

```
#pg 316
# Exa 9.20
#calculate the efficiency
# Given data
from math import acos,cos
V1 = 250.;# in V
V2 = 500.;# in V
Pcu = 100.;# in W
Pi = 80.;# in W
V = V2;# in V
A = 12.;# in A
#calculations
phi= acos(0.85);# in radians
# The efficiency of the transformer
Eta = ((V*A*cos(phi))/( V*A*cos(phi) + Pi+Pcu ))*100;# in %
print "The efficiency of the transformer in percent is",round(Eta,2)
```

In [21]:

```
#pg 317
# Exa 9.21
#calculate the iron and copper loss
# Given data
from math import acos,cos,sin
VA = 400.*10**3;# in Mean
Eta_fl = 98.77/100;# in %
phi1= acos(0.8);# in radians
phi2= acos(1);# in radians
Eta_hl = 99.13/100;# in %
n = 1/2.;
#calculations
#For full load, Eta_f1 = ((VA*cosd(phi1))/( VA*cosd(phi1) + Pi + Pcu_f1 )) or Pi+Pcu_f1 = VA*cosd(phi1)*(1-Eta_fl)/(Eta_f1) (i)
#For half load, Eta_hl = n*VA*cosd(phi2)/(n*VA*cosd(phi2)+Pi+n**2*Pcu_f1) or Pi+n**2*Pcu_f1 = n*VA*cosd(phi2)*( 1-Eta_hl)/Eta_hl (ii)
# From eq(i) and (ii)
Pcu_fl=(n*VA*cos(phi2)*( 1-Eta_hl)/Eta_hl-VA*cos(phi1)*(1-Eta_fl)/(Eta_fl))/(n**2-1);# in W
Pi=VA*cos(phi1)*(1-Eta_fl)/(Eta_fl)-Pcu_fl;# in W
# The copper loss on half load
C_loss_half_load=n**2*Pcu_fl;# in W
#results
print "The iron loss on full load and half load remain same in W which are : ",round(Pi,4)
print "The copper loss on full load in W is : ",round(Pcu_fl,3)
print "The copper loss on half load in W is : ",round(C_loss_half_load,3)
```

In [22]:

```
#pg 317
# Exa 9.22
#calculate the efficiency
# Given data
from math import cos,acos
VA = 100.*10**3;# in VA
Eta_max = 98.40/100;# in %
Eta_max1 = 90./100;# in %
phi= acos(1);# in radians
#Eta_max = (Eta_max1*VA*cos(phi)/(Eta_max1*VA*cos(phi) + 2*Pi);
Pi = (Eta_max1*VA*cos(phi)/Eta_max - Eta_max1*VA*cos(phi))/2;# in W
Pcu = Pi;# in W
n = 0.9;
# Pcu_fl/Pcu = (VA/(0.9*VA) )**2;
Pcu_fl = Pcu*(VA/(0.9*VA) )**2;# in W
Eta_fl = ( (VA*cos(phi))/( (VA*cos(phi)) + Pi + Pcu_fl ) )*100;# in %
#results
print "The efficiency of a transformer in percent is",round(Eta_fl,4)
```