In [10]:

```
#Given
N=100
A=10**-2 #m**2
B=0.5 #T
f=500/60.0
#Calculation
import math
w=2*math.pi*f
E0=N*A*B*w
E=E0*math.sin(60*3.14/180.0)
#Result
print"Maximum emf produced in the coil is", round(E0,2),"V"
print"Instantaneous value of e.m.f. is",round(E,1),"V"
```

In [17]:

```
#Given
N=50
A=2.5
B=0.3 #T
w=60
R=500 #ohm
#Calculation
E0=N*B*A*w
I0=E0/R
#Result
print"(i) Maximum current drawn from the gnerator is",I0,"A"
print"(ii) The maximum flux through the coil is zero and when induced current is maximum, flux through the coil is zero"
print"(iii) Yes the generator will work because the basic condition for induced an .m.f. is that there should be relative motion between coil and magnetic field"
```

In [23]:

```
#Given
N=150
A=2*10**-2 #m**2
B=0.15 #T
f=60
#Calculation
import math
w=2*math.pi*f
E0=N*A*B*w
#Result
print"Peak value of e.m.f is", round(E0,0),"V"
print"Average value of induced e.m.f is zero"
```

In [28]:

```
#Given
N=100
A=3
B=0.04 #T
w=60
R=500 #ohm
#Calculation
E0=N*A*B*w
I0=E0/R
P=E0*I0
#Result
print"Maximum power dissipated in the coil is", P,"W"
```

In [32]:

```
#Given
N=100
A=0.10 #m**2
f=0.5 #Hz
B=0.01 #T
#Calculation
import math
w=2*math.pi*f
E0=N*A*B*w
#Result
print"Maximum voltage generated in the coil is", round(E0,3),"V"
```

In [35]:

```
#Given
V=240 #V
I=5 #A
R=4 #ohm
#Calculation
Eb=V-(I*R)
#Result
print"Value of back e.m.f is", Eb,"V"
```

In [44]:

```
#Given
I=20 #A
R=2 #ohm
n=0.5
P=2000 #W
#Calculation
P1=P/n
V=P1/I
Eb=V-(I*R)
#Result
print"The back e.m.f is", Eb,"V \nSupply voltage is",V,"V"
```

In [49]:

```
#Given
V=100 #V
I=6 #A
V1=0.7
#Calculation
Pin=V*I
R=(V1*Pin)/I**2
#Result
print"Armature resistance is", round(R,2),"ohm"
```

In [61]:

```
#Given
V=200 #V
I=5 #A
R=8.5 #ohm
#Calculation
Eb=V-(I*R)
Pi=V*I
P0=Eb*I
n=(P0*100)/Pi
#Result
print"(i) Back e.m.f of motor is", Eb,"V"
print"(ii) Power input is",Pi,"W"
print"(iii) Output power is",P0,"W"
print"(iv) Efficiency of motor is",n,"%"
```

In [69]:

```
#Given
Vp=200 #V
n=200.0
Ip=2 #A
#Calculation
Vs=Vp*n
Is=(Ip*V)/Vs
#Result
print"(i) Voltage developed in the secondary is", Vs,"V"
print"(ii) The current in the secondary is",Is ,"A"
```

In [79]:

```
#Given
Vp=220.0 #V
Is=5 #A
n=20
#Calculation
Vs=Vp*n
Ip=(Vs*Is)/Vp
P=Vs*Is
#Result
print"(i) Voltage across secondary is",Vs,"V"
print"(ii) The current in primary is",Ip,"A"
print"(iii) The power output is",P*10**-3,"K W"
```

In [94]:

```
#Given
P=120*10**3 #W
R=0.4 #ohm
Ev=240.0 #V
Ev1=24000.0 #V
#Calculation
Iv=P/Ev
P=Iv**2*R
Iv1=P/Ev1
P1=Iv1**2*R
#Result
print"(i) Power loss at 240 V is", P*10**-3,"K W"
print"(ii) Power loss at 24000 V is", round(P1,0),"W"
```

In [99]:

```
#Given
Np=5000
Vp=2200 #V
Vs=220 #V
Pout=8 #K W
n=0.9
#Calculation
Ns=(Vs*Np)/Vp
Pin=Pout/n
#Result
print"(i) The number of turns in the secondary is", Ns
print"(ii) Input power is",round(Pin,1),"K W"
```

In [104]:

```
#Given
Vp=220.0 #V
Vs=22 #V
Z=220 #ohm
Is=0.1
#Calclation
Ip=(Vs*Is)/Vp
#Result
print"Current drawn is", Ip,"A"
```

In [116]:

```
#Given
Vs=24 #v
R=9.6 #ohm
Vp=120.0
#Calculation
Is=Vs/R
Ip=(Vs*Is)/Vp
P1=Vs*Is
#Result
print"(i) Current in the secondary coil is", Is,"A"
print"(ii) Current in primary coil is",Ip ,"A"
print"Power used is",P1,"W"
```