In [1]:

```
# Variables
R = 0.2 #total resistance of cable in ohms
I = 200. #current in A
t = 100. #time in hours
V = 240. #voltage in volts
c = 0.8 #math.cost of electrical energy in Rs per unit
# Calculations
V1 = I*R #voltage drop in the cable
#(i)consumer voltage
Vc = V-V1
#(ii)Power loss in the cable
P = I*I*R #in watts
E = P*t/1000 #energy loss in kWh
C = E*c #math.cost of energy loss in Rs.
# Results
print 'i)Consumer voltage is %3.1f Volts '%(Vc)
print 'ii)cost of energy loss is Rs %3.2f '%(C)
```

In [2]:

```
# Variables
Vi = 220. #voltage in volts supplied by dynamo
Vo = 200. #voltage in volts required for lighting
I = 40. #current in Amperes
# Calculations
Pi = Vi*I #power output of dynamo
Po = Vo*I #power consumed for lighting
L = Pi-Po #line losses
R = L/(I**2) #resistance of lines math.since line losses = I**2*R
t = 10 #time in hrs
N = (Po*t)/1000 #no of units of consumed in B.O.T units
Nw = (L*t)/1000 #No of units wasted in B.O.T units
# Results
print 'i)Resistance of lines is %3.1f Ohms '%(R)
print 'ii)No. of B.O.T units consumed in 10hrs is %3.2f B.O.T units'%(N)
print 'iii)No. of B.O.T units wasted in 10hrs is %3.2f B.O.T units'%(Nw)
```

In [3]:

```
# Variables
M = 250000. #weight of water lifted per hr in kg
h = 50. #height in metres
g = 9.81 #gravitational const.
WD = M*h*g #work done by pump per hr in watt-sec
P = WD/3600 #Power output of pump per sec in watts
V = 500. #supply voltage in volts
Ep = 0.8 #efficiency of pump
Em = 0.9 #efficiency of motor
# Calculations
E = Em*Ep #overall efficiency
I = P/(V*E) #current in amperes
# Results
print 'Current drawn by the motor is %3.2f Amperes'%(I)
```

In [4]:

```
import math
# Variables
P = 10. #Power developed by motor in H.P
N = 600. #Speed of motor in rpm
#1HP = 735.5Nw-m/sec = 75kgm/sec
a = 75.
b = 735.5
# Calculations
#Torque in kg-m
Tkgm = (P*a*60)/(2*math.pi*N) #math.since P = 2*pi*NT/60
#Torque in Nw-m
TNwm = (P*b*60)/(2*math.pi*N) #math.since P = 2*pi*NT/60
# Results
print 'i)Torque in kg.meter is %3.2f kg-m '%(Tkgm)
print 'ii)Torque in Newton.meter is %3.2f Nw-m'%(TNwm)
```

In [5]:

```
# Variables
P = 25. #Output of diesel engine in kW
s = 12500. #calorific value of fuel oil in k-cal/kgm
e = 0.35 #overall efficiency of diesel set
P1 = P/e #input energy required in 1 hour in kWh
# Calculations
P2 = P1*860 #input energy in kcal
m = P2/s #mass of oil needed per hr in kgm
w = 1000 #weight of 1 ton of oil in kgm
Eg = (P*w)/m #Energy generated by 1ton of oil in kWh
# Results
print 'i)Mass of oil required per hr is %3.3f kgm '%(m)
print 'ii)Eletrical energy generated per ton of fuel is %4.1f Kwh'%(Eg)
```

In [7]:

```
import math
# Variables
rho = 1.7*(10**-6) #resistivity of copper in ohm-cm
l = 5 #length in metres
t = 0.005 #thickness in m
D = 0.08 #external diameter in m
# Calculations
d = D-(2*t) #internal diameter in m
a = math.pi*(D**2-d**2)/4 #cross section area in cm**2
R = rho*l/a #resistance of copper tube in ohm
R1 = R/(10**-4) #resistance in micro-ohm
# Results
print 'Thus the resistance of copper tube is %3.2f micro-ohm'%(R1)
```

In [8]:

```
# Variables
rho = 1.7*(10**-8) #resistivity in ohm-m
K = 1/rho #conductivity in mho/m
# Calculations
a = 0.125*(10**-4) #cross sectional area of cable in m**2
l = 2000. #length of cable in meters
G = K*a/l #conducmath.tance
# Results
print 'Thus conductivity of cable is %e mho/metres '%(K)
print 'and conducmath.tance of cable is %3f mho'%(G)
```

In [9]:

```
# Variables
V = 0.05 #volume in m**3
l = 300 #length in m
R = 0.0306 #resistance of conductor in ohm
# Calculations
rho = R*V/(l**2) #resistivity of conducting material
# Results
print 'Thus resistivity of conducting material is %e ohm-m'%(rho)
```

In [10]:

```
# Variables
rho = 0.67*(10**-6) #resistivity in ohm-inch
m = 39.4 #1meter = 39.4inch
m2 = 1525. #1 meter2 = 1525 square inch
# Calculations
rhoc = rho*m/m2 #resistivity of copper in ohm/m**3
rho1 = rhoc/(10**-6)
# Results
print 'Thus resistivity of copper is %e ohm/m**3'%(rhoc)
print 'which is equal to %2.4f micro-ohm/m**3'%(rho1)
```

In [11]:

```
import math
# Variables
R1 = 0.12 #old conductor resistance in ohm
d1 = 15. #diameter of old conductor in cm
d2 = 0.4*d1 #diameter of new conductor in cm
# Calculations
a1 = math.pi*(d1**2)/4 #area of cross section of old conductor
a2 = math.pi*(d2**2)/4 #area of cross section of new conductor
#R = rho*l/a = rho*V/a**2
#Henec R is proportional to 1/a**2
R2 = R1*((a1/a2)**2) #resistance of new conductor
# Results
print 'Thus resistance of new conductor is %2.4f ohm'%(R2)
```

In [12]:

```
# Variables
lab = 10. #la = 10*lb ratio of length of A to length of B.
Aab = 1./2 #Aa = 1/2*Ab ratio of area of A to area of B
RHOab = 1./2 #RHOa = 2*RHOb ratio of resistivity of A to resistivity of B
Ra = 2. #resistance of A in ohm
# Calculations
Rb = (Ra*Aab)/(lab*RHOab) #resistance of B in ohm
#Since Ra = RHOa*la/Aa and Rb = RHOb*lb/Ab so from ratio of two we get Rb
# Results
print 'Thus resistance of resistor B is %2.2f ohm'%(Rb)
```

In [13]:

```
import math
# Variables
RHOo = 10.3*(10**-6) #resistivity of platinum wire at 0 degree in ohm-cm
d = 0.0074 #diameter of platinum wire
a = math.pi*(d**2)/4 #area of cross section of platinum wire in sq cm
Ro = 4. #resistance of wire in ohm
# Calculations
l = Ro*a/RHOo #length of wire in cm
alphao = 0.0038
t = 100 #temp in degree C
R100 = Ro*(1+(alphao*t))
# Results
print 'Thus length of wire required is %3.2f cms'%(l)
print 'and Resistance of wire at 100 degreeC is %2.2f ohms'%(R100)
```

In [14]:

```
# Variables
Ra = 1. #resistance of A in ohm
lab = 20. #ratio of length of A to length of B
Aab = 1./3 #ratio of area of A to area of B
# Calculations
#resistivity is same for both wires
Rb = Ra*(Aab/lab) #resistance of wire B in ohm
#math.since Ra = rho*la/Aa and Rb = rho*lb/Ab so from ratio of both we get Rb
# Results
print 'Thus resistance of wire B is %2.4f omhs'%(Rb)
```

In [15]:

```
# Variables
I1 = 2./(2+3) #current across 2V battery in circuit EBD in A
Vbe = 3*I1 #voltage dropp across BE in V
I2 = 4./(5+3) #current across 4V battery in circuit AFC in A
# Calculations
Vaf = 3*I2 #voltage dropp across AF in V
V = Vbe+4-Vaf #sum of potential drops starting from E and ending at F
# Results
#V is the P.D. between E and F
print 'Thus the P.D. between E and F is %2.1f Volts'%(V)
```

In [16]:

```
# Calculations
#Let current in XA = I, in XY = I1, in AY = I-40, in YB = I-40+I1-60, in BX = I+I1-150.
#By Kirchhoff's second law, in circuit XAYA I-I1 = 20
# and in circuit XAYBX 25I+15I1 = 1950
I1 = (1950-500.)/(15.+25) #in Amperes
# Results
print 'Thus the current in branch XY is I1 = %2.2f Amps'%(I1)
```

In [17]:

```
# Variables
A = 30. #area of hysteresis material in cm**2
s1 = 0.4 #scale is 1cm = 0.4Wb/m**2
s2 = 400. # and 1cm = 400AT/m
# Calculations
V = 1.2*(10**-3)
f = 50 #frequency in Hz
H = A*s1*s2 #hysteresis loss/m**3/cycle in joules
Hp = H*V*f #hysteresis power loss in Watts
# Results
print 'Thus hysteresis power loss is %3.2f Watts'%(Hp)
```

In [18]:

```
# Variables
d = 7500. #density of iron in kg/m**3
w = 12. #weight of iron in kgm
# Calculations
V = w/d #volume of iron in m**3
f = 25. #frequency in Hz
N = 3600.*f #number of cycle per hour
A = 300 #area in joules/m**3
E = A*V*N #Total energy loss per hour in joules
# Results
print 'Thus total energy loss per hour is %5.2f Joules'%(E)
```

In [19]:

```
# Variables
l = 0.5 #length of coil in meters
d = 0.1 #diameter of coil
N = 1500. #no of turns of coil
# Calculations
a = math.pi*(d**2)/4 #cross sectional area of coil in m**2
Ur = 1 #relative permeability
Uo = 4*math.pi*(10**-7) #permeability
I = 8 #current in A
L = ((N**2)*a*Uo*Ur)/l #self inducmath.tance of coil in H
E = (1./2)*L*(I**2) #Energy stored in Joules
# Results
print 'Thus Self Inducmath.tance of coil is %2.3f H'%(L)
print 'and Energy stored is %1.2f Joules'%(E)
```

In [20]:

```
# Variables
N = 600. #number of turns on the coil
I = 2. #current pasmath.sing through solenoid in A
l = 0.6 #length of solenoid in meter
H = N*I/l #magnetic field at the centre in AT/m
Ur = 1. #relative permeability
# Calculations
Uo = 4*math.pi*(10**-7) #permeability
d = 0.025 #diameter in meters
a = math.pi*(d**2)/4 #cross sectional area of coil in m**2
phi = Uo*Ur*H*a #flux in Wb
# Results
print 'Thus Magenetic field at centre is %3.2f AT/m'%(H)
print ' and Flux is %e Wb'%(phi)
```

In [21]:

```
# Variables
Ur = 1. #relative permeability
B = 1.257 #flux density in Wb/m**2
# Calculations
Uo = 4*math.pi*(10**-7) #permeability
H = B/(Uo*Ur) #magnetimath.sing force in AT/m
l = 0.004 #length of air gap in meter
AT = H*l #AT required for the air gap
# Results
print 'Thus AT required for the air gap is %3.1f '%(AT)
```

In [22]:

```
# Variables
D = 0.3 #diameter of anchor ring in m
l = math.pi*D #length of iron ring in m
N = 400. #number of turns on the iron ring
a = 0.0012 #area of cross section of iron path in m**2
Ur = 1000. #relative permeability
# Calculations
Uo = 4*math.pi*(10**-7) #permeability
I = 2 #current in A
phi = (N*I)/(l/(Uo*Ur*a)) #flux through iron path in WB
phi1 = phi/(10**-3) #flux in mWb
# Results
print 'Thus flux through iron path is %2.2f mWb'%(phi1)
```

In [23]:

```
# Variables
a = 0.01 #crosssectional area of ring in m**2
Uo = 4*(math.pi)*(10**-7) #absolute permeability
lf = 1.25 #leakage factor
Ur = 400. #permeability
N = 175. #no of turns
# Calculations
phig = 0.8*(10**-3) #flux through air gap in Wb
Bg = phig/a #Flux density in air gap in Wb/m**2
Hg = Bg/Uo #magnetimath.sing force in air gap in AT/m
Lg = 0.004 #length of air gap in m
ATg = Hg*Lg #AT required for air gap in AT
phii = phig*lf #flux through iron path in Wb
Bi = phii/a #Flux density in iron path in Wb/m**2
Hi = Bi/(Uo*Ur) #magnetimath.sing force in iron path in AT/m
Li = 1.5 #length of iron path in m
ATi = Hi*Li #At required for iron path in AT
AT = ATi+ATg #total AT required
I = ATg/N #Magnetimath.sing current required in A
# Results
print 'Thus the magnetimath.sing current required is %2.2f Amps'%(I)
```

In [24]:

```
# Variables
SI = 0.2 #steady current in A
t = 0.2 #time in sec
Q = SI*t #charge given to condenser in Coulomb
V = 220. #PD across condenser in Volts
# Calculations
C = Q/V #Capacitance of condenser in F
C1 = C*(10**6) #Capacitance in mircoF
# Results
print 'Thus the Charge of condenser is %2.2f Coulomb'%(Q)
print 'And the Capacitance of condenser is %3.2f microF'%(C1)
```

In [25]:

```
# Variables
C = 2*(10**-6) #capacitance of condenser in F
V = 10000 #PD across condenser in Volts
# Calculations
E = (1./2)*C*(V**2) #energy stored in condenser in Joules
H = E/4.2 #heat produced in the wire in calories
# Results
print 'Thus heat produced in the wire is %2.2f calories'%(H)
```

In [26]:

```
# Variables
V = 15*(10**3) #potential difference applied in V
A = 0.02 #surface area of plate in m**2
d = 0.001 #dismath.tance between plates in m
C = 4.5*(10**-10) #Capacitance of capacitor in F
Ko = 8.854*(10**-12) #consmath.tant
# Calculations
K = (C*d)/(Ko*A) #dielectric consmath.tant
q = C*V #charge on condenser in C
D = q/A #Electric flux density in C/m**2
# Results
print 'Thus the Charge of condenser is %e Coulomb'%(q)
print 'And the electric flux density of condenser is %e microF'%(D)
```

In [27]:

```
# Variables
m = 0.6 #mass of water in kgm
S = 4200. #specific heat of water
T1 = 100. #temperature in degreeC
T2 = 10. #temperature in degreeC
t = 5*60. #time in sec
V = 230. #Supply voltage in Volts
# Calculations
H = m*S*(T1-T2) #Heat required to raise the temp of water from 0 to 100 degree. in J
e = 0.78 #efficiency of kettle
Ei = H/e #Energy input in Joules
Ei1 = Ei/(100*3600) #Energy input in kWh
W = Ei/t #Rating of kettle in watts
R = (V*V)/W #Resistance of heating element in ohms
# Results
print 'Thus Resistance of heating element is %2.1f ohms'%(R)
```

In [28]:

```
# Variables
m1 = 120. #mass of water to be heated in kg
m2 = 20. #mass of copper math.tank in kg
S1 = 1. #specific heat of water
S2 = 0.095 #specific heat of copper
T1 = 10. #temp in degreeC
T2 = 60. #temp in degreeC
# Calculations
H = (m1*S1*(T2-T1))+(m2*S2*(T2-T1)) #heat required to raise the temp of water and math.tank in kcal
H1 = H*4200 #heat required in Joules
e = 0.8 #thermal efficiency
E = H1/e #Energy input in joules
E1 = E/(1000*3600) #energy input in kWh
r = 3 #rating of heater in kW
t = E1/r #time taken in hours
# Results
print 'Thus the time taken to raise the temp is %2.3f hours'%(t)
```

In [29]:

```
import math
# Variables
rho = 5*(10**-5) #specific resistance for steel in ohm-cm
U = 1. #relative permeability
d = 0.15 #depth of penetration in cm
# Calculations
f = (rho*(10**9))/(U*d*d*4*(math.pi**2)) #frequency required in cycles per sec
f1 = f/1000 #frquency in k.cycles/sec
# Results
print 'Thus the frequency required is %3.3f k.cycles/sec'%(f1)
```

In [30]:

```
import math
# Variables
v = 50.*20*2 #Volume of board to be heated in cm**3
Mw = 0.56 #weight of wood in gm/cm**3
m = Mw*v/1000 #mass of wood in kgm
S = 0.35 #specific heat of wood
t = 15./60 #time in hrs
f = 30*(10**6) #frequency in cycles/sec
t2 = 150.;t1 = 30. #temp in degreeC
H = m*S*(t2-t1) #heat required to raise the temp in kcal
Hw = H*1000/860 #heat required in kW
# Calculations
P = Hw/t #power required in Watts
e = 0.5 #efficiency of dielectric heating process
Pi = P/e #power input required in Watts
Ko = 8.854*(10**-12) #absolute permittivity
K = 5 #relative permittivity
A = 0.5*0.2 #area in m
i = 0.02
C = Ko*K*A/i #capacitance of parallel plate capacitor in F
Xc = 1/(2*math.pi*f*C) #capacitive reacmath.tance in ohms
cosx = 0.05
tanx = 19.97
R = Xc*tanx #resistance
V = math.sqrt(Pi*R) #voltage in volts
Ic = V/Xc #current through the board in Amps
# Results
print 'Thus the power required is %2.1f Watts'%(Pi)
print 'And Voltage across the board is %3.2f volts'%(V)
print 'And the current through the board is %2.3f Amps'%(Ic)
```

In [31]:

```
# Variables
m = 2. #quantity of aluminium to be melted in kg
t1 = 15.
t2 = 660. #temp in degreeC
S = 0.212 #specific heat of aluminium
L = 78.8 #latent heat of aluminium in kcal/kg
# Calculations
H = (m*S*(t2-t1))+(m*L) #total heat required to melt Al in kcal
i = 5 #input to furnace in kW
E = i*(1000*10*60) #Energy input to furnace in watt-sec
E1 = E/4180 #energy input in kcal
e = H*100/E1 #efficiency of furnace
# Results
print 'Thus the efficiency of furnace is %2.3f percent'%(e)
```

In [32]:

```
# Variables
O = 5*735.5 #output of motor in W
e = 0.85 #efficiency of motor
c = 2. #math.cost of energy per unit in Rs
# Calculations
I = O/e #input of motor in Watts
t = 4 #time in hrs
E = I*t/1000 #energy consumed in kWh
C = c*E #math.cost of umath.sing the motor in Rs
# Results
print 'Thus the math.cost of umath.sing the motor is %2.3f Rs'%(C)
```

In [33]:

```
# Variables
I = 2.5*(10**-3) #current in Amp
t = 30*(10**-3) #time in sec
Q = I*t #charge pasmath.sing through the person in Coulumbs
# Calculations
e = 1.602*(10**-19) #charge of 1 electron in C
N = Q/e #no of electrons pasmath.sing through the person
# Results
print 'Thus the no of electrons pasmath.sing through the person is %e electrons'%(N)
```

In [34]:

```
# Variables
#(a)Finding resistance between 2 ends
l = 1 #length in m
a = 2.5*(10**-2)*0.05*(10**-2) #area of cross section in m**2
rho = 1.724*(10**-8) #specific resistance of copper in ohm-m
# Calculations
R = rho*l/a #resistance of the strip in ohm
#(b) Finding resistance between 2 faces
l1 = 0.05*(10**-2) #length in m
a1 = 2.5*(10**-2)*1 #area of cross section in m**2
R1 = rho*l1/a1 #resistance in ohm
# Results
print 'Thus the resistance of the strip is %e ohms '%(R)
print 'And the resistance between the faces is %e ohms'%(R1)
```

In [35]:

```
# Variables
m = 2. #weight of water to be heated in kg
t2 = 98.
t1 = 15. #temp in degreeC
s = 1. #specific heat of water
V = 200. #voltage in volts
# Calculations
H = m*s*(t2-t1) #energy required to raise the temp of water in kcal
H1 = H*4200 #energy in Watt-sec or Joules
e = 0.85 #efficiency of kettle
E = H1/e #energy input required in watt-sec
E1 = E/(1000*3600) #energy input in kWh
c = 35. #math.cost per unit in paise
C = c*E1 #ocst of energy used in paise
t = 10./60 #time in hrs
W = E1*1000/t #wattage of kettle in watts
R = V*V/W #resistance of heating element in ohms
# Results
print 'Thus the resistance of heating element is %2.0f ohms'%(R)
print 'And the math.cost of energy used is %2.0f paisa'%(C)
```

In [1]:

```
import math
# Variables
phi = 70000./(10**8) #flux to be set up in Wb math.since 10**8lines = 1Wb
d = 0.03 #diameter in m
a = math.pi*d*d/4 #area of cross section in m**2
B = phi/a #flux density in Wb/m**2
Lg = 0.002 #length of air gap in m
Ls = (math.pi*0.2)-Lg #length of steel path
Uo = 4*math.pi*(10**-7) #absolute permitivity
Ur = 800. #relative permitivity of steel
# Calculations
Hg = B/Uo
Hs = B/(Uo*Ur)
AT = (Hg*Lg)+(Hs*Ls) #total ampere turns required
N = 500. # no of turns
I = AT/N #exciting current in amps
# Results
print 'Thus the value of exciting current is %2.3f A'%(I)
```