In [1]:

```
from __future__ import division
import math
#initializing the variables:
Vl = 415;# in Volts
Rp = 30;# in ohms
#calculation:
Vp = Vl/(3**0.5)
Ip = Vp/Rp
Il = Ip
#Results
print "\n\n Result \n\n"
print "\n (a)the system phase voltage is ",round(Vp,2)," V"
print "\n (b)phase current is ",round(Ip,2)," A"
print "\n (c)line current is ",round(Il,2)," A"
```

In [2]:

```
from __future__ import division
import math
#initializing the variables:
R = 30;# in ohms
L = 0.1273;# in Henry
Ip = 5.08;# in Amperes
f = 50;# in Hz
#calculation:
XL = 2*math.pi*f*L
Zp = (R*R + XL*XL)**0.5
Il = Ip
Vp = Ip*Zp
Vl = Vp*(3**0.5)
#Results
print "\n\n Result \n\n"
print "\n (a)line voltage is ",round(Vl,2)," V"
```

In [1]:

```
from __future__ import division
import math
#initializing the variables:
V = 415;# in Volts
PR = 24000;# in Watt
Py = 18000;# in Watt
Pb = 12000;# in Watt
VR = 240;# in Volts
Vy = 240;# in Volts
Vb = 240;# in Volts
#calculation:
#For a star-connected system VL = Vp*(3**0.5)
Vp = V/(3**0.5)
phir = 90*math.pi/180
phiy = 330*math.pi/180
phib = 210*math.pi/180
# I = P/V for a resistive load
IR = PR/VR
Iy = Py/Vy
Ib = Pb/Vb
Inh = IR*math.cos(phir) + Ib*math.cos(phib) + Iy*math.cos(phiy)
Inv = IR*math.sin(phir) + Ib*math.sin(phib) + Iy*math.sin(phiy)
In = (Inh**2 + Inv**2)**0.5
#Results
print "\n\n Result \n\n"
print "\n (a)cuurnt in R line is ",round(IR,2)," A, cuurnt in Y line is ",round(Iy,2)," A "
print "and cuurnt in B line is ",round(Ib,2)," A"
print "\n (b)cuurnt in neutral line is ",round(In,2)," A"
```

In [2]:

```
from __future__ import division
import math
#initializing the variables:
R = 30;# in ohms
L = 0.1273;# in Henry
VL = 440;# in Volts
f = 50;# in Hz
#calculation:
XL = 2*math.pi*f*L
Zp = (R*R + XL*XL)**0.5
Vp = VL
#Phase current
Ip = Vp/Zp
#For a delta connection,
IL = Ip*(3**0.5)
#Results
print "\n\n Result \n\n"
print "\n (a)the phase current ",round(Ip,2)," A"
print "\n (b)line current ",round(IL,2)," A"
```

In [3]:

```
from __future__ import division
import math
#initializing the variables:
IL = 15;# in Amperes
VL = 415;# in Volts
f = 50;# in Hz
#calculation:
#For a delta connection
Ip = IL/(3**0.5)#phase current
Vp = VL
#Capacitive reactance per phase
Xc = Vp/Ip
C = 1/(2*math.pi*f*Xc)
#Results
print "\n\n Result \n\n"
print "\n capacitance is ",round(C*1E6,2),"uF"
```

In [1]:

```
from __future__ import division
import math
#initializing the variables:
R = 3;# in ohms
XL = 4;# in ohms
VL = 415;# in Volts
#calculation:
#For a star connection:
#IL = Ip
#VL = Vp*(3**0.5)
VLs = VL
Vps = VLs/(3**0.5)
#Impedance per phase,
Zp = (R*R + XL*XL)**0.5
Ips = Vps/Zp
ILs = Ips
#For a delta connection:
#VL = Vp
#IL = Ip*(3**0.5)
VLd = VL
Vpd = VLd
Ipd = Vpd/Zp
ILd = Ipd*(3**0.5)
#Results
print "\n\n Result \n\n"
print "\n (a)the line voltage for star connection is ",round(VLs,2)," V "
print "and the phase voltage for star connection is ",round(Vps,2)," V "
print "and the line voltage for delta connection is ",round(VLd,2)," V "
print "and the phase voltage for delta connection is ",round(Vpd,2)," V"
print "\n (b)the line current for star connection is ",round(ILs,2)," A "
print "and the phase current for star connection is ",round(Ips,2)," A "
print "and the line current for delta connection is ",round(ILd,2)," A "
print "and the phase current for delta connection is ",round(Ipd,2)," A"
```

In [5]:

```
from __future__ import division
import math
#initializing the variables:
Rp = 12;# in ohms
VL = 415;# in Volts
#calculation:
#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
Vp = VL/(3**0.5)# since the resistors are star-connected
#Phase current, Ip
Zp = Rp
Ip = Vp/Zp
#For a star connection
IL = Ip
# For a purely resistive load, the power factor cos(phi) = 1
pf = 1
P = VL*IL*(3**0.5)*pf
#Results
print "\n\n Result \n\n"
print "\n (a)total power dissipated by the resistors is ",round(P,2)," W"
```

In [8]:

```
from __future__ import division
import math
#initializing the variables:
P = 5000;# in Watts
IL = 8.6;# in amperes
VL = 400;# in Volts
#calculation:
#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
pf = P/(VL*IL*(3**0.5))
#Results
print "\n\n Result \n\n"
print "\n power factor is ",round(pf,3)
```

In [9]:

```
from __future__ import division
import math
#initializing the variables:
R = 10;# in ohms
L = 0.042;# in Henry
VL = 415;# in Volts
f = 50;# in Hz
#calculation:
#For a star connection:
#IL = Ip
#VL = Vp*(3**0.5)
XL = 2*math.pi*f*L
Zp = (R*R + XL*XL)**0.5
VLs = VL
Vps = VLs/(3**0.5)
#Impedance per phase,
Ips = Vps/Zp
ILs = Ips
#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
pfs = R/Zp
Ps = VLs*ILs*(3**0.5)*pfs
#For a delta connection:
#VL = Vp
#IL = Ip*(3**0.5)
VLd = VL
Vpd = VLd
Ipd = Vpd/Zp
ILd = Ipd*(3**0.5)
#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
pfd = R/Zp
Pd = VLd*ILd*(3**0.5)*pfd
#Results
print "\n\n Result \n\n"
print "\n total power dissipated in star is ",round(Ps,2)," W and in delta is ",round(Pd,2)," W"
```

In [10]:

```
from __future__ import division
import math
#initializing the variables:
Po = 12750;# in Watts
pf = 0.77;# power factor
eff = 0.85;
VL = 415;# in Volts
#calculation:
#eff = power_out/power_in
Pi = Po/eff
#Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
IL = Pi/(VL*(3**0.5)*pf)# line current
#For a delta connection:
#IL = Ip*(3**0.5)
Ip = IL/(3**0.5)
#Results
print "\n\n Result \n\n"
print "\n (a)power input is ",round(Pi,2)," W"
print "\n (b)line current is ",round(IL,2)," A"
print "\n (c)phase current is ",round(Ip,2)," A"
```

In [1]:

```
from __future__ import division
import math
#initializing the variables:
R = 30;# in ohms
XL = 40;# in ohms
VL = 400;# in Volts
#calculation:
Zp = (R*R + XL*XL)**0.5
#a delta-connected load
Vp = VL
#Phase current
Ip = Vp/Zp
IL = Ip*(3**0.5)
#Alternator output power is equal to the power dissipated by the load.
#Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
pf = R/Zp
P = VL*IL*(3**0.5)*pf
#Alternator output kVA,
S = VL*IL*(3**0.5)
#Results
print "\n\n Result \n\n"
print "\n (a)the current supplied by the alternator is ",round(IL,2)," A"
print "\n (b)output power is ",round(P/1000,2),"KW and kVA of the alternator is ",round(S/1000,2),"kVA"
```

In [2]:

```
from __future__ import division
import math
#initializing the variables:
R = 30;# in ohms
C = 80E-6;# in Farads
f = 50;# in Hz
VL = 400;# in Volts
#calculation:
#Capacitive reactance
Xc = 1/(2*math.pi*f*C)
Zp = (R*R + Xc*Xc)**0.5
pf = R/Zp
#a delta-connected load
Vp = VL
#Phase current
Ip = Vp/Zp
IL = Ip*(3**0.5)
#Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
P = VL*IL*(3**0.5)*pf
#Alternator output kVA,
S = VL*IL*(3**0.5)
#Results
print "\n\n Result \n\n"
print "\n (a)the phase current is ",round(Ip,2)," A"
print "\n (b)the line current is ",round(IL,2)," A"
print "\n (c) power is ",round(P/1000,2),"kW"
print "\n (d)kVA of the alternator is ", round(S/1000,2),"kVA"
```

In [14]:

```
from __future__ import division
import math
#initializing the variables:
Pi1 = 8000;# in Watts
Pi2 = 4000;# in Watts
#calculation:
#Total input power
Pi = Pi1 + Pi2
phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))
#Power factor
pf = math.cos(phi)
#Results
print "\n\n Result \n\n"
print "\n (a)power input is ",round(Pi,2),"W"
print "\n (b)power factor is ",round(pf,2)
```

In [15]:

```
from __future__ import division
import math
#initializing the variables:
Pi = 12000;# in Watts
pf = 0.6;# power factor
#calculation:
#If the two wattmeters indicate Pi1 and Pi2 respectively
# Pit = Pi1 + Pi2
Pit = Pi
# Pid = Pi1 - Pi2
#power factor = 0.6 = cos(phi)
phi = math.acos(pf)
Pid = Pit*math.tan(phi)/(3**0.5)
#Hence wattmeter 1 reads
Pi1 = (Pid + Pit)/2
#wattmeter 2 reads
Pi2 = Pit - Pi1
#Results
print "\n\n Result \n\n"
print "\n reading in each wattameter are ",round(Pi1,2),"W and ",round(Pi2,2),"W"
```

In [16]:

```
from __future__ import division
import math
#initializing the variables:
Pi1 = 10000;# in Watts
Pi2 = -3000;# in Watts
#calculation:
#Total input power
Pi = Pi1 + Pi2
phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))
#Power factor
pf = math.cos(phi)
#Results
print "\n\n Result \n\n"
print "\n (a)power input is ",round(Pi,2),"W"
print "\n (b)power factor is ",round(pf,2)
```

In [17]:

```
from __future__ import division
import math
#initializing the variables:
R = 8; # in ohms
XL = 8; # in ohms
VL = 415; # in Volts
#calculation:
#For a star connection:
#IL = Ip
#VL = Vp*(3**0.5)
VLs = VL
Vps = VLs/(3**0.5)
#Impedance per phase,
Zp = (R*R + XL*XL)**0.5
Ips = Vps/Zp
ILs = Ips
#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
pf = R/Zp
Ps = VLs*ILs*(3**0.5)*pf
#If wattmeter readings are P1 and P2 then P1 + P2 = Pst
Pst = Ps
# Pid = Pi1 - Pi2
phi = math.acos(pf)
Psd = Pst*math.tan(phi)/(3**0.5)
#Hence wattmeter 1 reads
Ps1 = (Psd + Pst)/2
#wattmeter 2 reads
Ps2 = Pst - Ps1
#For a delta connection:
#VL = Vp
#IL = Ip*(3**0.5)
VLd = VL
Vpd = VLd
Ipd = Vpd/Zp
ILd = Ipd*(3**0.5)
#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
Pd = VLd*ILd*(3**0.5)*pf
#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt
Pdt = Pd
# Pid = Pi1 - Pi2
Pdd = Pdt*math.tan(phi)/(3**0.5)
#Hence wattmeter 1 reads
Pd1 = (Pdd + Pdt)/2
#wattmeter 2 reads
Pd2 = Pdt - Pd1
#results
print "\n\n Result \n\n"
print "(a)When the coils are star-connected the wattmeter readings are", round(Ps1,2),"W and ",round(Ps2,2),"W"
print "(b)When the coils are delta-connected the wattmeter readings are are", round(Pd1,2),"W and", round(Pd2,2),"W"
```