Chapter 26: Power in a.c. circuits

Example 1, page no. 466

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
RL  =  12j;#  in  ohm
R  =  5;#  in  ohm
rv  =  52;#  in  volts
thetav  =  30;#  in  degree

#calculation:
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #impedance,  Z
Z  =  R  +  RL
 #current
I  =  V/Z
 #Active  power,  P
Pa  =  V.real*I.real  +  V.imag*I.imag


#Results
print  "\n\n  Result  \n\n"
print  "\nthe  active  power  in  the  circuit  ",Pa,"  W\n"

  Result  



the  active  power  in  the  circuit   80.0   W

Example 2, page no. 467

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  120  +  200j;#  in  volts
I  =  15  +  8j;#  in  amperes

#calculation:
 #Active  power,  P
Pa  =  V.real*I.real  +  V.imag*I.imag
 #Reactive  power,  Q
Q  =  V.imag*I.real  -  V.real*I.imag

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)  the  active  power  in  the  circuit  ",Pa,"  W\n"
print  "\n  (b)  the  reactive  power  in  the  circuit  ",Q,"  var\n"

  Result  



  (a)  the  active  power  in  the  circuit   3400.0   W


  (b)  the  reactive  power  in  the  circuit   2040.0   var

Example 3, page no. 468

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Vm  =  141.4;#  in  volts
w  =  10000;#  in  rad/sec
phiv  =  math.pi/9;#  in  radian
Pd  =  1732;#  in  Watts
pf  =  0.866;#  power  fctr

#calculation:
 #the  rms  voltage,
Vrms  =  0.707*Vm
 #Power  P  =  V*I*cos(phi)
 #current  magnitude,  Irms
Irms  =  Pd/(Vrms*pf)
phid  =  math.acos(pf)
 #current  phase  angle
phii  =  phiv  +  phid
phiid  =  phii*180/math.pi#  in  degrees
 #Voltage,  V
V  =  Vrms*math.cos(phiv)  +  1j*Vrms*math.sin(phiv)
 #current,  I
I  =  Irms*math.cos(phii)  +  1j*Irms*math.sin(phii)
 #Impedance,  Z
Z  =  V/I
 #resistance,  R
R  =  Z.real
 #capacitive  reactance,  Xc
Xc  =  abs(Z.imag)
 #capacitance,  C
C  =  1/  (w*Xc)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  current  flowing  and  Circuit  phase  angle  is  ",round(Irms,2),"/_",round(phiid,2),"deg  A\n"
print  "\n  (b)  the  resistance  is  ",round(R,2),"  ohm\n"
print  "\n  (c)  the  capacitance  is  ",round(C*1E6,2),"uF\n"

  Result  



  (a)the  current  flowing  and  Circuit  phase  angle  is   20.01 /_ 50.0 deg  A


  (b)  the  resistance  is   4.33   ohm


  (c)  the  capacitance  is   40.02 uF

Example 4, page no. 468

In [4]:
from __future__ import division
import math
#initializing  the  variables:
rv  =  100;#  in  volts
thetav  =  0;#  in  degrees
R  =  5;#  in  ohm
R1  =  3;#  in  ohms
RL  =  4j;#  in  ohm
Rc  =  -10j;#  in  ohms

#calculation:
 #impedance,  Z1
Z1  =  R1  +  RL
 #impedance,  Zc
Zc  =  Rc
 #Circuit  impedance,  Z
Z  =  R  +  (Z1*Zc/(Z1  +  Zc))
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)
I  =  V/Z
Imag  =  ((I.real)**2  +  (I.imag)**2)**0.5
 #Active  power  developed  between  points  A  and  B
Pab  =  (Imag**2)*R
 #Active  power  developed  between  points  C  and  D
Pcd  =  (Imag**2)*Zc.real
 #Current,  I1
I1  =  I*Zc/(Zc  +  Z1)
I1mag  =  ((I1.real)**2  +  (I1.imag)**2)**0.5
 #active  power  developed  between  points  E  and  F
Pef  =  (I1mag**2)*Z1.real


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Active  power  developed  between  points  A  and  B  is  ",round(Pab,2),"  W\n"
print  "\n  (b)Active  power  developed  between  points  C  and  D  is  ",round(Pcd,2),"  W\n"
print  "\n  (c)Active  power  developed  between  points  E  and  F  is  ",round(Pef,2),"  W\n"

  Result  



  (a)Active  power  developed  between  points  A  and  B  is   339.62   W


  (b)Active  power  developed  between  points  C  and  D  is   0.0   W


  (c)Active  power  developed  between  points  E  and  F  is   452.83   W

Example 5, page no. 469

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Pa  =  400;#  in  Watts
rv  =  100;#  in  volts
thetav  =  30;#  in  degrees
R  =  4;#  in  ohm
pf  =  0.766;#  power  factor

 #calculation:
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #magnitude  of  apparent  power,S  =  V*I
S  =  Pa/pf
phi  =  math.acos(pf)
theta  =  phi*180/math.pi#  in  degrees
 #Reactive  power  Q
Q  =  S*math.sin(phi)
 #magnitude  of  current
Imag  =  S/rv
thetai  =  thetav  -  theta
I  =  Imag*math.cos(thetai*math.pi/180)  +  1j*Imag*math.sin(thetai*math.pi/180)
 #Total  circuit  impedance  ZT
ZT  =  V/I
 #impedance  Z
Z  =  ZT  -  R


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)apparent  power  is  ",round(S,2),"  VA\n"
print  "\n  (b)reactive  power  is  ",round(Q,1),"  var lagging\n"
print  "\n  (c)the  current  flowing  and  Circuit  phase  angle  is  ",round(Imag,2),"/_",round(thetai,2),"deg  A\n"
print  "\n  (d)impedance,  Z  is  ",round(Z.real,2),"  +  (",round(  Z.imag,2),")i  ohm\n"

  Result  



  (a)apparent  power  is   522.19   VA


  (b)reactive  power  is   335.7   var lagging


  (c)the  current  flowing  and  Circuit  phase  angle  is   5.22 /_ -10.0 deg  A


  (d)impedance,  Z  is   10.67   +  ( 12.31 )i  ohm

Example 6, page no. 471

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
S  =  300000;#  in  VA
pf1  =  0.70;#  in  power  factor
pf2  =  0.90;#  in  power  factor

#calculation:
 #active  power,  P
Pa  =  S*pf1
phi1  =  math.acos(pf1)
phi1d  =  phi1*180/math.pi
 #Reactive  power,  Q
Q  =  S*math.sin(phi1)
phi2  =  math.acos(pf2)
phi2d  =  phi2*180/math.pi
 #The  capacitor  rating  needed  to  improve  the  power  factor  to  0.90
 #the  capacitor  rating,
Pr  =  Q  -  (Pa*math.tan(phi2))


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  rating  (in  kilovars)  of  the  capacitors  is  ",round((Pr/1E3),2),"  kvar leading\n"

  Result  



  the  rating  (in  kilovars)  of  the  capacitors  is   112.54   kvar leading

Example 7, page no. 471

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Z  =  3  +  4j;#  in  ohms
rv  =  50;#  in  volts
thetav  =  30;#  in  Degrees
f  =  1500;#  in  Hz
pf1  =  0.966;#  in  power  factor

#calculation:
V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)
 #Supply  current,  I
I  =  V/Z
Istr  =  I.real  -  1j*I.imag
 #Apparent  power,  S
S  =  V*Istr
 #active  power,  Pa
Pa  =  S.real
#reactive  power,  Q
Q  =  abs(S.imag)
 #apparent  power,  S
S  =  (S.real**2  +  S.imag**2)**0.5
phi1  =  math.acos(pf1)
phi1d  =  phi1*180/math.pi
 #rating  of  the  capacitor  
Pr  =  Q  -  Pa*math.tan(phi1)
 #Current  in  capacitor,  Ic
Ic  =  Pr/rv
 #Capacitive  reactance,  Xc
Xc  =  rv/Ic
C  =  1/(2*math.pi*f*Xc)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)supply  current,  I  is  ",round(I.real,2),"  +  (",round(  I.imag,2),")i  A\n"
print  "\n  (b)active  power  is  ",round(Pa,2),"  W,  apparent  power  is  ",round(  S,2),"  W  "
print   "and  reactive  power  is  ",round(  Q,2),"  W lagging\n"
print  "\n  (c)the  rating  of  the  capacitors  is  ",round(Pr,2),"  var leading\n"
print  "\n  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is  ",round(  C*1E6,2),"uF\n"

  Result  



  (a)supply  current,  I  is   9.2   +  ( -3.93 )i  A


  (b)active  power  is   300.0   W,  apparent  power  is   500.0   W  and  reactive  power  is   400.0   W lagging


  (c)the  rating  of  the  capacitors  is   319.71   var leading


  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is   13.57 uF