Chapter 24: Application of complex numbers to series a.c. circuits

Example 1, page no. 433

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
z1  =  12  +  5j;
z2  =  -40j;
r3  =  30;
theta3  =  60;#  in  degrees
r4  =  2.20E6;  
theta4  =  -30;#  in  degrees
f  =  50;#  in  Hz

#calculation:
 #for  an  R-L  series  circuit,  impedance
 #  Z  =  R  +  iXL
Ra  =  z1.real
XLa  =  z1.imag
La  =  XLa/(2*math.pi*f)
 #for  a  purely  capacitive  circuit,  impedance  Z  =  -iXc
Xcb  =  abs(z2.imag)
Cb  =  1/(2*math.pi*f*Xcb)
z3  =  r3*cmath.cos(theta3*math.pi/180)  +  (r3*cmath.sin(theta3*math.pi/180))*1j
Rc  =  z3.real
XLc  =  z3.imag
Lc  =  XLc/(2*math.pi*f)
z4  =  r4*cmath.cos(theta4*math.pi/180)  +  (r4*cmath.sin(theta4*math.pi/180))*1j
Rd  =  z4.real
Xcd  =  abs(z4.imag)
Cd  =  1/(2*math.pi*f*Xcd)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of  ",round(  Ra,2),"  ohm  "
print   "in  series  with  an  inductance  of  ",round(La*1000,2),"mH"
print  "\n  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance  ",round(Cb*1E6,2),"uF"
print  "\n  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of  ",round(Rc,2),"  ohm  "
print   "in  series  with  an  inductance  of  ",round(Lc*1000,2),"mH"
print  "\n  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of  ",round(Rd/1000,2),"kohm "
print   " in  series  with  a  capacitor  of  capacitance  ",round(Cd*1E9,2),"nF"


  Result  



  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of   12.0   ohm  
in  series  with  an  inductance  of   15.92 mH

  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance   79.58 uF

  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of   15.0   ohm  
in  series  with  an  inductance  of   82.7 mH

  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of   1905.26 kohm 
 in  series  with  a  capacitor  of  capacitance   2.89 nF

Example 2, page no. 434

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
L  =  0.1592  ;#  in  Henry
V  =  250;#  in  Volts
f  =  50;#  in  Hz
R  =  0;#  in  ohms

#calculation:
 #for  an  R–L  series  circuit,  impedance
 #  Z  =  R  +  iXL
XL  =  2*math.pi*f*L
Z  =  R  +  1j*XL
I  =  V/Z
x  =  I.real
y  =  I.imag
r  =  (x**2  +  y**2)**0.5
if  ((x==0)&(y<0)):
    theta  =  -90
elif  ((x==0)&(y>0)):
    theta  =  +90
else:
    theta  =  cmath.phase(complex(x,y))*180/math.pi



#Results
print  "\n\n  Result  \n\n"
print  "\n  current  is  (",round(r,2),"/_",theta,"deg)  A"


  Result  



  current  is  ( 5.0 /_ -90 deg)  A

Example 3, page no. 435

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
C  =  3E-6  ;#  in  farad
f  =  1000;#  in  Hz
ri  =  2.83;
thetai  =  90;#  in  degrees

#calculation:
 #Capacitive  reactance  Xc
Xc  =  1/(2*math.pi*f*C)
 #  circuit  impedance  Z
Z  =  -1*1j*Xc
I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)
V  =  I*Z
x  =  V.real
y  =  V.imag


#Results
print  "\n\n  Result  \n\n"
print  "\n  supply  p.d.  is ",round(abs(V),0),"V"


  Result  



  supply  p.d.  is  150.0 V

Example 4, page no. 435

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  240;#  in  Volts
f  =  50;#  in  Hz
Z  =  30  -  50j;

#calculation:
 #Since  impedance  Z  =  30  -  i50,
 #resistance
R  =  Z.real
 #capacitive  reactance
Xc  =  abs(Z.imag)
 #capacitance
C  =  1/(2*math.pi*f*Xc)
 #modulus  of  impedance
modZ  =  (R**2  +  Xc**2)**0.5
I  =  V/Z
x  =  I.real
y  =  I.imag
r  =  (x**2  +  y**2)**0.5
theta  =  cmath.phase(complex(x,y))*180/math.pi



#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)resistance  is  ",round(  R,2),"  ohm"
print  "\n  (b)capacitance  is  ",round(C*1E6,2),"uFarad"
print  "\n  (c)modulus  of  impedance  is  ",round(modZ,2),"  ohm"
print  "\n  (d)current  flowing  and  its  phase  angle  is  (",round(  r,2),"/_",round(  theta,2),"deg)  A"


  Result  



  (a)resistance  is   30.0   ohm

  (b)capacitance  is   63.66 uFarad

  (c)modulus  of  impedance  is   58.31   ohm

  (d)current  flowing  and  its  phase  angle  is  ( 4.12 /_ 59.04 deg)  A

Example 5, page no. 436

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  200;#  in  Volts
f  =  50;#  in  Hz
R  =  32;#  in  ohms
L  =  0.15;#  in  Henry

#calculation:
 #Inductive  reactance  XL
XL  =  2*math.pi*f*L
 #impedance,  Z
Z  =  R  +  1j*XL
 #Current  I
I  =  V/Z
xi  =  I.real
yi  =  I.imag
ri  =  (xi**2  +  yi**2)**0.5
if  ((xi==0)&(yi<0)):
         thetai  =  -90
elif  ((xi==0)&(yi>0)):
         thetai  =  +90
else:
         thetai  =  cmath.phase(complex(xi,yi))*180/math.pi

 #P.d.  across  the  resistor
VR  =  I*R
xr  =  VR.real
yr  =  VR.imag
rr  =  (xr**2  +  yr**2)**0.5
thetar  =  cmath.phase(complex(xr,yr))*180/math.pi
 #P.d.  across  the  coil,  VL
VL  =  I*1j*XL
xl  =  VL.real
yl  =  VL.imag
rl  =  (xl**2  +  yl**2)**0.5
thetal  =  cmath.phase(complex(xl,yl))*180/math.pi


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)impedance  is  ",round(Z.real,2),"  + ",round(  Z.imag,2),")i  ohm"
print  "\n  (b)current  flowing  and  its  phase  angle  is lagging the voltage = (",round(  ri,2),"/_",round(  thetai,2),"deg)  A"
print  "\n  (c)P.d.  across  the  resistor  is  (",round(rr,2),"/_",round(thetar,2),"deg)  V"
print  "\n  (d)P.d.  across  the  coil,  VL  is  (",round(rl,2),"/_",round(thetal,2),"deg)  V"


  Result  



  (a)impedance  is   32.0   +  47.12 )i  ohm

  (b)current  flowing  and  its  phase  angle  is lagging the voltage = ( 3.51 /_ -55.82 deg)  A

  (c)P.d.  across  the  resistor  is  ( 112.36 /_ -55.82 deg)  V

  (d)P.d.  across  the  coil,  VL  is  ( 165.46 /_ 34.18 deg)  V

Example 6, page no. 436

In [2]:
from __future__ import division
import math
#initializing  the  variables:
V  =  120  +  200j;#  in  Volts
f  =  5E6;#  in  Hz
I  =  7  +  16j;#  in  amperes

#calculation:
 #impedance,  Z
Z  =  V/I
R  =  Z.real
X  =  Z.imag  
if  ((R>0)&(X<0)):
    C  =  -1/(2*math.pi*f*X)
#Results
    print  "\n\n  Result  \n\n"
    print  "\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  ",round(R,2),"  ohm  "
    print   "and  a  capacitor  of  capacitive reactance", round(X*-1,3),"ohm and capacitance is",round(C*1E9,2)," nFarad\n"
elif  ((R>0)&(X>0)):
    L  =  2*math.pi*f*X
#Results
    print  "\n\n  Result  \n\n"
    print  "\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  ",round(R,2),"  ohm "
    print   " and  a  inductor  of  insuctance  ",round(L*100,2)," mHenry\n"


  Result  



  The  series  circuit  thus  consists  of  a  resistor  of  resistance   13.25   ohm  
and  a  capacitor  of  capacitive reactance 1.705 ohm and capacitance is 18.67  nFarad

Example 7, page no. 437

In [9]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  70;#  in  volts
thetav  =  30;#  in  degrees
ri  =  3.5;#  in  amperes
thetai  =  -20;#  in  degrees
 #z1  consist  of  two  resistance
R1  =  4.36;#  in  ohms
R2  =  -2.1j;#  in  ohms

 #calculation:
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)
 #impedance,  Z
Z  =  V/I
 #Total  impedance  Z  =  z1  +  z2
Z1  =  R1  +  R2
Z2  =  Z  -  Z1
x  =  Z2.real
y  =  Z2.imag  


#Results
print  "\n\n  Result  \n\n"
print  "\n  impedance  Z2  is  ",round(x,2),"  +  (",round(y,2),")i  ohm\n"


  Result  



  impedance  Z2  is   8.5   +  ( 17.42 )i  ohm

Example 8, page no. 437

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  90;#  in  ohms
XL  =  150;#  in  ohms
ri  =  1.35;#  in  amperes
thetai  =  0;#  in  degrees

#calculation:
I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)
 #Circuit  impedance  Z
Z  =  R  +  1j*XL
 #Supply  voltage,  V
V  =  I*Z
 #Voltage  across  90  ohm  resistor
VR  =  V.real
#Voltage  across  inductance,  VL
VL  =  V.imag
xv  =  V.real
yv  =  V.imag
rv  =  (xv**2  +  yv**2)**0.5
thetav  = cmath.phase(complex(xv,yv))*180/math.pi
phi  =  thetav  -  thetai


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Supply  voltage,  V  is  ",xv,"  +  (",yv,")i  V\n"
print  "\n  (b)Voltage  across  90  ohm resistor,  VR  is  ",VR,"  V\n"
print  "\n  (c)Voltage  across  inductance,  VL  is  ",VL,"  V\n"
print  "\n  (d)Circuit  phase  angle  is  ",round(phi,2),"deg lagging\n"


  Result  



  (a)Supply  voltage,  V  is   121.5   +  ( 202.5 )i  V


  (b)Voltage  across  90  ohm resistor,  VR  is   121.5   V


  (c)Voltage  across  inductance,  VL  is   202.5   V


  (d)Circuit  phase  angle  is   59.04 deg lagging

Example 9, page no. 438

In [5]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  25;#  in  ohms
L  =  0.02;#  in  henry
Vm  =  282.8;#  in  volts
w  =  628.4;#  in  rad/sec
phiv  =  math.pi/3;#  phase  angle

#calculation:
 #rms  voltage
Vrms  =  0.707*Vm*math.cos(phiv)  +  0.707*Vm*math.sin(phiv)*1j
 #frequency
f  =  w/(2*math.pi)
 #Inductive  reactance  XL
XL  =  2*math.pi*f*L
 #Circuit  impedance  Z
Z  =  R  +  XL*1j
 #Rms  current
Irms  =  Vrms/Z
phii  =  cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi
phi  =  phiv*180/math.pi  -  phii


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  rms  value  of  voltage  is  ",round(Vrms.real,2),"  +  (",round(  Vrms.imag,2),")i  V\n"
print  "\n  (b)the  circuit  impedance  is  ",round(R,2),"  +  (",round(  XL,2),")i  ohm\n"
print  "\n  (c)the  rms  current  flowing  is  ",round(Irms.real,2),"  +  (",round(  Irms.imag,2),")i  A\n"
print  "\n  (d)Circuit  phase  angle  is  ",round(phi,2),"deg lagging\n"


  Result  



  (a)the  rms  value  of  voltage  is   99.97   +  ( 173.15 )i  V


  (b)the  circuit  impedance  is   25.0   +  ( 12.57 )i  ohm


  (c)the  rms  current  flowing  is   5.97   +  ( 3.92 )i  A


  (d)Circuit  phase  angle  is   26.69 deg lagging

Example 10, page no. 438

In [11]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  12;#  in  ohms
L  =  0.10;#  in  henry
C  =  120E-6;#  in  Farads
f  =  50;#  in  Hz
V  =  240;#  in  volts

#calculation:
 #Inductive  reactance,  XL
XL  =  2*math.pi*f*L
 #Capacitive  reactance,  Xc
Xc  =  1/(2*math.pi*f*C)
 #Circuit  impedance  Z
Z  =  R  +  1j*(XL  -  Xc)
I  =  V/Z
phii  =  cmath.phase(complex(I.real, I.imag))*180/math.pi
phiv  =  0#  in  degrees
phi  =  phiv  -  phii


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  current  flowing  is  ",round(abs(I),1),"/_",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),"deg  A\n"
print  "and  Circuit  phase  angle  is  ",round(phi,1),"deg lagging\n"


  Result  



  the  current  flowing  is   18.5 /_ -22.2 deg  A

and  Circuit  phase  angle  is   22.2 deg lagging

Example 11, page no. 439

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
C  =  50E-6;#  in  Farads
f  =  50;#  in  Hz
V  =  225;#  in  volts
ri  =  1.5;#  in  Amperes
thetai  =  -30;#  in  degrees

#calculation:
I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)
 #Capacitive  reactance,  Xc
Xc  =  1/(2*math.pi*f*C)
 #Circuit  impedance  Z
Z  =  V/I
R  =  Z.real
XL  =  Z.imag  +  Xc
 #inductance  L
L  =  XL/(2*math.pi*f)
 #Voltage  across  coil
Zcoil  =  R  +  1j*XL
Vcoil  =  I*Zcoil
 #Voltage  across  capacitor,
Vc  =  I*(-1j*Xc)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)resistance  is  ",round(R,2),"  ohm  and  inductance  is  ",round(  L,3)," H\n"
print  "\n  (b)voltage  across  the  coil  is  ",round(Vcoil.real,2),"  +  (",round(  Vcoil.imag,2),")i  V\n"
print  "\n  (c)voltage  across  the  capacitor  is  ",round(Vc.real,2),"  +  (",round(  Vc.imag,2),")i  V\n"


  Result  



  (a)resistance  is   129.9   ohm  and  inductance  is   0.441  H


  (b)voltage  across  the  coil  is   272.75   +  ( 82.7 )i  V


  (c)voltage  across  the  capacitor  is   -47.75   +  ( -82.7 )i  V

Example 12, page no. 440

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
C  =  2.653E-6;#  in  Farads
R1  =  8;#  in  ohms
R2  =  5;#  in  ohms
L  =  0.477E-3;#  in  Henry
f  =  4000;#  in  Hz
ri  =  6;#  in  Amperes
thetai  =  0;#  in  degrees

#calculation:
I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)
 #Capacitive  reactance,  Xc
Xc  =  1/(2*math.pi*f*C)
 #impedance  Z1
Z1  =  R1  -  1j*Xc
 #inductive  reactance  XL
XL  =  2*math.pi*f*L
 #impedance  Z2,
Z2  =  R2  +  1j*XL
 #voltage  V1
V1  =  I*Z1
 #voltage  V2
V2  =  I*Z2
 #Supply  voltage,  V
V  =  V1  +  V2
phiv  =  cmath.phase(complex(V.real, V.imag))*180/math.pi
phi  =  phiv  -  thetai


#Results
print  "\n\n  Result  \n\n"
print  "\n  supply  voltage  is  ",round(V.real,2),"  +  (",round(  V.imag,2),")i  V\n"
print  "and  Circuit  phase  angle  is  ",round(phi,2),"deg leading"


  Result  



  supply  voltage  is   78.0   +  ( -18.06 )i  V

and  Circuit  phase  angle  is   -13.03 deg leading