# Chapter 16: Single-phase parallel a.c. circuits

### Example 1, page no. 239

In :
from __future__ import division
import math
#initializing  the  variables:
R  =  20;#  in  Ohms
L  =  2.387E-3;#  in  Henry
V  =  60;#  in  Volts
f  =  1000;#  in  Hz

#calculation:
IR  =  V/R
XL  =  2*math.pi*f*L
IL  =  V/XL
I  =  (IR**2  +  IL**2)**0.5
phi  =  math.atan(IL/IR)
phid  =  phi*180/math.pi
Z  =  V/I
P  =  V*I*math.cos(phi)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Current  through  resistor  is  ",round(IR,2),"  A    and  current  through  Inductor  is  ",round(  IL,2),"  A"
print  "\n  (b)current,  I  =  ",round(I,2),"  A  "
print  "\n  (c)phase  angle  =  ",round(phid,2),"deg lagging"
print  "\n  (d)Impedance  Z  =  ",round(Z,2),"  Ohm  "
print  "\n  (e)Power  consumed  =  ",round(P,2),"  Watt  "


Result

(a)Current  through  resistor  is   3.0   A    and  current  through  Inductor  is   4.0   A

(b)current,  I  =   5.0   A

(c)phase  angle  =   53.13 deg lagging

(d)Impedance  Z  =   12.0   Ohm

(e)Power  consumed  =   180.0   Watt  

### Example 2, page no. 240

In :
from __future__ import division
import math
#initializing  the  variables:
R  =  80;#  in  Ohms
C  =  30E-6;#  in  Farads
V  =  240;#  in  Volts
f  =  50;#  in  Hz

#calculation:
IR  =  V/R
Xc  =  1/(2*math.pi*f*C)
Ic  =  V/Xc
I  =  (IR**2  +  Ic**2)**0.5
phi  =  math.atan(Ic/IR)
phid  =  phi*180/math.pi
Z  =  V/I
P  =  V*I*math.cos(phi)
S  =  V*I

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Current  through  resistor  is  ",round(IR,2),"  A    and  current  through  capacitor  is  ",round(  Ic,2),"  A"
print  "\n  (b)current,  I  =  ",round(I,2),"  A "
print  "\n  (c)phase  angle  =  ",round(phid,2),"deg leading"
print  "\n  (d)Impedance  Z  =  ",round(Z,2),"  Ohm  "
print  "\n  (e)Power  consumed  =  ",round(P,2),"  Watt  "
print  "\n  (f)apparent  Power  =  ",round(S,2),"  VA  "


Result

(a)Current  through  resistor  is   3.0   A    and  current  through  capacitor  is   2.26   A

(b)current,  I  =   3.76   A

(c)phase  angle  =   37.02 deg leading

(d)Impedance  Z  =   63.88   Ohm

(e)Power  consumed  =   720.0   Watt

(f)apparent  Power  =   901.72   VA  

### Example 3, page no. 241

In :
from __future__ import division
import math
#initializing  the  variables:
pf  =  0.6;#  power  factor
V  =  120;#  in  Volts
f  =  200;#  in  Hz
I  =  2;#  in  Amperes

#calculation:
phi  =  math.acos(pf)
phid  =  phi*180/math.pi
IR  =  I*math.cos(phi)
Ic  =  I*math.sin(phi)
R  =  V/IR
C  =  Ic/(2*math.pi*f*V)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Resistance  R  =  ",round(R,2),"  Ohm  "
print  "\n  (b)Capacitance,C  =  ",round((C/1E-6),2),"  uF  "


Result

(a)Resistance  R  =   100.0   Ohm

(b)Capacitance,C  =   10.61   uF  

### Example 4, page no. 242

In :
from __future__ import division
import math
#initializing  the  variables:
C  =  25E-6;#  in  Farads
L  =  120E-3;#  in  Henry
V  =  100;#  in  Volts
f  =  50;#  in  Hz

#calculation:
XL  =  2*math.pi*f*L
IL  =  V/XL
Xc  =  1/(2*math.pi*f*C)
Ic  =  V/Xc
#IL  and  Ic  are  anti-phase.  Hence  supply  current,
I  =  IL  -  Ic
#the  current  lags  the  supply  voltage  V  by  90Â°
phi  =  math.pi/2
phid  =  phi*180/math.pi
Z  =  V/I
P  =  V*I*math.cos(phi)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Current  through  Inductor  is  ",round(IL,2),"  A    and  current  through  capacitor  is  ",round(  Ic,2),"  A"
print  "\n  (b)current,  I  =  ",round(I,2),"  A  "
print  "\n  (c)phase  angle  =  ",round(phid,2),"deg lagging"
print  "\n  (d)Impedance  Z  =  ",round(Z,2),"  Ohm  "
print  "\n  (e)Power  consumed  =  ",round(P,2),"  Watt  "


Result

(a)Current  through  Inductor  is   2.65   A    and  current  through  capacitor  is   0.79   A

(b)current,  I  =   1.87   A

(c)phase  angle  =   90.0 deg lagging

(d)Impedance  Z  =   53.56   Ohm

(e)Power  consumed  =   0.0   Watt  

### Example 5, page no. 242

In :
from __future__ import division
import math
#initializing  the  variables:
C  =  25E-6;#  in  Farads
L  =  120E-3;#  in  Henry
V  =  100;#  in  Volts
f  =  150;#  in  Hz

#calculation:
XL  =  2*math.pi*f*L
IL  =  V/XL
Xc  =  1/(2*math.pi*f*C)
Ic  =  V/Xc
#IL  and  Ic  are  anti-phase.  Hence  supply  current,
I  =  Ic  -  IL
#the  current  leads  the  supply  voltage  V  by  90Â°
phi  =  math.pi/2
phid  =  phi*180/math.pi
Z  =  V/I
P  =  V*I*math.cos(phi)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Current  through  Inductor  is  ",round(IL,2),"  A    and  current  through  capacitor  is  ",round(  Ic,2),"  A"
print  "\n  (b)current,  I  =  ",round(I,2),"  A  "
print  "\n  (c)phase  angle  =  ",round(phid,2),"deg leading"
print  "\n  (d)Impedance  Z  =  ",round(Z,2),"  Ohm  "
print  "\n  (e)Power  consumed  =  ",round(P,2),"  Watt  "


Result

(a)Current  through  Inductor  is   0.88   A    and  current  through  capacitor  is   2.36   A

(b)current,  I  =   1.47   A

(c)phase  angle  =   90.0 deg leading

(d)Impedance  Z  =   67.93   Ohm

(e)Power  consumed  =   0.0   Watt  

### Example 6, page no. 244

In :
from __future__ import division
import math
#initializing  the  variables:
C  =  30E-6;#  in  Farads
R  =  40;#  in  Ohms
L  =  159.2E-3;#  in  Henry
V  =  240;#  in  Volts
f  =  50;#  in  Hz

#calculation:
XL  =  2*math.pi*f*L
Z1  =  (R**2  +  XL**2)**0.5
ILR  =  V/Z1
phi1  =  math.atan(XL/R)
phi1d  =  phi1*180/math.pi
Xc  =  1/(2*math.pi*f*C)
Ic  =  V/Xc
phi2  =  math.pi/2
phi2d  =  phi2*180/math.pi
Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)
Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)
I  =  (Ih**2  +  Iv**2)**0.5
phi  =  math.atan(abs(Iv)/Ih)
Z  =  V/I
P  =  V*I*math.cos(phi)
phid  =  phi*180/math.pi
S  =  V*I
Q  =  V*I*math.sin(phi)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Current  through  coil  is  ",round(ILR,2),"  A  and  lagged  by  phase  angle  is  ",round(phi1d,2),"deg"
print  "\n  (b)Current  through  capacitor  is  ",round(Ic,2),"  A  and  lead  by  phase  angle  is  ",round(phi2d,2),"deg"
print  "\n  (c)supply  Current  is  ",round(I,2),"  A  and  lagged  by  phase  angle  is  ",round(phid,2),"deg"
print  "\n  (d)Impedance  Z  =  ",round(Z,2),"  Ohm  "
print  "\n  (e)Power  consumed  =  ",round(P,2),"  Watt  "
print  "\n  (f)apparent  Power  =  ",round(S,2),"  VA  "
print  "\n  (g)reactive  Power  =  ",round(Q,2),"  var  "


Result

(a)Current  through  coil  is   3.75   A  and  lagged  by  phase  angle  is   51.35 deg

(b)Current  through  capacitor  is   2.26   A  and  lead  by  phase  angle  is   90.0 deg

(c)supply  Current  is   2.43   A  and  lagged  by  phase  angle  is   15.85 deg

(d)Impedance  Z  =   98.64   Ohm

(e)Power  consumed  =   561.76   Watt

(f)apparent  Power  =   583.97   VA

(g)reactive  Power  =   159.53   var  

### Example 7, page no. 246

In :
from __future__ import division
import math
#initializing  the  variables:
C  =  0.02E-6;#  in  Farads
R  =  3000;#  in  Ohms
L  =  120E-3;#  in  Henry
V  =  40;#  in  Volts
f  =  5000;#  in  Hz

#calculation:
XL  =  2*math.pi*f*L
Z1  =  (R**2  +  XL**2)**0.5
ILR  =  V/Z1
phi1  =  math.atan(XL/R)
phi1d  =  phi1*180/math.pi
Xc  =  1/(2*math.pi*f*C)
Ic  =  V/Xc
phi2  =  math.pi/2
phi2d  =  phi2*180/math.pi
Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)
Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)
I  =  (Ih**2  +  Iv**2)**0.5
phi  =  math.atan((Iv)/Ih)
phid  =  phi*180/math.pi
Z  =  V/I
P  =  V*I*math.cos(phi)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Current  through  coil  is  ",round(ILR*1000,2),"mA  and  lagged  by  phase  angle  is  ",round(phi1d,1),"deg"
print  "\n  (b)Current  through  capacitor  is  ",round(Ic*1000,2),"mA  and  lead  by  phase  angle  is  ",round(phi2d,2),"deg"
print  "\n  (c)supply  Current  is  ",round(I*1000,1),"mA  and  leaded  by  phase  angle  is  ",round(phid,2),"deg"
print  "\n  (d)Impedance  Z  =  ",round(Z/1000,3),"KOhm  "
print  "\n  (e)Power  consumed  =  ",round(P*1000,1),"mWatt  "


Result

(a)Current  through  coil  is   8.3 mA  and  lagged  by  phase  angle  is   51.5 deg

(b)Current  through  capacitor  is   25.13 mA  and  lead  by  phase  angle  is   90.0 deg

(c)supply  Current  is   19.3 mA  and  leaded  by  phase  angle  is   74.5 deg

(d)Impedance  Z  =   2.068 KOhm

(e)Power  consumed  =   206.8 mWatt  

### Example 8, page no. 249

In :
from __future__ import division
import math
#initializing  the  variables:
C  =  40E-6;#  in  Farads
R  =  0;#  in  Ohms
L  =  150E-3;#  in  Henry
V  =  50;#  in  Volts

#calculation:
fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)
Xc  =  1/(2*math.pi*fr*C)
Icirc  =  V/Xc

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Parallel  resonant  frequency,  fr  =  ",round(fr,2),"  Hz  "
print  "\n  (b)Current  circulating  in  L  and  C  at  resonance  =  ",round(Icirc,2),"  A  "


Result

(a)Parallel  resonant  frequency,  fr  =   64.97   Hz

(b)Current  circulating  in  L  and  C  at  resonance  =   0.82   A  

### Example 9, page no. 250

In :
from __future__ import division
import math
#initializing  the  variables:
C  =  20E-6;#  in  Farads
R  =  60;#  in  Ohms
L  =  200E-3;#  in  Henry
V  =  20;#  in  Volts

#calculation:
fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)
Rd  =  L/(R*C)
Ir  =  V/Rd
Q  =  2*math.pi*fr*L/R

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Parallel  resonant  frequency,  fr  =  ",round(fr,2),"  Hz  "
print  "\n  (b)the  dynamic  resistance,RD  =  ",round(Rd,2),"  ohm  "
print  "\n  (c)Current  at  resonance  =  ",round(Ir,2),"  A  "
print  "\n  (d)Q-factor  =  ",round(Q,2)


Result

(a)Parallel  resonant  frequency,  fr  =   63.66   Hz

(b)the  dynamic  resistance,RD  =   166.67   ohm

(c)Current  at  resonance  =   0.12   A

(d)Q-factor  =   1.33

### Example 10, page no. 251

In :
from __future__ import division
import math
#initializing  the  variables:
fr  =  5000;#  in  ohm
R  =  800;#  in  Ohms
L  =  100E-3;#  in  Henry
V  =  12;#  in  Volts

#calculation:
C  =  1/(L*((2*math.pi*fr)**2  +  R*R/(L*L)))
Rd  =  L/(R*C)
Ir  =  V/Rd
Q  =  2*math.pi*fr*L/R

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)capacitance,  C  =  ",round((C/1E-9),2),"  nF  "
print  "\n  (b)the  dynamic  resistance,RD  =  ",round(Rd,2),"  ohm  "
print  "\n  (c)Current  at  resonance  =  ",round((Ir/1E-3),2),"  mA  "
print  "\n  (d)Q-factor  =  ",round(Q,2)


Result

(a)capacitance,  C  =   9.52   nF

(b)the  dynamic  resistance,RD  =   13137.01   ohm

(c)Current  at  resonance  =   0.91   mA

(d)Q-factor  =   3.93

### Example 11, page no. 252

In :
from __future__ import division
import math
#initializing  the  variables:
f  =  50;#  in  ohm
V  =  240;#  in  Volts
pf  =  0.6;#  power  factor
Im  =  50;#  in  amperes

#calculation:
phi  =  math.acos(pf)
phid  =  phi*180/math.pi
Ic  =  Im*math.sin(phi)
I  =  Im*math.cos(phi)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  capacitor  current  Ic  must  be  ",round(Ic,2),"  A  for  the  power  factor  to  be  unity.  "
print  "\n  (b)Supply  current  I  =  ",round(I,2),"  A  "


Result

(a)the  capacitor  current  Ic  must  be   40.0   A  for  the  power  factor  to  be  unity.

(b)Supply  current  I  =   30.0   A  

### Example 12, page no. 253

In :
from __future__ import division
import math
#initializing  the  variables:
Pout  =  4800;#  in  Watt
eff  =  0.8#  effficiency
f  =  50;#  in  ohm
V  =  240;#  in  Volts
pf1  =  0.625;#  power  factor
pf2  =  0.95;#  power  factor

#calculation:
Pin  =  Pout/eff
Im  =  Pin/(V*pf1)
phi1  =  math.acos(pf1)
phi1d  =  phi1*180/math.pi
phi2  =  math.acos(pf2)
phi2d  =  phi2*180/math.pi
Imh  =  Im*math.cos(phi1)
#Ih  =  I*cos(phi2)
Ih  =  Imh
I  =  Ih/math.cos(phi2)
Imv  =  Im*math.sin(phi1)
Iv  =  I*math.sin(phi2)
Ic  =  Imv  -  Iv
C  =  Ic/(2*math.pi*f*V)
kvar  =  V*Ic/1000

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)current  taken  by  the  motor,  Im  =  ",round(Im,2),"  A"
print  "\n  (b)supply  current  after  p.f.  correction,  I  =  ",round(I,2),"  A  "
print  "\n  (c)magnitude  of  the  capacitor  current  Ic  =  ",round(Ic,0),"  A"
print  "\n  (d)capacitance,  C  =  ",round((C/1E-6),0),"  uF  "
print  "\n  (d)kvar  rating  of  the  capacitor  =  ",round(kvar,2),"  kvar  "


Result

(a)current  taken  by  the  motor,  Im  =   40.0   A

(b)supply  current  after  p.f.  correction,  I  =   26.32   A

(c)magnitude  of  the  capacitor  current  Ic  =   23.0   A

(d)capacitance,  C  =   305.0   uF

(d)kvar  rating  of  the  capacitor  =   5.52   kvar  

### Example 13, page no. 254

In :
from __future__ import division
import math
#initializing  the  variables:
S  =  3000;#  in  VA
f  =  50;#  in  ohm
V  =  250;#  in  Volts
Iil  =  10;#  in  Amperes
Ifl  =  8;#  in  Amperes
pfil  = 1; #  power  factor
pffl  =  0.7;#  power  factor
pfm  =  0.8;#  power  factor
pf0  =  0.975;#  power  factor

#calculation:
phiil  =  math.acos(pfil)
phiild  =  phiil*180/math.pi
phifl  =  math.acos(pffl)
phifld  =  phifl*180/math.pi
phim  =  math.acos(pfm)
phimd  =  phim*180/math.pi
phi0  =  math.acos(pf0)
phi0d  =  phi0*180/math.pi
Im  =  S/V
Ih  =  Iil*math.cos(phiil)  +  Ifl*math.cos(phifl)  +  Im*math.cos(phim)
Iv  =  Iil*math.sin(phiil)  -  Ifl*math.sin(phifl)  -  Im*math.sin(phim)
Il  =  (Ih**2  +  Iv**2)**0.5
phi  =  math.atan(abs(Iv)/Ih)
phid  =  phi*180/math.pi
pf  =  math.cos(phi)
P  =  V*Il*pf
I  =  Il*math.cos(phi)/math.cos(phi0)
Ic  =  Il*math.sin(phi)  -  I*math.sin(phi0)
C  =  Ic/(2*math.pi*f*V)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)total  current,  Il  =  ",round(Il,2),"  A"
print  "\n  (b)Power  factor  =  ",round(pf,2),"lagging"
print  "\n  (c)Total  power,  P  =  ",round(P/1000,2),"KWatt"
print  "\n  (d)capacitance,  C  =  ",round((C/1E-6),2),"uF  "


Result

(a)total  current,  Il  =   28.32   A

(b)Power  factor  =   0.89 lagging

(c)Total  power,  P  =   6.3 KWatt

(d)capacitance,  C  =   91.29 uF