Chapter 29: parallel resonance and Q-factor

Example 1, page no. 521

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  10;#  in  ohms
L  =  0.005;#  IN  Henry
C  =  0.25e-6;#  IN  fARADS
V  =  50;#in  volts

#calculation:
 #Resonant  frequency,  for  parallel
fr  =  ((1/(L*C)  -  ((R**2)/(L**2)))**0.5)/(2*math.pi)
 #dynamic  resistance
Rd  =  L/(C*R)
 #Current  at  resonance
Ir  =  V/Rd
wr  =  2*math.pi*fr
 #Q-factor  at  resonance,  Q  =  wr*L/R
Qr  =  wr*L/R


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Resonance  frequency  is  ",round(fr,2),"  Hz\n"
print  "\n  (b)dynamic  resistance  ",round(Rd,2),"  ohm\n"
print  "\n  (c)Current  at  resonance,  Ir  is  ",round(Ir,2),"  A\n"
print  "\n  (d)Q-factor  at  resonance  is  ",round(Qr,2),"\n"


  Result  



  (a)Resonance  frequency  is   4490.31   Hz


  (b)dynamic  resistance   2000.0   ohm


  (c)Current  at  resonance,  Ir  is   0.02   A


  (d)Q-factor  at  resonance  is   14.11

Example 2, page no. 521

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
RL1  =  0;#  in  ohms
RL2  =  30;#  in  ohms
L  =  0.100;#  IN  Henry
C  =  40e-6;#  IN  fARADS
V  =  50;#in  volts

#calculation:
 #for  RL1
 #Resonant  frequency,
wr1  =  (1/(L*C))**0.5
fr1  =  wr1/(2*math.pi)
 #for  RL2
 #Resonant  frequency,
wr2  =  (1/(L*C)  -  ((RL2**2)/(L**2)))**0.5
fr2  =  wr2/(2*math.pi)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Resonance  frequency  at  RL  =  0  is  ",round(fr1,2),"  Hz"
print  "\n  (b)Resonance  frequency  at  RL  =  30  ohm  is  ",round(fr2,2),"  Hz\n"


  Result  



  (a)Resonance  frequency  at  RL  =  0  is   79.58   Hz

  (b)Resonance  frequency  at  RL  =  30  ohm  is   63.66   Hz

Example 3, page no. 523

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  150;#  in  ohms
L  =  0.120;#  IN  Henry
V  =  20;#in  volts
fr  =  4000;#  in  Hz

#calculation:
 #capacitance,  C
C  =  1/(L*((2*math.pi*fr)**2  +  ((R**2)/(L**2))))
Rd  =  L/(C*R)
 #Current  at  resonance
Ir  =  V/Rd
wr  =  2*math.pi*fr
 #Q-factor  at  resonance,  Q  =  wr*L/R
Qr  =  wr*L/R
 #bandwidth,.(f2  −  f1)
bw  =  fr/Qr
 #upper  half-power  frequency,  f2
f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2
 #lower  half-power  frequency,  f1
f1  =  f2  -  bw
 #impedance  at  the  −3  dB  frequencies
Z  =  Rd/(2**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  capacitance  of  the  capacitor,C  is  ",round(C*1E6,2),"uF"
print  "\n  (b)dynamic  resistance  ",round(Rd,2),"ohm\n"
print  "\n  (c)Current  at  resonance,  Ir  is  ",round(Ir*1000,2),"mA\n"
print  "\n  (d)Q-factor  at  resonance  is  ",round(Qr,2),"\n"
print  "\n  (e)bandwidth  is  ",round(bw,2),"  Hz\n"
print  "\n  (f)the  upper  half-power  frequency,  f2  is  ",round(f2,2),"  Hz  and "
print   " the  lower  half-power  frequency,  f1  is  ",round(f1,2),"  Hz\n"
print  "\n  (g)impedance  at  the  -3  dB  frequencies  is  ",round(Z,2)," ohm\n"


  Result  



  (a)the  capacitance  of  the  capacitor,C  is   0.01 uF

  (b)dynamic  resistance   60788.85 ohm


  (c)Current  at  resonance,  Ir  is   0.33 mA


  (d)Q-factor  at  resonance  is   20.11 


  (e)bandwidth  is   198.94   Hz


  (f)the  upper  half-power  frequency,  f2  is   4100.71   Hz  and 
 the  lower  half-power  frequency,  f1  is   3901.76   Hz


  (g)impedance  at  the  -3  dB  frequencies  is   42984.21  ohm

Example 4, page no. 525

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
RL  =  5;#  in  ohms
L  =  0.002;#  IN  Henry
C  =  25e-6;#  IN  fARADS
Rc  =  3;#  in  ohms

#calculation:
 #Resonant  frequency,  for  parallel
fr  =  (1/(2*math.pi*((L*C)**0.5)))*((RL**2  -  (L/C))/(Rc**2  -  (L/C)))**0.5


#Results
print  "\n\n  Result  \n\n"
print  "\n  resonant  frequency,  fr  is  ",round(fr,2),"  Hz"


  Result  



  resonant  frequency,  fr  is   626.45   Hz

Example 5, page no. 525

In [6]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
RL  =  3;#  in  ohms
fr  =  1000;#  in  Hz
Xc  =  10;#  IN  ohms
Rc  =  4;#  in  ohms

#calculation:
XL1  =  (((Rc**2  +  Xc**2)/Xc)  +  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2
XL2  =  (((Rc**2  +  Xc**2)/Xc)  -  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2
wr  =  2*math.pi*fr
 #inductance
L1  =  XL1/wr
L2  =  XL2/wr


#Results
print  "\n\n  Result  \n\n"
print  "\n  inductance  is  either  ",round(L1*1000,2),"mH  or  ",round(L2*1000,2),"mH"


  Result  



  inductance  is  either   1.71 mH  or   0.13 mH

Example 6, page no. 526

In [7]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
QL  =  60;#  Q-factor
Qc  =  300;#  Q-factor

#calculation:
QT  =  QL*Qc/(QL  +  Qc)


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  overall  Q-factor  is  ",round(QT,2)


  Result  



  the  overall  Q-factor  is   50.0

Example 7, page no. 527

In [8]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
C  =  10.61E-9;#  in  Farad
bw  =  500;#  in  Hz
fr  =  150000;#  in  Hz
x  =  0.004

#calculation:
 #Q-factor
Q  =  fr/bw
wr  =  2*math.pi*fr
 #dynamic  resistance,  RD
Rd =  Q/(C*wr)
de =  x
Z  =  Rd/(1  +  (2*de*Q*1j))

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Q-factor  ",round(Q,2),""
print  "\n  (b)dynamic  resistance  ",round(Rd,2),"ohm"
print  "\n  (c)magnitude  of  the  impedance  ",round(abs(Z),2),"ohm"


  Result  



  (a)Q-factor   300.0 

  (b)dynamic  resistance   30000.93 ohm

  (c)magnitude  of  the  impedance   11538.82 ohm