Chapter 28: Series resonance and Q-factor

Example 1, page no. 492

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  10;#  in  ohms
C  =  40e-6;#  IN  fARADS
L  =  0.075;#  IN  Henry
V  =  200;#  in  Volts

#calculation:
 #Resonant  frequency,
fr  =  1/(2*math.pi*((L*C)**0.5))
 #Current  at  resonance,  I
I  =  V/R


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Resonant  frequency  =  ",round(fr,2),"  Hz\n"
print  "\n  (b)Current  at  resonance,  I  is  ",I,"  A\n"


  Result  



  (a)Resonant  frequency  =   91.89   Hz


  (b)Current  at  resonance,  I  is   20.0   A

Example 2, page no. 493

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  8;#  in  ohms
L  =  0.010;#  IN  Henry
f  =  1000;#  in  Hz

#calculation:
 #At  resonance
 #capacitance  C
C  =  1/(L*(2*math.pi*f)**2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance,  C  is  ",round(C*1E6,2),"uF\n"


  Result  



  capacitance,  C  is   2.53 uF

Example 3, page no. 493

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
C1  =  1000e-12;#  IN  fARADS
C2  =  500e-12;#  IN  fARADS
fr1  =  92500;#  in  Hz
fr2  =  127800;#  in  Hz

#calculation:
 #For  a  series  R–L–C  circuit  the  resonant  frequency  fr  is  given  by:
 #fr  =  1/(2pi*(L*C)**2)
Cs  =  ((C1  -  C2)/((fr2/fr1)**2  -  1))  -  C2
L  =  1/((C1  +  Cs)*(2*math.pi*fr1)**2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)stray  capacitance,  Cs  is  ",round(Cs*1E12,2),"pF\n"
print  "\n  (b)inductance,  L  is  ",round(L*1000,2),"mH\n"


  Result  



  (a)stray  capacitance,  Cs  is   50.13 pF


  (b)inductance,  L  is   2.82 mH

Example 4, page no. 497

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  10;#  in  ohms
C  =  5e-6;#  IN  fARADS
rv  =  20;#in  volts
thetav  =  0;#  in  degrees
f  =  318.3;#  in  Hz

#calculation:
wr  =  2*math.pi*f
 #The  maximum  voltage  across  the  resistance  occurs  at  resonance  when  the  current  is  a  maximum.  
    #At  resonance,L  =  1/c*wr**2
L  =  1/(C*wr**2)
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #Current  at  resonance  Ir
Ir  =  V/R
 #p.d.  across  resistance,  VR
VR  =  Ir*R
 #inductive  reactance,  XL
XL  =  wr*L
 #p.d.  across  inductance,  VL
VL  =  Ir*(1j*XL)
 #capacitive  reactance,  Xc
Xc  =  1/(wr*C)
 #p.d.  across  capacitor,  Vc
Vc  =  Ir*(-1j*Xc)
 #Q-factor  at  resonance,  Qr
Qr  =  VL.imag/V


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)inductance,  L  is  ",round(L*1000,2),"mH\n"
print  "\n  (b)p.d.  across  resistance,  VR  is  ",VR,"  V,  p.d.  across  inductance,  VL  ",round( VL.imag,2),"j V  "
print   "and  p.d.  across  capacitor,  VC  ",round(Vc.imag,2)," V\n"
print  "\n  (c)Q-factor  at  resonance,  Qr  is  ",round(abs(Qr),2),"  \n"


  Result  



  (a)inductance,  L  is   50.0 mH


  (b)p.d.  across  resistance,  VR  is   (20+0j)   V,  p.d.  across  inductance,  VL   200.01 j V  
and  p.d.  across  capacitor,  VC   -200.01  V


  (c)Q-factor  at  resonance,  Qr  is   10.0

Example 5, page no. 502

In [5]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  80;#  in  ohms
C  =  0.4e-6;#  IN  fARADS
L  =  0.020;#  IN  Henry
Vm  =  12;#in  volts

#calculation:
 #Resonant  frequency,
fr  =  1/(2*math.pi*((L*C)**0.5))
wr  =  2*math.pi*fr
 #Q  =  wr*L/R
Q  =  wr*L/R
Vc  =  Q*Vm
 #the  frequency  f  at  which  VC  is  a  maximum  value,
f  =  fr*(1  -  (1/(2*Q*Q)))**0.5
 #the  maximum  value  of  the  p.d.  across  the  capacitor  is  given  by:
Vcm  =  Vc/((1  -  (1/(2*Q*Q)))**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)The  resonant  frequency  is  ",round(fr,2),"  Hz\n"
print  "\n  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency  ",round(Vc,2),"  V\n"
print  "\n  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is  ",round(f,2),"  Hz\n"
print  "\n  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is  ",round(Vcm,2),"  V\n"


  Result  



  (a)The  resonant  frequency  is   1779.41   Hz


  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency   33.54   V


  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is   1721.52   Hz


  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is   34.67   V

Example 6, page no. 503

In [6]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
QL  =  60;#  Q-factor
Qc  =  390;#  Q-factor

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


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  overall  Q-factor  is  ",QT


  Result  



  the  overall  Q-factor  is   52.0

Example 7, page no. 505

In [7]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  5;#  in  ohms
L  =  0.010;#  IN  Henry
fr  =  10000;#  in  Hz

#calculation:
wr  =  2*math.pi*fr
 #Q-factor  at  resonance  is  given  by
Qr  =  wr*L/R
 #Since  Qr  =  fr/(f2  -  f1),
bw  =  fr/Qr


#Results
print  "\n\n  Result  \n\n"
print  "\n  bandwidth  of  the  filter  is  ",round(bw,2),"  Hz\n"


  Result  



  bandwidth  of  the  filter  is   79.58   Hz

Example 8, page no. 507

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Zr  =  50;#  in  ohms
fr  =  1200;#  in  Hz
Qr  =  30;#  Q-factor

#calculation:
 #At  resonance  the  circuit  impedance,  Z
R  =  Zr
wr  =  2*math.pi*fr
 #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is
L  =  Qr*R/wr
 #At  resonance  r*L  =  1/(wr*C)
 #capacitance,  C
C  =  1/(L*wr*wr)
 #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
 #At  the  half-power  frequencies,  current  I
 #I  =  0.707*Ir
 #Hence  impedance
Z  =  (2**0.5)*R


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)inductance,  L  is  ",round(L*1000,2),"mH\n"
print  "\n  (b)capacitance,  C  is  ",round(C*1E9,2),"nF\n"
print  "\n  (c)bandwidth  is  ",round(bw,2),"  Hz\n"
print  "\n  (d)the  upper  half-power  frequency,  f2  is  ",round(f2,2),"  Hz "
print   " and  the  lower  half-power  frequency,  f1  is  ",round(f1,2),"  Hz\n"
print  "\n  (e)impedance  at  the  half-power  frequencies  is  ",round(Z,2),"  ohm\n"


  Result  



  (a)inductance,  L  is   198.94 mH


  (b)capacitance,  C  is   88.42 nF


  (c)bandwidth  is   40.0   Hz


  (d)the  upper  half-power  frequency,  f2  is   1220.17   Hz 
 and  the  lower  half-power  frequency,  f1  is   1180.17   Hz


  (e)impedance  at  the  half-power  frequencies  is   70.71   ohm

Example 9, page no. 508

In [9]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  0.2;#  in  Volts
I  =  0.004;#  in  Amperes
fr  =  3000;#  in  Hz
Qr  =  100;#  Q-factor

#calculation:
wr  =  2*math.pi*fr
 #At  resonance,  impedance
Z  =  V/I
 #At  resonance  the  circuit  impedance,  Z
R  =  Z
 #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is
L  =  Qr*R/wr
 #At  resonance  r*L  =  1/(wr*C)
 #capacitance,  C
C  =  1/(L*wr*wr)
 #Q-factor  at  resonance  in  a  series  circuit  represents  the  voltage  magnification  Qr  =  Vc/V,  then  Vc  is
Vc  =  Qr*V


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  circuit  resistance  is  ",round(R,2),"  ohm\n"
print  "\n  (b)inductance,  L  is  ",round(L*1000,2),"mH\n"
print  "\n  (c)capacitance,  C  is  ",round(C*1E9,2),"nF\n"
print  "\n  (d)the  voltage  across  the  capacitor  is  ",round(Vc,2),"  V\n"


  Result  



  (a)the  circuit  resistance  is   50.0   ohm


  (b)inductance,  L  is   265.26 mH


  (c)capacitance,  C  is   10.61 nF


  (d)the  voltage  across  the  capacitor  is   20.0   V

Example 10, page no. 509

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  8.84;#  in  ohms
L  =  0.3518;#  IN  Henry
C  =  20e-6;#  IN  fARADS

#calculation:
 #Resonant  frequency,
fr  =  1/(2*math.pi*((L*C)**0.5))
wr  =  2*math.pi*fr
 #Q-factor  at  resonance,  Q  =  wr*L/R
Qr  =  wr*L/R
 #bandwidth,.(f2  −  f1)
bw  =  fr/Qr
 #the  lower  −3  dB  frequency
f1  =  fr  -  bw/2
 #the  upper  −3  dB  frequency
f2  =  fr  +  bw/2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Resonant  frequency,  fr  is  ",round(fr,2),"  Hz\n"
print  "\n  (b)Q-factor  at  resonance  is  ",round(Qr,2),"\n"
print  "\n  (c)Bandwidth  is  ",round(bw,2),"  Hz\n"
print  "\n  (d)the  lower  -3dB  frequency,  f1  is  ",round(f1,2),"  Hz "
print   " and  the  upper  -3dB  frequency,  f2  is  ",round(f2,2),"  Hz\n"


  Result  



  (a)Resonant  frequency,  fr  is   60.0   Hz


  (b)Q-factor  at  resonance  is   15.0 


  (c)Bandwidth  is   4.0   Hz


  (d)the  lower  -3dB  frequency,  f1  is   58.0   Hz 
 and  the  upper  -3dB  frequency,  f2  is   62.0   Hz

Example 11, page no. 511

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R  =  15;#  in  ohms
L  =  0.008;#  IN  Henry
C  =  0.3e-6;#  IN  fARADS
rv  =  7.56;#in  volts
thetav  =  0;#  in  degrees
x  =  0.03;

#calculation:
 #Resonant  frequency,
fr  =  1/(2*math.pi*((L*C)**0.5))
wr  =  2*math.pi*fr
 #At  resonance,
Zr  =  R
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #Current  at  resonance
Ir  =  V/Zr
 #Q-factor  at  resonance,  Q  =  wr*L/R
Qr  =  wr*L/R
 #If  the  frequency  is  3%  above    fr,  then
de  =  x
I  =  Ir/(1  +  (2*de*Qr*1j))
Z  =  V/I


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)Current  at  resonance,  Ir  is  ",round(abs(Ir),2),"  A\n"
print  "\n  (b)current  flowing  in  the  circuit  when  frequency  3  percent"
print   "      above  the  resonant  frequency  is  ",round(I.real,2),"  +  (",round(  I.imag,2),")i  A\n"
print  "\n  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent"
print   "      above  the  resonant  frequency  is  ",round(Z.real,2),"  +  (",round(Z.imag,2),")i  A\n"


  Result  



  (a)Current  at  resonance,  Ir  is   0.5   A


  (b)current  flowing  in  the  circuit  when  frequency  3  percent
      above  the  resonant  frequency  is   0.35   +  ( -0.23 )i  A


  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent
      above  the  resonant  frequency  is   15.0   +  ( 9.8 )i  A