# Chapter - 7 : Inductance and Capacitance Measurements¶

## Example : 7.1 - Page No : 178¶

In [9]:
#Given data
Z1= 100 # in Ω
theta1= 30 # in °
Z2= 50 # in Ω
theta2= 0 # in °
Z3= 200 # in Ω
theta3= -90 # in °
Z4= 100 # in Ω
theta4= 30 # in °
if Z1*Z4 == Z2*Z3 :
print "The first condition is satisfied"
if theta1+theta4 == theta2+theta3 :
print "The second condiiton is also satisfied, So it is possible to balance the bridge under the given condition"
else:
print "The second condition is not satisfied."
print "So balance is not possible under given condition"
The first condition is satisfied
The second condition is not satisfied.
So balance is not possible under given condition

## Example : 7.2 - Page No : 178¶

In [47]:
#Given data
Z1= 1000 # in Ω
theta1= -90 # in °
Z2= 500 # in Ω
theta2= 0 # in °
Z3= 1000 # in Ω
theta3= 0 # in °
R4= 100 # in Ω
XL4= 500 # in Ω
i_XL4= 500j # imaginary part
Z4=(R4+i_XL4) # in °
theta4= math.atan2(Z4.imag,Z4.real)*180/pi # in °
if theta1+theta4 ==  theta2+theta3 :
print "The first condiiton is satisfied."
else :
print "Balance is not possible with given configuration"

# 1/Z1=1/R1+j*omega*C1     (i)
# According to figure  1/Z1= R4/(Z2*Z3)+%i*XL4/(Z2*Z3)        (ii)
# Comparing real and j-components of Eqn (i) and (ii)
R1= Z2*Z3/R4 # in Ω
OmegaC1= Z2*Z3/XL4 # in Ω
print "\nSince X_C1 is already equal to ",int(OmegaC1)," Ω, the bridge can be balanced simply by placing a "
print "resistance of ",int(R1)," Ω across the capacitor arm 1"
# Z3= R3-j*X_C3        (iii)
#Z3= Z1*expm(%i*theta1*pi/180)*Z4*expm(%i*theta4*pi/180)/(Z2*expm(%i*theta2*pi/180)) #       (iv)
# Comparing real and j-components of Eqn (iii) and (iv)
R3= 1000 # in Ω
XC3= 200 # in Ω
print "\nSince R3 is already of ",int(R3)," Ω, the bridge can be balanced simply by adding a"
print "capacitor of reactance X_C3 of ",int(XC3)," Ω in series with the resistor R3 in arm 3."
Balance is not possible with given configuration

Since X_C1 is already equal to  1000  Ω, the bridge can be balanced simply by placing a
resistance of  5000  Ω across the capacitor arm 1

Since R3 is already of  1000  Ω, the bridge can be balanced simply by adding a
capacitor of reactance X_C3 of  200  Ω in series with the resistor R3 in arm 3.

## Example : 7.3 - Page No : 180¶

In [51]:
from __future__ import division
#Given data
C2= 0.2 # in micro F
Ratio21= 10/1 # resistance ratio R2/R1
C1= C2*Ratio21 # in micro F
Ratio21_desh= 1/10
C1_desh= C2*Ratio21_desh # in micro F
print "The range of measurement of unknown capacitance = ",round(C1_desh,2),"µF to",int(round(C1)),"µF"
The range of measurement of unknown capacitance =  0.02 µF to 2 µF

## Example : 7.4 - Page No : 182¶

In [5]:
from __future__ import division
from numpy import pi
#Given data
R2= 5 # in ohm
R3= 2000 # in ohm
R4= 2950 # in ohm
C2= 0.5 # in micro F
C2=C2*10**-6 # in F
r2=0.4 # in  ohm
f=450 # in Hz
omega= 2*pi*f
# Under Balace Condition Z1*Z4=Z2*Z3
# [r1+1/(j*omega*C1)]*R4= [r2+R2+1/(j*omega*C2)]*R3
# Equating the real parts we have, r1*R4= (r2+R2)*R3
r1= (r2+R2)*R3/R4 # in ohm
print "Value of r1 = %0.3f ohm" %r1
# Equating imaginary parts we have R4/(j*omega*C1)= R3/(j*omega*C2)
C1= R4/R3*C2 # in F
print "Value of C1 = %0.4f micro F" %(C1*10**6)
Tan_toh= omega*C1*r1
print "Dissipation Factor = %0.3e" %Tan_toh
Value of r1 = 3.661 ohm
Value of C1 = 0.7375 micro F
Dissipation Factor = 7.634e-03

## Example : 7.5 - Page No : 183¶

In [9]:
#Given data
f=1000 #in Hz
R1=1000 #in ohm
R2=1000 # in ohm
R3=2000 #in ohm
R4=2000 #in ohm
C1=1*10**-6 #in F
r1= 10 # in ohm
omega=2*pi*f
C2=C1*R1/R2 #in F
print "Unknown capacitance = %0.f µF "%(C2*10**6)

r2=(R2*(R3+r1)-R1*R4)/R1 #in ohm
del_1=del_1*180/pi # in °
print "Phase angle error1 = %0.1f degree" %del_1
del_2=del_2*180/pi # in degree
print "Phase angle error2 = %0.1f degree" %del_2
Unknown capacitance = 1 µF
Phase angle error1 = 3.6 degree
Phase angle error2 = 3.6 degree

## Example : 7.6 - Page No :183¶

In [12]:
#Given data
f=500 #in Hz
R2=4.8 #in ohm
R3=2*10**3 # in ohm
R4=2.85*10**3 #in ohm
C2=0.5*10**-6 #in F
r2= 0.4 # in ohm
omega=2*pi*f
C1=C2*R4/R3 #in F
print "The value of unknown capacitance = %0.4f micro F" %(C1*10**6)
r1=(R3*(R2+r2))/R4 #in ohm
print "Resistance of unknown capacitance = %0.3f ohm" %r1
Tan_del_1= omega*C1*r1
print "Dissipation factor = %0.5f" %Tan_del_1
The value of unknown capacitance = 0.7125 micro F
Resistance of unknown capacitance = 3.649 ohm
Dissipation factor = 0.00817

## Example : 7.7 - Page No : 184¶

In [13]:
#Given data
f=50 #in Hz
R2=330*10**3 #in ohm
R3=15*10**3 # in ohm
R4=22*10**3 #in ohm
C2=0.12*10**-6 #in F
omega=2*pi*f
R1= R2*R3/R4 # in ohm
print "Resistive component of unknown resistance = %0.f kohm" %(R1*10**-3)
C1= C2*R4/R3 # in F
print "Capacitive component of unknown capacitor = %0.3f micro F" %(C1*10**6)
D=1/(omega*C1*R1)
print "Dissipation factor = %0.2f" %D
Resistive component of unknown resistance = 225 kohm
Capacitive component of unknown capacitor = 0.176 micro F
Dissipation factor = 0.08

## Example : 7.8 - Page No : 191¶

In [14]:
#Given data
f=50 #in Hz
R4=309 #in ohm
R2=100 # in ohm
C3=109*10**-12 #in F
C4=0.5*10**-6 #in F
omega=2*pi*f
Cx= C3*R4/R2 # in F
print "Equivalent series capacitance = %0.2f µµF"  %(Cx*10**12)
Rx= C4*R2/C3 # in ohm
# Power factor of the specimen
Tan_delta= omega*Cx*Rx
print "Power factor of the specimen = %0.4f" %Tan_delta
Equivalent series capacitance = 336.81 µµF
Power factor of the specimen = 0.0485

## Example : 7.9 - Page No : 192¶

In [21]:
from math import cos, tan
from numpy import pi
#Given data
f=50 #in Hz
R4=1000 #in ohm
C3=50*10**-12 #in F
delta=9 # in °
epsilon_r= 2.3
epsilon_0= 8.854*10**-12
d= 0.3*10**-2 # in meter
A=314 # area of each electrode in square cm
A=A*10**-4 # in square meter
omega=2*pi*f
C1= epsilon_r*epsilon_0*A/d # in F
# Formula tan (delta)= 1/(omega*C1*R1)
R1= 1/(omega*C1*tan(delta*pi/180)) # in ohm
C4= 1/(omega**2*C1*R1*R4) # in F
print "Variable capacitor = %0.1f micro F" %(C4*10**6)
R2= C3*R4*(cos(delta*pi/180))**2/C1 # in ohm
print "Variable resistance = %0.f ohm" %R2

# Note: Calculation of R2 in the book is wrong
Variable capacitor = 0.5 micro F
Variable resistance = 229 ohm

## Example : 7.10 - Page No : 192¶

In [22]:
#Given data
f=25 #in kHz
f=f*10**3 # in Hz
R1=3.1*10**3 #in ohm
R2=25*10**3 #in ohm
R4=100*10**3 #in ohm
C1=5.2*10**-6 #in F
omega= 2*pi*f
# From C3/C1= R2/R4-R1/R3
# C3= C1*(R2/R4-R1/R3)         (i)
# and omega= 1/sqrt(R1*R3*C1*C3)
# R3= 1/(omega**2*R1*C1*C3), putting this value in (i)
C3= C1*R2/(R4*(1+R1**2*C1**2*omega**2)) # in F
print "Equivalent capacitance = %0.3f µµF" %(C3*10**12)
R3= 1/(omega**2*R1*C1*C3) # in ohm
print "Equivalent parallel resistance = %0.1f kohm" %(R3*10**-3)

# Note Evaluating the value of C3 in the book is wrong.
Equivalent capacitance = 0.203 µµF
Equivalent parallel resistance = 12.4 kohm

## Example : 7.11 - Page No :193¶

In [23]:
#Given data
R2= 5 # in ohm
R3= 2000 # in ohm
R4= 2950 # in ohm
C2= 0.5 # in miu F
C2=C2*10**-6 # in F
r2=0.4 # in  ohm
f=450 # in Hz
omega= 2*pi*f
# Under Balace Condition Z1*Z4=Z2*Z3
# [r1+1/(j*omega*C1)]*R4= [r2+R2+1/(j*omega*C2)]*R3
# Equating the real parts we have, r1*R4= (r2+R2)*R3
r1= (r2+R2)*R3/R4 # in ohm
print "Value of r1 = %0.3f ohm" %r1
# Equating imaginary parts we have R4/(j*omega*C1)= R3/(j*omega*C2)
C1= R4/R3*C2 # in F
print "Value of C1 = %0.4f micro F" %(C1*10**6)
Tan_toh= omega*C1*r1
print "Dissipation Factor = %0.3e" %Tan_toh
Value of r1 = 3.661 ohm
Value of C1 = 0.7375 micro F
Dissipation Factor = 7.634e-03

## Example : 7.12 - Page No : 195¶

In [24]:
#Given data
L1= 52.6 # in mH
r1= 28.5 # in ohm
R2= 1.68 # in ohm
R3= 80 # resistance in ohm
R4= 80 # resistance in ohm
r2= r1*R3/R4-R2 # in ohm
print "Resistance of coil = %0.2f ohm" %r2
L2=L1*R3/R4 # in mH
print "Inductance of coil = %0.1f mH" %L2
Resistance of coil = 26.82 ohm
Inductance of coil = 52.6 mH

## Example : 7.13 - Page No : 195¶

In [25]:
#Given data
L= 47.8 # in mH
R= 1.36 # in ohm
R2= 100 # in ohm
R3= 32.7 #in ohm
R4= 100 #in ohm
R1= R2*R3/R4-R # in ohm
print "Resistance of coil = %0.2f ohm" %R1
L1= R2/R4*L # in mH
print "Inductance of coil = %0.1f mH" %L1
Resistance of coil = 31.34 ohm
Inductance of coil = 47.8 mH

## Example : 7.14 - Page No : 198¶

In [28]:
#Given data
R2= 1000 # in ohm
R3= 1000 #in ohm
R4= 1000 #in ohm
C4= 0.5 # in miu F
C4= C4*10**-6 # in F
R1= R2*R3/R4 # in ohm
print "Resistance of inductor = %0.f ohm" %R1
L1= C4*R2*R3 # in H
print "Inductance of inductor = %0.1f H" %L1
Resistance of inductor = 1000 ohm
Inductance of inductor = 0.5 H

## Example : 7.15 - Page No : 200¶

In [29]:
#Given data
r= 469 # in ohm
R2= 1000 # in ohm
R3= 218 #in ohm
R4= 1000 #in ohm
C= 10 # in miu F
C= C*10**-6 # in F
R1= R2*R3/R4 # in ohm
print "Resistance of inductor = %0.f ohm" %R1
L1= C*R2/R4*(r*(R3+R4)+R3*R4) # in H
print "Inductance of inductor = %0.3f H" %L1

Resistance of inductor = 218 ohm
Inductance of inductor = 7.892 H

## Example : 7.16 - Page No : 200¶

In [30]:
#Given data
r= 500 # in ohm
R2= 400 # in ohm
R3= 400 #in ohm
R4= 400 #in ohm
C= 2 # in miu F
C= C*10**-6 # in F
R= R2*R3/R4 # in ohm
print "Resistance of AB = %0.f ohm" %R
L= C*R2/R4*(r*(R3+R4)+R3*R4) # in H
print "Inductance of AB = %0.2f H" %L
Resistance of AB = 400 ohm
Inductance of AB = 1.12 H

## Example : 7.17 - Page No : 200¶

In [32]:
#Given data
r= 100 # in ohm
R2= 1000 # in ohm
R3= 500 #in ohm
R4= 1000 #in ohm
C= 3 # in micro F
C= C*10**-6 # in F
Rx= R2*R3/R4 # in ohm
print "Value of Rx = %0.f ohm" %Rx
Lx= C*R2/R4*(r*(R3+R4)+R3*R4) # in H
print "Value of Lx = %0.2f H" %Lx
Value of Rx = 500 ohm
Value of Lx = 1.95 H

## Example : 7.18 - Page No : 203¶

In [33]:
#Given data
R2= 1000 # in ohm
R3= 16800 #in ohm
R4= 833 #in ohm
C4= 0.38 # in miu F
C4= C4*10**-6 # in F
f= 50 # in Hz
omega=2*pi*f
L1= R2*R3*C4/(1+(omega*C4*R4)**2) # in H
print "Unknown inductance = %0.2f H" %L1
R1= R2*R3*R4*omega**2*C4**2/(1+(omega*C4*R4)**2) # in ohm
print "Unknown resistance = %0.2f ohm" %R1
Unknown inductance = 6.32 H
Unknown resistance = 197.49 ohm

## Example : 7.19 - Page No : 203¶

In [35]:
#Given data
R1= 500 #in ohm
R2= 1000 # in ohm
R3= R2 #in ohm
L1= 0.18 # in H
f= 5000/(2*pi) # in Hz
omega= 2*pi*f
# L1= R2*R3*C4/(1+(omega*C4*R4)**2)     (i)
# and  R1= R2*R3*R4*omega**2*C4**2/(1+(omega*C4*R4)**2) or R1= omega**2*R4*C4*L1
R4C4= R1/(omega**2*L1)
# From eq (i)
C4= L1*(1+(omega*R4C4)**2)/(R2*R3) # in F
print "The value of C = %0.4f micro F" %(C4*10**6)
R4= R4C4/C4 # in ohm
print "The value of R4 = %0.f ohm" %R4
The value of C = 0.2356 micro F
The value of R4 = 472 ohm

## Example : 7.20 - Page No : 204¶

In [37]:
#Given data
R2= 1000 #in ohm
R3= 10000 # in ohm
R4= 2000 #in ohm
C4= 1*10**-6 # in F
L1= R2*R3*C4/(1+(omega*C4*R4)**2) # in H
print "Equivalent inductance of the network = %0.2f H" %L1
R1= R2*R3*R4*omega**2*C4**2/(1+(omega*C4*R4)**2) # in ohm
print "Equivalent resistance of the network = %0.3f kohm" %(R1*10**-3)
Equivalent inductance of the network = 0.27 H
Equivalent resistance of the network = 4.865 kohm

## Example : 7.21 - Page No : 204¶

In [38]:
#Given data
R2= 2410 #in ohm
R3= 750 # in ohm
R4= 64.5 #in ohm
C4= 0.35*10**-6 # in F
r4= 0.4 # series resistance of capacitor in ohm
f=500 #/ in Hz
R4= R4+r4 # in ohm
R1= R2*R3*R4*omega**2*C4**2/(1+(omega*C4*R4)**2) # in ohm
print "Resistance of the choke coil = %0.2f ohm" %R1
L1= R2*R3*C4/(1+(omega*C4*R4)**2) # in H
print "Inductance of the choke coil = %0.4f H" %L1

# Note: Calculation of finding the value of L1 in the book is wrong
Resistance of the choke coil = 141.11 ohm
Inductance of the choke coil = 0.6294 H

## Example : 7.22 - Page No :205¶

In [50]:
from math import atan2
#Given data
R2= 834 # in Ω
R3= 100 # in Ω
C2= 0.124 # in µF
C2= C2*10**-6 # in F
C4= 0.1 # in µF
C4= C4*10**-6 # in F
L1= R2*R3*C4 # in H
f= 2 # in kHz
f= f*10**3 # in kHz
print "The value of L1 = %0.2f mH" %(L1*10**3)
R1= R3*C4/C2 # in Ω
print "The value of R1 = %0.2f Ω" %R1
pi_2_f_L1= 2*pi*f*L1 # value of 2*pi*f*L1
i= 1j # complex number
i_XL= i*pi_2_f_L1 #imaginary part
Z= R1+i_XL # impedance in ohm
print "The magnitude of effective impedence = %0.2f Ω" %abs(Z)
theta= atan2(Z.imag,Z.real)*180/pi
print "The angle of effective impedence = %0.2f°" %theta
The value of L1 = 8.34 mH
The value of R1 = 80.65 Ω
The magnitude of effective impedence = 132.24 Ω
The angle of effective impedence = 52.42°

## Example : 7.23 - Page No : 211¶

In [51]:
#Given data
fr= 2 # in MHz
fr=fr*10**6 # in Hz
C=230+8 # in pF
C=C*10**-12 # in F
# Formula fr= 1/(2*pi*sqrt(L*C))
L= 1/((2*pi*fr)**2*C) # in H
print "Value of L = %0.1f µH" %(L*10**6)
# From the first set of data
fr= 1 # in MHz
fr=fr*10**6 # in Hz\
C= 1/((2*pi*fr)**2*L) # in F
print "Value of C = %0.f pF" %(C*10**12)
Value of L = 26.6 µH
Value of C = 952 pF

## Example : 7.24 - Page No : 212¶

In [52]:
#Given data
C1= 208 # in pF
C1=C1*10**-12 # in F
Q1= 80
C2= 184 # in pF
C2=C2*10**-12 # in F
Q2= 50
f=165 # in kHz
f=f*10**3 # in Hz
# Part (i)
Rm= 1/omega*(1/(C2*Q2)-1/(C1*Q1)) # in ohm
print "Resistive component of unknown impedence = %0.2f ohm" %Rm
# Part(ii)
Xm= 1/omega*(1/C2-1/C1) # in ohm
print "Reactive component of unknown impedence = %0.f ohm" %Xm
Resistive component of unknown impedence = 46.88 ohm
Reactive component of unknown impedence = 605 ohm

## Example : 7.25 - Page No : 212¶

In [53]:
#Given data
C1= 160*10**-12 # in F
C2= 36*10**-12 # in F
f1=250 # in kHz
f1=f1*10**3 # in Hz
f2=500 # in kHz
f2=f2*10**3 # in Hz
Cd= (C1-4*C2)/3 # in F
print "Self Capacitance of the coil = %0.2f µµF" %(Cd*10**12)
# Formula f1= 1/(2*pi*sqrt(L*(C1+Cd)))
L= 1/((2*pi*f1)**2*(C1+Cd)) # in H
print "Self inductance of the coil = %0.f µH" %(L*10**6)
Self Capacitance of the coil = 5.33 µµF
Self inductance of the coil = 2451 µH

## Example : 7.26 - Page No : 213¶

In [54]:
#Given data
C1= 251*10**-12 # in F
C2= 50*10**-12 # in F
f1=3 # in MHz
f1=f1*10**6 # in Hz
f2=6 # in MHz
f2=f2*10**6 # in Hz
Cd= (C1-4*C2)/3 # in F
print "Self Capacitance of the coil = %0.f pF" %(Cd*10**12)
Self Capacitance of the coil = 17 pF

## Example : 7.27 - Page No : 213¶

In [55]:
#Given data
C1= 1530 # in pF
C2= 162 # in pF
f1=1 # in MHz
f1=f1*10**6 # in Hz
f2=3 # in MHz
f2=f2*10**6 # in Hz
# f1= 1/(2*pi*sqrt(L*(C1+Cd)))
#  f1= 1/(2*pi*sqrt(L*(C2+Cd))) and f2= 3*f1 so
Cd= (C1-9*C2)/8 # in pF
print "Self capacitance of the coil = %0.f pF" %Cd
Self capacitance of the coil = 9 pF

## Example : 7.28 - Page No : 213¶

In [56]:
#Given data
f= 450 # in kHz
f=f*10**3 # in Hz
C=250 # in pF
C=C*10**-12 # in F
Rsh= 0.75 # in ohm
Q= 105
# Formula f= 1/(2*pi*sqrt(L*C))
L= 1/((2*pi*f)**2*C) # in H
print "Inductance of the coil = %0.f µH" %(L*10**6)
R= omega*L/Q-Rsh # in ohm
print "Resistance of the coil = %0.2f ohm" %R
Inductance of the coil = 500 µH
Resistance of the coil = 12.72 ohm

## Example : 7.29 - Page No : 213¶

In [57]:
#Given data
f= 500 # in kHz
f=f*10**3 # in Hz
C=120 # in pF
C=C*10**-12 # in F
R= 5 # in ohm
r=0.02 # resistance used across the oscillatory circuit in ohm
Q_True= 1/(omega*C*R)
Q_indicated= 1/(omega*C*(R+r))
PerError= (Q_True-Q_indicated)*100/Q_True # in %
print "Percentage Error = %0.1f %%" %PerError
Percentage Error = 0.4 %

## Example : 7.30 - Page No : 214¶

In [58]:
#Given data
f1= 800 # in kHz
f1=f1*10**3 # in Hz
f2= 2.5 # in MHz
f2=f2*10**6 # in Hz
C1=95 # in pF
C1=C1*10**-12 # in F
# L= 1/(omega1**2*(C1+Cd))        (i)
# L= 1/(omega2**2*Cd)                (ii)
# From eq(i) and eq(ii)
Cd= f1**2*C1/(f2**2-f1**2) # in F
print "Self capacitance of the radio coil = %0.2f pF" %(Cd*10**12)
Self capacitance of the radio coil = 10.84 pF

## Example : 7.31 - Page No : 214¶

In [60]:
#Given data
f1= 1 # in MHz
f1=f1*10**6 # in Hz
f2= 2 # in MHz
f2=f2*10**6 # in Hz
C1=480 # in pF
C1=C1*10**-12 # in F
C2=90 # in pF
C2=C2*10**-12 # in F
R=10 # in ohm

# Part (i)
Cd= (C1-4*C2)/3 # in F
print "(i)  : Self capacitance of the coil = %0.f pF" %(Cd*10**12)

# Part(ii)
Q_indicated1= 1/(omega1*(C1+Cd)*R)
print "(ii) : Indicated or effective Q for first measurement = %0.3f" %Q_indicated1
Q_True1=  1/(omega1*C1*R)
print "True Q for first measurement = %0.3f " %Q_True1
Q_indicated2= 1/(omega2*(C2+Cd)*R)
print "Indicated or effective Q for second measurement = %0.3f" %Q_indicated2
Q_True2=  1/(omega2*C2*R)
print "True Q for second measurement = %0.2f" %Q_True2
(i)  : Self capacitance of the coil = 40 pF
(ii) : Indicated or effective Q for first measurement = 30.607
True Q for first measurement = 33.157
Indicated or effective Q for second measurement = 61.213
True Q for second measurement = 88.42