# Chapter 6 : A.C. Bridges¶

## Example - 6.1 : Page No - 6.4¶

In :
from __future__ import division
from cmath import rect
#Given data
Z1 = 50 # in ohm
Z2 = 250 # in ohm
Z3 = 200 # in ohm
theta1 = 80 # in degree
theta2 = 0 # in degree
theta3 = 30 # in degree
#bridge balance equation, Z1*Z4 = Z2*Z3
Z4 = (Z2*Z3)/Z1 # in ohm
#phase angle condition, theta1+theta4 = theta2+theta3
theta4 = theta2+theta3-theta1 # in degree
theta4= theta4*pi/180 # in radian
Z4= rect(Z4,theta4)
print "The resistance part of Z4 = %0.2f Ω" %Z4.real
print "while it is in series with capacitive reactance of",abs(round(Z4.imag,1)),"Ω"

The resistance part of Z4 = 642.79 Ω
while it is in series with capacitive reactance of 766.0 Ω


## Example - 6.2 : Page No - 6.4¶

In :
#Given data
Z1 = 50 # in ohm
Z2 = 100 # in ohm
Z3 = 15 # in ohm
Z4 = 30 # in ohm
theta1 = 40 # in degree
theta2 = -90 # in degree
theta3 = 45 # in degree
theta4 = 30 # in degree
if abs(Z1*Z4)== abs(Z3*Z2) :
flag1=1
print "The condition of balance for magnitude is satisfied"
else :
flag1=0
print "The condition of balance for magnitude is not satisfied"
if theta1+theta4 == theta2+theta3 :
flag2=1
print "The condition of balance for phase is also satisfied"
else :
flag2=0
print "But the condition of balance for phase is not satisfied"

if flag1==1 :
if flag2==1 :
print "Hence the bridge is under balanced condition"
else :
print "Hence the bridge is not under balanced condition"
else :
print "Hence the bridge is not under balanced condition"

The condition of balance for magnitude is satisfied
But the condition of balance for phase is not satisfied
Hence the bridge is not under balanced condition


## Example - 6.3 : Page No - 6.6¶

In :
#Given data
C3 = 10 # in µF
C3 = C3*10**-6 # in F
R1 = 1.2 # in k ohm
R1 = R1 * 10**3 # in ohm
R2 = 100 # in k ohm
R2 = R2 * 10**3 # in ohm
R3 = 120 # in k ohm
R3 = R3 * 10**3 # in ohm
Rx = (R2*R3)/R1 #unknown resistance in ohm
Rx = Rx * 10**-6 # in M ohm
print "The value of Rx = %0.f MΩ " %Rx
Cx = (R1*C3)/R2 # in F
Cx = Cx * 10**6 #unknown capacitance in µF
print "The value of Cx = %0.2f µF " %Cx

The value of Rx = 10 MΩ
The value of Cx = 0.12 µF


## Example - 6.4 : Page No - 6.8¶

In :
#Given data
L3 = 8 # in mH
L3 = L3 * 10**-3 # in H
R1 = 1 # in k ohm
R1 = R1 * 10**3 # in ohm
R2 = 25 # in k ohm
R2 = R2 * 10**3 # in ohm
R3 = 50 # in  k ohm
R3 = R3 * 10**3 # in ohm
Rx = (R2*R3)/R1 #unknown resistance in ohm
Rx = Rx * 10**-6 # in M ohm
print "The value of Rx = %0.2f MΩ " %Rx
Lx = (R2*L3)/R1 #unknown inductance in H
Lx = Lx * 10**3 # in mH
print "The value of Lx = %0.f mH " %Lx

The value of Rx = 1.25 MΩ
The value of Lx = 200 mH


## Example - 6.5 : Page No - 6.13¶

In :
#Given data
C1 = 0.5 # in µF
C1 = C1 * 10**-6 # in µF
R1 = 1200 # in ohm
R2 = 700 # in ohm
R3 = 300 # in ohm
# From bridge balance equation
Rx = (R2*R3)/R1 # in ohm
print "Component of the brach BC :"
print "Rx = ",int(Rx),"Ω"
Lx = R2*R3*C1 # in H
Lx = Lx * 10**3 # in mH
print "Lx = ",int(Lx),"mH"

Component of the brach BC :
Rx =  175 Ω
Lx =  105 mH


## Example - 6.6 : Page No - 6.16¶

In :
#Given data
R2 = 1000 #resistance in ohm
R3 = 500 # resistance in ohm
R4 = 1000 # resistance in ohm
C = 3 #capacitance in µF
C = C * 10**-6 # in F
r = 100 # in ohm
Rx = (R2*R3)/R4 #value of Rx in ohm
print "The value of Rx = %0.f Ω " %Rx
Lx = ((C*R2)/R4)*( (R3*r) + (R4*r) + (R3*R4) ) #value of Lx in H
print "The value of Lx = %0.2f H " %Lx

The value of Rx = 500 Ω
The value of Lx = 1.95 H


## Example - 6.7 : Page No - 6.20¶

In :
#Given data
R1 = 5.1 # in k ohm
R1 = R1 * 10**3 # in ohm
R2 = 7.9 # in k ohm
R2 = R2 * 10**3 # in ohm
R3 = 790 # in ohm
C1 = 2 # in µF
C1 = C1 * 10**-6 # in F
omega = 1000 # in rad/sec
Rx = (((omega)**2)*R1*((C1)**2)*R2*R3)/( 1+(((omega)**2) * ((R1)**2)* ((C1)**2)) ) # unknown resistance in ohm
Rx = Rx * 10**-3 # in k ohm
print "The value of unknown resistance = %0.3f kΩ " %Rx
Lx = (R2*R3*C1)/( 1+(((omega)**2) * ((R1)**2)* ((C1)**2)) ) # unknown inductance in H
Lx = Lx * 10**3 # in mH
print "The value of unknown inductance = %0.2f mH " %Lx

The value of unknown resistance = 1.212 kΩ
The value of unknown inductance = 118.83 mH


## Example - 6.8 : Page No - 6.24¶

In :
from numpy import pi
#Given data
R1 = 1.2 # in k ohm
R1 = R1 * 10**3 # in ohm
R2 = 4.7 # in k ohm
R2 = R2 * 10**3 # in ohm
C1 = 1 # in µF
C1 = C1 * 10**-6 # in F
C3 = 1 # in µF
C3 = C3 * 10**-6 # in F
Rx = (R2*C1)/C3 # unknown resistance in ohm
Rx = Rx * 10**-3 # in k ohm
Cx = (R1*C3)/R2 # unknown capacitance in F
Cx = Cx * 10**6 # in µF
print "The unknown resistance = %0.1f kΩ is " %Rx
print "The unknown capacitance = %0.3f µF " %Cx
f = 0.5 # in kHz
f = f * 10**3 # in Hz
# omega = 2*pi*f
D = 2*pi*f*Cx*10**-6*Rx*10**3 # dissipation factor
print "The dissipation factor = %0.3f " %D

The unknown resistance = 4.7 kΩ is
The unknown capacitance = 0.255 µF
The dissipation factor = 3.770


## Example - 6.9 : Page No - 6.28¶

In :
#Given data
R1 = 2.7 # in k ohm
R1 = R1 * 10**3 # in ohm
R2 = 22 # in k ohm
R2 = R2 * 10**3 # in ohm
R4 = 100 # in k ohm
R4 = R4 * 10**3 # in ohm
C1 = 5 # in µF
C1 = C1 * 10**-6 # in F
f = 2.2 # in kHz
f = f * 10**3 # in Hz
#From omega**2 = 1/(R1*C1*R3*C3)
# C3 = 1/(R1*C1*R3*(omega**2))            (i)
# R2/R4 = R1/R3 + C3/C1                   (ii)
# From eq(i) and (ii)
R3 = R4*(R1+1/((2*pi*f)**2*R1*C1**2))/R2 # equivalent parallel resistance in ohm
R3= R3*10**-3 # in k ohm
print "The equivalent parallel resistance = %0.3f kΩ " %R3
R3= R3*10**3 # in ohm
C3 = 1/(R1*C1*R3*((2*pi*f)**2)) # equivalent parallel capacitance in F
C3 = C3 * 10**12 # in pF
print "The equivalent parallel capacitance = %0.2f pF " %C3

The equivalent parallel resistance = 12.273 kΩ
The equivalent parallel capacitance = 31.59 pF


## Example - 6.10 : Page No - 6.47¶

In :
from __future__ import division
#Given data
C1 = 550 # in pF
C2 = 110 # in pF
Cd = (C1-(4*C2))/3 # distributed capacitance in pF
print "The distributed capacitance = %0.2f pF " %Cd
Cd = Cd * 10**-12 # in F
C1 = C1 * 10**-12 # in F
f1 = 1.5 # in MHz
f1 = f1 * 10**6 # in Hz
# f1 = 1/(2*pi*(sqrt( L*(C1+Cd))))
L = ((1/(2*pi*f1))**2) * (1/(C1+Cd)) # distributed inductance in H
L = L * 10**6 # in µH
print "The distributed inductance = %0.2f µH " %L

The distributed capacitance = 36.67 pF
The distributed inductance = 19.19 µH


## Example - 6.11 : Page No - 6.48¶

In :
#Given data
f = 1.5 #frequency in MHz
f = f * 10**6 # in Hz
C = 60 # in pF
C = C * 10**-12 # in F
R = 8 # in ohm
R_SH = 0.02 # in ohm
omega = 2*pi*f
Qactual = 1/(omega*C*R) # true value of Q
Qobserved = 1/(omega*C*(R+R_SH)) # observed value of Q
PerError = ((Qactual-Qobserved)/Qactual) * 100 # Percentage error in %
print "The Percentage error = %0.2f %% " %PerError

The Percentage error = 0.25 %


## Example - 6.12 : Page No - 6.49¶

In :
#Given data
f1 = 2 #frequency in MHz
f1 = f1 * 10**6 # in Hz
C1 = 500 # in pF
C2 = 60 # in pF
# f1 = 1/(2*pi*sqrrt(L*(C1+Cd)))  (i)
# f2 = 1/(2*pi*sqrrt(L*(C2+Cd)))  (ii)
# and f2 = 2.5*f1      (iii)
#From eq(i),(ii) and (iii)
Cd = (C1 - (6.25*C2))/5.25 # value of self capacitance in pF
print "The value of self capacitance = %0.2f pF " %Cd

The value of self capacitance = 23.81 pF


## Example - 6.13 : Page No - 6.50¶

In :
#Given data
f = 1 # in MHz
f = f * 10**6 # in Hz
omega = 2*pi*f # in rad/sec
C = 65 # in pF
C = C * 10**-12 # in F
R = 10 # in ohm
R_SH = 0.02 # in ohm
# Q = X_L/R = X_C/R = 1/(omega*C*R)
Qactual = 1/(omega*C*R) # True value of Q
Qmeasured = 1/(omega*C*(R+R_SH)) # measured value of Q
PerError = ((Qactual-Qmeasured)/Qactual)*100 #percentage error in %
print  "The Percentage error = %0.1f %% " %PerError

The Percentage error = 0.2 %


## Example - 6.14 : Page No - 6.51¶

In :
#Given data
C1 = 450 #capacitance in pF
C1 = C1 * 10**-12 # in F
C2 = 60 #capacitance in pF
C2 = C2 * 10**-12 # in F
# f1 = 1/(2*pi*(sqrt(L*(C1+Cd))))     (i)
# f2 = 1/(2*pi*(sqrt(L*(C2+Cd))))     (ii)
# and f2 = 2.5*f1      (iii)
# from eq(i),(ii) and (iii)
Cd = (C1 - (6.25*C2))/5.25 # value of self capacitance in F
Cd = Cd * 10**12 # in pF
print "The value of self capacitance = %0.2f pF " %Cd

The value of self capacitance = 14.29 pF


## Example - 6.15 : Page No - 6.52¶

In :
from __future__ import division
#Given data
f1 = 8 #frequency in MHz
f1= f1*10**6 # in Hz
f2 = 12 #frequency in MHz
f2= f2*10**6 # in Hz
C1 = 120 #capacitance in pF
C1 = C1 * 10**-12 # in F
C2 = 40 #capacitance in pF
C2 = C2 * 10**-12 # in F
# f1 = 1/(2*pi*(sqrt(L*(C1+Cd))))     (i)
# f2 = 1/(2*pi*(sqrt(L*(C2+Cd))))     (ii)
# From eq(i) and (ii)
Cd= (f2**2*C2-f1**2*C1)/(f1**2-f2**2) # in F
# From eq(i)
C=C1+Cd
L=1/((C1+Cd)*(2*pi*f1)**2) # inductance in H
L= L*10**6 # in µH
Cd= Cd*10**12 # self capacitance in pF
print "The self capacitance = %0.1f pF " %Cd
print "The inductance = %0.3f µH " %L

The self capacitance = 24.0 pF
The inductance = 2.749 µH


## Example - 6.16 : Page No - 6.53¶

In :
#Given data
f1 = 1 #frequency in MHz
f1= f1*10**6 # in Hz
f2 = 2 #frequency in MHz
f2= f2*10**6 # in Hz
C1 = 500 #capacitance in pF
C1 = C1 * 10**-12 # in F
C2 = 110 #capacitance in pF
C2 = C2 * 10**-12 # in F
# f1 = 1/(2*pi*(sqrt(L*(C1+Cd))))     (i)
# f2 = 1/(2*pi*(sqrt(L*(C2+Cd))))     (ii)
# From eq(i) and (ii)
Cd= (f2**2*C2-f1**2*C1)/(f1**2-f2**2) # in F
# From eq(i)
C=C1+Cd
L=1/((C1+Cd)*(2*pi*f1)**2) # in H
L= L*10**6 #inductance in µH
Cd= Cd*10**12 # self capacitance in pF
print "The self capacitance = %0.f pF " %Cd
print "The inductance = %0.2f µH " %L

The self capacitance = 20 pF
The inductance = 48.71 µH


## Example - 6.17 : Page No - 6.54¶

In :
#Given data
f = 1 #frequency in kHz
f = f * 10**3 # in Hz
R1 = 400 #resistance in ohm
R2 = 150 #resistance in ohm
C2 = 0.2 #capacitance in µF
C2 = C2 * 10**-6 # in F
XC2= 1/(2*pi*f*C2)
R3 = 100 #resistance in ohm
L3 = 10 #inductance in mH
L3 = L3 * 10**-3 # in H
XL3= 2*pi*f*L3
Z1= complex(R1,0) # in Ω
Z2= R2-1j*XC2 # in Ω
Z3= R3+1j*XL3 # in Ω
Z4= Z2*Z3/Z1 # in Ω
R4= Z4.real#resistance in Ω
XC4= abs(Z4.imag) # in Ω
C4= 1/(2*pi*f*XC4) # in F
C4= C4*10**6 # in µF
print "The components of branch CD : "
print "R4 = ",round(R4,1)," Ω"
print "C4 = ",round(C4,4)," µF"

The components of branch CD :
R4 =  162.5  Ω
C4 =  0.9075  µF


## Example - 6.18 : Page No - 6.55¶

In :
#Given data
C3 = 10 #capacitance in µF
C3 = C3 * 10**-6 # in F
R1 = 1.2 #resistance in  k ohm
R1 = R1 * 10**3 # in ohm
R2 = 100 #resistance in k ohm
R2 = R2 * 10**3 # in ohm
R3 = 120 #resistance in k ohm
R3 = R3 * 10**3 # in ohm
Rx = (R2*R3)/R1 #resistance of unknown impedance  in ohm
Rx = Rx * 10**-6 # in M ohm
print "The resistance of unknown impedance = %0.f MΩ " %Rx
Cx = (R1*C3)/R2 #capacitance of unknown impedance  in F
Cx = Cx * 10**6 # in µF
print "The capacitance of unknown impedance = %0.2f µF " %Cx

The resistance of unknown impedance = 10 MΩ
The capacitance of unknown impedance = 0.12 µF


## Example - 6.19 : Page No - 6.56¶

In :
#Given data
f = 1000 #frequency in Hz
C1 = 0.2 #capacitance in µF
C1 = C1 * 10**-6 # in F
XC1= 1/(2*pi*f*C1)
R2= 500 # in Ω
R3= 300 # in Ω
C3= 0.1*10**-6 # in F
XC3= 1/(2*pi*f*C3)
omega = 2*pi*f # in rad/sec
Z1= 0-1j*XC1 # in Ω
Z2= R2 # in Ω
Y3= 1/R3+1j*1/XC3 # in Ω
Z3= R3*XC3/(R3+XC3) # in Ω
Z4= Z2/(Z1*Y3) # in Ω
R4= Z4.real # in Ω
XL4= abs(Z4.imag) # in Ω
L4= XL4/(2*pi*f) # in F
L4= L4*10**3 # in mH
print "The components of branch CD : "
print "Rx = ",round(R4,2),"Ω"
print "Lx = %0.f mH" %L4

The components of branch CD :
Rx =  34.31 Ω
Lx = 29 mH


## Example - 6.20 : Page No - 6.57¶

In :
#Given data
f = 2 # in kHz
f = f * 10**3 # in Hz
omega = 2*pi*f # in rad/sec
Z1 = 10 # in k ohm
Z2 = 50 # in k ohm
R3 = 100 # in k ohm
C3 = 100 # in µF
C3 = C3 * 10**-6 # in F
XC3= 1/(2*pi*f*C3)
Z3= R3-1j*XC3 # in Ω
# From balance equation, Z1*Z4= Z2*Z3
Z4= Z2*Z3/Z1 # in Ω
R4= Z4.real # in kΩ
XC4= abs(Z4.imag) # in kΩ
C4= 1/(2*pi*f*XC4) # in F
C4= C4*10**6 # in µF
print "The components of branch DC : "
print "Rx= %0.f kΩ" %R4
print "Cx= %0.f µF" %C4

The components of branch DC :
Rx= 500 kΩ
Cx= 20 µF


## Example - 6.21 : Page No - 6.59¶

In :
#Given data
f = 1 # in kHz
f = f * 10**3 # in Hz
omega = 2*pi*f # in rad/sec
Z1 = 200 # in ohm
R2 = 200 # in ohm
C2 = 5 # in µF
C2 = C2 * 10**-6 # in F
XC2= 1/(2*pi*f*C2)
Z2= R2-1j*XC2 # in Ω
R3 = 500 # in ohm
C3 = 0.2 # in µF
C3 = C3 * 10**-6 # in F
XC3= 1/(2*pi*f*C3)
Z3= R3-1j*XC3 # in Ω
# From balance equation, Z1*Z4= Z2*Z3
Z4= Z2*Z3/Z1 # in Ω
R4= Z4.real# in Ω
XC4= abs(Z4.imag) # in Ω
C4= 1/(2*pi*f*XC4) # in F
C4= C4*10**9 # in nF
print "The components of Zx : "
print "Rx = ",round(R4,3),"Ω"
print "Cx = ",round(C4,3),"nF"

The components of Zx :
Rx =  373.349 Ω
Cx =  181.818 nF


## Example - 6.22 : Page No - 6.60¶

In :
#Given data
f = 1 # in kHz
f = f * 10**3 # in Hz
omega = 2*pi*f # in rad/sec
Z1 = 1.65 # in k ohm
Z2 = 15.3 # in k ohm
R3 = 2.5 # in k ohm
C3 = 10 # in µF
C3 = C3 * 10**-6 # in F
XC3= 1/(2*pi*f*C3)
Z3= R3-1j*XC3 # in Ω
# From balance equation, Z1*Z4= Z2*Z3
Z4= Z2*Z3/Z1 # in Ω
R4= Z4.real # in kΩ
XC4= abs(Z4.imag) # in kΩ
C4= 1/(2*pi*f*XC4) # in F
C4= C4*10**6 # in µF
print "The components of branch DC : "
print "Rx = ",round(R4,3)," kΩ"
print "Cx = ",round(C4,2)," µF"

The components of branch DC :
Rx =  23.182  kΩ
Cx =  1.08  µF


## Example - 6.23 : Page No - 6.61¶

In :
#Given data
f = 1 # in kHz
f = f * 10**3 # in Hz
R1 = 600 # in ohm
C1 = 1 # in µF
C1 = C1 * 10**-6 # in F
XC1= 1/(2*pi*f*C1)
R2 = 100 # in ohm
R3 = 1 # in k ohm
R3 = R3 * 10**3 # in ohm
omega = 2*pi*f # in rad/sec
Y1= 1/R1+1j*1/XC1 # in Ω
Z2=R2 # in Ω
Z3= R3 # in Ω
# From balance equation, Z1*Z4= Z2*Z3
Z4= Z2*(Z3*Y1) # in Ω
R4= Z4.real # in Ω
XL4= abs(Z4.imag) # in Ω
L4= XL4/(2*pi*f) # in F
print "Rx= ",int(R4)," Ω"
print "Lx= %0.3f H" %L4

Rx=  166  Ω
Lx= 0.100 H


## Example - 6.24 : Page No - 6.62¶

In :
#Given data
R2 = 842 #resistance in ohm
C2 = 0.135 #capacitance in µF
C2 = C2 * 10**-6 # in F
f=1000 #frequency in Hz
XC2= 1/(2*pi*f*C2)
R3= 10 #resistance in ohm
C4= 1*10**-6 #capacitance in F
XC4= 1/(2*pi*f*C4)
Z2= R2-1j*XC2 #impedance in ohm
Z3= complex(R3,0) #impedance in ohm
Z4= -1j*XC4 #impedance in ohm
# From balance equation
Z1= Z2*Z3/Z4 # in Ω
R1= Z1.real # in Ω
XL1= abs(Z1.imag) # in Ω
L1= XL1/(2*pi*f) # in F
L1= L1*10**3 # in mH
print "The value of R1 = %0.3f Ω " %R1
print "The value of L1 = %0.2f mH " %L1

The value of R1 = 74.074 Ω
The value of L1 = 8.42 mH


## Example - 6.25 : Page No - 6.63¶

In :
#Given data
L2 = 47.8 #inductance in mH
R2 = 1.36 #resistance in ohm
r1 = 32.7 #resistance in ohm
R1 = 1.36 #resistance in ohm
#At balance, 100*(r1+J*oemga*L1) = 100*((R2+r2)+(J*omega*L2))
L1 = L2 # in mH (equating imaginary terms)
print "The inductance of coil = %0.1f mH " %L1
# R2+r2 = r1 (equating real terms)
r2 = r1-R1 #resistance of coil in ohm
print "The resistance of coil = %0.2f ohm " %r2

# Note: In the book the value of L1 is wrong.

The inductance of coil = 47.8 mH
The resistance of coil = 31.34 ohm


## Example - 6.26 : Page No - 6.64¶

In :
#Given data
R=1.36 #resistance in ohm
r2= 32.7 #resistance in ohm
L2= 47.8 #inductance in mH
L2= L2*10**-3 # in H
f=1000 #frequency in Hz
XL2=2*pi*f*L2 # in Ω
Z3 = 100 # in ohm
Z4 = 100 # in ohm
Z2= r2+1j*XL2 # in ohm
# Under balance condition
Z1= Z2*Z3/Z4 # in ohm
R1= Z1.real # in ohm
r1= R1-R #resistance of the coil in ohm
XL1= Z1.imag # in ohm
L1= XL1/(2*pi*f) #inductance of the coil in F
L1= L1*10**3 # in mH
print "The resistance of the coil = %0.2f Ω " %r1
print "The inductance of the coil = %0.1f mH " %L1

The resistance of the coil = 31.34 Ω
The inductance of the coil = 47.8 mH


## Example - 6.27 : Page No - 6.66¶

In :
#Given data
Z1 = 400 # in ohm
Z2 = 200 # in ohm
Z3 = 800 # in ohm
Z4 = 400 # in ohm
theta1 = 50 # in degree
theta2 = 40 # in degree
theta3 = -50 # in degree
theta4 = 20 # in degree
if abs(Z1*Z4) == abs(Z3*Z2) : # Applying the condition of balance for magnitude
flag1=1
print "The condition of balance for magnitude is satisfied"
else :
flag1=0
print "The condition of balance for magnitude is not satisfied"

if theta1+theta4==theta2+theta3 : # Applying the condition of balance for phases
flag2=1
print "The condition of balance for phase is also satisfied"
else :
flag2=0
print "But the condition of balance for phase is not satisfied"

if flag1==1 :
if flag2==1 :
print "Hence the bridge is under balanced condition"
else :
print "Hence the bridge is not under balanced condition"
else :
print "Hence the bridge is not under balanced condition"

The condition of balance for magnitude is satisfied
But the condition of balance for phase is not satisfied
Hence the bridge is not under balanced condition


## Example - 6.28 : Page No - 6.67¶

In :
#Given data
Z1 = 200 # in ohm
Z2 = 400 # in ohm
Z3 = 300 # in ohm
Z4 = 600 # in ohm
theta1 = 60 # in degree
theta2 = -60 # in degree
theta3 = 0 # in degree
theta4 = 30 # in degree
if abs(Z1*Z4)== abs(Z3*Z2) : # Applying the condition of balance for magnitude
flag1=1
print "The condition of balance for magnitude is satisfied"
else :
flag1=0
print "The condition of balance for magnitude is not satisfied"

if theta1+theta4==theta2+theta3 : # Applying the condition of balance for phases
flag2=1
print "The condition of balance for phase is also satisfied"
else :
flag2=0
print "But the condition of balance for phase is not satisfied"

if flag1==1 :
if flag2==1 :
print "Hence the bridge is under balanced condition"
else :
print "Hence the bridge is not under balanced condition"

else :
print "Hence the bridge is not under balanced condition"

The condition of balance for magnitude is satisfied
But the condition of balance for phase is not satisfied
Hence the bridge is not under balanced condition


## Example - 6.29 : Page No - 6.68¶

In :
#Given data
f = 1 #frequency in kHz
f = f * 10**3 # in Hz
C1 = 0.2 # in µF
C1 = C1 * 10**-6 # in F
XC1= 1/(2*pi*f*C1) # in Ω
C2 = 0.1 # in µF
C2 = C2 * 10**-6 # in F
XC2= 1/(2*pi*f*C2) # in Ω
R2= 300 # in Ω
R3= 500 # in Ω
Z1= 0-1j*XC1 # in Ω
Z2= R2*-1j*XC2/(R2-1j*XC2) # in Ω
Z3=R3 # in Ω
# For balanced condition
Z4= Z2*Z3/Z1 # in Ω
R4= Z4.real # in Ω
XL4= Z4.imag # in Ω
L4= XL4/(2*pi*f) # in H
L4= L4*10**3 # in mH
print "Components of arm CD : "
print "L4 =",round(L4,2),"mH"
print "R4 =",round(R4,4),"Ω"

Components of arm CD :
L4 = 28.97 mH
R4 = 34.3115 Ω


## Example - 6.30 : Page No - 6.69¶

In :
#Given data
R3 = 100 # in ohm
R4 = 200 # in ohm
R2 = 250 # in ohm
C = 1 # in µF
C = C * 10**-6 # in F
r = 229.7 # in ohm
r1 = 43.1 # in ohm
# Value of unknown resistance for Anderson's bridge
R1 = ((R2*R3)/R4) - r1 #resistance in ohm
print "The resistance = %0.1f ohm " %R1
L1 = ((C*R3)/R4) * ( ((R2+R4)*r) + (R2*R4) ) #inductance in H
L1 = L1 * 10**3 # in mH
print "The inductance = %0.4f mH " %L1

The resistance = 81.9 ohm
The inductance = 76.6825 mH


## Example - 6.31 : Page No - 6.70¶

In :
#Given data
f = 450 #frequency in Hz
omega = 2*pi*f # in rad/sec
R2 = 4.8 # in ohm
R3 = 200 # in ohm
R4 = 2850 # in ohm
C2 = 0.5 # in µF
C2 = C2*10**-6 # in F
XC2= 1/(2*pi*f*C2) # in Ω
r2 = 0.4 # in ohm
Z2= (R2+r2)-1j*XC2 # in Ω
Z3= R3 # in Ω
Z4= R4 # in Ω
# For balanced condition
Z1= Z2*Z3/Z4 # in Ω
r1= Z1.real # in Ω
XC1= abs(Z1.imag) # in Ω
C1= 1/(2*pi*f*XC1) # in F
Df= 2*pi*f*C1*r1 # dissipating factor
C1= C1*10**6 # in µF
print "The value of r1 = %0.4f Ω " %r1
print "The value of C1 = %0.3f µF " %C1
print "The dissipating factor = %0.6f " %Df

The value of r1 = 0.3649 Ω
The value of C1 = 7.125 µF
The dissipating factor = 0.007351


## Example - 6.32 : Page No - 6.72¶

In :
#Given data
f = 2 #frequency in kHz
f = f * 10**3 # in Hz
omega = 2*pi*f # in rad/sec
R2 = 834 #resistance in ohm
C2 = 0.124 #capacitance in µF
C2 = C2 * 10**-6 # in F
XC2= 1/(2*pi*f*C2) # in Ω
R3= 100 #resistane in Ω
C4= 0.1*10**-6 #capacitance in F
XC4= 1/(2*pi*f*C4) # in Ω
Z2= R2-1j*XC2 # in Ω
Z3=R3 # in Ω
Z4= 0-1j*XC4 # in Ω
# For balanced condition, effective impedance
Z1= Z2*Z3/Z4 #in Ω
print "The effective impedance = (",round(Z1.real,4),"+ j",round(Z1.imag,4),")Ω"

The effective impedance = ( 80.6452 + j 104.8035 )Ω


## Example - 6.33 : Page No - 6.73¶

In :
#Given data
R1= 20*10**3 #resistance in ohm
R2= 50*10**3 #resistance in ohm
C2= 0.003*10**-6 #capacitance in F
R4= 10*10**3 #resistance in ohm
C1= 150*10**-12 #capacitance in F
omega= 10**6 # in rad/sec
Z1= R1/(1+1j*omega*C1*R1) # in ohm
Z2= (1+1j*omega*C2*R2)/(1j*omega*C2) # in ohm
# At balance condition : Z1*R4 = Z2*(Rx+1j*omega*Lx) or
# R4= omega**2*R1*C2*(R1*R4*C1-Lx)      (i)
# R4= R1*(Rx*C2-R4*C1)/(R2*C2)             (ii)
Rx= R4*(R1*C1+R2*C2)/(R1*C2) # in Ω from eq(ii)
Lx= R4*(R2*C1-1/(omega**2*R1*C2)) # in H from eq (i)
Rx= Rx*10**-3 # in ohm
Lx= Lx*10**3 # in mH
print "The value of Rx = %0.1f Ω " %Rx
print "The value of Lx = %0.3f mH " %Lx

The value of Rx = 25.5 Ω
The value of Lx = 74.833 mH


## Example - 6.34 : Page No - 6.75¶

In :
#Given data
R2 = 1000 #resistance in Ω
R3 = 1000 #resistance in Ω
R4 = 1000 #resistance in Ω
C4 = 0.5 #capacitance in µF
C4 = C4 * 10**-6 # in F
#At balance, (R1+(%i*omega*L1))*(R4/( 1+(%i*omega*C4*R4) )) = R2*R3
# R1*R4 + (%i*omega*L1*R4) = (R2*R3) + (%i*omega*R2*R4*C4)
R1 = (R2*R3)/R4 # in Ω (equating real terms)
L1 = R2*R3*C4 # in H (equating imaginary terms)
print "The value of R1 = %0.f ohm " %R1
print "The value of L1 = %0.1f H " %L1

The value of R1 = 1000 ohm
The value of L1 = 0.5 H


## Example - 6.35 : Page No - 6.76¶

In :
#Given data
R3 = 260 #resistance in ohm
C4 = 0.5 # in µF
C4 = C4 * 10**-6 # in F
C2 = 106 # in pF
C2 = C2 * 10**-12 # in F
R4 = 1000/pi #resistance in ohm
r1 = (C4/C2)*R3 #resistance in ohm
C1 = (R4/R3)*C2 # in F
Epsilon_o = 8.854*10**-12
d = 4.5# in mm
d = d * 10**-3 # in m
D= 0.12 # in m
A= pi*D**2/4 # in m**2
print "The resistance = %0.2e Ω " %r1
C1= C1*10**12 # in pF
print "The capacitance = %0.2f pF " %C1
C1= C1*10**-12 # in F
f = 50 # in Hz
omega = 2*pi*f # in rad/sec
Pf= omega*C1*r1 # power factor
print "The power factor = %0.2f " %Pf
# C1 = Epsilon_r*Epsilon_o*(A/d)
Epsilon_r = (C1*d)/(Epsilon_o*A) # the relative permittivity
print "The relative permittivity = %0.1f " %Epsilon_r

# Note: The calculation of evaluating the value of C1 is wrong, so the answer of C1 in the book is wrong.
#       But they putted the correct value of C1 to find the value of relative permittivity

The resistance = 1.23e+06 Ω
The capacitance = 129.77 pF
The power factor = 0.05
The relative permittivity = 5.8


## Example - 6.36 : Page No - 6.76¶

In :
from math import atan
from numpy import pi
#Given data
C2 = 500 # capacitance in nF
C2 = C2 * 10**-9 # in F
f = 50 #frequency in Hz
omega = 2*pi*f # in rad/sec
C4 = 0.148 #capacitance in µF
C4 = C4 * 10**-6 # in F
R4 = 72.6 #resistance in ohm
R3 = 300 #resistance in ohm
C1 = C2*(R4/R3) # capacitance in F
C1 = C1 * 10**6 # in µF
print "The capacitance = %0.3f µF " %C1
delta = 90-(atan(omega*C4*R4)*180/pi) #dielectric loss angle of capacitance in degree
print "The dielectric loss angle of capacitance = %0.2f degree " %delta

# Note: The calculation in the book is wrong, so the answer in the book is wrong.

The capacitance = 0.121 µF
The dielectric loss angle of capacitance = 89.81 degree


## Example - 6.37 : Page No - 6.77¶

In :
#Given data
f1 = 3 #frequency in MHz
f1 = f1 * 10**6 # in Hz
C1 = 251 #capacitance in pF
C1 = C1  * 10**-12 # in F
f2 = 6 #frequency in MHz
f2 = f2 * 10**6 # in Hz
C2 = 50 #capacitance in pF
C2 = C2 * 10**-12 # in F
# f1 = 1/(2*pi*(sqrt(L*(C1+Cd))) )    (i)
# f2 = 1/(2*pi*(sqrt(L*(C2+Cd))) )    (ii)
# From eq(i) and (ii)
Cd = (C1 - (4*C2))/3 # self capacitance of the coil in F
Cd = Cd * 10**12 # in pF
print "The self capacitance of the coil = %0.f pF " %Cd

The self capacitance of the coil = 17 pF


## Example - 6.38 : Page No - 6.78¶

In :
#Given data
f=500 #frequency in Hz
R2 = 2410 #resistance in ohm
R3 = 750 #resistance in ohm
R4 = 64.5 #resistance in ohm
R_C4 = 0.4 #resistance in ohm
C4 = 0.35 #capacitance in µF
C4 = C4 * 10**-6 # in F
XC4= 1/(2*pi*f*C4) # in Ω
Z4= R4+R_C4-1j*XC4 # in Ω
Z2= R2 # in Ω
Z3= R3 # in Ω
Z1= Z2*Z3/Z4 # in Ω
R1= Z1.real #resistance of choke coil  in Ω
XL1= Z1.imag # in Ω
L1= XL1/(2*pi*f) #inductance of choke coil  in H
print "The resistance of choke coil = %0.4f Ω " %R1
print "The inductance of choke coil = %0.4f H " %L1

The resistance of choke coil = 141.1084 Ω
The inductance of choke coil = 0.6294 H


## Example - 6.39 : Page No - 6.79¶

In :
#Given data
f = 50 # in Hz
omega = 2*pi*f # in rad/sec
R1 = 50 # in ohm
L1 = 0.1 # in H
XL1= 2*pi*f*L1 # in Ω
R2= 100 # in Ω
R3= 1000 # in Ω
Z1= R1+1j*XL1 # in Ω
Z2= R2 # in Ω
Z3= R3 # in Ω
# The bridge balance condition
Zx= Z2*Z3/Z1 # in Ω
# Comparing real part
Rx= Zx.real # in Ω
# Comparing imaginary part
XCx= abs(Zx.imag) # in Ω
Cx= 1/(2*pi*f*XCx) # in F
print "The value of Rx = %0.4f Ω " %Rx
print "The value of Cx = %0.4f µF " %(Cx*10**6)

The value of Rx = 1433.9136 Ω
The value of Cx = 3.5330 µF


## Example - 6.40 : Page No - 6.80¶

In :
from math import atan2
from numpy import pi
#Given data
f = 2 # in kHz
f = f * 10**3 # in Hz
R2= 834 # in Ω
C2= 0.124*10**-6 # in F
XC2= 1/(2*pi*f*C2) # in Ω
R3= 100 # in Ω
C4 = 0.1 # in µF
C4 = C4*10**-6 # in F
XC4= 1/(2*pi*f*C4) # in Ω
Z2= R2+1j*XC2 # in Ω
Z3= R3 # in Ω
Z4= -1j*XC4 # in Ω
# The bridge balance condition
Z1= Z2*Z3/Z4 # in Ω
mag= abs(Z1) # magnitude of effective impedence in Ω
theta= atan2(Z1.imag,Z1.real)*180/pi # phase angle of effective impedence in °
print "The magnitude of effective impedence = %0.2f Ω " %mag
print "The phase angle of effective impedence = %0.2f ° " %theta

The magnitude of effective impedence = 132.24 Ω
The phase angle of effective impedence = 127.58 °


## Example - 6.41 : Page No - 6.81¶

In :
#Given data
L1 = 52.6 #inductance in mH
R2 = 1.68 #resistance in ohm
# 80*(r1+(J*omega*L1)) = 80*( (R2+r2) + (J*omega*L2) )
L2 = L1 #inductance of the coil in mH
print "The inductance of the coil = %0.1f mH " %L2
r1 = 28.5 # in ohm
r2 = r1-R2 #resistance of the coil in ohm
print "The resistance of the coil = %0.2f ohm " %r2

The inductance of the coil = 52.6 mH
The resistance of the coil = 26.82 ohm


## Example - 6.42 : Page No - 6.83¶

In :
#Given data
Q = 1 # in k ohm
Q = Q * 10**3 # in  ohm
S = Q # in ohm
P = 500 # in ohm
r = 100 # in ohm
C = 0.5 # in µF
C = C * 10**-6 # in F
#Using standard condition, Rx = (R2*R3)/R4
Rx = (P*Q)/S # in ohm
print "The value of Rx = %0.f Ω " %Rx
#Lx = ((C*R2)/R4) * ( (R3*r) + (R4*r) + (R3*R4) )
Lx = ((C*P)/S) * ( (Q*r) + (S*r) + (Q*S) ) # in H
print "The value of Lx = %0.1f H " %Lx

The value of Rx = 500 Ω
The value of Lx = 0.3 H