# Chapter 10 : Bridge Circuits¶

## Example10_1,pg 292¶

In [2]:
# wheatstone bridge

import math
#Variable declaration
Vs=12.0                  #source voltage
R=120.0                  #resistance of arms
delv=0.3                 #variation in output voltage(+/-)0.3
Rm=100.0                 #meter resistance

#Calculations
delRbyR=(4.0/Vs)*(delv)*100
delIm=(delRbyR/100.0)/(4.0*R*(1+(Rm/R)))

#Result
print("percent change in resistance:")
print("delRbyR = %.f%% \n"%delRbyR)
print("current variation:")
print("delIm = %.6f A"%delIm)
# Answer current variation is not matchhing with the book

percent change in resistance:
delRbyR = 10%

current variation:
delIm = 0.000114 A


## Example10_2,pg 295¶

In [5]:
# high resistance measurement bridge

import math
#variable declaration
#in absence of the guard point arrangement, two 10^10 ohm resistances in series become parallel
#to the 10^9 ohm resistance, making the effective unknown resistance

#case-1
Rh=10.0**9
Ra1=10.0**10
Rb1=10.0**10
#case-2
Ra2=10.0**9
Rb2=10.0**9

#Calculations
Rue1=((Rh*2*Ra1)/(Rh+(2*Ra1)))        #effective resistance
err1=((Rh-Rue1)/Rh)*100               #percentage error
Rue2=((Rh*2*Ra2)/(Rh+(2*Ra2)))        #effective resistance
err2=((Rh-Rue2)/Rh)*100               #percentage error

#Result
print("percentage error case-1:")
print("err1 = %.0f%% \n"%err1)
print("percentage error case-2:")
print("err2 = %.1f%%"%err2)

percentage error case-1:
err1 = 5%

percentage error case-2:
err2 = 33.3%


## Example10_3,pg 297¶

In [15]:
# capacitance and resistance of AC bridge
import math
#Variable declaration
Z1=20.0+80.0j            #impedance in first arm
Z2=200.0                 #impedance in second arm
Z3=100.0+200.0j          #impedance in third arm
f=50.0                   #excitation frequency

#Calculations
Zu=((Z2*Z3)/Z1)                 #impedance of fourth arm
Cu=(1.0/(2*math.pi*f*Zu.real))  #capacitance in fourth arm
Ru=-Zu.imag                     #resistance in fourth arm

#Result
print("capacitance in fourth arm:")
print("Cu = %f F\n"%(Cu*10**6))
print("resistance in fourth arm:")
print("Ru = %.2f ohm"%Ru)
#Answer is slightly different than book

capacitance in fourth arm:
Cu = 6.012520 F

resistance in fourth arm:
Ru = 117.65 ohm


## Example10_4,pg 301¶

In [25]:
# schering bridge

import math
#Variable declaration
C3=0.001*10**-6                    #capacitor
Fd=6.0*10**-4                      #dissipation factor
f=1.0*10**3                        #schering bridge frequency
R1=10.0*10**3
R2=10.0*10**3

#Calculations
Ru=(Fd/(2*math.pi*f*C3))           #standard resistor
C1=C3*(1/R2)*Ru

#Result
print("standard resistor:")
print("Ru = %.3f ohm\n"%Ru)
print("capacitor:")
print("C1 = %.1f pF"%(C1*10**12))
#Answer do not match with the book

standard resistor:
Ru = 95.493 ohm

capacitor:
C1 = 9.5 pF


## Example10_5,pg 303¶

In [27]:
# wein bridge

import math
#Variable declaration
R=10*10**3                    #resistor
C=0.001*10**-6                #capacitor
R3=10.0*10**3                 #reistance in third arm

#Calculations
f=(1.0/(2*math.pi*R*C))       #supply frequency
R4=(R3/2)                     #reistance in fourth arm

#Result
print("supply frequency:")
print("f = %.2f kHz\n"%(f/1000))
print("reistance in fourth arm:")
print("R4 = %.1f k-ohm"%(R4/1000))

supply frequency:
f = 15.92 kHz

reistance in fourth arm:
R4 = 5.0 k-ohm


## Example10_6,pg 303¶

In [8]:
# balance condition in wein bridge

import math
#Variable declaration
f=47.76*10**3                         #supplu frequency
C=10**-9                              #assume

#Calculations
CR=(1.0/(2*math.pi*f))                #resistor capacitor product
R=(CR/C)                              #resistor

#Result
print("for (R3/R4) = 2\nR3 and R4 may be maintained at earlier values")

for (R3/R4) = 2
R3 and R4 may be maintained at earlier values


## Example10_7,pg 309¶

In [11]:
# relation between Vo and t for Vi given

import math
#Variable declaration
a1=3.81*10**-3
a2=-6.17*10**-7
#R1=(R2/2),i.e R2/R1=2
R1=10*10**3
R2=20*10**3
R5=4*10**3
R6=20*10**3

#Calculations
B=(R5/(R5+R6))
#using relation 10.68(b)

#Result
print("(Vo/Vi)= (-3.05*10^-3)t/(1+0.76*10^-3)t")
print("thus for, t<=130 C, Vo is approx. linear. This however can be extended with proper choice")
print("i.e R5 and R6 in relation to R1,R3 and R4")

(Vo/Vi)= (-3.05*10^-3)t/(1+0.76*10^-3)t
thus for, t<=130 C, Vo is approx. linear. This however can be extended with proper choice
i.e R5 and R6 in relation to R1,R3 and R4


## Example10_8,pg 503¶

In [49]:
# find deflection in galvanometer

import math
#Variable declaration
R1=120.0                    #resistance of arm-1
R2=120.0                    #resistance of arm-2
R3=120.0                    #resistance of arm-3
R4=121.0                    #resistance of arm-4
Rm=100.0                    #meter resistance
Vs=6.0                      #source voltage
n=1*10**-3                  #meter sensitivity

#Calculations
k=(R3/(R3+R4))
k = math.floor(k*10**3)/10**3
Vm=Vs*((R1/(R1+R2))-k)                  #voltage across meter
Rb=((R1*R2)/(R1+R2))+((R3*R4)/(R3+R4))  #thevenised bridge resistance
Rb=math.floor(Rb*1000)/1000
Ig=(Vm/(Rb+Rm))                         #current through galvanometer

D=Ig*10**6

#Result
print("deflection in meter:")
print("D = %f mm"%D)
#Calcualtions in the book are not correct

deflection in meter:
D = 81.726054 mm


## Example10_9,pg 503¶

In [61]:
# find insulating post resistance

import math
#Variable declaration
err=0.5/100.0                 #(+/-)0.5%
R=100.0*10**6               #test resistance

#Calculations
#Re=((R*2*Rip)/(R+(2*Rip)))
Re1=R-(err*R)               #err=+0.5
Re2=R+(err*R)               #err=-0.5
Rip1=((R*Re1)/(2*(R-Re1)))  #err=+0.5
Rip2=((R*Re2)/(2*(Re2-R)))  #err=-0.5

#Result
print("resistance of each insulating post-1:")
print("Rip1 = %.2f M-ohm\n"%(Rip1*10**-6))
print("resistance of each insulating post-2:")
print("Rip2 = %.2f M-ohm"%(Rip2*10**-6))
# Answer in the book are not matching

resistance of each insulating post-1:
Rip1 = 9950.00 M-ohm

resistance of each insulating post-2:
Rip2 = 10050.00 M-ohm


## Example10_10,pg 504¶

In [64]:
# maxwell bridge

import math
#Variable declaration
Ru=130.0                     #resistance
Lu=31*10**-3                 #inductance
R2=10*10**3                  #resistance in arm-2
C1=0.01*10**-6               #capacitance in arm

#Calculations
R3=(Lu/(C1*R2))              #resistance in arm-3
R1=((R2*R3)/Ru)              #resistance in arm-1

#Result
print("R1 = %.2f k-ohm"%(R1/1000))
print("R3 = %.f ohm"%R3)
print("yes values are unique")

R1 = 23.85 k-ohm
R3 = 310 ohm
yes values are unique


## Example10_11,pg 504¶

In [74]:
# hay bridge

import math
#Variable declaration
f=1000.0                 #supply frequency
C1=0.04*10**-6           #capacitance
R1=220.0                 #resistance in arm-1
R2 = 10*10**3            #resistance in arm-2
Lu=22.0*10**-3           #inductance

#Calculations
pi= math.floor(math.pi*100)/100
Ru=((2*pi*f)**2)*C1*R1*Lu #resistance
R3=((R1*Ru)+(Lu/C1))/R2        #resistance in arm-3

#Result
print("resistance of inductor:")
print("Ru = %.3f ohm\n"%Ru)
print("resistance of arm-3:")
print("R3 = %.2f ohm"%R3)
#Answer for R3 is not matching.

resistance of inductor:
Ru = 7.635 ohm

resistance of arm-3:
R3 = 55.17 ohm


## Example10_12,pg 505¶

In [7]:
# find C1 C3 and dissipation factor

import math
#Variable declaration
C4=0.0033*10**-6              #lossy capacitor
R2=12.0*10**3                 #arm-2 resistance
R1=10.0*10**3                 #arm-1 resistance
f = 50.0                      # frequency

#Calculations
C3=((C4*R2)/R1)               #standard capacitance
R4=0.1
C1=((R4*C3)/R2)
Fd=2*math.pi*f*C4*R4         #dissipation factor

#Result
print("capacitance set value:")
print("C1 = %.5f * 10^12 F\n"%(C1*10**12))
print("value of standard capacitance:")
print("C3 = %.5f *10^6 F\n"%(C3*10**6))
print("dissipation factor:")
print("Fd = %.4f * 10^-6"%(math.floor(Fd*10**10)/10**4))
#Answer for C1 is wrong

capacitance set value:
C1 = 0.03300 * 10^12 F

value of standard capacitance:
C3 = 0.00396 *10^6 F

dissipation factor:
Fd = 0.1036 * 10^-6


## Example10_13,pg 505¶

In [12]:
# wein bridge

import math
#Variable declaration
f=10.0*10**3                     #supply frequency
R1=10.0*10**3                    #reistance of arm-1
C1=0.01*10**-6
C2=0.01*10**-6
R3=20*10**3                    #resistance of arm-3

#Calaculations
R2=(1/(f**2))*(1/(C1*C2*R1))   #resistance of arm-2
R4=(R3/((R1/R2)+(C2/C1)))      #resistance of arm-4

#Result
print("resistance of arm-2:")
print("R4 = %.f k-ohm\n"%(R2/1000))
print("resistance of arm-4:")
print("R2 = %.f k-ohm\n"%(R4/1000))

resistance of arm-2:
R4 = 10 k-ohm

resistance of arm-4:
R2 = 10 k-ohm