# Chapter 11 Bridges¶

## Example 11.1 Page no 323¶

In [3]:
#Given
R2=15*10**3                      #ohm
R3=40*10**3                      #ohm
R1=10.0*10**3                      #ohm

#Calculation
Rx=(R2*R3)/R1

#Result
print"Value of unknown resistance is ",Rx*10**-3,"Kohm"

Value of unknown resistance is  60.0 Kohm


## Example 11.2 Page no 325¶

In [1]:
#Given
E=6                             #Volts
R1=1.0*10**3                      #ohm
R2=2.5*10**3                    #ohm
R3=3.5*10**3                    #ohm
R4=10*10**3                     #ohm
Rg=300.0                          #ohm

#Calculation
Eth=E*((R4/(R2+R4))-(R3/(R1+R3)))
Rth=((R1*R3)/(R1+R3))+((R2*R4)/(R2+R4))
Ig=Eth/(Rth+Rg)

#Result
print"Current through galvanometer is ", round(Ig*10**6,0),"micro A"

Current through galvanometer is  43.0 micro A


## Example 11.3 Page no 327¶

In [30]:
#Given
E=10                              #Volts
r=35
R=700                             #ohm
Rth=700                            #ohm
Rg=125                              #ohm

#Calculation
Eth=E*r/(4.0*R)
Ig=Eth/(Rth+Rg)
print"current through the galvanometer is ", round(Ig*10**6,1),"micro A"

current through the galvanometer is  151.5 micro A


## Example 11.4 Page no 331¶

In [33]:
#Given
R1=5                                   #ohm
Rb=1                                   #ohm
Ra=1000.0                               #ohm

#Calculation
R2=R1/0.5
Rx=(Rb*R2)/Ra

#Result
print"Value of Rx is ",Rx,"ohm"

Value of Rx is  0.01 ohm


## Example 11.5 Page no 333¶

In [4]:
#Given
R1=5.0*10**3                                 #ohm
R2=5.0*10**3                                  #ohm
R3=5.0*10**3                                  #ohm
E=6                                         #Volts
RV=4.5*10**3                                #ohm
r=500

#Calculation
Rv=(R2*R3)/R1
es=E*((R3/(R1+R3))-(RV/(R2+RV)))

#Result
print"(i) The bridge is balanced when the temperature is 80 degree."
print"The amplitude of the error signal at 60 degree ", round(es,3),"V"

(i) The bridge is balanced when the temperature is 80 degree.
The amplitude of the error signal at 60 degree  0.158 V


## Example 11.6 (a) Page no 337¶

In [15]:
#Given
R1=10*10**3                                   #ohm
R2=50*10**3                                    #ohm
R3=100*10**3                                     #ohm

#Calculation
Rx=(R2*R3)/R1
Cx=(R1*C3)/R2

#Result
print"The unknown capacitance is ", Cx*10**6,"micro F"

The unknown capacitance is  20.0 micro F


## Example 11.6 (b) Page no 338¶

In [6]:
#Given
R1=2000.0                        #ohm
R2=2850.0                        #ohm
R4=52                          #ohm
f=400                           #Hz

#Calculation
import math
Rx=(R1*R4)/R2
Cx=(R2*C4)/R1
A=2*math.pi*f*Cx*Rx

#Result
print"Loss angle of capacitor is ", round(A,3)

Loss angle of capacitor is  0.052


## Example 11.7 Page no 339¶

In [11]:
#Given
L3=10*10**-3                      #H
R1=10.0*10**3                        #ohm
R2=40*10**3                        #ohm
R3=100*10**3                        #ohm

#Calculation
Rx=(R2*R3)/R1
Lx=(R2*L3)/R1

#Result
print"Equivalent unknown resistance is ", Lx*10**3,"mH"

Equivalent unknown resistance is  40.0 mH


## Example 11.8 (a) Page no 341¶

In [8]:
#Given
R1=470.0*10**3                                 #ohm
R2=5.1*10**3                                 #ohm
R3=100*10**3                                 #ohm

#Calculation
Rx=(R2*R3)/R1
Lx=R2*R3*C1

#Result
print"The value of unknown impedence ", Lx,"H"

The value of unknown impedence  5.1 H


## Example 11.8 (b) Page no 341¶

In [22]:
#Given
R1=32.7                           #ohm
L1=50*10**-3                       #H
R2=1.36                            #ohm
R3=100.0                             #ohm
R4=100                             #ohm

#Calculation
r=((R4*R1)/R3)-R2
L2=L1*(R4/R3)

#Result
print"Resistance of the coil is ",r,"ohm \nInductance of the coil is " ,L2*10**3,"mH"

Resistance of the coil is  31.34 ohm
Inductance of the coil is  50.0 mH


## Example 11.9 (a) Page no 343¶

In [27]:
#Given
R2=10*10**3                         #ohm
R1=2*10**3                          #ohm
R3=1*10**3                          #ohm

#Calculation
Rx=(w**2*R1*R2*R3*C1**2)/(1+w**2*R1**2*C1**2)
Lx=(R2*R3*C1)/(1+w**2*R1**2*C1**2)

#Result
print" series equivalent impedence is ", round(Lx*10**3,0),"mH","\n Series equivqlent resistance is ",round(Rx*10**-3,2),"Kohm"

 series equivalent impedence is  270.0 mH
Series equivqlent resistance is  4.86 Kohm


## Example 11.9 (b) Page no 343¶

In [35]:
#Given
R2=1000                          #ohm
R3=16500                         #ohm
R4=800                           #ohm
C4=2*10**-6                      #F
f=50                             #Hz

#Calculation
import math
w=2*math.pi*f
Lx=(R2*R3*C4)/(1+(w**2*C4**2*R4**2))
Rx=(w**2*C4**2*R4*R2*R3)/(1+w**2*C4**2*R4**2)

#Result
print"Resistance of the coil is ", round(Rx*10**-3,1),"Kohm"
print"Inductance of the coil is ",round(Lx,1),"H"

Resistance of the coil is  4.2 Kohm
Inductance of the coil is  26.3 H


## Example 11.9 (c) Page no 345¶

In [3]:
#Given
C4=1*10**-6                        #F
R2=1000                            #ohm
R3=1000
R4=1000

#Calculation
L1=(R3*R4*C4)/(1+(w**2*C4**2*R4**2))
Rx=(w**2*C4**2*R4*R2*R3)/(1+w**2*C4**2*R4**2)

#Result
print"unknown resistance is" ,round(L1,2),"H"
print"unknown inductance is", round(Rx,2),"ohm"

unknown resistance is 0.91 H
unknown inductance is 89.75 ohm


## Example 11.10 (a) Page no 346¶

In [4]:
#Given
R2=2*10**3                             #ohm
R1=1*10**3                              #ohm
f=1000                                  #Hz

#Calculation
import math
Rx=(C1*R2)/C3
Cx=(R1*C3)/R2
D=2*math.pi*f*Cx*Rx

#Result
print" The unknown capacitance is ",Cx*10**6,"micro farad", "\n Dissipation factor is ",round(D,4)

 The unknown capacitance is  0.25 micro farad
Dissipation factor is  3.1416


## Example 11.10 (b) Page no 347¶

In [19]:
#Given
R3=100.0                                   #ohm
R4=300                                     #ohm
C4=0.5*10**-6                              #F
C2=100.0                                   #pf
f=50                                       #Hz

#Calculation
import math
W=2*math.pi*f
C1=(R4/R3)*C2
R1=(C4/C2)*R3
Pf=W*R4*C4

#Result
print"Capacitance is ", C1,"pF"
print"Equivalent series resistance is ",R1*10**6,"Mohm"
print"PF of insulation is ",round(Pf,4)

Capacitance is  300.0 pF
Equivalent series resistance is  0.5 Mohm
PF of insulation is  0.0471


## Example 11.10 (c) Page no 348¶

In [40]:
#Given
R4=100                          #ohm
C4=0.1                   #micro F
C2=500                    #pf
R3=300.0                     #ohm
f=50                         #Hz

#Calculation
import math
W=2*math.pi*f
C1=(R4/R3)*C2
A=W*C4*R4*10**-5

#Result
print"The Capacitance is ",round(C1,1),"pF"
print"Dielectric loss of angle is ",round(A*180/3.14,1),"degree"

The Capacitance is  166.7 pF
Dielectric loss of angle is  1.8 degree


## Example 11.10 (d) Page no 348¶

In [63]:
#Given
d=4.5*10**-3
f=50                    #Hz
C2=105*10**-12
C4=0.5*10**-6               #F
R3=260.0                   #ohm
R4=1000/3.14
K0=8.854*10**-12

#Calculation
import math
W=2*math.pi*f
C1=(R4/R3)*C2
A=W*C4*R4
a=math.pi*(6*10**-2)**2
K=(C1*d)/(K0*a)

#Result
print"The capacitance is ",round(C1*10**12,1),"pF"
print"Value of PF is ",round(A,3)
print"Relative permittivity is ",round(K,3)

The capacitance is  128.6 pF
Value of PF is  0.05
Relative permittivity is  5.78


## Example 11.10 (e) Page no 349¶

In [71]:
#Given
R3=100               #ohm
R4=300               #ohm
C2=100.0                #pF
C4=0.5                #micro F

#Calculation
import math
C1=(R4/R3)*C2
r1=R3*(C4/C2)
W=2*math.pi*f
C1=(R4/R3)*C2
A=W*C4*R4*10**-6

#Result
print"The value of capacitor is ",C1,"pF"
print"PF of the capacitor is ",round(A,4)

The value of capacitor is  300.0 pF
PF of the capacitor is  0.0471


## Example 11.10 (f) Page no 350¶

In [80]:
#Given
R1=300                    #ohm
R2=100.0                    #ohm
C1=0.6*10**-6                   # F
C3=100.0*10**-12                   #F

#Calculation
Rx=R2*(C1/C3)
C=(R1/R2)*C3

#Result
print"The value of series resistance is ", Rx*10**-5,"Mohm"
print"The value of capacitance is ",C*10**12,"pF"

The value of series resistance is  6.0 Mohm
The value of capacitance is  300.0 pF


## Example 11.10 (g) Page no 350¶

In [102]:
#Given
C1=0.1*10**-6                    #F
C2=0.25*10**-6                    #F
R2=5000.0                           #ohm
C3=0.25*10**-6                    #F
f=2000                           #Hz

#Calculation
import math
Rx=R2*(C1/C2)
C=(R1/R2)*C3
W=2*math.pi*f
A=W*C*Rx*10**-6
#Result
print"The value of capacitance is ", round(C*10**7,1),"micro F"
print"Dissipation factor is ", round(A*10**7,3)

The value of capacitance is  0.1 micro F
Dissipation factor is  3.77


## Example 11.11 Page no 352¶

In [6]:
#Given
R1=4.7*10**3                                     #ohm
R2=20*10**3                                      #ohm
R3=10*10**3                                      #ohm
R4=100*10**3                                     #ohm

#Calculation
import math
f=1/(2.0*math.pi*(math.sqrt(C1*R1*R3*C3)))
print"Frequency of the circuit is ", round(f*10**-3,3),"KHz"

Frequency of the circuit is  3.283 KHz


## Example 11.12 Page no 353¶

In [20]:
#Given
R1=3.1*10**3                                       #ohm
R2=25.0*10**3                                        #ohm
f=2.5*10**3                                        #ohm
R4=100.0*10**3                                       #ohm

#Calculation
import math
w=2*math.pi*f
R3=R4/R2*(R1+(1/w**2*R1*C1**2))
C3=R2/R4*(C1/(1+w**2*R1**2*C1**2))

#CResult
print" Equivalent parallel resistance is ",R3*10**-3,"Kohm ","\n Equivalent parallel capaciatnce is ", round(C3*10**12,1),"pf"

 Equivalent parallel resistance is  12.4 Kohm
Equivalent parallel capaciatnce is  20.3 pf


## Example 11.13 Page no 353¶

In [30]:
#Given
R1=800.0                                  #ohm
R2=400                                   #ohm
R4=1000                                  #ohm

#Calculation
import math
f=1/(2.0*math.pi*math.sqrt(R1*C1*R2*C2))
R3=R4/((R2/R1)+(C1/C2))

#Result
print"(i) The value of frequency for which the bridge is balanced ", round(f*10**-3,3),"KHz"
print"(ii) The value of R required to produce balance ",R3,"ohm"

(i) The value of frequency for which the bridge is balanced  0.398 KHz
(ii) The value of R required to produce balance  1000.0 ohm


## Example 11.14 Page no 359¶

In [108]:
#Given
R2=200                                 #ohm
R3=1000                                  #ohm
R4=1000.0

The value of inductance is  7.952 H