Chapter - 6 : Resistance Measurements

Example : 6.1 - Page No : 156

In [1]:
#Given data
V= 100 # in volt
I=5 # in mA
I=I*10**-3 # in amp
VS= 1000 # sensitivity of voltmeter in ohm
VR= 150 # voltmeter range in volt
Rv= VS*VR # in ohm
# Part (i)
Rm= V/I # in ohm
Rm= Rm*10**-3 # in kohm
print "Apparent value of unknown resistor = %0.f kohm" %Rm 

# Part (ii)
Rx= V/(I*(1-V/(I*Rv))) # in ohm
Rx= Rx*10**-3 #/ in kohm
print "Actual value of unknown resistor = %0.3f kohm" %Rx

# Part (iii)
epsilon_r= (Rm-Rx)/Rx*100 # in %
print "Error percentage due to loading effect of voltmeter = %0.2f %%" %epsilon_r 

Apparent value of unknown resistor = 20 kohm
Actual value of unknown resistor = 23.077 kohm
Error percentage due to loading effect of voltmeter = -13.33 %

Example : 6.2 - Page No : 156

In [12]:
from __future__ import division
from math import sqrt
#Given data
V=38.4 # in volt
I=0.4 # in amp
ammeterRange= 1 # in amp
voltmeterRange= 50 # in volt
inst_acc= 1/2 # instrument accurcy in %
R= 100 # resistance in ohm

R_A= 2.5 # in ohm
R_V= 6000 # in ohm
Rx= sqrt(R_A*R_V) # in ohm
print "Value of unknown resistance = %0.1f ohm" %Rx
print "Since the unknown resistance is of value smaller than ",round(Rx,1)," ohm, the voltmeter should be connected"
print "directly across the unknown resistance as it will give more accurate result"
Rm= V/I # in ohm
Rx= V/(I*(1-V/(I*R_V))) # in ohm
ErrorAmmeter= inst_acc*ammeterRange/R # Error in ammeter reading in amp
ErrorVoltmeter= inst_acc*voltmeterRange/R # Error in voltmeter reading in volt
# Percentage error at 0.4 A reading 
E1= ErrorAmmeter/0.4*100 #in %
# Percentage error at 38.4 V reading 
E2= ErrorVoltmeter/38.4*100 #in %
#Error due to ammeter and voltmeter
E= sqrt(E1**2+E2**2) 
#Absolute error due to ammeter and voltmeter
Error_ammeter_voltmeter= E/R*Rx # in pos and neg
print "\nAbsolute error due to ammeter and voltmeter = %0.3f ohm" %Error_ammeter_voltmeter
print "So the resistance is specified as (",round(Rx,2),"±",round(Error_ammeter_voltmeter,3),") ohm"

Value of unknown resistance = 122.5 ohm
Since the unknown resistance is of value smaller than  122.5  ohm, the voltmeter should be connected
directly across the unknown resistance as it will give more accurate result

Absolute error due to ammeter and voltmeter = 1.375 ohm
So the resistance is specified as ( 97.56 ± 1.375 ) ohm

Example : 6.3 - Page No : 157

In [13]:
 #Given data
V=120 # in volt
I=8 # in amp
R_A= 0.3 # in ohm
ammeterReading= 0.01 # in A
voltmeterReading= 0.1 # in V
AmmeterRange= 10 #in A
VoltmeterRange= 150 #in V
EA= 0.25 # constructional error of the ammeter in %
EV= 0.5 # constructional error of the voltmeter in %

Rm= V/I # in ohm
Rx= Rm-R_A # in ohm
ErrorAmmeter= ammeterReading/AmmeterRange*100 # in %
ErrorVoltmeter= voltmeterReading/VoltmeterRange*100 # in %
del_I= ErrorAmmeter+EA # in %
del_V= ErrorVoltmeter+EV # in %
# since R=V/I
TotalError= del_I+del_V # in % in neg and pos
print "Total systematic error in measurement = ± %0.3f %%" %TotalError
print "So the value of Rx is specified as : (",round(Rx,1),"±",round(Rx*TotalError/100,3),") ohm"
    

Total systematic error in measurement = ± 0.917 %
So the value of Rx is specified as : ( 14.7 ± 0.135 ) ohm

Example : 6.4 - Page No : 164

In [17]:
 #Given data
P=100 # in ohm
Q=10 # in ohm
S=46 # in ohm
R= P*S/Q #in ohm
print "The value of unknown resistance = %0.f ohm" %R

The value of unknown resistance = 460 ohm

Example : 6.5 - Page No : 164

In [18]:
 #Given data
S=6 # in ohm
AB= 25 # in cm
BC= 75 # in cm
R= S*AB/BC # in ohm
print "The value of unknown resistance = %0.f ohm" %R

The value of unknown resistance = 2 ohm

Example : 6.6 - Page No : 164

In [21]:
 #Given data
resistor= 5000 # in ohm
LVR1= resistor-resistor*0.1/100 # Limiting value of 5000 ohm resistor in negative error
LVR2= resistor+resistor*0.1/100  #Limiting value of 5000 ohm resistor in positve error
print "Limiting value of 5000 ohm resistance =",int(LVR1),"to",int(LVR2),"ohm"
print "Thus dials of 1000 , 100 , 10 and 1 ohm would to be adjusted" 

Limiting value of 5000 ohm resistance = 4995 to 5005 ohm
Thus dials of 1000 , 100 , 10 and 1 ohm would to be adjusted

Example : 6.7 - Page No : 164

In [24]:
 #Given data
P=100 # in ohm
Q=100 # in ohm
S=230 # in ohm
R=P*S/Q # in ohm
del_P_BY_P= 0.02 # in %
del_Q_BY_Q= 0.02 # in %
del_S_BY_S= 0.01 # in %
del_R_BY_R= del_P_BY_P + del_Q_BY_Q + del_S_BY_S # in %
print "Relative limiting error of unknown resistance = %0.2f %%" %del_R_BY_R
print "So limiting values of unknown resistance =",round(R-R*del_R_BY_R/100,3),"to",round(R+R*del_R_BY_R/100,3),"ohm"

Relative limiting error of unknown resistance = 0.05 %
So limiting values of unknown resistance = 229.885 to 230.115 ohm

Example : 6.8 - Page No : 165

In [27]:
 #Given data
P=1000 # in ohm
Q=1000 # in ohm
S=100 # in ohm
E=2 # in volt
Rg=50 # in ohm
R_desh= 101 # in ohm
R=Q*S/P # in ohm
del_R= R_desh-R # in ohm
E_Th= E*((R+del_R)/(R+del_R+S)- P/(P+Q)) # in volt
R_Th= ((R+del_R)*S/(R+del_R+S)+ P*Q/(P+Q)) #in ohm
Ig= E_Th/(R_Th+Rg) # in amp
Ig=Ig*10**+6 # in micro amp
print "The galvanometer current = %0.3f µA" %Ig

The galvanometer current = 8.288 µA

Example : 6.9 - Page No : 165

In [30]:
 #Given data
P=100 # in ohm
Q=1000 # in ohm
S=2000 # in ohm
E=5 # in volt
Si= 5 # in mm/miuA
Rg=200 # in ohm
R_desh= 202 # in ohm
R=P*S/Q # in ohm
del_R= R_desh-R # in ohm
E_Th= E*((R+del_R)/(R+del_R+S)- P/(P+Q)) # in volt
R_Th= ((R+del_R)*S/(R+del_R+S)+ P*Q/(P+Q)) #in ohm
Ig= E_Th/(R_Th+Rg) # in amp
Ig=Ig*10**+6 # in micro amp
theta= Si*Ig # in mm
print "Deflection of the galvanometer = %0.1f mm" %theta
S_B= theta/del_R # in mm/ohm
print "Sensitivity of the bridge = %0.2f mm/ohm" %S_B

Deflection of the galvanometer = 43.5 mm
Sensitivity of the bridge = 21.76 mm/ohm

Example : 6.10 - Page No : 166

In [31]:
 #Given data
P=1000 # in ohm
Q=100 # in ohm
R=200 # in ohm
E=5 # in volt
Si1= 10 # in mm/miuA
Si2= 5 # in mm/miuA
Rg1= 400 # in ohm
Rg2= 100 # in ohm
S=R*Q/P # in ohm
R_Th= R*S/(R+S)+ P*Q/(P+Q) # in ohm
# theta=Si1*E*S*del_R/((R+S)**2*(R_Th+Rg))
# RatioTheta21= theta2/theta1  
RatioTheta21= Si2/Si1*(R_Th+Rg1)/(R_Th+Rg2) 
print "Ratio of deflection on two galvanometers = %0.3f" %RatioTheta21

Ratio of deflection on two galvanometers = 1.217

Example : 6.11 - Page No : 167

In [33]:
 #Given data
P=500 # in ohm
S=P 
Q=S 
R=P 
R_Th=R # in ohm
Rg=100 # in ohm
E=10 # in volt
Ig= 1 # in nA
Ig=Ig*10**-9 #in amp
# Formula Ig= E_Th/(R_Th+Rg) and E_Th= E*del_R/(4*R) so
# Ig= (E*del_R/(4*R))/(R_Th+Rg) and
del_R= Ig*(R_Th+Rg)*4*R/E # in ohm
del_R= del_R*10**3 #in mohm
print "The smallest change in resistance = %0.2f m ohm" %del_R

The smallest change in resistance = 0.12 m ohm

Example : 6.12 - Page No : 167

In [35]:
 #Given data
R=200 # in ohm
S=R 
P=S 
Q=P 
r=2 # in ohm
E=24 # in volt
Power= 0.5 # in W
# Formula Power= I**2/R
I= sqrt(Power/R) # in A
print "Maximum power dissipation = %0.2f A" %I
V=I*2*R # in volt
# Formula E= V+2*I*(r+R)
R= (E-V)/(2*I)-r # in ohm
print "Series resistance = %0.f ohm" %R

Maximum power dissipation = 0.05 A
Series resistance = 38 ohm

Example : 6.13 - Page No : 167

In [38]:
 #Given data
P=10000 # in ohm
Q=10 # in ohm
S=5 # in kohm
S=S*10**3 # in ohm
E=12 # in volt
R=P*S/Q # in ohm
print "(i) : The maximum value of resistance that can be measurement with the given argument = %0.f Mohm" %(R*10**-6)
R_Th= R*S/(R+S)+ P*Q/(P+Q) # in ohm

# Part (ii)
theta= 2.5 # in mm
Rg=100 # in ohm
Si=100 # in mm/miuA
Si=Si*10**6 # in mm/amp
del_R= theta*(R_Th+Rg)*(R+S)**2/(Si*E*S) # in ohm
print "(ii) : Change in resistance  = %0.2f kohm" %(del_R*10**-3)

(i) : The maximum value of resistance that can be measurement with the given argument = 5 Mohm
(ii) : Change in resistance  = 53.28 kohm

Example : 6.14 - Page No : 171

In [41]:
 #Given data
p=200.62 # in ohm
q=400 # in ohm
P=200.48 # in ohm
Q=400 # in ohm
S=100.03 # in micro ohm
S=S*10**-6 # in ohm
r=700 # in micro ohm
r=r*10**-6 # in ohm
X= P*S/Q+q*r/(p+q)*(P/Q-p/q) # in ohm
print "Unknown resistance = %0.4f micro ohm" %(X*10**+6)

Unknown resistance = 49.9719 micro ohm