# Chapter 6: Measuring instruments¶

## Example 1: pg 235¶

In [1]:
#pg 235
#calculate the shunt resistance
# Given data
Rm = 8.;# in ohm
Im = 20.;# in mA
Im = Im * 10**-3;# in A
I = 1.;# in A
#calculations
# Multiplying factor
N = I/Im;
# Shunt resistance
Rsh = Rm/(N-1);# in ohm
#results
print "The shunt resistance required in (ohm) = ",round(Rsh,3)


The shunt resistance required in (ohm) =  0.163


## Example 2: pg 235¶

In [2]:
#pg 235
#calculate the multiplying factor
# Given data
Rm = 6;# in ohm
Rsh = 0.025;# in ohm
#calculations
N = 1 + (Rm/Rsh);# multiplying factor
#results
print "The multiplying factor is",N

The multiplying factor is 241.0


## Example 3: pg 235¶

In [3]:
#pg 235
#calculate the resistances to be connected in parallel and series
# Given data
Rm = 5.;# in ohm
Im = 15.;# in mA
Im = Im * 10**-3;# in A
I = 1.;# in A
#calculations
N = I/Im;# multiplying factor
Rsh = Rm/(N-1);# in ohm
print "The resistance to be connected in parallel in (ohm) = ",round(Rsh,4)
V = 10;# in V
Rs = (V/Im)-Rm;# in ohm
print "The resistance to be connected in series in (ohm) = ",round(Rs,2)

The resistance to be connected in parallel in (ohm) =  0.0761
The resistance to be connected in series in (ohm) =  661.67


## Example 4: pg 236¶

In [4]:
#pg 236
#calculate the current range of the instrument
# Given data
V=250.;# full scale voltage reading in V
Rm = 2.;# in ohm
Rsh = 2.;# in m ohm
Rsh = Rsh * 10**-3;# in ohm
R = 5000.;# in ohm
#calculations
Im = V/(Rm+R);# in A
Ish = (Im*Rm)/Rsh;# in A
# Current range of instrument
I = Im+Ish;# in A
#results
print "The current range of instrument in A is",round(I)

The current range of instrument in A is 50.0


## Example 5: pg 236¶

In [5]:
#pg 236
#calculate the percentage error
import math
from math import acos,pi,cos
# Given data
V = 230.;# in V
I = 35.;# in A
N = 200.;
t = 64.;# in sec
kwh = 500.;
#calculations
Er = N/kwh;# in kWh
Et =  V*I*cos(phi)*t;# in Joules
Et = Et/3600.;# in W hour
Et = Et * 10**-3;# in kWh
# percentage error
PerError = ((Er-Et)/Et)*100;# in %
#results
print "The percentage error in (percentage) is",round(PerError,2)
print 'The answer is a bit different from textbook due to rounding off error'

The percentage error in (percentage) is 249.38
The answer is a bit different from textbook due to rounding off error


## Example 6: pg 237¶

In [6]:
#pg 237
#calculate the percentage error
from math import acos,cos,pi
# Given data
I = 50.;# in A
V = 230.;# in V
N = 61.;
t = 37.;# in sec
KWh = 500.;
#calculations
Er = N/KWh;# in kWh
Et = V*I*cos(phi)*t;# in Joules
Et = Et/3600.;# in Wh
Et = Et*10**-3;# in kWh
# Percentage error
PerError = ((Er-Et)/Et)*100;# in %
#results
print "The percentage error in (percentage) is ",round(PerError,2)
print 'The answer is a bit different from textbook due to rounding off error'

The percentage error in (percentage) is  3.22
The answer is a bit different from textbook due to rounding off error


## Example 7: pg 237¶

In [7]:
#pg 237
#calculate the series resistance
# Given data
Im = 20.;# in mA
Im =  Im * 10**-3;# in A
Vm = 50.;# in mV
Vm = Vm * 10**-3;# in V
V = 500.;# in V
#calculations
Rm = Vm/Im;# in ohm
Rs = (V/Im)-Rm;# in ohm
#results
print "The series resistance in ohm is",Rs

The series resistance in ohm is 24997.5


## Example 8: pg 238¶

In [8]:
#pg 238
#calculate the values of resistances
# Given data
Rm = 50;# in ohm
Im = 10;# in mA
Im = Im * 10**-3;# in A
V = 100;# in V
#calculations
Rs = (V/Im)-Rm;# in ohm
print "The value of Rs in (ohm) is",Rs
N = 1/Im;
Rsh = Rm/(N-1);# in ohm
print "The value of Rsh in (ohm) is",round(Rsh,3)

The value of Rs in (ohm) is 9950.0
The value of Rsh in (ohm) is 0.505


## Example 9: pg 238¶

In [9]:
#pg 238
#calculate the percentage error
# Given data
from math import acos,cos
I = 40.;# in A
V = 230.;# in V
N = 600.;
t = 46.;# in sec
#calculations
P = V*I*cos(phi);# in W
P = P * 10**-3;# in kW
# 1 kWh = 500 revolution
P = P * 500.;# in revolution
T = (3600./t)*60;# in revolution
# Percentage error
PerError = ((T-P)/P)*100;# in %
#results
print "The percentage error in (percentage) is",round(PerError,2)

The percentage error in (percentage) is 2.08


## Example 10: pg 238¶

In [11]:
#pg 238
#calculate the percentage error
# Given data
N = 100.;
I = 20.;# in A
V = 210.;# in V
Er = 350.;# in rev
a = 3.36;# assumed
#calculations
Et = (a*3600.)/3600;# in kWh
# 1 kWh = 100;#  revolution
Et = Et*N;# revolution
# Percentage error
PerError = ((Er-Et)/Et)*100;# in %
#results
print "The percentage error in (percentage) is",round(PerError,3)

The percentage error in (percentage) is 4.167


## Example 11: pg 239¶

In [12]:
#pg 239
#calculate the percentage error
# Given data
I = 5.;# in A
V = 230.;# in V
N = 61.;# number of revolution
t = 37.;# in sec
# speed of the disc
discSpeed= 500.;# in rev/kWh
#calculations
Er = N/discSpeed;
Et = (V*I*t)/(3600*100);
# percentage error
PerError = ((Er-Et)/Et)*100;# in %
#results
print "The percentage error in (percentage) is",round(PerError,2)

The percentage error in (percentage) is 3.22