Chapter 18: ELECTRICAL MEASURING INSTRUMENTS

Example 18.1,Page number: 598

In [2]:
#Question:
"""Finding the deflecting torque in Newton-metres for a PMMC instrument."""

#Variable Declaration:
I=15e-03              #Current flowing through the coil(in Amperes) 
B=0.2                 #Flux density in the air gap(in Tesla)
l=2e-02               #Length of the magnetic field(in m)
d=2.5e-02             #Mean width of the coil(in m) 
r=d/2                 #Radius of the coil(in cm) 
n1=42                 #Number of turns(lower limit) 
n2=43                 #Number of turns(upper limit)


#Calculations:
F1=I*B*l*n1
F2=I*B*l*n2
net_torque=(F1+F2)*r


#Result:
print "The deflecting torque is %e Nm." %(net_torque)

The deflecting torque is 6.375000e-05 Nm.

Example 18.2,Page Number: 604

In [4]:
#Question:
"""Finding the shunt resistance for measuring a maximum current of 10 mA."""     

#Variable Declaration:
Ifsd=10e-03           #Maximum current(in Amperes)
Im=100e-06            #Full-scale deflection current(in Amperes)  
Rm=100                #Meter Resistance(in Ohms)  


#Calculations:
Ish=Ifsd-Im
Rsh=(Im*Rm)/Ish


#Result:
print "The shunt resistance needed is %.6f Ohms." %(Rsh)

The shunt resistance needed is 1.010101 Ohms.

Example 18.3,Page number: 605

In [4]:
#Question:
"""Designing an universal shunt for a multi-range ammeter."""

#Variable Declaration:
Im=100e-06            #Full-scale deflection current(in Amperes)
Rm=100.0              #Internal resistance(in Ohms)


#Calculations:
"""For 1-mA range,the required shunt can be calculated as follows."""
Ifsd=1e-03
Rsh=(Im*Rm)/(Ifsd-Im)

Rm=900.0
R=100.0
"""(a)Range-switch at 1 mA:"""
Rm1=Rm
Ish1=(1e-03)-(0.1e-03)
Rsh1=(Rm1*Im)/(Ish1)

"""(b)Range-switch at 10 mA:"""
"""Rm2=Rm+R1"""
Ish2=(10e-03)-(0.1e-03)
"""Rsh2=R2+R3+R4+R5=R-R1=100-R1;"""
"""Rsh2=(Rm2*Im)/Ish2;"""
R1=(9900.0-900.0)/100.0

"""(c)Range-switch at 100 mA:"""
"""Rm3=Rm+R1+R2"""
Ish3=(100e-03)-(0.1e-03)
"""Rsh3=R3+R4+R5=R-R1-R2=100-90-R2=90-R2;"""
"""Rsh3=(Rm3*Im)/Ish3;"""
R2=(9990.0-990.0)/1000.0

"""(d)Range-switch at 500 mA:"""
"""Rm4=Rm+R1+R2+R3"""
Ish3=(500e-03)-(0.1e-03)
"""Rsh4=R4+R5=R-R1-R2-R3=100-90-9-R3=1-R3;"""
"""Rsh4=(Rm4*Im)/Ish4;"""
R3=(4999.0-999.0)/5000.0


"""(e)Range-switch at 1 A:"""
"""Rm5=Rm+R1+R2+R3+R4"""
Ish3=(1000e-03)-(0.1e-03)
"""Rsh5=R5=R-R1-R2-R3-R4=100-90-9-0.8-R4=0.2-R4;"""
"""Rsh5=(Rm5*Im)/Ish5;"""
R4=(1999.8-999.8)/10000.0
R5=R-R1-R2-R3-R4


#Result:
print "The resistor values are: "
print "R1=%.2f Ohms." %(R1)
print "R2=%.2f Ohms." %(R2)
print "R3=%.2f Ohms." %(R3)
print "R4=%.2f Ohms." %(R4)
print "R5=%.2f Ohms." %(R5)

The resistor values are: 
R1=90.00 Ohms.
R2=9.00 Ohms.
R3=0.80 Ohms.
R4=0.10 Ohms.
R5=0.10 Ohms.

Example 18.4,Page number: 609

In [5]:
#Question:
"""To convert a d'Arsonval meter movement into a voltmeter.""" 

#Variable Declaration:
Im=100e-06            #Current sensitivity(in Amperes)
Rm=100                #Resistance of the coil(in Ohms)
Vfsd=100              #Full-scale deflection of voltmeter(in Volts) 


#Calculations:
Rs=(Vfsd/Im)-Rm


#Result:
print "To convert the d'Arsonval meter movement into a voltmeter of range 100V, we connect a resistor Rs in series."
print "Rs = %.3f kilo Ohms. " %(Rs/1000)

To convert the d'Arsonval meter movement into a voltmeter of range 100V, we connect a resistor Rs in series.
Rs = 999.900 kilo Ohms.

Example 18.5,Page number: 609

In [6]:
#Question:
"""Finding the multipiler resistance and the voltage mutiplyiing factor of a dc voltmeter."""

#Variable Declaration:
Im=50e-06             #Full-scale deflection current(in Amperes)  
Rm=1e03               #Meter Resistance(in Ohms) 
Vfsd=50               #Full-scale deflection of voltmeter(in Volts)     


#Calculations:
Rs=(Vfsd/Im)-Rm
n=Vfsd/(Im*Rm)

#Result:
print "(a)The multiplier resistance needed is %.2f kilo Ohms." %(Rs/1000)
print "(b)The voltage multiplying factor is %d." %(round(n,0))

(a)The multiplier resistance needed is 999.00 kilo Ohms.
(b)The voltage multiplying factor is 1000.

Example 18.6,Page number: 612

In [1]:
#Question:
"""Finding the reading and error in measurement of voltmeter."""
    
""" NOTE: All resistances expressed in kilo Ohms."""

#Variable Declaration:
range_A=50            #Range of voltmeter-A(in Volts)
range_B=50            #Range of voltmeter-B(in Volts)
sens_A=1000           #Sensitivity of voltmeter-A(in Ohm/Volts)  
sens_B=20000          #Sensitivity of voltmeter-B(in Ohm/Volts)


#Calculations:
Vt=150.0*(50.0/(100.0+50.0))
R_i1=(range_A*sens_A)/1000.0
Req=1/((1/R_i1)+(1.0/50))
V1=(150.0)*(Req/(100+Req))
R_i2=(range_B*sens_B)/1000.0
Req=1/((1/R_i2)+(1.0/50))
V2=(150.0)*(Req/(Req+100))
err_A=((Vt-V1)/Vt)*100.0
err_B=((Vt-V2)/Vt)*100.0


#Result:
print "The reading of voltmeter-A is %.2f V\nThe reading of voltmeter-B is %.2f V." %(V1,V2)
print "\nThe error in the reading of voltmeter-A is %.2f percent." %(err_A)
print "The error in the reading of voltmeter-B is %.2f percent." %(err_B)

The reading of voltmeter-A is 30.00 V
The reading of voltmeter-B is 48.39 V.

The error in the reading of voltmeter-A is 40.00 percent.
The error in the reading of voltmeter-B is 3.23 percent.

Example 18.7,Page number: 617

In [18]:
#Question:
"""Finding the deflection in an ammeter."""

from math import sin,asin,degrees,pow

#Variable Declaration:
I1=20.0               #Initial current(in Amperes) 
I2=12.0               #Final current(in Amperes)
angle1=60             #Initial deflection(in degrees)
""" Given: Deflecting torque is directly proportional to the current."""


#Calculations:
"""For spring control:  Controlling torque is directly proportional to deflection.
                        For steady state deflection, controlling torque=deflecting torque.
                        Therefore,deflection is directly proportional to current. """
angle2_a=(I2/I1)*angle1
"""For gravity control: Controlling torque is directly proportional to sine of the deflection angle.
                        For steady state deflection, controlling torque=deflecting torque.
                        Therefore,sine of the angle of deflection is directly proportional to the current. """  
angle2_b=asin((I2/I1)*sin(radians(angle1)))


#Result:
print "(a)The deflection for a current of 12A when the instrument is spring controlled is %.2f degrees." %(angle2_a) 
print "(b)The deflection for a current of 12A when the instrument is gravity controlled is %.2f degrees." %(degrees(angle2_b)) 

(a)The deflection for a current of 12A when the instrument is spring controlled is 36.00 degrees.
(b)The deflection for a current of 12A when the instrument is gravity controlled is 31.31 degrees.

Example 18.8,Page number: 618

In [19]:
#Question:
"""Finding the deflection in degrees in a gravity-controlled instrument."""

from math import asin,degrees

#Variable Declaration:
W=0.005               #Controlling weight(in kilograms)
l=2.4e-02             #Distance of controlling weight from the axis(in metres)
torque=1.05e-04       #Deflecting torque(in kg-m)


#Calculations:
angle=asin(torque/(W*l))


#Result:
print "The deflection corresponding to a deflecting torque of 1.05e-04 kg-m is %.2f degrees." %(degrees(angle))

The deflection corresponding to a deflecting torque of 1.05e-04 kg-m is 61.04 degrees.

Example 18.9,Page number: 618

In [20]:
#Question:
"""Finding the deflection in an ammeter."""

from math import sin,asin,degrees,pow

#Variable Declaration:
I1=10.0               #Initial current(in Amperes) 
I2=5.0                #Final current(in Amperes)
angle1=90             #Initial deflection(in degrees)
""" Given: Deflecting torque is directly proportional to square of the current."""


#Calculations:
"""For spring control:  Controlling torque is directly proportional to deflection.
                        For steady state deflection, controlling torque=deflecting torque.
                        Therefore,deflection is directly proportional to square of the current. """
angle2_a=pow((I2/I1),2)*angle1
"""For gravity control: Controlling torque is directly proportional to sine of the deflection angle.
                        For steady state deflection, controlling torque=deflecting torque.
                        Therefore,sine of the angle of deflection is directly proportional to square of the current. """  
angle2_b=asin((pow((I2/I1),2))*sin(radians(angle1)))


#Result:
print "(a)The deflection for a current of 5A when the instrument is spring controlled is %.2f degrees." %(angle2_a) 
print "(b)The deflection for a current of 5A when the instrument is gravity controlled is %.2f degrees." %(degrees(angle2_b))   

(a)The deflection for a current of 5A when the instrument is spring controlled is 22.50 degrees.
(b)The deflection for a current of 5A when the instrument is gravity controlled is 14.48 degrees.

Example 18.10,Page number: 619

In [8]:
#Question:
"""Finding the current required to produce a deflection of 60 degrees in a moving coil instrument."""

#Variable Declaration:
w=4e-02                 #Width of the coil(in metres)
l=5e-02                 #Length of the coil(in metres)
N=80.0                  #Number of turns in the coil
torque_control=0.5e-07  #Controlling torque per degree deflection of the coil(in Nm)
B=0.1                   #Magnetic flux density in the air gap(in Wb per square-metre)


#Calculations:
A=w*l
torque_c=torque_control*60.0
I=torque_c/(B*N*A)


#Result:
print "The current required to give a deflection of 60 degrees is %e A." %(I)

The current required to give a deflection of 60 degrees is 1.875000e-04 A.