# Chapter 7: Measurement of High Voltages and Currents¶

## Example 1: pg 277¶

In [1]:
#example 7.1
#calculation of capacitance of generating voltmeter
from math import pi,sqrt
#given data
Irms=2.*10**-6#current(in A)
V1=20.*10**3#applied voltage(in V)
V2=200.*10**3#applied voltage(in V)
rpm=1500.#assume synchronous speed(in rpm) of motor

#calculation
Cm=Irms*sqrt(2)/(V1*(rpm/60)*2*pi)
Irmsn=V2*Cm*2*pi*(rpm/60)/sqrt(2)
#results
print 'The capacitance of the generating voltmeter is (pF) = ',round(Cm*10**12,1)

The capacitance of the generating voltmeter is (pF) =  0.9


## Example 2: pg 277¶

In [2]:
#example 7.2
#Design of a peak reading voltmeter

#given data
r=1000.#ratio is 1000:1
R=10**7#value of resistance(in ohm)

#calculation
#take range as 0-10 microampere
Vc2=V/r#voltage  at C2 arm
#Cs * R = 1 to 10 s
Cs=10./R
#results
print 'The value of Cs is (microfarad) = ',round(Cs*10**6)
print '%s %.1e %s' %('\nThe value of R is ',R,'ohm')

The value of Cs is (microfarad) =  1.0

The value of R is  1.0e+07 ohm


## Example 3: pg 278¶

In [4]:
#example 7.3
#calculation of correction factors for atmospheric conditions

#given data
t=37#temperature(in degree celsius)
p=750.#atmospheric pressure(in mmHg)

#calculation
d=p*293./(760*(273+t))

#results
print 'The air density factor is ',round(d,4)

The air density factor is  0.9327


## Example 4: pg 278¶

In [5]:
#example 7.4
#calculation of divider ratio

#given data
R1=16.*10**3#high voltage arm resistance(in ohm)
n=16.#number of members
R=250.#resistance(in ohm) of each member in low voltage arm
R2dash=75.#terminating resistance(in ohm)

#calculation
R2=R/n
a=1+(R1/R2)+(R1/R2dash)
#results
print 'The divider ratio is ',round(a,1)

The divider ratio is  1238.3


## Example 5: pg 278¶

In [8]:
#example 7.5
#calculation of capacitance needed for correct compensation

#given data
n=15.#number of capacitors
r=120.#resistance(in ohm)
R2=5.#resistance(in ohm) of LV arm

#calculation
Ce=(2./3)*n*Cgdash
R1=n*r/2
T=R1*Ce/2
C2=T/R2
#results
print '%s %.1e %s %d %s' %('The value of capacitance needed for correct compensation is ',C2,' F or ',C2*10**9,' nf')

The value of capacitance needed for correct compensation is  1.8e-08  F or  17  nf


## Example 6: pg 279¶

In [9]:
#example 7.6
#calculation of ohmic value of shunt an its dimensions
from math import sqrt,pi
#given data
I=50.*10**3#impulse current (in A)
Vm=50.#voltage(in V) drop across shunt
B=10.*10**6#bandwidth(in Hz) of the shunt
mu0=4.*pi*10**-7#magnetic permeability(in H/m) of free space

#calculation
R=Vm/I#resistance of shunt
L0=1.46*R/B
mu=mu0#in this case ...mu = mu0 * mur ~mu0
rho=30*10**-8#resistivity(in ohm m) of the tube material
d=sqrt((1.46*rho)/(mu*B))#thickness of the tube(in m)
l=10**-1#length(in m) (assume)
r=(rho*l)/(2*pi*R*d)
#results
print 'The value of resistance is (milliohm) = ',round(R*10**3)
print '\nThe length of shunt is (cm) = ',round(l*100)
print '\nThe radius of shunt is (mm) = ',round(r*10**3,1)
print '\nThe thickness of shunt is (mm) = ',round(d*10**3,3)

The value of resistance is (milliohm) =  1.0

The length of shunt is (cm) =  10.0

The radius of shunt is (mm) =  25.6

The thickness of shunt is (mm) =  0.187


## Example 7: pg 280¶

In [10]:
#example 7.7
#Estimation of values of mutual inductance,resistance and capacitance
from math import pi
#given data
It=10.*10**3#impulse current(in A)
Vmt=10.#meter reading(in V) for full scale deflection
dibydt=10.**11#rate of change of current(in A/s)

#calculation
MbyCR=Vmt/It
t=It/dibydt
f=1/(4.*t)
omega=2*pi*f
CR=10*pi/omega
M=10**-3*CR
R=2*10**3#assume resistance(in ohm)
C=CR/R
#results
print 'The value of mutual inductance is (nH) = ',round(M*10**9)
print '%s %.1e %s' %('\nThe value of resistance is',R,'ohm')
print '\nThe value of capacitance is (pF) = ',round(C*10**12)

The value of mutual inductance is (nH) =  2.0

The value of resistance is 2.0e+03 ohm

The value of capacitance is (pF) =  1000.0


## Example 8: pg 281¶

In [11]:
#example 7.8
#calculation of resistance and capacitance
from math import pi
#given data
t1=8.*10**-6#fronttime(in s)
t2=20.*10**-6#tailtime(in s)

#calculation
f2=1/t2#frequency corresponding to tail time
fl=f2/5
omega=2*pi*fl
CR=10*pi/omega
M=10**-3*(1/CR)
R=2*10**3#assume resistance(in ohm)
C=CR/R

print '%s %.1e %s' %('The value of resistance is ',R,'ohm')
print '\nThe value of capacitance is (microfarad) = ',round(C*10**6,2)

The value of resistance is  2.0e+03 ohm

The value of capacitance is (microfarad) =  0.25