Chapter2 - Calculation of Capacitance & Inductance

Exa 2.1 page 69

In [1]:
from __future__ import division
from numpy import pi
N=400 
a=4*10**-4 
MUo=4*pi*10**-7 
MUr=800 
l=0.3 
L=(MUo*MUr*a*N**2)/l 
print "Self inductance of the coil=%.3f H"%(L)

Self inductance of the coil=0.214 H

Exa 2.2 page 69

In [2]:
P0=8.854*10**-12 
Pr1=5.5 
d1=10**-3 
b1=d1/Pr1 
Pr2=2.2 
d2=10**-3 
b2=d2/Pr2 
Pr3=1.5 
d3=10**-3 
b3=d3/Pr3 
A=100*10**-4 
C=P0*A/(b1+b2+b3) 
V=5000 
Q=C*V*10**6 
print "stored charge in the capacitor=%.2f coulombs"%(Q)
D=Q/A 
D=146*10**-6 
g1=D*10**-3/(P0*Pr1) 
print "potential gradient g1=%.2f kV/m"%(g1)
g2=D*10**-3/(P0*Pr2) 
print "potential gradient g2=%.2f kV/m"%(g2)
g3=D*10**-3/(P0*Pr3) 
print "potential gradient g3=%.2f kV/m"%(g3)

stored charge in the capacitor=0.34 coulombs
potential gradient g1=2998.13 kV/m
potential gradient g2=7495.33 kV/m
potential gradient g3=10993.15 kV/m

Exa 2.3 page 69

In [3]:
from math import log10
a=0.5/2 
b=0.25+0.4 
Pr=4.5 
C=(0.024*Pr)/(log10(b/a))
C_total=300*C
print "capacitance of the cable=%.2f uF"%(C_total)

capacitance of the cable=78.08 uF