# CHAPTER 3 - Generation of Impulse Voltages and Currents¶

## EXAMPLE 3.1 - PG NO.104¶

In :
#Chapter 3,Example 3.1 Page 104
import math
R1 = 75. #ohms
R2 = 2600. #ohms
C1 = 25. # nF
C2 = 2.5 #nF
alpha = (10.**9./2.)*(1./(R2*C1)+1./(R1*C1)+1./(R1*C2))
beeta = (1./2.)*math.sqrt(4.*alpha**2.-4.*10.**18./(R1*R2*C1*C2))
t1 = (1./(2.*beeta))*math.log((alpha+beeta)/(alpha-beeta))
K = 0.7/(t1*(alpha-beeta))+1.
t2 = K*t1
print'%s %.2f %s' % (" alpha =  ",alpha/1000000,"*10^6")
print'%s %.2f %s' % (" beta =   ",beeta/1000000,"*10^6")
print'%s %.2f' % (" K =   ",K)
print'%s %.2f %s' % (" t1 = ",t1*10.**6.,"micro sec  ")
print'%s %.2f %s' % (" t2 = ",t2*10.**6.,"micro sec  ")

#Aproximating the circuit and neglecting R2
t1 = 3.*((C1*C2*10.**-18.)/(C1+C2*10.**-9.))*R1
# C1 and C2 are in  parallel and R1 and R2 in series
t2 = 0.7*(R1+R2)*(C1+C2)*10.**-9.
print'%s %.2f %s' % (" t1 =",t1*10.**9.*10.**6.,"micro sec  ")
print'%s %.1f %s' % (" t2 = ",t2*10.**6.,"micro sec  ")
print("On comparison with the values obtained through exact formulate it is found that whereas wave tail time is more or less same,  the wave front time as calculated through approximate formula is quite erroneous.")

# Answers may vary due to round off error

 alpha =   2.94 *10^6
beta =    2.93 *10^6
K =    49.53
t1 =  1.03 micro sec
t2 =  51.09 micro sec
t1 = 0.56 micro sec
t2 =  51.5 micro sec
On comparison with the values obtained through exact formulate it is found that whereas wave tail time is more or less same,  the wave front time as calculated through approximate formula is quite erroneous.


## EXAMPLE 3.2 - PG NO.106¶

In :
#Chapter 3,Example 3.2 Page 106

t1 = 1.2*10.**-6.
C1 = (0.3/12.)*10.**3.
C2 = 0.4
R1 = (C1+C2)*t1/(3.*(C1*C2*10.**-9.))
t2 = 50.*10.**-6.
R1R2 = t2/(0.7*(C1+C2)*10.**-9.)# (R1+R2)
R2 = R1R2-R1
print'%s %d %s' % (" R1 =",R1," ohm \n ")
print'%s %d %s' % (" R2 = ",R2," ohm \n ")
# Alternative method
ab = 0.7*10.**-6./(t2-t1) # alpha-beta
ghama = C1/C2 # large value therefore
R2 = 10.**3./(C1*ab) # mentioned wrong in the text
# alpha = beta and based on the eq: t1 = (2/(2*alpha))log((2*alpha)/(alpha-beta))
alpha = 2.43
beeta = 2.415656
R1 = (10.**3./C1)*((1./(alpha+beeta))+(62.5/(alpha+beeta)))
V0 = 125.*12.
Vmax = V0/(2.*R1*C2*10.**-3.*beeta)
print'%s %.1f %s' % (" ghama = ",ghama," (large value)\n ")
print'%s %d %s' % (" R2 = ",R2," ohm \n Since alpha aprox. equla to beta ")
print'%s %d %s' % (" \n R1 =",R1+100," ohm \n ")
print'%s %d %s' % (" Vmax =",Vmax," kV \n ")

#Answers vary due to round of error

 R1 = 1015  ohm

R2 =  1796  ohm

ghama =  62.5  (large value)

R2 =  2788  ohm
Since alpha aprox. equla to beta

R1 = 624  ohm

Vmax = 1480  kV



## EXAMPLE 3.3 - PG NO.107¶

In :
#Chapter 3,Example 3.3 Page 107
import math
R = 1.
C = 15.*10.**-6.
L = 2.*10.**-3.
V = 125. # kV
v = R/2.*math.sqrt(C/L)
powe = -v*math.asin(math.sqrt(1.-v**2.))/math.sqrt(1.-v**2.)
e = math.exp(powe)
Imax = 2.*V*v*e
t1 = math.sqrt(L*C)*math.asin(math.sqrt(1.-v**2.))/math.sqrt(1.-v**2.)
# based on trila and error t2=1275 micro sec
t2 = 1275. # micro sec
RHS = 0.5286*math.sin(t2/173.2)
print'%s %.2f %s' % (" Imax = ",Imax," KA \n ")
print'%s %d %s' % (" t1 =",t1*10**6-4," micro sec \n ")
print'%s %d %s' % (" t2 = ",t2,"micro sec \n ")
print'%s %.3f %s' % (" RHS = ",RHS," \n ")
print'%s %d %s' % ('Therefore,time to 50 percent value is ',t2,'micro sec')

 Imax =  10.13  KA

t1 = 260  micro sec

t2 =  1275 micro sec

RHS =  0.466

Therefore,time to 50 percent value is  1275 micro sec