In [1]:

```
import math
#Variable declaration
f = 50.0 #Generator frequency(Hz)
kV = 7.5 #emf to neutral rms voltage(kV)
X = 4.0 #Reactance of generator & connected system(ohm)
C = 0.01*10**-6 #Distributed capacitance(F)
#Calculation
#Case(a)
v = 2**0.5*kV #Active recovery voltage i.e phase to neutral(kV)
V_max_restrike = v*2 #Maximum restriking voltage i.e phase to neutral(kV)
#Case(b)
L = X/(2.0*math.pi*f) #Inductance(H)
f_n = 1/(2.0*math.pi*(L*C)**0.5*1000) #Frequency of transient oscillation(kHZ)
#Case(c)
t = 1.0/(2.0*f_n*1000) #Time(sec)
avg_rate = V_max_restrike/t #Average rate of rise of voltage upto first peak of oscillation(kV/s)
#Result
print('Case(a): Maximum re-striking voltage(phase-to-neutral) = %.1f kV' %V_max_restrike)
print('Case(b): Frequency of transient oscillation, f_n = %.1f kHz' %f_n)
print('Case(c): Average rate of rise of voltage upto first peak of oscillation = %.f kV/s' %avg_rate)
print('\nNOTE: Changes in the obtained answer from that of textbook is due to more approximation in the textbook')
```

In [1]:

```
import math
#Variable declaration
kV = 132.0 #Voltage(kV)
pf = 0.3 #Power factor of the fault
K3 = 0.95 #Recovery voltage was 0.95 of full line value
f_n = 16000.0 #Natural frequency of the restriking transient(Hz)
#Calculation
kV_phase = kV/3**0.5 #System voltage(kV)
sin_phi = math.sin(math.acos(pf)) #Sinφ
K2 = 1.0
v = K2*K3*kV/3**0.5*2**0.5*sin_phi #Active recovery voltage(kV)
V_max_restrike = 2*v #Maximum restriking voltage(kV)
t = 1.0/(2.0*f_n) #Time(sec)
RRRV = V_max_restrike/(t*10**6) #Rate of rise of restriking voltage(kV/µ-sec)
#Result
print('Rate of rise of restriking voltage, R.R.R.V = %.2f kV/µ-sec' %RRRV)
```

In [1]:

```
#Variable declaration
kV = 132.0 #Voltage(kV)
C = 0.01*10**-6 #Phase to ground capacitance(F)
L = 6.0 #Inductance(H)
i = 5.0 #Magnetizing current(A)
#Calculation
V_pros = i*(L/C)**0.5/1000 #Prospective value of voltage(kV)
R = 1.0/2*(L/C)**0.5/1000 #Resistance to be used across the contacts to eliminate the restriking voltage(k-ohm)
#Result
print('Voltage across the pole of a CB = %.1f kV' %V_pros)
print('Resistance to be used across the contacts to eliminate the restriking voltage, R = %.2f k-ohm' %R)
print('\nNOTE: ERROR: Unit of final answer R is k-ohm, not ohm as in the textbook solution')
```

In [1]:

```
#Variable declaration
I = 1200.0 #Rated normal current(A)
MVA = 1500.0 #Rated MVA
kV = 33.0 #Voltage(kV)
#Calculation
I_breaking = MVA/(3**0.5*kV) #Rated symmetrical breaking current(kA)
I_making = I_breaking*2.55 #Rated making current(kA)
I_short = I_breaking #Short-time rating(kA)
#Result
print('Rated normal current = %.f A' %I)
print('Breaking current = %.2f kA' %I_breaking)
print('Making current = %.f kA' %I_making)
print('Short-time rating = %.2f kA for 3 secs' %I_short)
```

In [1]:

```
#Variable declaration
kVA = 7500.0 #Rated kVA
X_st = 9.0 #Sub-transient reactance(%)
X_t = 15.0 #Transient reactance(%)
X_d = 100.0 #Direct-axis reactance(%)
kV = 13.8 #Voltage(kV). Assumption
#Calculation
kVA_base = 7500.0 #Base kVA
kVA_sc_sustained = kVA_base/X_d*100 #Sustained S.C kVA
I_sc_sustained = kVA_base/(3**0.5*kV) #Sustained S.C current(A). rms
I_st = kVA*100/(X_st*3**0.5*kV) #Initial symmetrical rms current in the breaker(A)
I_max_dc = 2**0.5*I_st #Maximum possible dc component of the short-circuit(A)
I_moment = 1.6*I_st #Momentary current rating of the breaker(A)
I_interrupt = 1.1*I_st #Current to be interrupted by the breaker(A)
I_kVA = 3**0.5*I_interrupt*kV #Interrupting kVA
#Result
print('Case(a): Sustained short circuit in the breaker = %.1f A (rms)' %I_sc_sustained)
print('Case(b): Initial symmetrical rms current in the breaker = %.f A' %I_st)
print('Case(c): Maximum possible dc component of the short-circuit in the breaker = %.f A' %I_max_dc)
print('Case(d): Momentary current rating of the breaker = %.f A (rms)' %I_moment)
print('Case(e): Current to be interrupted by the breaker = %.f A (rms)' %I_interrupt)
print('Case(f): Interrupting kVA = %.f kVA' %I_kVA)
```