CHAPTER 1: REVIEW OF ELECTRIC CIRCUITS

Example 1.1, Page number 2-3

In [1]:
#Variable declaration
Q = 4.0      #Charge(C) 
t = 0.54     #Time(sec) 

#Calculation
I = Q/t      #Current(A) 

#Result
print('Value of Current is , I = %.2f A' %I) 

Value of Current is , I = 7.41 A

Example 1.2, Page number 4-5

In [1]:
#Variable declaration
V = -24.0     #Voltage(V)
I = 3.0       #Current(A)

#Calculation 
P = V*I       #Power supplied by the element A(W) 

#Result
print('Power supplied by the element A is , P = %.1f W' %P)

Power supplied by the element A is , P = -72.0 W

Example 1.3, Page number 7-9

In [1]:
#Variable declaration
R1 = 5.0   #Resistance(ohm)
R2 = 4.0   #Resistance(ohm)
R3 = 9.0   #Resistance(ohm)
R4 = 6.0   #Resistance(ohm)
V1 = 10.0  #Resistance(ohm)
V2 = 6.0   #Resistance(ohm)


#Calculation
R_th = (R1*R4/(R1+R4))+R2    #Thevenin resistance(ohm) by removing R3 & short-circuiting voltage sources
I = (V1-V2)/(R1+R4)          #Current(A) by applying KVL
V_th = 6*I+V2                #Thevenin voltage(V) by applying KVL
I_9ohm = V_th/(R_th+R3)      #Current through 9 ohm resistor(A)

#Result
print('Current through 9 ohm resistor , I_9Ω = %.2f A' %I_9ohm) 

Current through 9 ohm resistor , I_9Ω = 0.52 A

Example 1.4, Page number 10-11

In [1]:
from scipy.integrate import quad

#Variable declaration
V_t1 = 30.0            #Magnitudes of voltages(V)  0 < t1 < 2
V_t2 = -10.0           #Magnitudes of voltages(V)  2 < t2 < 4
T = 4.0                #Time period(sec) from figure

#Calculation
def integrand(V):
    return V**0

a, err = quad(integrand, 0, 2)

def integrand(V):
    return V**0

b, err = quad(integrand, 2, 4)

V_rms = ((a*V_t1**2+b*V_t2**2)/4)**0.5   #RMS value of voltage waveform(V)

#Result
print('RMS value , V_rms = %.2f V' %V_rms)

RMS value , V_rms = 22.36 V

Example 1.5, Page number 15-16

In [1]:
import math
import cmath

#Variable declaration
V_P = 200.0                                 #Magnitude of each phase(V) 

#Calculation
V_an = V_P*cmath.exp(1j*0*math.pi/180)      #Magnitude of 3-phase voltage(V)
V_bn = V_P*cmath.exp(1j*-120*math.pi/180)   #Magnitude of 3-phase voltage(V)
V_cn = V_P*cmath.exp(1j*120*math.pi/180)    #Magnitude of 3-phase voltage(V)
V_L = 3**0.5*V_P                            #Magnitude of line voltage(V)

#Result
print('Expression of phase voltages are,')
print('\t\t\t  V_an = %.f∠%.f° V' %(abs(V_an),cmath.phase(V_an)))
print('\t\t\t  V_bn = %.f∠%.f° V' %(abs(V_bn),cmath.phase(V_bn)*180/math.pi))
print('\t\t\t  V_cn = %.f∠%.f° V' %(abs(V_cn),cmath.phase(V_cn)*180/math.pi))
print('Magnitude of the line voltage , V_L = %.1f V' %V_L) 

Expression of phase voltages are,
			  V_an = 200∠0° V
			  V_bn = 200∠-120° V
			  V_cn = 200∠120° V
Magnitude of the line voltage , V_L = 346.4 V

Example 1.6, Page number 16-17

In [1]:
import math
import cmath

#Variable declaration
R = 10.0        #Resistance of each coil(ohm)
X = 15.0        #Inductive reactance of each coil(ohm)
V_L = 420.0     #Line voltage(V)
f = 50.0        #Frequency of supply(Hz)

#Calculation
V_an = (V_L/3**0.5)*cmath.exp(1j*(0-30)*math.pi/180)       #Phase voltage(V)
V_bn = (V_L/3**0.5)*cmath.exp(1j*(-120-30)*math.pi/180)    #Phase voltage(V)
V_cn = (V_L/3**0.5)*cmath.exp(1j*(120-30)*math.pi/180)     #Phase voltage(V)
Z_P = complex(R,X)                                         #Phase impedance(ohm)
#For case(i)
I_L1 = V_an/Z_P         #Line current(A)
I_L2 = V_bn/Z_P         #Line current(A)
I_L3 = V_cn/Z_P         #Line current(A)
#For case(ii)
pf = R/abs(Z_P)         #Power factor

#Result
print('(i) Values of line currents are,')
print('\t I_L1 = I_an = %.2f∠%.2f° A' %(abs(I_L1),cmath.phase(I_L1)*180/math.pi))
print('\t I_L2 = I_bn = %.2f∠%.2f° A' %(abs(I_L2),cmath.phase(I_L2)*180/math.pi))
print('\t I_L3 = I_cn = %.2f∠%.2f° A' %(abs(I_L3),cmath.phase(I_L3)*180/math.pi))
print('(ii) Power factor is , pf = %.1f lag' %pf)
print('\nNOTE : I_L2 has an angle -206.31° in textbook which is same as 153.69° i.e (360-206.31)° obtained here')

(i) Values of line currents are,
	 I_L1 = I_an = 13.45∠-86.31° A
	 I_L2 = I_bn = 13.45∠153.69° A
	 I_L3 = I_cn = 13.45∠33.69° A
(ii) Power factor is , pf = 0.6 lag

NOTE : I_L2 has an angle -206.31° in textbook which is same as 153.69° i.e (360-206.31)° obtained here

Example 1.7, Page number 19-20

In [1]:
import math
import cmath

#Variable declaration
Z_P = complex(10,15)     #Per phase impedance(ohm)
V_L = 420.0              #Voltage(V)

#Calculation
#For case(i)
V_ab = V_L*cmath.exp(1j*0*math.pi/180)        #Phase voltage(V)
V_bc = V_L*cmath.exp(1j*-120*math.pi/180)     #Phase voltage(V)
V_ca = V_L*cmath.exp(1j*120*math.pi/180)      #Phase voltage(V)
I_ab = V_ab/Z_P                               #Phase current(A)
I_bc = V_bc/Z_P                               #Phase current(A)
I_ca = V_ca/Z_P                               #Phase current(A)
#For case(ii)
I_P = abs(I_ab)                               #Phase current magnitude(A)
I_L = 3**0.5*I_P                              #Line current magnitude(A)

#Result
print('(i) Phase currents are,')
print('\t\t I_ab = %.2f∠%.2f° A' %(abs(I_ab),cmath.phase(I_ab)*180/math.pi))
print('\t\t I_bc = %.2f∠%.2f° A' %(abs(I_bc),cmath.phase(I_bc)*180/math.pi))
print('\t\t I_ca = %.2f∠%.2f° A' %(abs(I_ca),cmath.phase(I_ca)*180/math.pi))
print('(ii) Magnitude of line current , I_L = %.2f A' %I_L)

(i) Phase currents are,
		 I_ab = 23.30∠-56.31° A
		 I_bc = 23.30∠-176.31° A
		 I_ca = 23.30∠63.69° A
(ii) Magnitude of line current , I_L = 40.35 A

Example 1.8, Page number 22-23

In [1]:
import math
import cmath

#Variable declaration
V_P = 280.0          #Generator Phase voltage(V)
Z_P = complex(2,3)   #Line impedance per phase(ohm)
Z_L = complex(4,5)   #Load impedance per phase(ohm)

#Calculation
V_An = V_P*cmath.exp(1j*0*math.pi/180)        #Phase voltage(V)
V_Bn = V_P*cmath.exp(1j*-120*math.pi/180)     #Phase voltage(V)
V_Cn = V_P*cmath.exp(1j*120*math.pi/180)      #Phase voltage(V)
Z_t = Z_P+Z_L                                 #Total impedance(ohm)
I_Aa = V_An/Z_t                               #Magnitude of line current for phase A(A)
I_Bb = V_Bn/Z_t                               #Magnitude of line current for phase B(A)
I_Cc = V_Cn/Z_t                               #Magnitude of line current for phase C(A)
V_an = I_Aa*Z_L                               #Phase voltage of load(V)
V_bn = I_Bb*Z_L                               #Phase voltage of load(V)
V_cn = I_Cc*Z_L                               #Phase voltage of load(V)

#Result
print('Line currents are,')
print('\t\t I_Aa = %.f∠%.f° A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))
print('\t\t I_Bb = %.f∠%.f° A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))
print('\t\t I_Cc = %.f∠%.f° A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))
print('\nLoad phase voltages are,')
print('\t\t V_an = %.1f∠%.1f° V' %(abs(V_an),cmath.phase(V_an)*180/math.pi))
print('\t\t V_bn = %.1f∠%.1f° V' %(abs(V_bn),cmath.phase(V_bn)*180/math.pi))
print('\t\t V_cn = %.1f∠%.1f° V' %(abs(V_cn),cmath.phase(V_cn)*180/math.pi))
print('\nNOTE : ERROR : Z_L = 6.4∠38.6°Ω is taken in textbook solution instead of 6.4∠51.34°Ω = (4+j5)Ω')

Line currents are,
		 I_Aa = 28∠-53° A
		 I_Bb = 28∠-173° A
		 I_Cc = 28∠67° A

Load phase voltages are,
		 V_an = 179.3∠-1.8° V
		 V_bn = 179.3∠-121.8° V
		 V_cn = 179.3∠118.2° V

NOTE : ERROR : Z_L = 6.4∠38.6°Ω is taken in textbook solution instead of 6.4∠51.34°Ω = (4+j5)Ω

Example 1.9, Page number 23-24

In [1]:
import math
import cmath

#Variable declaration
Z = complex(6,8)                            #Per phase impedance of load(ohm)
V_AN = 340.0*cmath.exp(1j*0*math.pi/180)    #Phase voltage(V)

#Calculation
V_P = abs(V_AN)                             #Voltage(V)
V_BN = V_P*cmath.exp(1j*-120*math.pi/180)   #Phase voltage(V)
V_CN = V_P*cmath.exp(1j*120*math.pi/180)    #Phase voltage(V)
I_an = V_AN/Z                               #Load current(A)
I_bn = V_BN/Z                               #Load current(A)
I_cn = V_CN/Z                               #Load current(A)
I_n = I_an+I_bn+I_cn                        #Neutral current(A)

#Result
print('Phase current in each load = Line current in each load are,')
print('\t\t\t\t I_an = I_Aa = %.f∠%.f° A' %(abs(I_an),cmath.phase(I_an)*180/math.pi))
print('\t\t\t\t I_bn = I_Bb = %.f∠%.f° A' %(abs(I_bn),cmath.phase(I_bn)*180/math.pi))
print('\t\t\t\t I_cn = I_Cc = %.f∠%.f° A' %(abs(I_cn),cmath.phase(I_cn)*180/math.pi))
print('Neutral current is , I_n = %.f A' %abs(I_n))

Phase current in each load = Line current in each load are,
				 I_an = I_Aa = 34∠-53° A
				 I_bn = I_Bb = 34∠-173° A
				 I_cn = I_Cc = 34∠67° A
Neutral current is , I_n = 0 A

Example 1.10, Page number 25-26

In [1]:
import math
import cmath

#Variable declaration
Z = complex(3,4)                                 #Per phase impedance of load(ohm)
V_AN = 200.0*cmath.exp(1j*0*math.pi/180)         #Phase voltage(V)

#Calculation
V_P = abs(V_AN)                                   #Voltage(V)
V_AB = 3**0.5*V_P*cmath.exp(1j*30*math.pi/180)    #Line voltage(V)
V_BC = 3**0.5*V_P*cmath.exp(1j*-90*math.pi/180)   #Line voltage(V)
V_CA = 3**0.5*V_P*cmath.exp(1j*150*math.pi/180)   #Line voltage(V)
#For case(i)
I_ab = V_AB/Z                               #Load current(A)
I_bc = V_BC/Z                               #Load current(A)
I_ca = V_CA/Z                               #Load current(A)
#For case(ii)
I_Aa = I_ab-I_ca                            #Line current(A)
I_Bb = I_bc-I_ab                            #Line current(A)
I_Cc = I_ca-I_bc                            #Line current(A)

#Result
print('(i) Magnitude of load currents are,')
print('\t\t I_ab = %.1f∠%.2f° A' %(abs(I_ab),cmath.phase(I_ab)*180/math.pi))
print('\t\t I_bc = %.1f∠%.2f° A' %(abs(I_bc),cmath.phase(I_bc)*180/math.pi))
print('\t\t I_ca = %.1f∠%.2f° A' %(abs(I_ca),cmath.phase(I_ca)*180/math.pi))
print('\n(ii) Magnitude of line currents are,')
print('\t\t I_Aa = %.2f∠%.2f° A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))
print('\t\t I_Bb = %.2f∠%.2f° A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))
print('\t\t I_Cc = %.2f∠%.2f° A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))
print('\nNOTE : ERROR : Calculation mistakes in textbook')

(i) Magnitude of load currents are,
		 I_ab = 69.3∠-23.13° A
		 I_bc = 69.3∠-143.13° A
		 I_ca = 69.3∠96.87° A

(ii) Magnitude of line currents are,
		 I_Aa = 120.00∠-53.13° A
		 I_Bb = 120.00∠-173.13° A
		 I_Cc = 120.00∠66.87° A

NOTE : ERROR : Calculation mistakes in textbook

Example 1.11, Page number 28-29

In [1]:
import math
import cmath

#Variable declaration
Z = complex(3,4)                                 #Per phase impedance of load(ohm)
V_AN = 150.0*cmath.exp(1j*0*math.pi/180)         #Phase voltage(V)

#Calculation
V_P = abs(V_AN)                                   #Voltage(V)
V_BN = V_P*cmath.exp(1j*-120*math.pi/180)         #Phase voltage(V)
V_CN = V_P*cmath.exp(1j*120*math.pi/180)          #Phase voltage(V)
I_Aa = V_AN/Z                                     #Line current(A)
I_Bb = V_BN/Z                                     #Line current(A)
I_Cc = V_CN/Z                                     #Line current(A)
pf = Z.real/abs(Z)                                #Power factor
I = abs(I_Aa)                                     #Magnitude of line current(A)
P = V_P*I*pf*10**-3                               #Power supplied to each phase(kW)
P_t = 3*P                                         #Total power supplied(kW)

#Result
print('Line currents are,')
print('  I_Aa = %.f∠%.2f° A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))
print('  I_Bb = %.f∠%.2f° A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))
print('  I_Cc = %.f∠%.2f° A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))
print('Power factor , pf = %.1f ' %pf)
print('Power supplied to each phase , P = %.1f kW' %P)
print('Total Power supplied to the load , P_t = %.1f kW' %P_t)

Line currents are,
  I_Aa = 30∠-53.13° A
  I_Bb = 30∠-173.13° A
  I_Cc = 30∠66.87° A
Power factor , pf = 0.6 
Power supplied to each phase , P = 2.7 kW
Total Power supplied to the load , P_t = 8.1 kW

Example 1.12, Page number 32

In [1]:
import math

#Variable declaration
P = 120.0         #Total power(kW)
pf = 0.6          #Power factor 

#Calculation
teta = math.acos(pf)                       #Power factor angle(radians)
teta_deg = teta*180/math.pi                #Power factor angle(degree)
P_2 = 1.0/2*((math.tan(teta)*P/3**0.5)+P)  #Second wattmeter reading(kW)

#Result
print('Second wattmeter reading , P_2 = %.1f kW' %P_2)

Second wattmeter reading , P_2 = 106.2 kW

Example 1.13, Page number 35

In [1]:
import math

#Variable declaration
P = 5000.0         #Power(W)
pf_1 = 0.8         #Initial Power factor
V = 110.0          #rms Voltage(V)
f = 50.0           #Frequency(Hz)
pf_2 = 0.9         #Final Power factor

#Calculation
phi_1 = math.acos(pf_1)               #Initial Power factor angle(radians)
phi_1_deg = phi_1*180/math.pi         #Initial Power factor angle(degree)
phi_2 = math.acos(pf_2)               #Final Power factor angle(radians)
phi_2_deg = phi_2*180/math.pi         #Final Power factor angle(degree)
C = P*(math.tan(phi_1)-math.tan(phi_2))/(2*math.pi*f*V**2)*10**6   #Parallel capacitance(µF)

#Result
print('Capacitance , C = %.1f µF' %C)

Capacitance , C = 349.5 µF

Example 1.14, Page number 35-36

In [1]:
import math

#Variable declaration
pf_1 = 0.85        #Initial Power factor
kVA = 20.0         #Load(kVA)
f = 50.0           #Frequency(Hz)
pf_2 = 0.95        #Final Power factor
V = 200.0          #Voltage(V)
R = 0.05           #Resistance(ohm)
X = 0.2            #Inductive reactance(ohm)

#Calculation
phi_1 = math.acos(pf_1)               #Initial Power factor angle(radians)
phi_1_deg = phi_1*180/math.pi         #Initial Power factor angle(degree)
phi_2 = math.acos(pf_2)               #Final Power factor angle(radians)
phi_2_deg = phi_2*180/math.pi         #Final Power factor angle(degree)
P = kVA*pf_1                          #Load power(kW)
C = P*1000*(math.tan(phi_1)-math.tan(phi_2))/(2*math.pi*f*V**2)*10**6   #Parallel capacitance(µF)
#Before adding capacitor
I_1 = P*1000/(pf_1*V)                 #Line current(A)
P_1 = I_1**2*R                        #Power loss in line(W)
#After adding capacitor\n",
S = P*1000/pf_2                       #Apparent power(VA)
I_2 = S/V                             #Line current(A)
P_2 = I_2**2*R                        #Power loss in line(W)

#Result
print('Capacitance , C = %.1f µF' %C)
print('Power loss in the line before adding capacitor , P_1 = %.1f W' %P_1)
print('Power loss in the line after adding capacitor , P_2 = %.1f W' %P_2)

Capacitance , C = 393.8 µF
Power loss in the line before adding capacitor , P_1 = 500.0 W
Power loss in the line after adding capacitor , P_2 = 400.3 W