Chapter 5 - Steady state sinusoidal analaysis¶
Pg: 181 Ex:5.1¶
In [1]:
from __future__ import division
from numpy import arange,cos,pi,sqrt
%matplotlib inline
from matplotlib.pyplot import plot,subplot,title,xlabel,ylabel,show
R=50#
t=arange(0,0.05+0.000001,0.000001)
V_t=[]
for tt in t:
V_t.append(100*cos(100*pi*tt))
V_m=100# #peak value
V_rms=V_m/sqrt(2)#
P_avg=(V_rms**2)/R#
P_t=[]
for vv in V_t:
P_t.append(vv**2/R)
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'RMS value of voltage = %0.2f volts'%V_rms
print 'average power = %0.2f watts'%P_avg
subplot(211)
plot([tt*10**3 for tt in t],V_t)#
title('voltage vs time')
xlabel('time in ms')
ylabel('voltage in volts') #ms-milli seconds(10**-3)
subplot(212)
plot([tt*10**3 for tt in t],P_t)
title('power vs time')
xlabel('time in ms')
ylabel('power in watts') #ms-milli seconds(10**-3)
show()
Pg: 182 Ex:5.2¶
In [2]:
from __future__ import division
from numpy import arange,cos,pi,sqrt
%matplotlib inline
from matplotlib.pyplot import plot,subplot,title,xlabel,ylabel,show
from sympy.mpmath import quad
#plot of V and t(already given with the question but to get clarity we plot it)
t_1=arange(0,1+0.001,0.001)
t_2=arange(1.001,2+0.001,0.001)
t=[]
for tt in t_1:
t.append(tt)
for tt in t_2:
t.append(tt)
V_1=[];V_2=[]
for tt in t_1:
V_1.append(3*tt)
for tt in t_2:
V_2.append(6-3*tt)
V=[]
for vv in V_1:
V.append(vv)
for vv in V_2:
V.append(vv)
plot(t,V)
title('voltage vs time')
xlabel('time in seconds')
ylabel('voltage in volts')
show()
#now find V_rms
T=2# #from the plot of V vs t
V_rms=sqrt((1/T)*((quad(lambda t:(3*t)**2,[0,1]))+(quad(lambda t:(6-3*t)**2,[1,2]))))
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'RMS value = %0.2f volts'%V_rms
Pg: 183 Ex:5.3¶
In [3]:
from __future__ import division
from math import pi,sqrt,sin,cos,atan
#V_1 and V_2 are phasors of given voltages
theta_1=-pi/4# #for V_1
theta_2=-pi/6# #for V_2 (in cos form)
V_1=complex(20*cos(theta_1),20*sin(theta_1))#
V_2=complex(10*cos(theta_2),10*sin(theta_2))#
V_s=V_1+V_2#
V=sqrt(((V_s.real)**2)+((V_s.imag)**2))# #peak voltage of resultant summation
phi=atan((V_s.imag)/(V_s.real))# #phase angle of resultant sum voltage
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'Peak value of resultant voltage = %0.2f volts'%V
print 'phase of resulting voltage = %0.2f degrees'%(phi*180/pi) #converting phi in radians to degrees
#result : V_t=Vcos(wt+phi)
Pg: 184 Ex:5.4¶
In [4]:
from __future__ import division
from math import pi,sqrt,sin,cos,atan
L=0.3#
C=40*10**-6#
R=100#
V_s_max=100# #peak value of given voltage
W=500# #angular frequency
V_s_phi=pi/6# #phase angle in degrees
V_s=complex(V_s_max*cos(V_s_phi),V_s_max*sin(V_s_phi))# #phasor for voltage source
Z_L=1J*W*L# #complex impedance of inductance
Z_C=-1J/(W*C)# #complex impedance of capacitance
Z_eq=R+Z_L+Z_C# #R,Z_L and Z_C in series
I=V_s/Z_eq# #phasor current
V_R=R*I#
V_L=Z_L*I#
V_C=Z_C*I#
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
#for resistance R
print 'For resistance R'
V_R_max=sqrt(((V_R.real)**2)+((V_R.imag)**2))
V_R_phi=(atan((V_R.imag)/(V_R.real)))*180/pi#
print 'peak value of voltage = %0.2f volts'%V_R_max
print 'phase angle = %0.2f degrees'%V_R_phi
#result : V_R=Vcos(wt+phi) V-peak voltage
#for inductance L
print 'For inductance L'
V_L_max=sqrt(((V_L.real)**2)+((V_L.imag)**2))#
V_L_phi=(atan((V_L.imag)/(V_L.real)))*180/pi#
print 'peak value of voltage = %0.2f volts'%V_L_max
print 'phase angle = %0.2f degrees'%V_L_phi
#result : V_L=Vcos(wt+phi) V-peak voltage
#for capacitor C
print 'For capacitor C'
V_C_max=sqrt(((V_C.real)**2)+((V_C.imag)**2))#
V_C_phi=(atan((V_C.imag)/(V_C.real)))*180/pi#
print 'peak value of voltage = %0.2f volts'%V_C_max
print 'phase angle = %0.2f degrees'%V_C_phi #cos(t)=cos(t-180) (we get 75 instead of -105 given in textbook)
#result : V_C=Vcos(wt+phi) V-peak voltage
print 'The phasor diagram cannot be plotted'
Pg: 185 Ex: 5.5¶
In [5]:
from __future__ import division
from math import pi,sqrt,sin,cos,atan
V_s_max=10# #peak value of source voltage
phi=-pi/2# #phase of source voltage
V_s=complex(10*cos(pi/2),10*sin(pi/2))# #phasor of source voltage
W=1000# #angular frequency
R=100#
L=0.1#
C=10*10**-6#
Z_L=1J*W*L# #impedance of inductance
Z_C=-1J/(W*C)# #impedance of capacitance
Z_RC=1/((1/R)+(1/Z_C))# #R and Z_C in parallel combination
V_C=V_s*Z_RC/(Z_L+Z_RC)# #voltage division principle
I=V_s/(Z_L+Z_RC)# #current through source and inductor
I_R=V_C/R# #current through resistance
I_C=V_C/Z_C# #current through capacitor
#cos(t)=cos(180-t)
print 'peak value of Vc = %0.2f volts'%(sqrt(((V_C.real)**2)+((V_C.imag)**2)),)
print 'phase angle of Vc = %0.f degrees'%((atan((V_C.imag)/(V_C.real)))*180/pi)
##result : V_C=Vcos(wt+phi) V-peak voltage
print 'current through source and inductor =',I,'amperes'
print 'current through resistance = ',I_R,'amperes'
print 'current through capacitance = ',I_C,'amperes'
Pg: 187 Ex:5.6¶
In [6]:
from __future__ import division
from math import pi,sqrt,sin,cos,atan
from numpy import mat
V_s_max=2# #peak value of source voltage
V_s_phi=-pi/2# #phase angle of source voltage
V_s=complex(V_s_max*cos(V_s_phi),V_s_max*sin(V_s_phi))#
R=10#
Z_C=-1J*5# #impedance of capacitance
Z_L=1J*10# #impedance of inductance
I_s_max=1.5# #peak value of current source
I_s_phi=0# #phase angle of current source
I_s=complex(I_s_max*cos(I_s_phi),I_s_max*sin(I_s_phi))#
#we write the standard equations of V_1 and V_2 in matrix form
#from node-voltage relation
A=[[0.1+1J*0.2,-1J*0.2],[-1J*0.2,1J*0.1]] #coefficient matrix
B=[[-1J*2],[1.5]]# #constant matrix
#As in A*X=B form
A=mat(A);B=mat(B)
V=(A**-1)*B#
V_1=sqrt(((V[0,0].real))**2+((V[0,0].imag))**2)# #peak value of V_1
V_1_phi=atan((V[0,0].imag)/(V[0,0].real))# #phase angle of V_1
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'peak value of V1 = %0.2f volts'%V_1
print 'phase angle of V1 = %0.2f degrees'%(V_1_phi*180/pi)
Pg: 188 Ex:5.7¶
In [7]:
from __future__ import division
from math import pi,sqrt,sin,cos,atan
from numpy import mat,conj
phi_v=-pi/2# #angle of voltage source
phi_i=-3*pi/4# #angle of current source
phi=phi_v-phi_i# #power angle
V_s_max=10# #peak value of voltage source
V_s_phi=phi_v# #phase angle of voltage source
R=100#
V_s=complex(V_s_max*cos(V_s_phi),V_s_max*sin(V_s_phi))# #phasor of voltage source
X_L=1J*100#
X_C=-1J*100#
I_max=0.1414# #peak value of current
I_phi=phi_i# #phase angle of current
I=complex(I_max*cos(I_phi),I_max*sin(I_phi))# #phasor of current
V_s_rms=V_s_max/sqrt(2)# #rms value of voltage
I_rms=I_max/sqrt(2)# #rms value of current
I_R_max=0.1# #peak value
I_R_phi=-2*pi# #phase angle
I_R=complex(I_R_max*cos(I_R_phi),I_R_max*sin(I_R_phi))# #phasor of current
I_R_rms=I_R_max/sqrt(2)# #rms value
I_C_max=0.1# #peak value
I_C_phi=-pi/2# #phase angle
I_C=complex(I_C_max*cos(I_C_phi),I_C_max*sin(I_C_phi))# #phasor current in capacitor
I_C_rms=I_C_max/sqrt(2)# #rms value
P=V_s_rms*I_rms*cos(phi)# #power by source
Q=V_s_rms*I_rms*sin(phi)# #reactive power by source
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'power delivered by source = %0.2f watts'%P
print 'reactive power delivered by source = %0.2f VARs'%Q
#using complex power method
print 'Using complex power method:'
S=(1/2)*V_s*conj(I)# #complex power
P=(S.real)#
Q=(S.imag)#
print 'power delivered by source = %0.2f watts'%P
print 'reactive power delivered by source = %0.2f VARs'%Q
print 'we see that, in both the methods answers are the same'
Q_L=I_rms**2*X_L/1J# #reactive power to inductance
Q_C=I_C_rms**2*X_C/1J# #reactive power to capacitance
P_R=I_R_rms**2*R# #power to resistance
print 'reactive power delivered to inductance = %0.2f VARs'%abs(Q_L)
print 'reactive power delivered to capacitance = %0.2f VARs'%abs(Q_C)
print 'power delivered to resistance = %0.2f watts'%P_R
Pg: 189 Ex:5.8¶
In [8]:
from __future__ import division
from math import degrees,pi,sqrt,acos,tan,atan,cos
Vrms = 10**2 #V
Irms= 10**2 #amp
pf= 0.5
pf1= 0.7
r= 1.41
#CALCULATIONS
PA= Vrms*Irms*pf
QA= -sqrt((Vrms*Irms)**2-PA**2)/1000
a= acos(pf1)*180/pi
QB= PA*tan(pi/180*a)/1000
P= 2*PA/1000
Q= QA+QB
o= atan(Q/P)
pf2= cos(o)
A= degrees(o)+69.18
S= sqrt(P**2+Q**2)
I= S*r
#RESULTS
print 'Phasor Current = %.3f A'%I
print '\n Angle = %.2f degrees'%A
Pg: 190 Ex: 5.9¶
In [9]:
from __future__ import division
from math import degrees,pi,sqrt,acos,tan,atan,cos
#L is load
P_L=50*10**3# #power of load
f=60# #frequency
V_rms=10*10**3# #rms voltage
PF_L=0.6# #power factor
phi_L=acos(PF_L)# #power angle
Q_L=P_L*tan(phi_L)# #reactive power of load
#when capacitor is added, power angle changes
PF_L_new=0.9#
phi_L_new=acos(PF_L_new)#
Q_new=P_L*tan(phi_L_new)#
Q_C=Q_new-Q_L# #reactive power of capacitance
X_C=-V_rms**2/Q_C# #reactance of capacitor
W=2*pi*f# #angular frequency
C=1/(W*abs(X_C))# #capacitance
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'Required capacitance = %0.2f micro-farads'%(C*10**6)
Pg: 191 Ex: 5.10¶
In [10]:
from __future__ import division
from math import degrees,pi,sqrt,acos,tan,atan,cos
R=100#
V_s_max=100# #peak value of voltage
V_s_phi=0# #phase angle of voltage
V_s=complex(V_s_max*cos(V_s_phi),V_s_max*sin(V_s_phi))# #phasor of voltage
Z_C=-1J*100# #impedance of capacitance
I_s_max=1# #peak value of current
I_s_phi=pi/2# #phase angle of current
I_s=complex(I_s_max*cos(I_s_phi),I_s_max*sin(I_s_phi))# #phasor of current
#zeroing sources to find Z_t i.e., thevenin impedance
Z_t=1/((1/R)+(1/Z_C))# #R and Z_C are in parallel combination
#apply short-circuit to find I_sc i.e., short-circuit current
I_R=abs(V_s)/R# #ohm's law
I_sc=I_R-I_s# #applying KCL
V_t=I_sc*Z_t# #thevenin voltage
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'FOR THEVENIN CIRCUIT:'
print 'thevenin voltage'
print 'peak value of voltage = %0.2f volts'%abs(V_t)
#cos(t)=cos(t-180)
print atan((V_t.imag)/(V_t.real))*180/pi,'phase angle in degrees'
print 'thevenin resistance'
print 'peak value of resistance = %0.2f ohms'%abs(Z_t)
print 'phase angle = %0.2f degrees'%(atan((Z_t.imag)/(Z_t.real))*180/pi)
print 'FOR NORTON CIRCUIT:'
print 'norton current'
print 'peak value of norton current = %0.2f amperes'%(abs(I_sc))
print 'phase angle = %0.2f degrees'%(atan((I_sc.imag)/(I_sc.real))*180/pi)
print 'resistance'
print 'peak value of resistance = %0.2f ohms'%abs(Z_t)
print 'phase angle = %0.2f degrees'%(atan((Z_t.imag)/(Z_t.real))*180/pi)
Pg: 192 Ex: 5.11¶
In [11]:
from __future__ import division
from math import degrees,pi,sqrt,acos,tan,atan,cos
from numpy import conj
#thevenin voltage
V_t_max=100#
V_t_phi=-pi/2#
V_t=complex(V_t_max*cos(V_t_phi),V_t_max*sin(V_t_phi))#
#thevenin resistance
Z_t_max=70.71#
Z_t_phi=-pi/4#
Z_t=complex(Z_t_max*cos(Z_t_phi),Z_t_max*sin(Z_t_phi))#
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
#a) Any complex load
print 'FOR ANY COMPLEX LOAD'
Z_load=conj(Z_t)#
I_a=V_t/(Z_t+Z_load)# #ohm's law
I_a_rms=I_a/sqrt(2)# #rms value
P_1=abs(I_a_rms)**2*(Z_load.real)# #power
print 'required complex load impedance : {0:0.2f}+j*{0:0.2f}'.format(Z_load.real,Z_load.imag)
print 'power delivered to load = %0.2f watts'%P_1
#b) purely resistive load
print 'FOR PURE RESISTIVE LOAD'
R_load=abs(Z_t)#
I_b=V_t/(Z_t+R_load)#
I_b_rms=I_b/sqrt(2)#
P_2=abs(I_b_rms)**2*R_load#
print 'required pure resistive load : ',R_load
print 'power delivered to load: %0.2f'%P_2
Pg: 193 Ex: 5.12¶
In [12]:
from __future__ import division
from math import degrees,pi,sqrt,acos,tan,atan,cos
V_Y=1000# #line to neutral voltage
f=60# #frequency
L=0.1# #inductance
R=50#
W=2*pi*f# #angular frequency
Z=complex(R,W*L)# #complex impedance
phi=atan((Z.imag)/(Z.real))#
#Balanced wye-wye calculations
V_an=complex(1000*cos(0),1000*sin(0))#
V_bn=complex(1000*cos(-2*pi/3),1000*sin(-2*pi/3))#
V_cn=complex(1000*cos(2*pi/3),1000*sin(2*pi/3))#
I_aA=V_an/Z#
I_bB=V_bn/Z#
I_cC=V_cn/Z#
#line-line phasors
V_ab=V_an*sqrt(3)*complex(cos(pi/6),sin(pi/6))#
V_bc=V_bn*sqrt(3)*complex(cos(pi/6),sin(pi/6))#
V_ca=V_cn*sqrt(3)*complex(cos(pi/6),sin(pi/6))#
I_L=abs(I_aA)#
P=(3/2)*V_Y*I_L*cos(phi)# #power
Q=(3/2)*V_Y*I_L*sin(phi)# #reactive power
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'LINE CURRENTS'
print 'IaA= {0:0.3f}+j*{1:0.3f}'.format(I_aA.real,I_aA.imag)
print 'IbB= {0:0.3f}+j*{1:0.3f}'.format(I_bB.real,I_bB.imag)
print 'IcC= {0:0.3f}+j*{1:0.3f}'.format(I_cC.real,I_cC.imag)
print 'LINE-LINE VOLTAGES'
print 'Vab= {0:0.3f}+j*{1:0.3f}'.format(V_ab.real,V_ab.imag)
print 'Vbc= {0:0.3f}+j*{1:0.3f}'.format(V_bc.real,V_bc.imag)
print 'Vca= {0:0.3f}+j*{1:0.3f}'.format(V_ca.real,V_ca.imag)
print 'POWER = %0.2f WATTS'%P
print 'REACTIVE POWER = %0.2f VARs'%Q
print 'the phasor diagram cannot be plotted'
Pg: 194 Ex: 5.13¶
In [13]:
from __future__ import division
from math import degrees,pi,sqrt,acos,tan,atan,cos
Z_line=complex(0.3,0.4)# #impedance of wire
Z_d=complex(30,6)# #load impedance
R=(Z_d.real)#
R_line=(Z_line.real)#
#source voltages
V_ab=complex(1000*cos(pi/6),1000*sin(pi/6))#
V_bc=complex(1000*cos(-pi/2),1000*sin(-pi/2))#
V_ca=complex(1000*cos(5*pi/6),1000*sin(5*pi/6))#
#choosing A phase of wye-equivalent circuit
V_an=V_ab/(sqrt(3)*complex(cos(pi/6),sin(pi/6)))#
Z_Y=Z_d/3#
I_aA=V_an/(Z_line+Z_Y)# #line current
I_aA_rms=abs(I_aA)/sqrt(2)#
V_An=I_aA*Z_Y# #line to neutral voltage
V_AB=V_An*sqrt(3)*complex(cos(pi/6),sin(pi/6))# #line to line voltage at the load
I_AB=V_AB/Z_d# #current through phase AB
I_AB_rms=abs(I_AB)/sqrt(2)# #rms value
P_AB=I_AB_rms**2*R# #power delivered to phase AB
#power delivered in other two phases is same
P=3*P_AB# #total power
P_A=I_aA_rms**2*R_line# #power lost in line A
#power lost in other two lines is same
P_line=3*P_A#
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
IbB= I_aA*complex(cos(-2*pi/3),sin(-2*pi/3))
IcC= I_aA*complex(cos(2*pi/3),sin(2*pi/3))
print 'LINE CURRENTS'
print 'IaA= {0:0.3f}+j*{1:0.3f}'.format(I_aA.real,I_aA.imag)
print 'IbB= {0:0.3f}+j*{1:0.3f}'.format(I_bB.real,I_bB.imag)
print 'IcC= {0:0.3f}+j*{1:0.3f}'.format(I_cC.real,I_cC.imag)
VBB=V_AB*complex(cos(-2*pi/3),sin(-2*pi/3)),'VBB='
VCC=V_AB*complex(cos(2*pi/3),sin(2*pi/3)),'VCC='
print 'LINE-LINE VOLTAGES'
print 'VAB= {0:0.3f}+j*{1:0.3f}'.format(V_ab.real,V_ab.imag)
print 'VBC= {0:0.3f}+j*{1:0.3f}'.format(V_bc.real,V_bc.imag)
print 'VCA= {0:0.3f}+j*{1:0.3f}'.format(V_ca.real,V_ca.imag)
print 'power delivered to load = %0.2f watts'%P
print 'total power dissipated in the line = %0.2f VArs'%P_line
Pg: 194 Ex: 5.14¶
In [14]:
from numpy import mat,conj
from __future__ import division
from math import sqrt,pi,sin,cos
V_1=10**3*2.2*sqrt(2)*complex(cos(0),sin(0))#
V_2=10**3*2*sqrt(2)*complex(cos(-pi/18),sin(-pi/18))#
#writing matrix form of mesh current equaions obtained by KVL
Z=[[5+3*1J+50*complex(cos(-pi/18),sin(-pi/18)),-50*complex(cos(-pi/18),sin(-pi/18))],[-50*complex(cos(-pi/18),sin(-pi/18)),4+1J+50*complex(cos(-pi/18),sin(-pi/18))]] #coefficient matrix
V=[[2200*sqrt(2)],[-2000*sqrt(2)*complex(cos(-pi/18),sin(-pi/18))]]# #voltage matrix
Z=mat(Z);V=mat(V)
I=Z/V# #current matrix
S_1=(1/2)*V_1*conj((I[0,0]))# #complex power
P_1=(S_1.real)# #power
Q_1=(S_1.imag)# #reactive power
print "All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'real power supplied by V1 = %0.2f watts'%P_1
print 'reactive power supplied by V1 = %0.2f VARs'%Q_1