# Appendix - Solved problems¶

## Ex 2 Page 340¶

In [49]:
from __future__ import division
from math import pi
# Given
t1=2# # mm
t2=5# # mm
t3=7# # mm
eps1=3  #  dielectric constant
eps2=4  #  dielectric constant
eps3=6# # dielectric constant
eps0=1/36/pi*10**-9# # dielectric constant
d=10/100   #  m
V=1500 # V
#E1*t1+E2*t2+E3*t3=V
# using eps1*E1=eps2*E2=eps3*E3
E1=V/(t1+(eps1/eps2)*t2+(eps1/eps3)*t3)# V/mm
E1=E1*10 # V/cm
E2=eps1*E1/eps2 # V/cm
E3=eps1*E1/eps3 # V/cm
print "Voltage Gradients :\n   for A : %.f V/cm\n   for B : %.f V/cm\n   for C : %.f V/cm"%(E1,E2,E3)

A = pi*d**2
W1=1/2*eps0*eps1*E1**2*10**4*A*t1*10**-3# # J
W2=1/2*eps0*eps2*E2**2*10**4*A*t2*10**-3# # J
W3=1/2*eps0*eps3*E3**2*10**4*A*t3*10**-3# # J
W1=W1*10**6 # # uJ
W2=W2*10**6 # # uJ
W3=W3*10**6 # # uJ
print "\nEnergy stored :\n   for A : %.1f uJ\n   for B : %.1f uJ\n   for C : %.1f uJ"%(W1,W2,W3)

# Answer in the textbook are not accurate

Voltage Gradients :
for A : 1622 V/cm
for B : 1216 V/cm
for C : 811 V/cm

Energy stored :
for A : 21.9 uJ
for B : 41.1 uJ
for C : 38.3 uJ


## Ex 3 Page 341¶

In [50]:
from __future__ import division
from math import pi

# Given
N=680# # turns
fi=1.6*10**3   #  Wb
d1=4/100   #  m
d2=24/100  # m
l=0.6  # m
mu0=4*pi/10**7   #  constant

# For air gap :
A=d1**2   #  m**2
Bg=fi/A   # weber/m**2
Hg=Bg/mu0  # AT/m
mmf1=0.001/mu0   #  AT

# For central limb :
A=d1**2  #  m**2
Bc=fi/A   # weber/m**2
Hc=900  # AT/m (from magnetization curve)
mmf2=Hc*d2   #  AT

# For side limb :
fi=1/2*fi   #  Wb
A=d1**2  #  m**2
Bc=fi/A   # weber/m**2
Hc=520  # AT/m (from magnetization curve)
mmf3=Hc*l   #  AT

mmf_total = mmf1+mmf2+mmf3  #  AT
i=mmf_total/N   #  A
print "Current required  = %0.2f A"%(i)

# Answer in the textbook are not accurate

Current required  = 1.95 A


## Ex 4 Page 343¶

In [51]:
from __future__ import division
from math import pi

# Given

D=15/100   #  m
A=10/10**-4  # m**2
N=200# # turns
fi=1.6*10**3   #  Wb
B=1   # weber/m**2
mu0=4*pi/10**7   #  constant
mur=500   #  constant
lg=2/1000  # m

# without air gap
l=pi*D  # m
R=l/mu0/mur/A  # A/Wb
fi=B*A  # Wb
mmf=fi*R  # AT
I=mmf/N  # A
L=N**2/R/10**6  # mH
E=1/2*L*I**2/100  # J

# with air gap
Rg=lg/mu0/A  # A/Wb
Rt=R+Rg  # A/Wb
fi=B*A  # Wb
mmf=fi*Rt  # AT
I2=mmf/N  # A
L2=N**2/Rt/10**6  # mH
E2=1/2*L2*I2**2/100  # J

print "\t\t\tWithout air gap       With air gap"
print "\nExciting current        %.2f A                %.1f A"%(I,I2)
print "\nInductance              %.1f mH                %.1f mH"%(L,L2)
print "\nStored Energy           %.3f J               %.2f J"%(E,E2)

# Answer in the textbook are not accurate

			Without air gap       With air gap

Exciting current        3.75 A                11.7 A

Inductance              5.3 mH                1.7 mH

Stored Energy           0.375 J               1.17 J


## Ex 5 Page 344¶

In [52]:
from __future__ import division
from numpy import array

# Given

VA=60  # V
I=0.6  # A
# (VB-VA)/20+(VB-VC)/20+VB/20-I=0
#3*VB-VC=72 for node B eqn(1)
#(VC-VA)/50+(VC-VB)/30+(VC-12)/50+VC/100=0
#-5*VB+10*VC=144 eqn(2)
A=array([[3, -1],[-5 ,10]])
B=array([[72],[144]])
X=A**-1*B#
VB=X[0,0]  # V
VC=X[0,1]  # V
print "Voltage acroos 100 ohm  = %.1f V"%(VC)
VC=24  # V
VB=(72+VC)/3   #  from eqn(1)
# Node C
# (VC-VA)/50+(VC-VB)/20+(VC-12)/50+VC/100+VC/R=0 eqn(3)
R=100*VC/(144+5*VB-10*VC)  # ohm
print "\nR=%.1f ohm"%(R)

Voltage acroos 100 ohm  = -72.0 V

R=37.5 ohm


## Ex 6 Page 346¶

In [53]:
from __future__ import division
from math import pi,cosh

# Given

Ro=600  # ohm
fc=2*1000  # Hz
alfa=10  # dB

L=Ro/pi/fc*1000  # mH
C=1/(pi*Ro*fc)*10**6  # uF
alfa=alfa/8.686  # nepers
f=fc*cosh(alfa/2)/1000  # kHz
print "\nat f = %.2f kHz, the above filter will have required attenuation."%(f)

at f = 2.34 kHz, the above filter will have required attenuation.


## Ex 7 Page 347¶

In [54]:
from __future__ import division
from math import pi,atan,sqrt,asin,sin

# Given
#v=100*sin(314*t)
R=25  # ohm
C=80  # uF
Vm=100  # V

Xc=1/omega/(C*10**-6)  # ohm
Z=sqrt(R**2+Xc**2)  # ohm
Im=Vm/Z  # A
print "equation for instant current:"
print "i=%.2f*sin(%d*t+%.2f)"%(Im,omega,theta)
P=(Im/sqrt(2))**2*R  # W
print "Power consumed = %.1f W"%(P)
Vcm=Im*Xc  # V
#(when i=Im/2)
i=0.5*Im  # A
#vc=Vcm*sin(314*t+theta-pi/2)
#i=Im*sin(314*t+theta)
vcm=Vcm*sin(tt-pi/2)
print "Voltage across capacitor = %.1f V(+ve & -ve)"%(abs(vcm))

equation for instant current:
i=2.13*sin(314*t+1.01)
Power consumed = 56.6 W
Voltage across capacitor = 73.3 V(+ve & -ve)


## Ex 8 Page 348¶

In [55]:
from __future__ import division
from numpy import pi,roots

# Given

Z1=(6.25+1J*1.25)  # ohm
Z2=(5+1J*0)  # ohm
#Z3=(5-1J*XC)  # ohm
V=100  # V
f=50  # Hz
#Z23=(250+5*Xc**2)/(100+Xc**2)-1J*(25*Xc)/(100+Xc**2)
#for in phase condition imag part must be zero
#5*Xc**2-100*Xc+5*100=0
A=[5 ,-100, 500]  # polynomal
XC=roots(A)#
XC=XC[0]  # ohm
C=1/(2*pi*f*XC)*10**6  # uF
print "Capacitance of XC = %.f uF"%(C)
Z=XC  # ohm
I=V/Z  # A
P=I**2*Z/1000  # kW
print "Circuit current = %.f A and power = %.f kW"%(I,P)

Capacitance of XC = 318 uF
Circuit current = 10 A and power = 1 kW


## Ex 9 Page 349¶

In [56]:
from __future__ import division
from math import pi

# Given

R=3  # ohm
IBYIo=1/2  # ratio

fo=omega_o/2/pi  # Hz
f=omega/2/pi  # Hz
#I/Io=1/(sqrt(1+Q**2*(f/fo-fo/f)**2))
Q=sqrt(1/IBYIo**2-1)/(fo/f-f/fo)
#Q=1/omega_0/R/C
C=1/omega_o/R/Q*10**6  # uF
#Q=omega_0*L/R
L=Q*R/omega_o*1000  # mH
print "L = %.1f mH\n C=%.f uF"%(L,C)

L = 10.4 mH
C=267 uF


## Ex 10 Page 350¶

In [57]:
from __future__ import division
from math import pi,sqrt

# Given

V=400  # V
f=50  # Hz
n=3  # no of phase
R=100  # ohm

#Star connection
Vph=V/sqrt(n)  # V
Iph=Vph/R  # A
IL=Iph  # A
cos_fi=1  #  for only resitor load
P=sqrt(3)*V*IL*cos_fi/1000  # kW
print "Star Connection : P=%.1f kW"%(P)
#Delta Connection

Vph=V  # V
Iph=Vph/R  # A
IL=sqrt(3)*Iph  # A
VL=Vph  # V
P=sqrt(3)*VL*IL*cos_fi/1000  # kW
print "Delta Connection : P=%.1f kW"%(P)

Star Connection : P=1.6 kW
Delta Connection : P=4.8 kW


## Ex 11 Page 351¶

In [58]:
from __future__ import division
from numpy import exp,pi
import cmath as cm
# Given
Rab=6;Rbc=8;Rca=4  # ohm
Vab=100*exp(1J*0)  # V
Vbc=100*exp(1J*-120*pi/180)  # V
Vca=100*exp(1J*120*pi/180)  # V
Zab=6+1J*8  # ohm
Zbc=8+1J*6  # ohm
Zca=4-1J*3  # ohm

#Phase current
Iab=Vab/Zab  # A
Ibc=Vbc/Zbc  # A
Ica=Vca/Zca  # A
print "Phase Current:"
print "Iab=%.f angle=%.2f degree "%(abs(Iab),cm.phase(Iab)*180/pi)
print "Ibc=%.f angle=%.2f degree "%(abs(Ibc),cm.phase(Ibc)*180/pi)
print "Ica=%.f angle=%.2f degree "%(abs(Ica),cm.phase(Ica)*180/pi)
#Line current
Iaa=Iab-Ica  # A
Ibb=Ibc-Iab  # A
Icc=Ica-Ibc  # A
print "\n\n  Line Current:"
print "Iaa=%.f angle=%.2f degree "%(abs(Iaa),cm.phase(Iaa)*180/pi)
print "Ibb=%.f angle=%.2f degree "%(abs(Iab),cm.phase(Ibb)*180/pi)
print "Icc=%.f angle=%.2f degree "%(abs(Icc),cm.phase(Icc)*180/pi)
#Power Consumed
Wab=abs(Iab)**2*Rab  # W
Wbc=abs(Ibc)**2*Rbc  # W
Wca=abs(Ica)**2*Rca  # W
W=Wab+Wbc+Wca  # W
W=W/1000  # kW
print "\n\n Total Power, W=%.f kW"%(W)
#Answer wrong for line current in the textbook.

Phase Current:
Iab=10 angle=-53.13 degree
Ibc=10 angle=-156.87 degree
Ica=20 angle=156.87 degree

Line Current:
Iaa=29 angle=-33.03 degree
Ibb=10 angle=165.00 degree
Icc=15 angle=127.97 degree

Total Power, W=3 kW


## Ex 12 Page 353¶

In [59]:
from __future__ import division
from numpy import exp,pi
import cmath as cm

# Given
VRY=200*exp(1J*0)  # V
VYB=200*exp(1J*-120*pi/180)  # V
VBR=200*exp(1J*120*pi/180)  # V

ZA=10*exp(1J*60*pi/180)  # ohm
ZB=10*exp(1J*0*pi/180)  # ohm
ZC=10*exp(1J*60*pi/180)  # ohm

#Phase current
IRY=VRY/ZA  # A
IYB=VYB/ZB  # A
IBR=VBR/ZC  # A

IR=IRY-IBR  # A
PVA=(VRY).conjugate()*IR  # W
IB=IBR-IYB  # A
PVB=(-VYB).conjugate()*IB  # W
print "Wattmeter W2 reading=%.f W or %.f kW"%((PVB.real),(PVB.real)/1000)

Wattmeter W1 reading=0 W
Wattmeter W2 reading=8000 W or 8 kW


## Ex 13 Page 354¶

In [60]:
from __future__ import division
from numpy import exp,pi,sqrt
import cmath as cm
# Given
#v=250*sin(omega*t)+50*sin(3*omega*t+pi/3)+20*sin(5*omega*t+5*pi/6)
V1=250;V3=50;V5=20  # V
fi1=0;fi3=60;fi5=150  # degree
R=20  # omh
L=0.05  # H

X1=omega*L  # ohm
Z1=R+1J*X1  # ohm
X3=3*omega*L  # ohm
Z3=R+1J*X3  # ohm
X5=5*omega*L  # ohm
Z5=R+1J*X5  # ohm
r1=abs(Z1);t1=cm.phase(Z1)
r3=abs(Z3);t3=cm.phase(Z3)
r5=abs(Z5);t5=cm.phase(Z5)
print "expression for current:"
print "i = %.2f*sin(omega*t-%.1f)+%.1f*sin(3*omega*t%.1f)+%.2f*sin(5*omega*t-%.1f)"%(V1/r1,fi1-t1*180/pi,V3/r3,fi3-t3*180/pi,V5/r5,fi5-t5*180/pi)

I1m=V1/r1  # A
I3m=V3/r3  # A
I5m=V5/r5  # A
Irms=sqrt(I1m**2/2+I3m**2/2+I5m**2/2)  # A
Vrms=sqrt(V1**2/2+V3**2/2+V5**2/2)  # A
print "Irms=%.f A\n Vrms=%.f V"%(Irms,Vrms)
P=Irms**2*R  # W
print "Total Power, P=%.f W"%(P)
cosfi=P/Vrms/Irms  # Power factor
print "Power factor = %.2f"%(cosfi)

expression for current:
i = 9.83*sin(omega*t--38.1)+1.0*sin(3*omega*t-7.0)+0.25*sin(5*omega*t-74.3)
Irms=7 A
Vrms=181 V
Total Power, P=977 W
Power factor = 0.77


## Ex 14 Page 355¶

In [61]:
from __future__ import division
from math import pi,exp,sin,atan

# Given
f=50  # Hz
Vm=400  # V
R=10  # ohm
L=0.1  # H
t=0.02  # sec
XL=2*pi*f*L  # ohm
Z=R+1J*XL  # ohm
Im=Vm/abs(Z)  # A
fi=atan(XL/R)  # degree
lamda=L/R  # sec
print "expression for current:"
print "i = %.1f*sin(314*t-%.3f)+0.95*e**(-100*t)"%(Im,fi)
i = Im*sin(314*t-fi)+0.95*exp(-100*t)  # A
print "current after 0.02 sec is : %0.1f A"%(i)
i2=Im*(0.95*exp(-100*t))  # A
print "transient component is : %0.2f A"%(i2)

expression for current:
i = 12.1*sin(314*t-1.263)+0.95*e**(-100*t)
current after 0.02 sec is : -11.4 A
transient component is : 1.56 A


## Ex 15 Page 356¶

In [62]:
from __future__ import division
from math import pi,sqrt,asin,exp,sin
from sympy import Symbol,solve
from scipy.misc import derivative

# Given
C=5*10**-6  # F
L=2  # H
R=200  # ohm

if R<2*sqrt(L/C):
print "Since R<2sqrt(L/C), the circuit is originally oscillatory."

a=R/2/L
#i=Im*exp(-a*t)*sin(omega*t+fi)
#at t=0 sec
i0=0  # A
vc=10  # V
fi=asin(i0)  # degree
#L*di/dt=vc at t=0
Im=Symbol('Im')
def current(t):
i=Im*exp(-a*t)*sin(omega*t+fi)
return i
#i=Im*expm(-a*t)*sin(omega*t+fi)

LdiBYdt=L*derivative(current,0)
#temp = coeff(LdiBYdt)
Im=solve(LdiBYdt,Im)[0]
#vc/temp(2)
print "Expression for current :\n  i = %.3f*exp(-%dt)*sin(%.1ft)"%(Im,a,omega)
Rn=2*sqrt(L/C)  # ohm

Since R<2sqrt(L/C), the circuit is originally oscillatory.
Expression for current :
i = 0.000*exp(-50t)*sin(312.2t)

Additional resistance required = 1064 ohm


## Ex 16 Page 357¶

In [63]:
from __future__ import division
from math import pi,sqrt

# Given
#i=0.5+0.3*sin(omega*t)-0.2*sin(2*omega*t)
I0=0.5;I1m=0.3;I2m=-0.2  # from above expression
Iav=I0  # A
R=1000  # ohm
L=1/1000  # H

Irms=sqrt(I0**2+(I1m/sqrt(2)**2+(I2m/sqrt(2)**2)))  # A
print "Reading of hot wire instrument = %.3f A"%(Irms)
VR=Irms*R  # V
print "Reading of electrostatic voltmeter acroos 1000 ohm = %d V"%(VR)
#vl_dash=L*di/dt=300*cos(w*t)-400*cos(2*w*t)
vl1=300;vl2=4  # V
vl=sqrt((300/sqrt(2))**2+(400/sqrt(2))**2)
print "Reading of electrostatic voltmeter acroos 1 mH inductor = %d V"%(vl)

Reading of hot wire instrument = 0.548 A
Reading of electrostatic voltmeter acroos 1000 ohm = 547 V
Reading of electrostatic voltmeter acroos 1 mH inductor = 353 V


## Ex 17 Page 358¶

In [64]:
from __future__ import division

# Given
R=80  # ohm
V=2  # V
l=50  # cm
vd=.1  # V
emf=1.43  # V
Rc=850  # ohm
sg=17.5  # mm/uA
df=1  # mm

I=R/V  # A
Rw=vd/I  # ohm (Resistance of side wire)
Id=df/sg*10**-6  # A (current for 1mm deflection)
el=1/sg*Rc  # uV
print "The limit of error = %.1f uV"%(el)
Rw1=0.2/l*Rw  # ohm (for 2cm slide wire)
dV=I*Rw1*1000  # mV (Change in voltage from null point)
r1=emf/I  # ohm (tapped portion)
r2=r1*22.8/R  # ohm
Ig=dV/1000/(Rc+r2)  # A
d=dV/1000/(Rc+r2)/Id  # mm
print "Deflection  = %.1f mm"%(d)

The limit of error = 48.6 uV
Deflection  = 8.2 mm


## Ex 18 Page 359¶

In [65]:
from __future__ import division
from math import pi,sqrt

# Given
R=50  # ohm
Vrms=100  # V
Rd1=50  # ohm
Rd2=100  # ohm

Vm=Vrms/sqrt(2)  # V
#v=Vm*sin(theta)
Rf=R+Rd1  # ohm
Rb=R+Rd2  # ohm
#i_f=v/Rf  # A
#i_b=v/Rb  # A
print "reading of hot wire ammeter = %.2f A"%(Irms)
print "reading of moving coil ammeter = %.2f A"%(Iav)
P=1/2*(Vrms**2/Rf+Vrms**2/Rb)  # W
print "\n\n Power taken from the mains = %.1f W"%(P)
Pc=Irms**2*R  # W
Pd=P-Pc  # W
print "Power dissipated in rectifying device = %d W"%(Pd)

reading of hot wire ammeter = 0.56 A
reading of moving coil ammeter = 0.30 A

Power taken from the mains = 83.3 W
Power dissipated in rectifying device = 67 W


## Ex 19 Page 361¶

In [66]:
from __future__ import division
from math import pi,sqrt,sin

# Given
d=1/100  # m
S=-1/100  # m
Ve=2  # kV
theta=30  # degree
e=1.6*10**-19  # C
m=9.67*10**-31  # kg

u=sqrt(2*e*Ve*1000/m)  # m/s
uy=u*sin(pi/180*theta)  # m/s
vy=0  # since final velocity =0
#vy**2-uy**2=2*ay*S
ay=(vy**2-uy**2)/2/S  # m**2/s
#ay=e/m*V/d
V=ay*m*d/e  # V
print "Required potential difference = %.f V"%(V)

Required potential difference = 500 V


## Ex 20 Page 361¶

In [67]:
from __future__ import division

# Given
#Ia=0.0004*(Va+40*Vg)**(3/2)  #  mA --eqn(1)
Va=250  # V
Vg=-3  # V
#mu=delVa/delVg
#differenciation wrt Vg eqn(1)
#(4*10**-6*3/2*(Va+40*Vg))**(1/2)*(mu+40)=0
mu=-40  # constant
print "Amplification factor, mu = %.f"%(mu)
#differenciation wrt Va eqn(1)
#delIa/delVa=(4*10**-6*3/2*(Va+40*Vg))**(1/2)*(0+40)
gm=(4*10**-6*3/2)*(Va+40*Vg)**(1/2)*(0+40)*1000  #  mA/V or S
print "Mutual conductance, gm = %.2f S"%(gm)
#differenciation wrt Ia eqn(1) keeping Vg constant
#1=(4*10**-6)*3/2*(Va+40*Vg)**(1/2)*(delVa/delIa+0)
#ra=delVa/delIa
ra=1/((4*10**-6)*3/2*(Va+40*Vg)**(1/2))
print "Plate resistance, ra = %.1f kohm"%(ra/1000)

Amplification factor, mu = -40
Mutual conductance, gm = 2.74 S
Plate resistance, ra = 14.6 kohm


## Ex 21 Page 363¶

In [68]:
from __future__ import division
from math import pi

# Given
fc=25*10**6  # Hz
fm=400  # Hz
Vm=4  # V
Del=10*10**3  # Hz
m=Del/fm  # modulation index
mf=m;mp=m  # modulation index
print "General equation of FM wave is:"
print "v=%d*sin(%.2e*t+%d*sin(%d*t)"%(Vm,wc,mf,wm)
print "\n\n General equation of PM wave is:"
print "v=%d*sin(%.2e*t+%d*sin(%d*t)"%(Vm,wc,mp,wm)
# Changing modulating frequency
fm_new=2*10**3  # Hz
a=fm_new/fm  # increase in angular frequency
print "\n\n Now equation of FM wave is:"
print "v=%d*sin(%.2e*t+%d*sin(%d*t)"%(Vm,wc,mf,a*wm)
print "\n\n Now equation of PM wave is:"
print "v=%d*sin(%.2e*t+%d*sin(%d*t)"%(Vm,wc,mp,a*wm)

General equation of FM wave is:
v=4*sin(1.57e+08*t+25*sin(2513*t)

General equation of PM wave is:
v=4*sin(1.57e+08*t+25*sin(2513*t)

Now equation of FM wave is:
v=4*sin(1.57e+08*t+25*sin(12566*t)

Now equation of PM wave is:
v=4*sin(1.57e+08*t+25*sin(12566*t)


## Ex 22 Page 363¶

In [69]:
from __future__ import division

# Given
Ebb=300  # V
Ibb=20  # A
Emm=150  # V
Po=4.5*10**3  # W

m=Emm/Ebb  # modulation index
Pbb=Ebb*Ibb
eta=Po/Pbb*100  # %
P=Po*(1+m**2/2)  # W
Pdo=Pbb-Po  # W
Pd=Pdo*(1+m**2/2)  # W
print "modulation index = %.1f"%(m)
print "carrier power under modulated condition = %0.2f kW"%(P/1000)
print "plate circuit efficiency = %.f percent"%(eta)
print "plate dissipation under unmodulated condition = %.1f kW"%(Pdo/1000)
print "plate dissipation under modulated condition = %.2f kW"%(Pd/1000)

modulation index = 0.5
carrier power under modulated condition = 5.06 kW
plate circuit efficiency = 75 percent
plate dissipation under unmodulated condition = 1.5 kW
plate dissipation under modulated condition = 1.69 kW


## Ex 23 Page 364¶

In [70]:
from __future__ import division
from math import cos,pi,sin
# Given
Zo=50  # ohm
VSWR=2  # ratio
#lm=0.2*lamda
lmBYlamda=0.2
betaINTOlamda=2*pi
rho=(VSWR-1)/(VSWR+1)  # reflection coefficient
#exp(j*theta)=cos(theta)+1J*sin(theta)
ZL=Zo*(1-rho*(cos(theta)+1J*sin(theta)))/(1+rho*(cos(theta)+1J*sin(theta)))  # ohm
Rs=(ZL.real)  # ohm
Xs=abs((ZL.imag))  # ohm(capacitive)
print "Series equivalent circuit:"
print "Rs = %0.1f ohm"%(Rs)
print "Xs = %0.1f ohm"%(Xs)
YL=(1/ZL)*1000  # mS
Rp=1000/(YL.real)  # ohm
Xp=1000/(YL.imag)  # ohm
print "\n\n Parallel equivalent circuit:"
print "Rp = %0.1f ohm"%(Rp)
print "Xp = %0.f ohm"%(Xp)

Series equivalent circuit:
Rs = 77.7 ohm
Xs = 34.3 ohm

Parallel equivalent circuit:
Rp = 92.8 ohm
Xp = 211 ohm


## Ex 24 Page 366¶

In [71]:
from __future__ import division
from math import pi,sqrt

# Given
b=3  # cm
a=4.5  # cm
f=9*10**9  # Hz
v=3*10**10  # cm/s
lamda=v/f  # cm

print "For TE10 mode:"
m=1  #  for TE10 mode
lamda_c = 2*a/m  # cm
rho=sqrt(1-(lamda/lamda_c)**2)
lamda_g=lamda/rho  # cm
vg=rho*v  # cm/s
vp=v/rho  # cm/s
ZTE=120*pi/rho  # ohm

print "cutoff wavelength = %.f cm"%(lamda_c)
print "guide wavelength = %.2f cm"%(lamda_g)
print "Group velocity = %.1e m/s"%(vg/100)
print "Phase velocity = %.1e m/s"%(vp/100)
print "Characteristic wave impedence = %.f ohm"%(ZTE)

print "\n\n For TM11 mode:"
m=1;n=1# for TE10 mode
lamda_c = 2/sqrt((m/a)**2+(n/b)**2)  # cm
rho=sqrt(1-(lamda/lamda_c)**2)
lamda_g=lamda/rho  # cm
vg=rho*v  # cm/s
vp=v/rho  # cm/s
ZTM=120*pi*rho  # ohm
print "cutoff wavelength = %.f cm"%(lamda_c)
print "guide wavelength = %.2f cm"%(lamda_g)
print "Group velocity = %.1e m/s"%(vg/100)
print "Phase velocity = %.1e m/s"%(vp/100)
print "Characteristic wave impedence = %.f ohm"%(ZTM)

For TE10 mode:
cutoff wavelength = 9 cm
guide wavelength = 3.59 cm
Group velocity = 2.8e+08 m/s
Phase velocity = 3.2e+08 m/s
Characteristic wave impedence = 406 ohm

For TM11 mode:
cutoff wavelength = 5 cm
guide wavelength = 4.48 cm
Group velocity = 2.2e+08 m/s
Phase velocity = 4.0e+08 m/s
Characteristic wave impedence = 281 ohm


## Ex 25 Page 367¶

In [72]:
from __future__ import division

# Given
fs_max=1600  # kHz
fs_min=500  # kHz
IF=465  # kHz

fr=fs_max/fs_min  # ratio
C_ratio = fr**2  # Cs_max/Cs_min
#Part (a)
fo_min=IF+fs_min  # kHz
fo_max=IF+fs_max  # kHz
fr_o=fo_max/fo_min  # frequency ratio for oscillator
C_ratio_o = fr_o**2  # Cs_max/Cs_min
print "part(a):"
print "For fo>fs, the range of fo : %.f to %.f kHz"%(fo_min,fo_max)
print "frequency ratio = %.2f"%(fr_o)
print "capacitance ratio = %.2f"%(C_ratio_o)

#Part (b)
fo_min=-IF+fs_min  # kHz
fo_max=-IF+fs_max  # kHz
fr_o=fo_max/fo_min  # frequency ratio for oscillator
C_ratio_o = fr_o**2  # Cs_max/Cs_min
print "\n\n part(b):"
print "For fo>fs, the range of fo : %.f to %.f kHz"%(fo_min,fo_max)
print "frequency ratio = %.1f"%(fr_o)
print "capacitance ratio = %.1f"%(C_ratio_o)
#ans wrong for part b in the book.

part(a):
For fo>fs, the range of fo : 965 to 2065 kHz
frequency ratio = 2.14
capacitance ratio = 4.58

part(b):
For fo>fs, the range of fo : 35 to 1135 kHz
frequency ratio = 32.4
capacitance ratio = 1051.6


## Ex 26 Page 368¶

In [73]:
from __future__ import division

# Given
Ic=3  # mA
hfe=45  # unitless
Vcc=12  # V
VBE=0.5  # V
S=0.05  # stability factor
Beta=45  # unitless

RR=Vcc/2/(Ic*10**-3)  # ohm (let RL+Re=RR)
#Re=20% of (Re+Rl)
Re=RR*20/100  # ohm
RL=RR-Re  # ohm
Ve=(Ic+Ic/Beta)*10**-3*Re  # V
Vb=Ve+VBE  # V
#S=Re/Rb=0.5 => Rb=Re/S
R1=Vcc*Re/S/Vb/1000  # kohm
# Vb/Vcc = R2/(R2+R1)
R2=Vb*R1/(Vcc-Vb)  # kohm
print "Resistor values are : "
print "RL=%.2f kohm"%(RL/1000)
print "Re=%.2f kohm"%(Re/1000)
print "R1=%.2f kohm"%(R1)
print "R2=%.2f kohm"%(R2)

Resistor values are :
RL=1.60 kohm
Re=0.40 kohm
R1=55.60 kohm
R2=9.34 kohm


## Ex 27 Page 369¶

In [74]:
from __future__ import division
from math import pi

# Given
Vcc=50  # V
Vmin=10  # V
Pd=40  # W

Vo=Vcc-Vmin  # V
K=Vo/Vcc  # constant
Rdash=2*Vcc**2/pi/Pd*(K-pi*K**2/4)  # ohm
Po=K**2*Vcc**2/2/Rdash  # W
eta=pi*K/4*100  # %

print "load presented by transformer = %.1f ohm"%(Rdash)
print "load power output = %.1f W"%(Po)
print "conversion efficiency = %.1f percent"%(eta)

load presented by transformer = 11.8 ohm
load power output = 67.6 W
conversion efficiency = 62.8 percent


## Ex 28 Page 370¶

In [75]:
from __future__ import division

# Given
Rs=1000  # ohm
Rc1=2*1000  # ohm
Re2=2*1000  # ohm
#CE configuration
hie=1100  # ohm
hre=2.5*10**-4#
hfe=50#
hoe=25*10**-6  # s
#CC configuration
hic=1.1  # kohm
hrc=1#
hfc=-51#
hoc=25*10**-6  # s

print "for 2nd stage(CC stage)"
AI2=-hfc/(1+hoe*Re2)  # current gain
Ri2=hic+hrc*AI2*Re2  # kohm
Av2=AI2*Re2/Ri2  # Voltage Gain
print "current gain = %0.2f"%(AI2)
print "Input impedence = %0.2f kohm"%(Ri2/1000)
print "Voltage gain = %0.2f"%(Av2)

print "\n\n for 1st stage(CE stage)"
RL1=Rc1*Ri2/(Rc1+Ri2)  # kohm
AI1=-hfe/(1+hoe*RL1)  # current gain
print "current gain = %.2f"%(AI1)
Ri1=hie+hre*AI1*RL1  # kohm
print "Input impedence = %0.2f kohm"%(Ri1/1000)
Av1=AI1*RL1/Ri1  # Voltage gain
print "Voltage gain = %0.2f"%(Av1)
Ro1=1/(hoe-hfe*hre/(hie+100))  # ohm
print "Output impedence = %.2f kohm"%(Ro1/1000)
Ro1dash=Ro1*Rc1/(Ro1+Rc1)  # /ohm
print "Output impedence taking Rc1 into account = %.2f kohm"%(Ro1dash/1000)

print "\n\n for overall amplifier"
Ro=1/(hoc*100-hfc*hrc/(hic+Ro1dash))  # ohm
print "Output impedence = %.2f ohm"%(Ro)
Rodash=Ro*Re2*1000/(Ro1+Re2*1000)  # /ohm
print "Output impedence taking Re2 into account = %.2f ohm"%(Rodash)
AI=AI1*AI2*Rc1/(Ri2+Rc1)  #  current gain
print "current gain = %.2f"%(AI)
Av=Av1*Av2  # voltage gain
print "Voltage gain = %.2f"%(Av)
#answer is wrong for overall amplifier in the book.

for 2nd stage(CC stage)
current gain = 48.57
Input impedence = 97.14 kohm
Voltage gain = 1.00

for 1st stage(CE stage)
current gain = -47.66
Input impedence = 1.08 kohm
Voltage gain = -86.76
Output impedence = 68.57 kohm
Output impedence taking Rc1 into account = 1.94 kohm

for overall amplifier
Output impedence = 34.81 ohm
Output impedence taking Re2 into account = 33.65 ohm
current gain = -46.70
Voltage gain = -86.76


## Ex 29 Page 372¶

In [76]:
from __future__ import division
from math import pi

# Given
fT=6*10**6  # Hz
hfe=50#
Rs=500  # ohm
gm=0.04  # S
rbb_dash=100  # ohm
Cc=10*10**-12  # F
RL=1  # kohm

rbe=hfe/gm  # ohm
Ce=gm/2/pi/fT  # F
C=Ce+Cc*(1+gm*RL)  # F
hie=rbb_dash+rbe  # ohm
R=(Rs+rbb_dash)*rbe/((Rs+rbb_dash)+rbe)  # ohm
f2=1/2/pi/R/C  # Hz
print "Voltage gain upper BW frequency = %.2f MHz"%(f2/10**6)
AIS=-hfe*Rs/(Rs+hie)  # current gain
print "Current gain = %.2f"%(AIS)
AVS=-hfe*RL*1000/(Rs+hie)  # voltage gain
print "Voltage gain = %.2f"%(AVS)
AVSf2=AVS*f2  # Hz
print "Voltage gain BW product = %.2f MHz"%(abs(AVSf2/10**6))
AISf2=AIS*f2  # Hz
print "Current gain BW product = %.2f MHz"%(abs(AISf2/10**6))

#answer in the textbook are wrong.

Voltage gain upper BW frequency = 0.37 MHz
Current gain = -13.51
Voltage gain = -27.03
Voltage gain BW product = 9.90 MHz
Current gain BW product = 4.95 MHz


## Ex 30 Page 373¶

In [77]:
from __future__ import division
from math import pi,sqrt

# Given
VP=-2  # V
IDSS=1.75/1000  # A
VDD=24  # V
ID=1/1000  # A
VP=0.2  # V (pinch off voltage)

VGS=(1-sqrt(ID/IDSS))*VP  # V
gmo=-2*IDSS/VP  # S
gm=gmo*(1-VGS/VP)  # S
Rs=-VGS/ID  # ohm

print "VGS=%.2f V"%(VGS)
print "gm = %.2f mS"%(gm*1000)
print "Rs = %.f ohm"%(Rs)
#Ans are wrong in the book.

VGS=0.05 V
gm = -13.23 mS
Rs = -49 ohm


## Ex 31 Page 374¶

In [78]:
from math import sqrt
from __future__ import division
# Given
G=37  # dB
f1=50  # Hz
f2=18.7*1000  # Hz
BW1=f2  # Hz (f2-f1~=f2)

A1=10**(G/20)  # Gain
A3=sqrt(A1)  # Gain
RL1BYRL2=A1/A3  # ratio
RL2BYRL1=A3/A1  # ratio
#BW=2*pi*Cd*RL
BW1BYBW2=RL2BYRL1#
BW2BYBW1=RL1BYRL2#
f2dash=f2*sqrt(sqrt(2)-1)#
BW2=BW2BYBW1*f2dash  # Hz
print "Bandwidth of redesigned aplifier, BW=%.f kHz"%(BW2/1000)

Bandwidth of redesigned aplifier, BW=101 kHz


## Ex 32 Page 375¶

In [79]:
from math import sqrt
from __future__ import division

# Given
L=30  # H
C=10*10**-6  # F
RL=8*10**3  # ohm
f=50  # Hz

r=sqrt(2)/12/(2*pi*f)**2/L/C*100  # %
Lc=2*RL/6/(2*pi*f)  # H
print "ripple factor = %.1f percent"%(r)
print "Critical inductance, Lc=%.1f H"%(Lc)

ripple factor = 0.4 percent
Critical inductance, Lc=8.5 H


## Ex 33 Page 376¶

In [80]:
from __future__ import division
# Given
V=500  # V
Pp=1500*10**3  # W (+ve side)
Pn=2000*10**3  # W (-ve side)

P=Pp+Pn  # W
I=P/V  # A
Ip=Pp/(V/2)  # A
In=Pn/(V/2)  # A
Iob=In-Ip  # A
Ia=Iob/2  # A
print "Current supplied by the main generator = %.f A"%(I)
print "Current supplied on +ve side = %.f A"%(Ip)
print "Current supplied on -ve side = %.f A"%(In)
print "out-off balance Current = %.f A"%(Iob)
print "Current in each armature = %.f A"%(Ia)

Current supplied by the main generator = 7000 A
Current supplied on +ve side = 6000 A
Current supplied on -ve side = 8000 A
out-off balance Current = 2000 A
Current in each armature = 1000 A


## Ex 34 Page 377¶

In [81]:
from __future__ import division
from math import sqrt
# Given
l=20  # km
P=10000  # kW
V=11  # kV
pf=0.707  # lagging
R=0.02  # ohm/km/phase
X=0.07  # ohm/km/phase

#for pf = 0.707

IL=P*10**3/sqrt(3)/(V*1000)/pf  # A
VRphase=V*1000/sqrt(3)  # V
R_phase=l*R  # ohm
X_phase=l*X  # ohm
Z_phase=R_phase+1J*X_phase  # ohm
Vd_phase=IL*(pf-1J*pf)*Z_phase  # V
VS=(Vd_phase+VRphase)  # V
regulation=(VS-VRphase)/VRphase*100  # %
print "regulation = %.f percent"%abs(regulation)
PL=3*IL**2*R_phase/1000  # kW
Po=P+PL  # kW
eta=P/Po*100  # %
print "Efficiency = %.f percent"%(eta)

#for pf2=0.9  # lagging
pf=0.9  # lagging
IL=P*10**3/sqrt(3)/(V*1000)/pf  # A
VRphase=V*1000/sqrt(3)  # V
R_phase=l*R  # ohm
X_phase=l*X  # ohm
Z_phase=R_phase+1J*X_phase  # ohm
Vd_phase=IL*(pf-1J*.435)*Z_phase  # V
VS=(Vd_phase+VRphase)  # V
regulation=(VS-VRphase)/VRphase*100  # %
print "\n\n regulation = %.1f percent"%abs(regulation)
PL=3*IL**2*R_phase/1000  # kW
Po=P+PL  # kW
eta=P/Po*100  # %
print "Efficiency = %.f percent"%(eta)
#ans wrong for 2nd part in the book.

regulation = 17 percent
Efficiency = 94 percent

regulation = 13.4 percent
Efficiency = 96 percent


## Ex 35 Page 378¶

In [82]:
from __future__ import division

# Given
C1=1.2  # Rs
C2=5  # Rs
P1=100  # W
P2=30  # W
t=1000  # hours
Ce=60  # Rs/kW/annum for max demand
Cee=3   # paise/unit for extra

#first lamp
Cl1=C1*100/t  #  paise / hour
dmax1=P1/1000  # kW/hour
Cmax1=Ce*100*dmax1/(24*365)  # paise/hour
CE1=Cee*dmax1  # /paise / hour
CT1=Cl1+Cmax1+CE1  # paise (total cost per hour)
print "lamp1 : Total cost/hour = %.3f paise"%(CT1)
#second lamp
Cl2=C2*100/t  #  paise / hour
dmax2=P2/1000  # kW/hour
Cmax2=Ce*100*dmax2/(24*365)  # paise/hour
CE2=Cee*dmax2  # /paise / hour
CT2=Cl2+Cmax2+CE2  # paise (total cost per hour)
print "lamp2 : Total cost/hour = %.2f paise"%(CT2)
print "on comparing cost it is clear lamp1 will be more economical. "
#Cl1+Cmax1/x+CE1=Cl2+Cmax2/x+CE2
print "\n\n load factor = %.f percent"%(x)

lamp1 : Total cost/hour = 0.488 paise
lamp2 : Total cost/hour = 0.61 paise
on comparing cost it is clear lamp1 will be more economical.



## Ex 36 Page 379¶

In [83]:
from __future__ import division

# Given
p=4  # pole
I1=143  # A
Z=492  # armature conductors
theta=10  # degree
I2=10  # A

Ia=I1+I2  # A
Iw=Ia/2  # A for wave sound
Il=Ia/4  # for lap sound
#part(a)
ATw=Z*Iw*theta/360  # AT
nw=ATw/theta  # turns
print "(a) extra shunt field turns required = %d"%(nw)

#part(b)
ATl=Z*Il*theta/360  # AT
nl=ATl/theta  # turns
print "(b) extra shunt field turns required = %d"%(nl)
#answer wrong for part(a) in the boko.

(a) extra shunt field turns required = 104
(b) extra shunt field turns required = 52


## Ex 37 Page 380¶

In [84]:
from __future__ import division
from numpy import roots
# Given
Wh=250  # W
We=100  # W
N=1000/60  # rps

A=Wh/N  # constant
B=We/N**2  # constant
Wnew=1/2*(Wh+We)  # W
#Wnew=A*N1+B*N1**2
p=[B,A,-Wnew]  # polynomial for N1
N1=roots(p)  # rps
print "New speed will be %.f rpm"%(N1)

New speed will be 570 rpm


## Ex 38 Page 381¶

In [85]:
from __future__ import division

# Given
P=50  # kW
V=250#V
N1=400  # rpm
Ra=0.02  # ohm
Rf=50  # ohm
Pi=50  # kW
Vin=250  # V
Vd=1  # V per Brush

I=P*10**3/V#A
Ish=V/Rf  # A
Ia=I+Ish  # A
Va=Ia*Ra  # V
Vbd=2*Vd  # V
Eb1=V+Va+Vbd  # V

#as motor
I=P*10**3/V#A
Ish=V/Rf  # A
Ia=I-Ish  # A

Va=Ia*Ra  # V
Vbd=2*Vd  # V
Eb2=V-Va-Vbd  # V
N2=(Eb2/Eb1)*N1  # rpm
print "Speed of the machine running as shunt motor = %.f rpm"%(N2)

Speed of the machine running as shunt motor = 381 rpm


## Ex 39 Page 383¶

In [86]:
from __future__ import division

# Given
VL=250  # V
Is=50  # A
Ia=380  # A
If1=5  # A
If2=4.2  # A
ra=0.02  # ohm

#Machine Losses:
ma_cu_loss=Ia**2*ra  # W (motor armature cu loss)
ga_cu_loss=(Ia-Is)**2*ra  # W (generator armature cu loss)
P=VL*Is  # W
stray_loss=P-ma_cu_loss-ga_cu_loss  # W
stray_loss_per_machine=stray_loss/2  # W

#Motor efficiency
field_cu_loss=VL*If2  # W
total_loss=ma_cu_loss+field_cu_loss+stray_loss_per_machine  # W
Pin_motor=(VL*Ia)+(VL*ra)  # W
Pout_motor=Pin_motor-total_loss  # W
Eff=Pout_motor/Pin_motor*100  # %
print "Motor efficiency = %.f percent"%(Eff)

#Generator efficiency
field_cu_loss=VL*If1  # W
total_loss=ga_cu_loss+field_cu_loss+stray_loss_per_machine  # W
Pout_generator=VL*(Ia-Is)  # W
Pin_generator=Pout_generator+total_loss  # W
Eff=Pout_motor/Pin_motor*100  # %
print "Generator efficiency = %.f percent"%(Eff)

Motor efficiency = 92 percent
Generator efficiency = 92 percent


## Ex 40 Page 384¶

In [87]:
from __future__ import division
from math import sqrt
# Given
KVA=4  # kVA
V1=200#V
V2=400#V
f=50  # Hz
Io1=0.8  # A
P1=70  # W
Vs2=17.5  # V
Is2=9  # A
P2=50  # W

I_loss=P1  # W
I2=KVA*1000/V2  # A
Cu_loss=(I2/Is2)**2*P2  # W
Zo2=Vs2/Is2  # ohm
Ro2=P2/Is2**2  # ohm
Xo2=sqrt(Zo2**2-Ro2**2)  # ohm

#(a)
print "Full load efficiency : "
#unity pf
pf=1  # power factor
Output=KVA*pf  # kW
Losses=Cu_loss+I_loss  # W
eta=Output*1000/(Output*1000+Losses)*100  # %
print "at unity power factor = %.1f percent"%(eta)
#0.8 pf
pf=0.8  # power factor
Output=KVA*pf  # kW
eta=Output*1000/(Output*1000+Losses)*100  # %
print "at 0.8 power factor = %.1f percent"%(eta)

#(b)
#(i) unity pf
Vd=I2*Ro2  # V
V22=V2-Vd  # V
print "\n\n Voltage drop at unity pf = %.1f V"%(V22)
#(i) 0.8 pf lagging
pf=0.8  # power factor
Vd=I2*(Ro2*pf+Xo2*sqrt(1-pf**2))  # V
V22=V2-Vd  # V
print "Voltage drop at 0.8 pf lagging = %.1f V"%(V22)
pf=0.8  # power factor
Vd=I2*(Ro2*pf-Xo2*sqrt(1-pf**2))  # V
V22=V2-Vd  # V
print "Voltage drop at 0.8 pf leading = %.f V"%(V22)

Full load efficiency :
at unity power factor = 96.8 percent
at 0.8 power factor = 96.0 percent

Voltage drop at unity pf = 393.8 V
Voltage drop at 0.8 pf lagging = 384.0 V
Voltage drop at 0.8 pf leading = 406 V


## Ex 41 Page 385¶

In [88]:
from __future__ import division

# Given
KVA=15  # kVA
pf=1#
eff=98/100  # efficiency

L1=2  # kW
pf1=0.5  # lagging
t1=12  # hours
L2=12  # kW
pf2=0.8  # lagging
t2=6  # hours
L3=18  # kW
pf3=0.9  # lagging
t3=6  # hours

Po=KVA*pf  # kW
Pin=Po/eff  # kW
Losses=Pin-Po  # kW
Cu_loss=Losses/2  # kW
I_loss=Losses/2  # kW

KVA1=L1/pf1  # kVA
KVA2=L2/pf2  # kVA
KVA3=L3/pf3  # kVA
Cu_loss1=Cu_loss*(KVA1/KVA2)**2  # kW
Cu_loss2=Cu_loss*(KVA2/KVA2)**2  # kW
Cu_loss3=Cu_loss*(KVA3/KVA2)**2  # kW

Cu_loss_t1=Cu_loss1*t1  # kW
Cu_loss_t2=Cu_loss2*t2  # kW
Cu_loss_t3=Cu_loss3*t3  # kW
Cu_loss_total=Cu_loss_t1+Cu_loss_t2+Cu_loss_t3  # kW
I_loss24=I_loss*24  # Wh

Po24=L1*t1+L2*t2+L3*t3  # kWh
Pi24=Po24+Cu_loss_total+I_loss24  # kWh
eff_allday=Po24/Pi24*100  # %
print "All day efficiency = %.f percent"%(eff_allday)

All day efficiency = 97 percent


## Ex 42 Page 386¶

In [89]:
from __future__ import division
from math import sqrt
# Given
V1=250  # V
V2=150  # V
Vs1=200  # V
pf1=0.8  # lagging
pf2=1  # unity
Vs2=100  # V

t1_ratio=V1/Vs1  #
Ip1=I1*t1_ratio  # A at 0.8 pf (current drawn by primary)

t2_ratio=Vs2/V2  #
Ip2=I2*t2_ratio  # A at 0.8 pf (current drawn by primary)

Ipx=Ip1*pf1+Ip2  # A
Ipy=Ip1*sqrt(1-pf1**2)  # A
I3=sqrt(Ipx**2+Ipy**2)  # A
print "Current drawn by the transformer=%.2f A"%(I3)
pf=Ipx/I3  # power factor
print "power factor = %.1f lagging"%(pf)

Current drawn by the transformer=42.67 A
power factor = 0.9 lagging


## Ex 43 Page 387¶

In [90]:
from __future__ import division
from math import sqrt,atan,cos
# Given
R2=0.03  # ohm
X2=0.18  # ohm
Ns=100  # rpm
s1=3  # %

Nfl=(100-s1)  # rpm (full load speed)
N2=Nfl/2  # rpm
s2=(Ns-N2)/Ns*100  # %
V1BYV2=sqrt(s2/s1*(R2**2+(s1/100*X2)**2)/(R2**2+(s2/100*X2)**2))  # from torque equation
#let V1=V12BYV1 V2=1
V1=V1BYV2  # V
V2=1  # V
V12BYV1=(V1-1)/V1*100  #  % reduction in the stator (V12=V1-V2)
print "Percentage reduction in stator voltage = %.f percent"%(V12BYV1)
pf=cos(fi)  # power factor
print "power factor = %.1f"%(pf)

Percentage reduction in stator voltage = 23 percent
power factor = 0.3


## Ex 44 Page 388¶

In [91]:
from __future__ import division
from math import sqrt

# Given
zo=1+1J*1  # ohm
zi=0.2+1J*4  # ohm
Ri=zi.real   # ohm
Ro=(zo).real   # ohm

#at standstill
s=1  # % at standstill
Zo=sqrt((zo).real**2+(zo).imag**2)  # ohm
Zi=sqrt((zi).real**2+(zi).imag**2)  # ohm
ToBYTi=Ro/Ri*(Zi/Zo)**2  # torque ratio
print "(a) at standstill, To:Ti = %d:1"%(ToBYTi)

#at s=0.5
s=0.05  # %
Zo=sqrt((zo).real**2/s**2+(zo).imag**2)  # ohm
Zi=sqrt((zi).real**2/s**2+(zi).imag**2)  # ohm
ToBYTi=Ro/Ri*(Zi/Zo)**2  # torque ratio
print "(b) at s=0.05, To:Ti = %.1f:1"%(ToBYTi)

(a) at standstill, To:Ti = 40:1
(b) at s=0.05, To:Ti = 0.4:1


## Ex 45 Page 389¶

In [92]:
from __future__ import division
from math import sqrt
# Given
Edc=500  # V
fim=.085  # Wb
f=50  # Hz
E1=11000  # V
P=1500  # kW
p=8  # pole
pf=0.9
V=500  # V
J=3  # A/mm**2

E2=Edc/sqrt(2)#V
N2=E2/4.44/f/fim  # no. of turns
N1=E1/E2*N2  # no. of turns
print "no. of turns in primary = %d"%(N1)
print "no. of turns in secondary = %d"%(N2)
Idc=P*10**3/V  # A
eta=1  # because of no loss
ISR=0.472*Idc/(eta*pf)
A1=ISR/J*10**-6  # m**2 (cross section area)
I1=N2/N1*ISR  # A
A2=I1/J*10**-6  # m**2 (cross section area of primary winding)
print "\ncross section of primary winding=%.2e m**2"%(A1)
print "cross section of secondary winding=%.1e m**2"%(A2)
#ans in the book are not accurate.

no. of turns in primary = 582
no. of turns in secondary = 18

cross section of primary winding=5.24e-04 m**2
cross section of secondary winding=1.7e-05 m**2


## Ex 46 Page 391¶

In [93]:
from __future__ import division
from math import sqrt

# Given
IscBYIfl=5  #  as Isc=5*Ifl
ILByIfl=3  #  as IL <= 3*Ifl
sf=5  # %

#IL=K**2*ISC
#dividing by Ifl
K=sqrt(ILByIfl/IscBYIfl)*100  # %
TstBYTf=(K/100)**2*IscBYIfl*sf/100*100  #  % #ratio of torque
print "Suitable auto transformation ratio = %.1f"%(K)
print "Starting torque Tst = %.f percent of full-load torque"%(TstBYTf)
#ans wrong in the textbook.

Suitable auto transformation ratio = 77.5
Starting torque Tst = 15 percent of full-load torque


## Ex 47 Page 391¶

In [94]:
from __future__ import division
from math import log

# Given
V=500  # V
ns=60  # slots
nc=20  # conductor/slot
ra=1.31  # ohm
Tmax=218  # N-m
fi=23*10**-3  # Wb

Tmin=Tmax/1.5#N-m
Z=ns*nc  # no of conductors
Ia=Tmax/(.159*fi*Z)  # A
Imax=1.5*Ia  # A
I1=Imax  # A
I2=Ia  # A
R1=V/I1  # ohm
n= log(R1/ra)/log(I1/I2)+1  # no of studs
N=n-1  # no of section
print "no of studs = %d and no. of section = %d"%(n,N)
R2=R1*(I2/I1)  # ohm
R3=R2*(I2/I1)  # ohm
R4=R3*(I2/I1)  # ohm
R1s=R1-R2  # ohm
R2s=R2-R3  # ohm
R3s=R3-R4  # ohm
R4s=R4-ra  # ohm
print "\nResistance of 1st section = %.2f ohm"%(R1s)
print "Resistance of 2nd section = %.2f ohm"%(R2s)
print "Resistance of 3rd section = %.2f ohm"%(R3s)
print "Resistance of 4th section = %.2f ohm"%(R4s)

no of studs = 5 and no. of section = 4

Resistance of 1st section = 2.24 ohm
Resistance of 2nd section = 1.49 ohm
Resistance of 3rd section = 0.99 ohm
Resistance of 4th section = 0.68 ohm


## Ex 48 Page 393¶

In [95]:
from __future__ import division
from math import log,exp

# Given
theta1=20  # degree C
theta2=35  # degree C
t1=0.5  # hour
t2=1  # hour
theta_m_dashBYthetam=80/100  # temperature rise

#theta=theta_m*(1-e**(-t/alfa))
#theta1/theta2=(1-exp(-t1/alfa))/(1-exp(-t2/alfa))
ee1=theta2/theta1-1  # let ee1=exp(-1/2/alfa)
theta_m=theta1/(1-ee1)  # degree C
theta_2=theta_m*(1-ee1**4)  #  degree C (after 2 hours)
print "temperature rise after 2 hours full load = %.f degree C"%(theta_2)
alfa=-1/2/log(ee1)  # hour
alfa_dash=theta_m_dashBYthetam*alfa  # hour
theta_m_dash=theta_m_dashBYthetam*theta_m
theta_dash=theta_m_dash*(1-exp(-t2/alfa))
print "temperature rise of cold water after 1 hour = %.f degree C"%(theta_dash)
#ans of 2nd part is wrong in the book.

temperature rise after 2 hours full load = 55 degree C
temperature rise of cold water after 1 hour = 28 degree C


## Ex 49 Page 394¶

In [96]:
from __future__ import division
from math import sqrt,cos,pi,sin

# Given
u=30  # degree
m=3  # no of phases

#Id=sqrt(2)*Vs*X*(1-cosd(u))*sin(pi/m)
#where IdBYVs_dash = m/pi*IdX/2
EdoBYVs=sqrt(2)*m/pi*sin(pi/m)  # dc output voltage/Vs with no overlap
EduBYVs=EdoBYVs-IdBYVs_dash  # dc output voltage/Vs with overlap
#part (a)
Reg1=(EdoBYVs-EduBYVs)/EdoBYVs*100  # % (regulation)
print "Regulation at no load voltage = %.1f percent"%(Reg1)
#part (b)
Reg2=(EdoBYVs-EduBYVs)/EduBYVs*100  # % (regulation)
print "Regulation at full load voltage = %.1f percent"%(Reg2)

Regulation at no load voltage = 6.7 percent
Regulation at full load voltage = 7.2 percent


## Ex 50 Page 395¶

In [97]:
from __future__ import division
from numpy import array
# Given
I12=2000  # A (I12=I1+I2)
R1=0.04  # ohm
R2=0.025  # ohm
rf1=25  # ohm
rf2=20  # ohm
E1=440  # V
E2=420  # V

#V*(1+R1/rf1)+R1*I1=E1#eqn(1)
#V*(1+R2/rf2)-I1*R2=E2-I12*R2# eqn(2)
A=array([[(1+R1/rf1),R1],[(1+R2/rf2),-R2]])# # matrix for solution
B=array([[E1],[E2-I12*R2]])  # matrix for solution
X=A**-1*B  # solution
V=X[0,0]  # V
I1=X[1,0]  # A
I2=I12-I1  # A
print "Current for each machine  = %.f A & %.f A "%(I1,I2)
Po1=V*I1  # W
Po2=V*I2  # W
print "Power output for each machine = %.1f kW & %.1f kW"%(Po1/1000,Po2/1000)
#ans in the book are wrong.

Current for each machine  = 370 A & 1630 A
Power output for each machine = 162.3 kW & 716.3 kW


## Ex 51 Page 396¶

In [98]:
from __future__ import division
from math import atan,pi,cos,degrees
# Given
ZA=0.15+0.5*1J  # ohm
ZB=0.1+0.6*1J  # ohm
EA=207  # V
EB=205  # V
ZL=2+1.5*1J  # ohm

IA=(EA*ZB+(EA-EB)*ZL)/(ZA*ZB+ZL*(ZA+ZB))  # A
IB=(EB*ZA-(EA-EB)*ZL)/(ZA*ZB+ZL*(ZA+ZB))  # A
V2=(IA+IB)*ZL  # V
fi_A=degrees(atan((V2).imag/(V2).real)-(atan((IA.imag)/(IA.real))))
pf_A=cos(pi/180*fi_A)  # lag
print "pf transformer A = %.2f lag"%(pf_A)
fi_B=degrees(atan((V2.imag)/(V2.real))-(atan((IB.imag)/(IB.real))))
pf_B=cos(pi/180*fi_B)  # lag
print "pf transformer B = %.2f lag"%(pf_B)
PA=abs(V2*IA*pf_A)  # W
print "power output transformer A = %.f W"%(PA)
PB=abs(V2*IB*pf_B)  # W
print "power output transformer B = %.f W"%(PB)
#Power output ans are wrong in the book.

pf transformer A = 0.82 lag
pf transformer B = 0.78 lag
power output transformer A = 6535 W
power output transformer B = 4925 W


## Ex 52 Page 397¶

In [99]:
from __future__ import division
# Given
d1=0.05*10**-3  # mm
l1=100*10**-2  # m
i2BYi1=1/4  # current ratio
#(d2/d1)**(3/2)=i2BYi1
d2=(i2BYi1)**(2/3)*d1*10**6  # um
l2=1/2*l1*d1/d2*10**6  # m
print "filament length = %.2f m"%(l2)
print "filament diameter = %.f um"%(d2)

filament length = 1.26 m
filament diameter = 20 um


## Ex 53 Page 398¶

In [100]:
from __future__ import division
# Given
d1=0.10*10**-3  # mm
l1=150*10**-2  # m
i2BYi1=1/3  # current ratio
#(d2/d1)**(3/2)=i2BYi1
d2=(i2BYi1)**(2/3)*d1*10**6  # um
l2=1/2*l1*d1/d2*10**6  # m
print "filament length = %.1f m"%(l2)
print "filament diameter = %.f um"%(d2)

filament length = 1.6 m
filament diameter = 48 um


## Ex 54 Page 398¶

In [101]:
from __future__ import division
from math import sqrt
# Given
zo=2+1J*2  # ohm
zi=0.5+1J*4  # ohm
Ri=(zi).real   # ohm
Ro=(zo).real   # ohm

#at standstill
s=1  # % at standstill
Zo=sqrt((zo.real)**2+(zo.imag)**2)  # ohm
Zi=sqrt((zi.real)**2+(zi.imag)**2)  # ohm
ToBYTi=Ro/Ri*(Zi/Zo)**2  # torque ratio
print "(a) at standstill, To:Ti = %d:1"%(ToBYTi)

#at s=0.5
s=0.05  # %
Zo=sqrt((zo.real)**2/s**2+(zo.imag)**2)  # ohm
Zi=sqrt((zi.real)**2/s**2+(zi.imag)**2)  # ohm
ToBYTi=Ro/Ri*(Zi/Zo)**2  # torque ratio
print "(b) at s=0.05, To:Ti = %.f:10"%(ToBYTi*10)

(a) at standstill, To:Ti = 8:1
(b) at s=0.05, To:Ti = 3:10


## Ex 55 Page 400¶

In [102]:
from __future__ import division
from math import sqrt

# Given
Edc=250  # V
fim=.065  # Wb
f=50  # Hz
E1=6000  # V
P=1500  # kW
p=8  # pole
pf=0.9
V=400  # V
J=3  # A/mm**2

E2=Edc/sqrt(2)#V
N2=E2/4.44/f/fim  # no. of turns
N1=E1/E2*N2  # no. of turns
print "no. of turns in primary = %d"%(N1)
print "no. of turns in secondary = %d"%(N2)
Idc=P*10**3/V  # A
eta=1  # because of no loss
ISR=0.472*Idc/(eta*pf)
A1=ISR/J*10**-6  # m**2 (cross section area)
I1=N2/N1*ISR  # A
A2=I1/J*10**-6  # m**2 (cross section area of primary winding)
print "\ncross section of primary winding=%.2e m**2"%(A1)
print "cross section of secondary winding=%.1e m**2"%(A2)

no. of turns in primary = 415
no. of turns in secondary = 12

cross section of primary winding=6.56e-04 m**2
cross section of secondary winding=1.9e-05 m**2


## Ex 56 Page 400¶

In [103]:
from __future__ import division
from math import sqrt
# Given
IscBYIfl=4  #  as Isc=5*Ifl
ILByIfl=3  #  as IL <= 3*Ifl
sf=4  # %

#IL=K**2*ISC
#dividing by Ifl
K=sqrt(ILByIfl/IscBYIfl)*100  # %
TstBYTf=(K/100)**2*IscBYIfl*sf/100*100  #  % #ratio of torque
print "Suitable auto transformation ratio = %.1f"%(K)
print "Starting torque Tst = %.f percent of full-load torque"%(TstBYTf)

Suitable auto transformation ratio = 86.6
Starting torque Tst = 12 percent of full-load torque


## Ex 57 Page 401¶

In [104]:
from __future__ import division
from math import log,exp
# Given
theta1=30  # degree C
theta2=45  # degree C
t1=0.5  # hour
t2=1  # hour
theta_m_dashBYthetam=60/100  # temperature rise

#theta=theta_m*(1-e**(-t/alfa))
#theta1/theta2=(1-exp(-t1/alfa))/(1-exp(-t2/alfa))
ee1=theta2/theta1-1  # let ee1=exp(-1/2/alfa)
theta_m=theta1/(1-ee1)  # degree C
theta_2=theta_m*(1-ee1**4)  #  degree C (after 2 hours)
print "temperature rise after 2 hours full load = %.f degree C"%(theta_2)
alfa=-1/2/log(ee1)  # hour
alfa_dash=theta_m_dashBYthetam*alfa  # hour
theta_m_dash=theta_m_dashBYthetam*theta_m
theta_dash=theta_m_dash*(1-exp(-t2/alfa))
print "temperature rise of cold water after 1 hour = %.f degree C"%(theta_dash)

temperature rise after 2 hours full load = 56 degree C
temperature rise of cold water after 1 hour = 27 degree C


## Ex 58 Page 401¶

In [105]:
from __future__ import division
from math import sqrt
# Given
zo=2+1J*3  # ohm
zi=0.5+1J*5  # ohm
Ri=(zi.real)   # ohm
Ro=(zo.real)   # ohm

#at standstill
s=1  # % at standstill
Zo=sqrt((zo.real)**2+(zo.imag)**2)  # ohm
Zi=sqrt((zi.real)**2+(zi.imag)**2)  # ohm
ToBYTi=Ro/Ri*(Zi/Zo)**2  # torque ratio
print "at slip=0, To:Ti = %d:1"%(ToBYTi)

#at s=0.5
s=0.05  # %
Zo=sqrt((zo.real)**2/s**2+(zo.imag)**2)  # ohm
Zi=sqrt((zi.real)**2/s**2+(zi.imag)**2)  # ohm
ToBYTi=Ro/Ri*(Zi/Zo)**2  # torque ratio
print "at s=0.05, To:Ti = %.f:10"%(ToBYTi*10)

at slip=0, To:Ti = 7:1
at s=0.05, To:Ti = 3:10


## Ex 59 Page 402¶

In [106]:
from __future__ import division
from math import sqrt,pi,sin,cos

# Given
u=45  # degree
m=3  # no of phases

#Id=sqrt(2)*Vs*X*(1-cosd(u))*sin(pi/m)
#where IdBYVs_dash = m/pi*IdX/2
EdoBYVs=sqrt(2)*m/pi*sin(pi/m)  # dc output voltage/Vs with no overlap
EduBYVs=EdoBYVs-IdBYVs_dash  # dc output voltage/Vs with overlap
#part (a)
Reg1=(EdoBYVs-EduBYVs)/EdoBYVs*100  # % (regulation)
print "part(a) Regulation at no load voltage = %.f percent"%(Reg1)
#part (b)
Reg2=(EdoBYVs-EduBYVs)/EduBYVs*100  # % (regulation)
print "part(b) Regulation at full load voltage = %.f percent"%(Reg2)

part(a) Regulation at no load voltage = 15 percent
part(b) Regulation at full load voltage = 17 percent


## Ex 60 Page 402¶

In [107]:
from __future__ import division

# Given
d1=0.15*10**-3  # mm
l1=150*10**-2  # m
i2BYi1=1/4  # current ratio
#(d2/d1)**(3/2)=i2BYi1
d2=(i2BYi1)**(2/3)*d1*10**6  # um
l2=1/2*l1*d1/d2*10**6  # m
print "length of filament = %.2f m"%(l2)
print "diameter of filament = %.f um"%(d2)

length of filament = 1.89 m
diameter of filament = 60 um


## Ex 61 Page 403¶

In [108]:
from __future__ import division
from math import sqrt,sin,pi
# Given
d=5/100  # m
S=-4/100  # m
Ve=3  # kV
theta=45  # degree
e=1.6*10**-19  # C
m=9.67*10**-31  # kg

u=sqrt(2*e*Ve*1000/m)  # m/s
uy=u*sin(pi/180*theta)  # m/s
vy=0  # since final velocity =0
#vy**2-uy**2=2*ay*S
ay=(vy**2-uy**2)/2/S  # m**2/s
#ay=e/m*V/d
V=ay*m*d/e  # V
print "Potential difference = %.f V"%(V)

Potential difference = 1875 V


## Ex 62 Page 403¶

In [109]:
from __future__ import division
from math import pi
# Given
R=150  # ohm
Vrms=200  # V
Rd1=65  # ohm
Rd2=140  # ohm

Vm=Vrms/sqrt(2)  # V
#v=Vm*sin(theta)
Rf=R+Rd1  # ohm
Rb=R+Rd2  # ohm
#i_f=v/Rf  # A
#i_b=v/Rb  # A
print "reading of ammeter 1= %.2f A"%(Irms)
print "reading of ammeter 2 = %.2f A"%(Iav)
P=1/2*(Vrms**2/Rf+Vrms**2/Rb)  # W
print "\n\nPower taken from the mains = %.1f W"%(P)
Pc=Irms**2*R  # W
Pd=P-Pc  # W
print "Power dissipated in rectifying device = %d W"%(Pd)

reading of ammeter 1= 0.56 A