Chapter 10: AC CIRCUITS

Example 10.1,Page number: 274

In [1]:
#Question:
"""Finding the power factor and the average power dissipated."""

from math import sqrt,pi,cos,degrees
from cmath import rect,phase

#Variable Declaration:
Vm=141.0              #Peak value of supply voltage(in Volts)
ang_freq=100.0*pi     #Angular frequency(in radians per second)        
R=3.0                 #Resistance(in Ohms) 
L=0.0127              #Self-inductance of the coil(in Henry)


#Calcultaions:
Vrms=Vm/sqrt(2.0)
V=rect(Vrms,0.0)
X_L=ang_freq*L
Z=R+ (X_L*1j)
I=V/Z
ang_I=degrees(phase(I))
P=Vrms*abs(I)*cos(phase(I))
pf=cos(phase(I))


#Result:
print "(a)The rms value of the steady-state current is %.2f A and the relative phase angle is %.2f degrees." %(abs(I),-ang_I)
print "(b)The expression for the instantaneous current is i=%.3f sin(100*pi*t%.2f degrees) A." %((sqrt(2)*abs(I)),ang_I)
print "(c)The average power dissipated in the circuit is %.2f W." %(P)
print "(d)The power factor is %.3f lagging." %(pf)

(a)The rms value of the steady-state current is 19.97 A and the relative phase angle is 53.06 degrees.
(b)The expression for the instantaneous current is i=28.246 sin(100*pi*t-53.06 degrees) A.
(c)The average power dissipated in the circuit is 1196.75 W.
(d)The power factor is 0.601 lagging.

Example 10.2,Page number: 276

In [2]:
#Question:
"""Finding the capacitance required,the phase angle,the power factor,the apparent power,and the reactive power."""

from math import sqrt,pi,degrees,acos,cos,sin

#Variable Declaration: 
V=230.0               #Rms value of supply voltage(in Volts)
f=50.0                #Frequency of supply(in Hertz) 
P_lamp=750.0          #Power rating of the lamp(in Watts) 
V_lamp=100.0          #Voltage rating of the lamp(in Volts)   


#Calculations:
I_rated=P_lamp/V_lamp
Vc=sqrt((V*V)-(V_lamp*V_lamp))
Xc=Vc/I_rated
C=1/(2*pi*f*Xc)
phase_angle=acos(V_lamp/V)
pf=cos(phase_angle)
app_P=V*I_rated
rea_P=V*I_rated*sin(phase_angle)


#Result:
print "(a)The required capacitance is %e F." %(C)
print "(b)The phase angle is %.3f degrees." %(degrees(phase_angle))
print "(c)The power factor is %.3f leading." %(pf)
print "(d)The apparent power is %.3f VA." %(app_P)
print "(e)The reactive power is %.3f VAr." %(rea_P)

(a)The required capacitance is 1.152611e-04 F.
(b)The phase angle is 64.229 degrees.
(c)The power factor is 0.435 leading.
(d)The apparent power is 1725.000 VA.
(e)The reactive power is 1553.424 VAr.

Example 10.3,Page number: 277

In [4]:
#Question: 
"""Finding the impedance,the power factor,supply voltage,voltage across resistor,apparent power and reactive power."""

from cmath import rect,phase
from math import pi,cos,sin,degrees

#Variable Declaration:
I=rect(0.9,0)         #Current in the circuit(in Amperes)
R=120.0               #Resistance of the resistor(in Ohms) 
Xc=250.0              #Reactance of the capacitor(in Ohms)


#Calculations:
Z=R- Xc*1j
pf=cos(phase(Z))
V=I*Z
V_R=I*R
V_C=I*Xc
act_P=abs(V)*abs(I)*cos(phase(Z))
app_P=abs(V)*abs(I)
rea_P=abs(V)*abs(I)*sin(phase(Z))


#Result:
print "The impedance is %.3f ohm at a phase angle of %.3f degrees." %(abs(Z),degrees(phase(Z)))
print "The power factor is %.3f leading." %(pf)
print "The supply voltage is %.3f V at a phase angle of %.3f degrees." %(abs(V),degrees(phase(V)))
print "The voltage across resistor is %.3f V at a phase angle of %.3f degrees." %(abs(V_R),degrees(phase(V_R)))
print "The voltage across capacitor is %.3f V at a phase angle of %.3f degrees." %(abs(V_C),-90)
print "The apparent power is %.3f VA." %(app_P)
print "The active power is %.3f W." %(act_P)
print "The reactive power is %.3f VAr." %(rea_P)
print "Note: Negative sign indicates the capacitor supplies reactive power."

The impedance is 277.308 ohm at a phase angle of -64.359 degrees.
The power factor is 0.433 leading.
The supply voltage is 249.578 V at a phase angle of -64.359 degrees.
The voltage across resistor is 108.000 V at a phase angle of 0.000 degrees.
The voltage across capacitor is 225.000 V at a phase angle of -90.000 degrees.
The apparent power is 224.620 VA.
The active power is 97.200 W.
The reactive power is -202.500 VAr.
Note: Negative sign indicates the capacitor supplies reactive power.

Example 10.4,Page number: 279

In [5]:
#Question:
"""Finding the impedance of the circuit and state whether it is inductive or capacitive."""

from cmath import phase,rect
from math import pi,degrees,radians,cos,sin

#Variable Declaration:
V=160.0 +120.0*1j     #ac sinusoidal voltage(in Volts) 
I=-4.0+10.0*1j        #Current in circuit(in Amperes) 


#Calculations:
Z=V/I

if Z.imag>0:
            print("The nature of circuit is inductive\n")
            pf_type="lagging"

elif Z.imag<0:
              print("The nature of circuit is capacitive\n")
              pf_type="leading"   
                                                
else:
      print("The nature of circuit is resistive\n")

pf=cos(phase(Z))
act_P=abs(V)*abs(I)*pf
rea_P=abs(V)*abs(I)*sin(pi+phase(Z))



#Result:
print "The power factor is %.3f %s." %(pf,pf_type)
print "The active power is %.2f W." %(act_P)
print "The reactive power is %.2f VAr." %(rea_P)     

The nature of circuit is capacitive

The power factor is 0.260 leading.
The active power is 560.00 W.
The reactive power is 2080.00 VAr.

Example 10.5,Page number: 280

In [6]:
#Question:
"""Finding the values of the two circuit elements."""

from math import pi

#Variable Declaration:
f=50                  #Frequency of the source(in Hertz)
Z=10+10*1j            #Impedance of the circuit(in Ohm) 


#Calculations:
ang_freq=2*pi*f
R=Z.real
X_L=Z.imag
L=X_L/ang_freq


#Result:
print "The nature of the impedance indicates that the circiut is inductive.\n"
print "The values of the two elements are: \n(a)Resistance(R)=%.2f ohm \n(b)Inductance(C)=%e H." %(R,L)

The nature of the impedance indicates that the circiut is inductive.

The values of the two elements are: 
(a)Resistance(R)=10.00 ohm 
(b)Inductance(C)=3.183099e-02 H.

Example 10.6,Page number: 280

In [7]:
#Question:
"""Finding the values of the two circuit elements."""

from math import pi

#Variable Declaration:
f=50.0                #Frequency of the source(in Hertz)
Z=10.0-10.0*1j        #Impedance of the circuit(in Ohm) 


#Calculations:
ang_freq=2.0*pi*f
Y=1.0/Z
R=1/Y.real
C=Y.imag/ang_freq


#Result:
print "The nature of the impedance indicates that the circiut is capacitive.\n"
print "The values of the two elements are: \n(a)Resistance(R)=%.2f ohm \n(b)Capacitance(C)=%e F." %(R,C)

The nature of the impedance indicates that the circiut is capacitive.

The values of the two elements are: 
(a)Resistance(R)=20.00 ohm 
(b)Capacitance(C)=1.591549e-04 F.

Example 10.7,Page number: 281

In [9]:
#Question:
"""Finding the impedance,the current,the phase angle,the voltage across each element and power factor."""

from cmath import rect,phase
from math import cos,pi,degrees

#Variable Declaration:
R=12.0                #Resistance of the resistor(in Ohms)
L=0.15                #Self-inductance of the inductor(in Henry)
C=100e-06             #Capacitance of the capacitor(in Farad)
f=50                  #Frequency of the source(in Hertz) 
V=rect(100,0)         #Supply voltage(in Volts) 


#Calculations:
ang_freq=2*pi*f
X_L=ang_freq*L
X_C=1/(ang_freq*C)
Z=R + (X_L-X_C)*1j

if Z.imag>0:
            pf_type="lagging"

elif Z.imag<0:
              pf_type="leading"   
                                                
else:
      print("The nature of circuit is resistive\n")

I=V/Z
V_R=abs(I)*R
V_C=abs(I)*X_C
V_L=abs(I)*X_L
pf=cos(phase(I))
app_P=abs(V)*abs(I)
avg_P=abs(V)*abs(I)*pf


#Result:
print "(a)The impedance is %.3f ohm at a phase angle of %.3f degrees." %(abs(Z),degrees(phase(Z)))
print "(b)The current is %.3f A at a phase angle of %.3f degrees." %(abs(I),degrees(phase(I)))
print "(c)The phase angle is %.3f degrees." %(degrees(phase(I)))
print "(d)The voltage across the resistor is %.2f V." %(V_R) 
print "   The voltage across the capacitor is %.2f V.\n   The voltage across the inductor is %.2f V." %(V_C,V_L)
print "(e)The power factor is %.3f %s." %(pf,pf_type)
print "(f)The apparent power is %.3f VA." %(app_P)     
print "(g)The average power is %.3f W." %(avg_P)

(a)The impedance is 19.439 ohm at a phase angle of 51.880 degrees.
(b)The current is 5.144 A at a phase angle of -51.880 degrees.
(c)The phase angle is -51.880 degrees.
(d)The voltage across the resistor is 61.73 V.
   The voltage across the capacitor is 163.75 V.
   The voltage across the inductor is 242.42 V.
(e)The power factor is 0.617 lagging.
(f)The apparent power is 514.431 VA.
(g)The average power is 317.567 W.

Example 10.8,Page number: 285

In [11]:
#Question:
"""Finding the voltage across the capacitor by applying Thevenin's theorem."""

from math import sqrt,degrees
from cmath import rect,phase

#Variable Declaration:
""" Note: All the impedances are expresssed in kilo ohm.""" 
ang_freq=3000.0   #Angular frequency(in radians per second)
Vs_m=40.0         #Peak value of the supply voltage(in Volts)


#Calculations:
Vs_rms=Vs_m/sqrt(2.0)
Vs=rect(Vs_rms,0)
Zeq=1.5 + ((1.0-(2.0*1j))/(1j+1.0-2.0*1j))*1.0j
I=Vs/Zeq
Im=abs(I)*sqrt(2)
V_Th=Vs*(1j/(1.5+1j))
Z_Th=1+ ((1.5*1j)/(1.5+1j))
Vc=V_Th*((-2*1j)/(Z_Th-2*1j))
Z_L=Z_Th.real-Z_Th.imag*1j 
C=1/(ang_freq*abs(Z_L.imag)*1000)


#Result:
print "(a)The expression for i(t) can be written as i(t)=%.1f sin(%dt%.2fdegrees) mA." %(Im,ang_freq,degrees(phase(I)))
print "(b)The voltage across the capacitor by applying Thevenin's theorem is %.3f V at a phase angle of %.3f degrees." %(abs(Vc),degrees(phase(Vc)))
print "(c)The values of the two elements of the load impedance that consumes maximum power are R=%e kilo Ohms and C=%e F." %(Z_L.real,C)

(a)The expression for i(t) can be written as i(t)=16.0 sin(3000t-36.87degrees) mA.
(b)The voltage across the capacitor by applying Thevenin's theorem is 16.000 V at a phase angle of 8.130 degrees.
(c)The values of the two elements of the load impedance that consumes maximum power are R=1.461538e+00 kilo Ohms and C=4.814815e-07 F.

Example 10.9,Page number: 286

In [1]:
#Question:
"""Finding the value of resistance for a desired power factor."""

from math import sqrt
from cmath import atan

#Variable Declaration:
pf=0.8                #Power factor of the circuit
X_C=60.0              #Capacitive reactance(in Ohms)


#Calculations:
sin_ang=sqrt(1-(pf*pf))
tan_ang=sin_ang/pf
R=X_C/tan_ang


#Result:
print "The value of R for which the power factor of the circuit is 0.8 is %.2f Ohms." %(R)

The value of R for which the power factor of the circuit is 0.8 is 80.00 Ohms.

Example 10.10,Page number: 286

In [3]:
#Question:
"""Finding the resistance and the inductance of the coil."""

from math import sqrt,pi

#Variable Declaration:
Vdc=20.0              #Supply dc voltage(in Volts)
Idc=4.0               #Current drawn by the coil(in Amperes)
f=50.0                #Frequency of supply voltage(in Hertz)
Vs=65.0               #Ac supply voltage(in Volts)
I=5.0                 #Current drawn by the choke when connected ac supply(in Amperes)


#Calculations:
R=Vdc/Idc
Z=Vs/I
X_L=sqrt((Z*Z)-(R*R))
L=X_L/(2*pi*f)
pf=R/Z
P=Vs*I*pf


#Result:
print "(a)The resistance of the coil is %.2f Ohms and the inductance of the coil is %.5f H." %(R,L)
print "(b)The power factor is %.3f lagging." %(pf)
print "(c)The power(real) drawn by the coil is %.3f W." %(P)  

(a)The resistance of the coil is 5.00 Ohms and the inductance of the coil is 0.03820 H.
(b)The power factor is 0.385 lagging.
(c)The power(real) drawn by the coil is 125.000 W.

Example 10.11,Page number: 286

In [7]:
#Question:
"""Finding the resistance and inductance of the coil."""

from math import pi,pow,sqrt

#Variable Declaration:
Vs=240.0              #AC supply voltage(in Volts)
f1=50.0               #Frequency of the ac supply voltage(in Hertz) 
I1=60.0               #First reading of the ammeter(in Amperes)
f2=100.0              #Frequency of the ac supply voltage(in Hertz)
I2=40.0               #Second reading of the ammeter(in Amperes)


#Calculations:
Z1=Vs/I1
Z2=Vs/I2
L=sqrt(((Z2*Z2)-(Z1*Z1))/((pow((200*pi),2.0))-(pow((100*pi),2.0))))
X1=2*pi*f1*L
R=sqrt((Z1*Z1)-(X1*X1))


#Result:
print "The resistance of the coil is %.2f Ohms." %(R)
print "The inductance of the coil is %.5f H." %(L)

The resistance of the coil is 3.06 Ohms.
The inductance of the coil is 0.00822 H.

Example 10.12,Page number: 287

In [1]:
#Question:
"""Finding the power consumed by the choke coil."""

#Variable Declaration:
R=100.0               #Resistance of the resistor(in Ohms)
V_R=200.0             #Voltage across resistor(in Volts)
V_Ch=300.0            #Voltage across choke coil(in Volts)
Vs=440.0              #Supply voltage(in Volts)

 
#Calculations:
pf=((Vs*Vs)-(V_R*V_R)-(V_Ch*V_Ch))/(2*V_R*V_Ch)
I=V_R/R
P=V_Ch*I*pf


#Result:
print "The power consumed by the choke coil is %.2f W." %(P) 

The power consumed by the choke coil is 318.00 W.

Example 10.13,Page number: 287

In [5]:
#Question:
"""Finding the power dissipated in each coil."""

from math import pi
from cmath import phase

#Variable Declaration:
R1=15.0               #Resistance of the first coil(in Ohms)
L1=0.2                #Inductance of the first coil(in Henry)
R2=25.0               #Resistance of the second coil(in Ohms)
L2=0.04               #Inductance of the second coil(in Henry)
V=230.0               #Supply voltage(in Volts)
f=50.0                #Frequency of supply voltage(in Hertz)


#Calculations:
X1=2*pi*f*L1
X2=2*pi*f*L2
Z1=R1+(1j*X1)
Z2=R2+(1j*X2)
Z=Z1+Z2
I=V/Z
V1=abs(I)*abs(Z1)
V2=abs(I)*abs(Z2)
P1=abs(I)*abs(I)*R1
P2=abs(I)*abs(I)*R2
pf=cos(phase(I))


#Result:
print "(a)The voltage across the first coil is %.2f V and the voltage across the second coil is %.2f V." %(V1,V2)
print "(b)The power dissipated by the first coil is %.2f W and power dissipated by the second coil is %.2f W." %(P1,P2)
print "(c)The power factor of the whole circuit is %.3f lagging." %(pf) 

(a)The voltage across the first coil is 174.07 V and the voltage across the second coil is 75.40 V.
(b)The power dissipated by the first coil is 108.92 W and power dissipated by the second coil is 181.54 W.
(c)The power factor of the whole circuit is 0.469 lagging.

Example 10.14,Page number: 288

In [8]:
#Question:
"""Finding the current and power drawn from the source."""

from math import sqrt,pi

#Variable Declaration:
I_A=8.0               #Current through coil A(in Amperes)
I_B=10.0              #Current through coil B(in Amperes)
V=100.0               #Voltage of the source(in Volts)
f=50.0                #Frequency of the supply(in Hertz)
P_A=120.0             #Power delivered to coil A(in Watts)  
P_B=500.0             #Power delivered to coil B(in Watts)


#Calculations:
Z_A=V/I_A
Z_B=V/I_B
R_A=P_A/(I_A*I_A)
R_B=P_B/(I_B*I_B)
X_A=sqrt((Z_A*Z_A)-(R_A*R_A))
X_B=sqrt((Z_B*Z_B)-(R_B*R_B))
R=R_A+R_B
X=X_A+X_B
Z=sqrt((R*R)+(X*X))
I=V/Z
P=I*I*R


#Result:
print "The current when the two coils are in series is %.2f A and the power taken from the source is %.2f W." %(I,P)

The current when the two coils are in series is 4.52 A and the power taken from the source is 140.58 W.

Example 10.15,Page number: 288

In [11]:
#Question:
"""Finding the voltage drop across each coil."""

from math import sqrt,pi

#Variable Declaration:
V=240.0               #Voltage of the source(in Volts)
f=50.0                #Frequency of the supply(in Hertz)
R_A=5.0               #Resistance of coil A(in Ohms)
L_B=0.015             #Inductance of the coil B(in Henry)
P=3e03                #Active power(in Watts)
Q=2e03                #Reactive power(in VAr)


#Calculations:
S=sqrt((P*P)+(Q*Q))
I=S/V
R_B=(P/(I*I))-R_A
X_B=2*pi*f*L_B
X_A=(Q/(I*I))-X_B
L_A=X_A/(2*pi*f)
Z_A=sqrt((R_A*R_A)+(X_A*X_A))
Z_B=sqrt((R_B*R_B)+(X_B*X_B))
V_A=Z_A*I
V_B=Z_B*I


#Result:
print "(a)The resistance of coil B is %.2f Ohms." %(R_B)
print "(b)The inductance of coil A is %.5f Henry." %(L_A)
print "(c)The voltage drop across coil A is %.2f V and across coil B is %.2f V." %(V_A,V_B)

(a)The resistance of coil B is 8.29 Ohms.
(b)The inductance of coil A is 0.01321 Henry.
(c)The voltage drop across coil A is 97.61 V and across coil B is 143.29 V.

Example 10.16,Page number: 289

In [4]:
#Question:
"""Finding the total power consumed by the circuit and the power factor of the circuit."""

from math import pi

#Variable Declaration:
P=100.0               #Power rating of the bulb(in Watts)
V=120.0               #Voltage rating of the bulb(in Volts)
Vs=240.0              #Supply voltage(in Volts)
f=50.0                #Frequency of the supply(in Hertz)


#Calculations:
I=P/V
V_R=Vs-V
R=V_R/I
pf_a=1.0
Pt_a=Vs*I
V_C=sqrt((Vs*Vs)-(V*V))
X_C=V_C/I
C=1/(2*pi*X_C*f)
pf_b=V/Vs
Pt_b=Vs*I*pf_b
Vr=I*10.0
V_L=sqrt((Vs*Vs)-((V+Vr)*(V+Vr)))
X_L=V_L/I
L=X_L/(2*pi*f)
V_R_c=V+Vr
pf_c=V_R_c/Vs
Pt_c=Vs*I*pf_c


#Result:
print "(a)The value of the component used R=%.2f Ohms." %(R)
print "   The total power consumed is %.2f W." %(Pt_a)
print "   The power factor is %.2f." %(pf_a)
print "(b)The value of the component used C=%e Farad." %(C)
print "   The total power consumed is %.2f W." %(Pt_b)
print "   The power factor is %.2f leading." %(pf_b)
print "(c)The value of the component used R=10 Ohms and L=%.3f Henry." %(L)
print "   The total power consumed is %.2f W." %(Pt_c)
print "   The power factor is %.3f lagging." %(pf_c)

(a)The value of the component used R=144.00 Ohms.
   The total power consumed is 200.00 W.
   The power factor is 1.00.
(b)The value of the component used C=1.276224e-05 Farad.
   The total power consumed is 100.00 W.
   The power factor is 0.50 leading.
(c)The value of the component used R=10 Ohms and L=0.775 Henry.
   The total power consumed is 106.94 W.
   The power factor is 0.535 lagging.

Example 10.17,Page number: 289

In [14]:
#Question:
"""Finding the value of capacitance C in the cirucit."""

#Variable Declaration:
R=20.0                #Resistance connected in series(in Ohms)
L=15e-03              #Pure inductance in parallel with the capacitor(in Henry)
ang_fre=1000.0        #Angular frequency of voltage source(in radians per second)


#Calculations:
X_L=ang_fre*L
X=R*tan(pi/4.0)
"""Case 1:(X is inductive)"""
X_C1=1.0/((1.0/X_L)-(1.0/X))
C1=1.0/(ang_fre*X_C1)
"""Case 2:(X is capacitive)"""
X_C2=1.0/((1.0/X_L)+(1.0/X))
C2=1.0/(ang_fre*X_C2)


#Result:
print "The value of capacitance is:"
print "Case 1: If net reactance X is inductive, C=%e Farad." %(C1) 
print "Case 2: If net reactance X is capacitive, C=%e Farad." %(C2)

The value of capacitance is:
Case 1: If net reactance X is inductive, C=1.666667e-05 Farad.
Case 2: If net reactance X is capacitive, C=1.166667e-04 Farad.

Example 10.18,Page number: 290

In [12]:
#Question:
"""Finding total current in the circuit."""

from cmath import phase
from math import degrees

#Variable Declaration:
Z1=(12+1j*15)         #Impedance of the first branch(in Ohms)  
Z2=(8-1j*4)           #Impedance of the second branch(in Ohms)
V=(230+1j*0)          #Potential difference across the parallel combination(in Volts)


#Calculations:
I1=V/Z1
I2=V/Z2
I=I1+I2
P=abs(V)*abs(I)*cos(phase(I))
P1=abs(I1)*abs(I1)*(Z1.real)
P2=abs(I2)*abs(I2)*(Z2.real)
pf1=Z1.real/abs(Z1)
pf2=Z2.real/abs(Z2)
pf=cos(phase(I))


#Result:
print "(a)The current supplied to branch 1 is %.2f A at a phase angle of %.2f degrees." %(abs(I1),degrees(phase(I1))) 
print "   The current supplied to branch 2 is %.2f A at a phase angle of %.2f degrees." %(abs(I2),degrees(phase(I2))) 
print "   The total current is %.2f A at a phase angle of %.2f degrees.\n" %(abs(I),degrees(phase(I)))
print "(b)The power consumed by branch 1 is %.2f W." %(P1)
print "   The power consumed by branch 2 is %.2f W." %(P2)
print "   The total power consumed is %.2f W.\n" %(P)
print "(c)The power factor of branch 1 is %.4f lagging." %(pf1)
print "   The power factor of branch 2 is %.4f leading." %(pf2)
print "   The overall power factor of the circuit is %.4f leading." %(pf)

(a)The current supplied to branch 1 is 11.97 A at a phase angle of -51.34 degrees.
   The current supplied to branch 2 is 25.71 A at a phase angle of 26.57 degrees.
   The total current is 30.56 A at a phase angle of 4.04 degrees.

(b)The power consumed by branch 1 is 1720.33 W.
   The power consumed by branch 2 is 5290.00 W.
   The total power consumed is 7010.33 W.

(c)The power factor of branch 1 is 0.6247 lagging.
   The power factor of branch 2 is 0.8944 leading.
   The overall power factor of the circuit is 0.9975 leading.

Example 10.19,Page number: 290

In [7]:
#Question:
"""Finding the resistance and inductance of a coil."""

from math import acos

#Variable Declaration:
V=230.0               #Voltage of the supply(in Volts)
f=50.0                #Frequency of the supply ac voltage(in Hertz)
R=50.0                #Reistance in series with the coil(in Ohms) 
V_coil=180.0          #Voltage across coil(in Volts)
V_R=130.0             #Voltage across resistance(in Volts) 


#Calculations:
I=V_R/R
"""From parallelogram OACB,the angle theta is calculated."""
cos_theta=((V*V)-(V_R*V_R)-(V_coil*V_coil))/(2*V_R*V_coil)
theta=acos(cos_theta)
V_L=V_coil*sin(theta)
V_r=V_coil*cos(theta)
L=V_L/(I*2*pi*f)
r=V_r/I
P=I*V_r


#Result:
print "(a)The resistance of the coil is %.2f Ohms and the inductance is %.2f H." %(r,L)
print "(b)The power dissipated in the coil is %.2f W." %(P)

(a)The resistance of the coil is 5.33 Ohms and the inductance is 0.22 H.
(b)The power dissipated in the coil is 36.00 W.

Example 10.20,Page number: 291

In [19]:
#Question:
"""Finding the component values of the circuit."""

from math import pi,sqrt,radians,cos

#Variable Declaration:
Vm=141.4              #Peak value of supply voltage(in Volts)  
Im=7.07               #Peak value of current in the series circuit(in Amperes)
ang_fre=2000.0        #Angular frequency of the ac signal(in radians per second) 


#Calculations:
V=Vm/sqrt(2.0)
I=Im/sqrt(2.0)
Z=V/I
""" We know that V=I*Z; 
                
    100=5*sqrt((R*R)+(X_C*X_C));
    
    (R*R)+(X_C*X_C)=400; 
    
    From the triangle OAB,
    
    cos(36.87)=V_R/V;          """
R=(V*cos(radians(36.87)))/I
X_C=sqrt((Z*Z)-(R*R))
C=1.0/(ang_fre*X_C)


#Result:
print "The elements of the circuit are R=%.2f Ohms and C=%e Farad." %(R,C)

The elements of the circuit are R=16.00 Ohms and C=4.166657e-05 Farad.

Example 10.21,Page number: 292

In [20]:
#Question:
"""Finding the voltage and current by Thevenin's theorem."""

from cmath import rect,phase
from math import degrees,radians

#Variable Declaration:
Vs=rect(100,radians(20))


#Calculations:
"""Using voltage divider rule,"""
Z2=((1j*20)*(15-1j*30))/((1j*20)+(15-1j*30))
Z=10+Z2
I=Vs/Z
V2_div=Vs*(Z2/(10+Z2))
V1_div=V2_div*((-1j*30)/(15-1j*30))
"""Using Thevenin's theorem,"""
V_Th=Vs*((15-1j*30)/(10+15-1j*30))
Z_Th=(10*(15-1j*30))/(10+(15-1j*30))
V2=V_Th*((1j*20)/(Z_Th+1j*20))
V_Th=Vs*((1j*20)/(10+1j*20))
Z_Th=15+(1.0/((1.0/10)+(1.0/(1j*20))))
V1=V_Th*((-1j*30)/(Z_Th-1j*30))


#Result:
print "(a)By voltage divider rule,"
print "(i)The current I is %.2f A at an angle of %.2f degrees." %(abs(I),degrees(phase(I))) 
print "(ii)The voltage V1 is %.2f V at an angle of %.2f degrees." %(abs(V1_div),degrees(phase(V1_div)))
print "(b)By applying Thevenin's theorem,"
print "(i)The voltage V1 is %.2f V at an angle of %.2f degrees." %(abs(V1),degrees(phase(V1)))
print "(ii)The voltage V2 is %.2f V at an angle of %.2f degrees." %(abs(V2),degrees(phase(V2)))

(a)By voltage divider rule,
(i)The current I is 2.32 A at an angle of -28.62 degrees.
(ii)The voltage V1 is 77.30 V at an angle of 5.07 degrees.
(b)By applying Thevenin's theorem,
(i)The voltage V1 is 77.30 V at an angle of 5.07 degrees.
(ii)The voltage V2 is 86.42 V at an angle of 31.63 degrees.

Example 10.22,Page number: 293

In [19]:
#Question:
"""Finding the current by Norton's theorem."""

from cmath import rect,phase
from math import degrees,radians

#Variable Declaration:
Is=rect(20,radians(45))  #Current supplied by current source(in Amperes)


#Calculations:
"""Using current divider rule,"""
Z2=(1j*3)+1.0/((1.0/4)+(1.0/(-1j*5)))
I=Is*(2/(Z2+2))
I_R_div=I*((-1j*5)/(4-1j*5))
"""Using Norton's theorem,"""
I_N=Is*(2/(2+1j*3))
Z_N=1.0/((1.0/(2+1j*3))+(1.0/(-1j*5)))
I_R=I_N*(Z_N/(Z_N+4))


#Result:
print "(a)By current divider rule,"
print "   The current I_R is %.2f A at an angle of %.2f degrees." %(abs(I_R_div),degrees(phase(I_R_div)))
print "(a)By Norton's theorem,"
print "   The current I_R is %.2f A at an angle of %.2f degrees." %(abs(I_R),degrees(phase(I_R)))

(a)By current divider rule,
   The current I_R is 6.85 A at an angle of -6.95 degrees.
(a)By Norton's theorem,
   The current I_R is 6.85 A at an angle of -6.95 degrees.

Example 10.23,Page number: 294

In [4]:
#Question:
"""Finding the voltage by Thevenin's theorem."""

from cmath import rect,phase
from math import radians,degrees

#Variable Declaration:
Vs=rect(200,radians(30.0))


#Calculations:
Z2=((40)*(20+1j*50))/(40+(20+1j*50))
V2=(Vs*Z2)/(Z2-1j*20)
V_div=V2*(20/(20+1j*50))
V_Th=Vs*(40/(40-1j*20))
Z_Th=(1j*50)+1.0/((1.0/(-1j*20))+(1.0/40))
V=(V_Th*20)/(Z_Th+20)


#Result:
print "(a)V using voltage divider rule is %.2f V at an angle of %.2f degrees." %(abs(V_div),degrees(phase(V_div)))
print "(b)The voltage V using Thevenin's theorem is %.2f V at an angle of %.2f degrees." %(abs(V),degrees(phase(V)))

(a)V using voltage divider rule is 81.23 V at an angle of 6.04 degrees.
(b)The voltage V using Thevenin's theorem is 81.23 V at an angle of 6.04 degrees.