Chapter 3: NETWORK ANALYSIS

Example 3.1,Page number: 53

In [1]:
#Question:
"""Finding the value of capacitance parameter."""

#Variable Declaration:
per_free_space=8.854e-12   #Permittivity of free space(in Farad per metre)
A=0.113                    #Total area(in square metre)  
rel_per=10.0               #Relative permittivity of mica layer
d=0.1e-03                  #Thickness of mica layer(in metre)


#Calculations:
C1=(per_free_space*rel_per*A)/d
W=0.05
V=100.0
C2=(2*W)/(V*V)
i=5e-03
dv=100
dt=0.1
C3=i/(dv/dt)


#Result:
print "(i)The value of capacitance is %.2f micro Farad." %(C1*1000000)
print "(ii)The value of capacitance is %.2f micro Farad." %(C2*1000000)
print "(iii)The value of capacitance is %.2f micro Farad." %(C3*1000000)

(i)The value of capacitance is 0.10 micro Farad.
(ii)The value of capacitance is 10.00 micro Farad.
(iii)The value of capacitance is 5.00 micro Farad.

Example 3.2,Page number: 54

In [2]:
#Question:
"""Finding the inductance of the coil for different cases."""

#Variable Declaration:
W=0.2                  #Energy stored in magnetic field(in Joules)    
i=0.2                  #Current producing magnetic field(in Amperes) 


#Calculations:
L1=(2*W)/(i*i)
v=10
di=0.1
dt=0.2
L2=v/(di/dt)
p=2.5
di=0.5
dt=1
i=0.1
L3=p/(i*(di/dt))


#Result:
print "(i)The inductance of the coil is %.2f H." %(L1)
print "(ii)The inductance of the coil is %.2f H." %(L2)
print "(iii)The inductance of the coil is %.2f H." %(L3)

(i)The inductance of the coil is 10.00 H.
(ii)The inductance of the coil is 20.00 H.
(iii)The inductance of the coil is 50.00 H.

Example 3.3,Page number: 55

In [5]:
#Question:
"""Finding the inductance of 2 coils."""

#Variable Declaration:
Leq=0.7                #Equivalent self-inductance(in Henry)


#Calculations:
""" Leq=0.7 H; 
    
    Leq=((L1*L2)/(L1+L2))+0.5 ;

    L1=(2*L2); """
L2=(Leq-0.5)*(3.0/2)
L1=2*L2


#Result:
print "L1=%.2f H. \nL2=%.2f H." %(L1,L2)

L1=0.60 H. 
L2=0.30 H.

Example 3.4,Page number: 55

In [7]:
#Question:
"""Finding the equivalent capacitance of the network and the voltage drop across each capacitor. """

#Variable Declaration:
V=220.0                #Voltage of the DC source(in Volts)
C1=0.05e-06            #Capacitance of the capacitor 1(in Farad)
C2=0.10e-06            #Capacitance of the capacitor 2(in Farad)
C3=0.20e-06            #Capacitance of the capacitor 3(in Farad)
C4=0.05e-06            #Capacitance of the capacitor 4(in Farad) 


#Calculations:
Cs=1/((1/C1)+(1/C2)+(1/C3)+(1/C4))
Q=Cs*V
V1=Q/C1
V2=Q/C2
V3=Q/C3
V4=Q/C4


#Result:
print "The equivalent capacitance of the network is %.4f micro Farad." %(Cs*1000000)
print "The voltage across capacitor C1= %.2f V." %(V1)
print "The voltage across capacitor C2= %.2f V." %(V2)
print "The voltage across capacitor C3= %.2f V." %(V3)
print "The voltage across capacitor C4= %.2f V." %(V4)

The equivalent capacitance of the network is 0.0182 micro Farad.
The voltage across capacitor C1= 80.00 V.
The voltage across capacitor C2= 40.00 V.
The voltage across capacitor C3= 20.00 V.
The voltage across capacitor C4= 80.00 V.

Example 3.5,Page number: 55

In [8]:
#Question:
"""Finding the voltage across each capacitor."""

#Variable Declaration:
Q1=400e-06            #Charge on the capacitor 1(in Coulombs)
Q2=200e-06            #Charge on the capacitor 2(in Coulombs) 
C1=2e-06              #Capacitance of the capacitor 1(in Farad)
C2=10e-06             #Capacitance of the capacitor 2(in Farad) 


#Calculations:
Q=Q1+Q2
C=C1+C2
V=Q/C


#Result:
print "After the switch is closed, the voltage that exists across each capacitor is %.2f V." %(V)

After the switch is closed, the voltage that exists across each capacitor is 50.00 V.

Example 3.6,Page number: 56

In [9]:
#Question:
""" A series combination of 2 capacitances C1=2 micro-F and C2=8 micro-F is connected across a dc-supply of 300-V.Determine(a)the charge,
    (b)the voltage,and(c)the energy stored in each capacitor. """

#Variable Declaration:
C1=2e-06               #Capacitance of the capacitor 1(in Farad) 
C2=8e-06               #Capacitance of the capacitor 2(in Farad)
V=300.0                #Voltage of Dc supply(in Volts)


#Calculations:
C=(C1*C2)/(C1+C2)
Q=C*V
V1=Q/C1
V2=Q/C2
W1=(1.0/2)*C1*(V1*V1)    
W2=(1.0/2)*C2*(V2*V2)

#Result:
print "(a)When connected in series,the charge on each capacitor is the same.The charge on each capacitor is %.2f micro Coulomb." %(Q*1000000)
print "(b)The voltage across the capacitor C1= %.2f V.\n   The voltage across the capacitor C2= %.2f V." %(V1,V2)
print "(c)The energy stored in the capacitor C1= %.2f milli Joule." %(W1*1000)
print "   The energy stored in the capacitor C2= %.2f milli Joule." %(W2*1000)

(a)When connected in series,the charge on each capacitor is the same.The charge on each capacitor is 480.00 micro Coulomb.
(b)The voltage across the capacitor C1= 240.00 V.
   The voltage across the capacitor C2= 60.00 V.
(c)The energy stored in the capacitor C1= 57.60 milli Joule.
   The energy stored in the capacitor C2= 14.40 milli Joule.

Example 3.7,Page number: 56

In [10]:
#Question:
"""Finding the value of capacitance."""

#Variable Declaration:
C_eq=1e-06            #Equivalent capacitance of the network(in Farad)
C1=1e-06              #Capacitance of the capacitor 1(in Farad)
C2=8e-06              #Capacitance of the capacitor 2(in Farad)
C3=2e-06              #Capacitance of the capacitor 3(in Farad)
C4=2e-06              #Capacitance of the capacitor 4(in Farad)
C5=6e-06              #Capacitance of the capacitor 5(in Farad)
C6=12e-06             #Capacitance of the capacitor 6(in Farad)
C7=4e-06              #Capacitance of the capacitor 7(in Farad) 


#Calculations:
C8=C3+C4
C9=1/((1/C5)+(1/C6))
C10=1/((1/C2)+(1/C8))
C11=C7+C9
C12=1/((1/C1)+(1/C11))
C13=C12+C10
C=1/((1/C_eq)-(1/C13))


#Result:
print "The value of capacitance C is %.2f micro Farad." %(C*1e06)

The value of capacitance C is 1.39 micro Farad.

Example 3.8,Page number: 59

In [30]:
#Question:
"""Finding the % change in V_L and I_L."""

#Variable Declaration:
E=3.0                  #EMF of the battery(in Volts) 
Ri=1.0                 #Internal resistance of the battery(in Ohms)
R_L_min=100.0          #Minimum value of variable load resistance(in Ohms)
R_L_max=1000.0         #Maximum value of variable load resistance(in Ohms) 


#Calculations and Result: 
I_L1=E/(R_L_min+Ri)
I_L2=E/(R_L_max+Ri)
V_L1=E-(I_L1*Ri)
V_L2=E-(I_L2*Ri)

if I_L1>I_L2 :
               change_I_L= ((I_L1-I_L2)/I_L1)*100
               ans="(a)The percentage decrease in I_L is %.2f percent" %(round(change_I_L,2))
               print(ans)
elif I_L2>IL1 : 
               change_I_L= ((I_L2-I_L1)/I_L1)*100
               ans="(a)The percentage increase in I_L is %.2f percent" %(round(change_I_L,2))
               print(ans)
else: print("(a)The percentage change in I_L is 0 percent.")

if V_L2>V_L1 :
              change_V_L= ((V_L2-V_L1)/V_L1)*100
              ans="   The percentage increase in V_L is %.2f percent" %(round(change_V_L,2))
              print(ans)
elif V_L1>V_L2 :
                change_V_L= ((V_L1-V_L2)/V_L1)*100
                ans="   The percentage decrease in V_L is %.2f percent" %(round(change_V_L,2))
                print(ans)
else : print("   The percentage change in V_L is 0","%")

R_L_min=1e-03
R_L_max=10e-03
I_L1=E/(R_L_min+Ri)
I_L2=E/(R_L_max+Ri)
V_L1=round((E-(I_L1*Ri)),3)
V_L2=round((E-(I_L2*Ri)),2)

if I_L1>I_L2 :
               change_I_L= ((I_L1-I_L2)/I_L1)*100
               ans="(b)The percentage decrease in I_L is %.2f percent" %(round(change_I_L,2))
               print(ans) 

elif I_L2>I_L1 :
                change_I_L= ((I_L2-I_L1)/I_L1)*100
                ans="(b)The percentage increase in I_L is %.2f percent" %(round(change_I_L,2))
                print(ans)

else : print("(b) The percentage change in I_L is 0 percent")

if V_L2>V_L1 :
               change_V_L= ((V_L2-V_L1)/V_L1)*100
               ans="   The percentage increase in V_L is %.3f percent" %(change_V_L)
               print(ans)

elif V_L1>V_L2 :
                change_V_L= ((V_L1-V_L2)/V_L1)*100
                ans="   The percentage decrease in V_L is %.2f percent" %(change_V_L)
                print(ans)

else: print("    The percentage change in V_L is 0 percent")

(a)The percentage decrease in I_L is 89.91 percent
   The percentage increase in V_L is 0.90 percent
(b)The percentage decrease in I_L is 0.89 percent
   The percentage increase in V_L is 900.000 percent

Example 3.9,Page number: 61

In [1]:
#Question:
"""Finding the power delivered by the ideal part of the sources."""

#Variable Declaration:
R_S=2.0                #Resistance of the current source(in Ohms)
I_S=3.0                #Current from the current source(in Amperes) 
R_L=4.0                #Resistance of the load(in Ohms)


#Calculations:
V_S=R_S*I_S
I_L1=I_S*(R_S/(R_S+R_L))
V_L1=I_L1*R_L
R1=1/((1/R_S)+(1/R_L))
P_S1=I_S*I_S*R1
I_L2=V_S/(R_S+R_L)
V_L2=V_S*(R_L/(R_S+R_L))
R2=R_S+R_L
P_S2=(V_S*V_S)/R2


#Result:
print "Case 1:"
print "I_L=%.2f A. \nV_L=%.2f V. \nP_S=%.2f W." %(I_L1,V_L1,P_S1)
print "\nCase 2:"
print "I_L=%.2f A. \nV_L=%.2f V. \nP_S=%.2f W." %(I_L2,V_L2,P_S2) 
print "\nWe find that the two energy sources are equivalent as regards the terminal relationships. But, the total power delivered by the ideal part of the two sources is different." 

Case 1:
I_L=1.00 A. 
V_L=4.00 V. 
P_S=12.00 W.

Case 2:
I_L=1.00 A. 
V_L=4.00 V. 
P_S=6.00 W.

We find that the two energy sources are equivalent as regards the terminal relationships. But, the total power delivered by the ideal part of the two sources is different.

Example 3.10,Page number: 62

In [5]:
#Question:
"""Finding the currents through the two resistors in the circut."""

#Calculations:
I_2_ohm1=16.0*(6.0/(6+2))
I_6_ohm1=16-I_2_ohm1
V_eq=16*2
I_2_ohm2=V_eq/(2+6)
I_6_ohm2=I_2_ohm2


#Result:
print "The current through the 2 ohm resistor is %.2f A." %(I_2_ohm1) 
print "The current through the 6 ohm resistor is %.2f A. \n" %(I_6_ohm1) 
print "After transformation,\nThe current through the 2 ohm resistor is %.2f A." %(I_2_ohm2) 
print "The current through the 6 ohm resistor is %.2f A." %(I_6_ohm2)

The current through the 2 ohm resistor is 12.00 A.
The current through the 6 ohm resistor is 4.00 A. 

After transformation,
The current through the 2 ohm resistor is 4.00 A.
The current through the 6 ohm resistor is 4.00 A.

Example 3.11,Page number: 64

In [6]:
#Question:
"""Finding the reduced network using source transformation."""

#Calculations:
"""First all the voltage sources are transformed to equivalent current sources."""
I1=8.0/2.0
I2=6.0/1.0
I3=6.0/2.0
I4=3.0/1.0
I5=I1+I2
I6=I3+I4
R1=1.0/((1.0/2.0)+(1.0/1.0))
R2=1.0/((1.0/2.0)+(1.0/1.0))
V5=I5*R1
V6=I6*R2
V_tot=V5+V6
R_tot=R1+R2


#Result:
print "The network is equivalent to a voltage source of %.3f V with an internal resistance of %.2f Ohms." %(V_tot,R_tot)

The network is equivalent to a voltage source of 10.667 V with an internal resistance of 1.33 Ohms.

Example 3.12,Page number: 65

In [3]:
#Question:
"""Finding the voltage across 3 Ohms resistor."""

#Variable Declaration:
I1=4.0                #Current delivered by source 1(in Amperes)
R1=1.0                #Resitance of resistor parallel to current source 1(in Ohms) 
V2=6.0                #Voltage of the voltage source 2(in Volts)
I2=5.0                #Current delivered by source 2(in Amperes)
R2=2.0                #Resistance of resistor 2(in Ohms) 
R3=3.0                #Resistance of resistor 3(in Ohms)


#Calculations:
V1=I1*R1
V3=V1+V2
I3=V3/R1
I4=I3-I2
V4=I4*R1
V_res=V4*(R3/(R1+R2+R3))


#Result:
print "The voltage across 3 Ohms resistor is %.2f Volts." %(V_res)

The voltage across 3 Ohms resistor is 2.50 Volts.

Example 3.13,Page number: 67

In [1]:
#Question:
"""Finding the current in the circuit."""

#Variable Declaration:
V=24.0                #Supply voltage(in Volts)
R1=4.0                #Resistance of resistor 1(in Ohms) 
R2=2.0                #Resistance of resistor 2(in Ohms)


#Calculations:
""" V1=4*I;
    Applying Kirchoff's Voltage Law to the closed loop, 24-(4*I)-(2*I)-(4.5*4*I)=0 ; """
I=24/((R1+R2)-(4.5*R1))
P=-4.5*R1*I*I
R=V/I


#Result:
print "(a)The value of the current I is %.2f A." %(abs(I))
print "(b)The power absorbed by the dependent source is %.2f W." %(P)
print "(c)The resistance 'seen' by the independent voltage source is %.2f Ohms." %(R) 
print "\nNOTE: The negative sign of the resistance is a result of the action of the dependent source."     

(a)The value of the current I is 2.00 A.
(b)The power absorbed by the dependent source is -72.00 W.
(c)The resistance 'seen' by the independent voltage source is -12.00 Ohms.

NOTE: The negative sign of the resistance is a result of the action of the dependent source.

Example 3.14,Page number: 68

In [7]:
#Question:
"""Finding the voltage in the circuit."""

#Variable Declaration:
V=100.0               #Supply voltage(in Volts)


#Calculations:
"""Appling KVL in the closed loop, +100-(40*I)-(60*I)=0."""
I=V/(40.0+60.0)
V1=I*60.0
"""Appling KVL in the open loop, -10+V1+(0*10)+30-Vab=0."""
Vab=30+V1-10


#Result:
print "The voltage Vab in the circuit is %.2f V." %(Vab)

The voltage Vab in the circuit is 80.00 V.

Example 3.15,Page number: 69

In [8]:
#Question:
"""Finding the unknown voltages Vx and Vcd."""

#Variable Declaration:
V=10.0                #Supply voltage(in Volts)
Vca=4.0               #Voltage between points c and a(in Volts) 
V1=6.0                #Voltage across resistor R1(in Volts) 
V2=-4.0               #Voltage across resistor R2(in Volts) 


#Calculations:
Vx=Vca-V2
Vcd=Vca-V1


#Result:
print "The voltage Vx in the circuit is %.2f V." %(Vx)
print "The voltage Vcd in the circuit is %.2f V." %(Vcd)

The voltage Vx in the circuit is 8.00 V.
The voltage Vcd in the circuit is -2.00 V.

Example 3.16,Page number: 71

In [20]:
#Question:
"""Finding the currents in the network."""

from sympy import *;

#Calculations:
"""KVL equations for loops 1,2 and 3 are written and the equations are solved using SymPy.""" 
print "Note: All currents are expressed in Amperes."
Ix, Iy, I1 = symbols('Ix Iy I1');
solve([(5*Ix)+(0*Iy)+(10*I1)-100,(7*Ix)+(2*Iy)+(-2*I1)+50,(3*Ix)+(-5*Iy)+(-3*I1)+50], [Ix,Iy,I1])

Note: All currents are expressed in Amperes.
Out[20]:
{I1: 585/49, Ix: -190/49, Iy: 25/49}

Example 3.17,Page number: 71

In [42]:
#Question:
"""Finding the currents using Kirchoff's laws."""

from sympy import *;

#Calculations:
"""KCL equations at nodes B and C,we get
    
   I1+I2=20;
   
   I3-I2=30;"""
"""KVL equation for the outer loop ABCDEFGHA, I1-(3*I2)-(2*I3)=-100; """
print "All the currents are expressed in Amperes."""
print "Resistances R1 and R2 are in Ohms."""
I1, I2,I3,R1,R2 = symbols('I1 I2 I3 R1 R2')
solve([I1+I2+(0*I3)-20,(0*I1)-I2+I3-30,I1+(-3*I2)+(-2*I3)+100,(110-(0.1*I1))-(20*R1),(120-(0.2*I3))-(30*R2)],[I1,I2,I3,R1,R2])

All the currents are expressed in Amperes.
Resistances R1 and R2 are in Ohms.
Out[42]:
{I1: 10.0000000000000,
 I2: 10.0000000000000,
 I3: 40.0000000000000,
 R1: 5.45000000000000,
 R2: 3.73333333333333}

Example 3.18,Page number: 74

In [11]:
#Question:
"""Finding the current through resistor in the circuit."""

#Variable Declaration:
I=2.0                 #Loop current(in Amperes)


#Calculations:
"""KVL Equation: 10-(5*(I1-2))-(8*I1)=0."""
I1=(10.0+10.0)/(8.0+5.0)


#Result:
print "The current through the 8 Ohms resistor is %.3f A." %(I1)

The current through the 8 Ohms resistor is 1.538 A.

Example 3.19,Page number: 75

In [13]:
#Question:
"""Finding the unknown voltage."""

#Calculations:
"""Writing the KVL equation around the loop of I, -(2*I)+(3*I)+6-(1*(I+5-4))=0."""
I=(6.0-1.0)/(2+3+1)
v=3*I


#Result:
print "The unknown voltage in the circuit is %.2f V." %(v)

The unknown voltage in the circuit is 2.50 V.

Example 3.20,Page number: 79

In [4]:
#Question:
"""Finding the current drawn by the source."""

from sympy import *;

#Calculations:
"""Apply KVL for the loops 1, 2 and 3 and solving for currents."""
print "All the currents are expressed in Amperes."""
print "The current delivered  by the source is I1."
I1, I2,I3,Is = symbols('I1 I2 I3 Is')
solve([(19*I1)+(-12*I2)+(0*I3)-60,(-12*I1)+(18*I2)+(-6*I3),(0*I1)+(-6*I2)+(18*I3),Is-I1], [I1,I2,I3,Is]) 

All the currents are expressed in Amperes.
The current delivered  by the source is I1.
Out[4]:
{I1: 6, I2: 9/2, I3: 3/2, Is: 6}

Example 3.21,Page number: 80

In [34]:
#Question:
"""Finding the mesh currents using mesh analysis."""

from sympy import *;

#Calculations:
"""Apply KVL for the loops 1, 2 and 3 and solving for currents."""
print "All the currents are expressed in Amperes."""
I1, I2, I3 = symbols('I1 I2 I3')
solve([(7*I1)+(-4*I2)+(0*I3)-13,(-4*I1)+(15*I2)+(-6*I3)-12,(0*I1)+(-6*I2)+(13*I3)-1], [I1,I2,I3]) 

All the currents are expressed in Amperes.
Out[34]:
{I1: 3, I2: 2, I3: 1}

Example 3.22,Page number: 81

In [43]:
#Question:
"""Finding the voltage across resistor using node-voltage analysis."""

from sympy import *;

#Calculations:
"""Apply KCL at nodes a and b and finding the nodal voltages."""
print "All the voltages are expressed in Volts."""
Va,Vb,Vab = symbols('Va Vb Vab')
solve([(Va/2)+((Va-Vb)/3)-5,((Vb-Va)/3)+(Vb/4)+6,Va-Vb-Vab], [Va,Vb,Vab])

All the voltages are expressed in Volts.
Out[43]:
{Va: 22/9, Vab: 34/3, Vb: -80/9}

Example 3.23,Page number: 82

In [15]:
#Question:
"""Finding the current through resistor."""

#Calculations:
"""Applying KCL at node 1,

   ((V1-0)/12.0)+((V1-60.0)/7.0)+((V1-0)/4.0)=0; """
V1=(60.0/7.0)/((1.0/12.0)+(1.0/7.0)+(1.0/4.0))
I1=V1/12.0


#Result:
print "The current I1 through the 12 Ohms resistor is %.2f A." %(I1)

The current I1 through the 12 Ohms resistor is 1.50 A.

Example 3.24,Page number: 82

In [47]:
#Question:
"""Finding the current through a resistor."""

from sympy import *;

#Calculations:
"""Apply KCL for supernode (a,b) and for node c and finding the nodal voltages."""
print "All the voltages are expressed in Volts."""
print "The current I is in Amperes."
Va,Vb,Vc,I = symbols('Va Vb Vc I')
solve([Va-Vb-6,(Va/3)+((Vb-Vc)/4)-2,((Vc-Vb)/4)+(Vc/5)+7,((Vb-Vc)/4)-I], [Va,Vb,Vc,I])

All the voltages are expressed in Volts.
The current I is in Amperes.
Out[47]:
{I: 35/12, Va: -11/4, Vb: -35/4, Vc: -245/12}

Example 3.25,Page number: 83

In [8]:
#Question:
"""Finding the voltage across resistor by nodal-voltage method."""

#Calculations:
"""Applying KCL at node a,
  
   ((Va-6.0)/1.0)+((Va-0.0)/5.0)=-4.0+5.0 ; """

Va=(6.0-1.0)/1.2
v=Va*(3.0/(2.0+3.0))


#Result:
print "The voltage across 3 Ohms resistor is %.2f V." %(v)

The voltage across 3 Ohms resistor is 2.50 V.

Example 3.26,Page number: 84

In [16]:
#Question:
"""Finding the current through resistor."""

from sympy import *;

#Calculations:
"""Applying KCL at nodes 1 and 2 and solving for nodal voltages."""
print "All the voltages are expressed in Volts."""
print "The current I is in Amperes."
V1,V2,I = symbols('V1 V2 I')
solve([(0.7*V1)+(-0.2*V2)-3,(-0.2*V1)+(1.2*V2)-2,((V1-V2)/5.0)-I],[V1,V2,I])

All the voltages are expressed in Volts.
The current I is in Amperes.
Out[16]:
{I: 0.500000000000000, V1: 5.00000000000000, V2: 2.50000000000000}

Example 3.27,Page number: 85

In [12]:
#Question:
"""Finding the current through a resistor."""

#Calculations:
"""Applying KCL at node 1,

   ((V-10.0)/2.0)+((V-0.0)/(1.0+3.0))+((V-8.0)/6.0)=0 ; """

V=((10.0/2.0)+(8.0/6.0))/((1.0/2.0)+(1.0/4.0)+(1.0/6.0))
I=V/(1.0+3.0)


#Result:
print "The current I is %.2f A." %(I)

The current I is 1.73 A.

Example 3.28,Page number: 85

In [31]:
#Question:
"""Finding the values of capacitances."""

from sympy import *;

#Variable Declaration:
Vs=200.0              #Voltage of the dc supply(in Volts)
V1=120.0              #Potential difference across C1(in Volts)
V_new=140.0           #Potential difference across C1 when a capacitor is added(in Volts)


#Calculations:
"""Case 1: The charge on each capacitor is the same.

           (C1*V1)=(C2*V2);
           (120*C1)=(80*C2);"""
"""Case 2: The capacitance of 3 micro Farad in parallel with C2 gives an equivalent capacitance of (C2+3) micro Farad.
            
           (140*C1)=(60*(C2+3)) ;"""
print "Note:The capacitances are expressed in Farads."          
C1,C2 = symbols('C1 C2')
solve([(120*C1)-(80*C2),(140*C1)-(60*(C2+(3e-06)))],[C1,C2])

Note:The capacitances are expressed in Farads.
Out[31]:
{C1: 3.60000000000000e-6, C2: 5.40000000000000e-6}

Example 3.29,Page number: 86

In [50]:
#Question:
"""Finding the current in each branch."""

#Calculations:
I1=20.0/10.0
I4=10.0/5.0
I5=10.0/2.0
I2=(20.0-(10.0+9.0))/1.0
I3=(20.0-50.0-10.0)/20.0
Ia=I1+I2+I3
Ib=I4+I5-I2-I3


#Result:
print "The currents are:" 
print " I1=%.2f A \n I2=%.2f A \n I3=%.2f A \n I4=%.2f A \n I5=%.2f A \n Ia=%.2f A \n Ib=%.2f A" %(I1,I2,I3,I4,I5,Ia,Ib)

The currents are:
 I1=2.00 A 
 I2=1.00 A 
 I3=-2.00 A 
 I4=2.00 A 
 I5=5.00 A 
 Ia=1.00 A 
 Ib=8.00 A

Example 3.30,Page number: 86

In [54]:
#Question:
"""Finding the currents through resistors."""

from sympy import *;

#Calculations:
"""Apply KVL for the loops 1 and 2 and solving for currents."""
print "All the currents are expressed in Amperes."""
print "Current through R1= Current through R2=I1"
print "Current through R3=I3"
print "Current through R4=I2"
I1, I2, I3 = symbols('I1 I2 I3')
solve([(4.5*I1)+(-3*I2)-2,(-3*I1)+(4*I2)-5,I1-I2-I3], [I1,I2,I3]) 

All the currents are expressed in Amperes.
Current through R1= Current through R2=I1
Current through R3=I3
Current through R4=I2
Out[54]:
{I1: 2.55555555555556, I2: 3.16666666666667, I3: -0.611111111111111}

Example 3.31,Page number: 87

In [3]:
#Question:
"""Finding the current through each resistor."""

#Variable Declaration:
E1=4.0                #EMF of cell 1(in Volts)
E2=8.0                #EMF of cell 2(in Volts)
R1=0.5                #Internal resistance of cell 1(in Ohms)
R2=1.0                #Internal resistance of cell 2(in Ohms)


#Calculations:
R3=1.0/((1.0/3.0)+(1.0/6.0))
Rt=R1+R2+R3+4.5
E=E2-E1
I=E/Rt
I_3=I*(6.0/(3.0+6.0))
I_6=I*(3.0/(3.0+6.0))
V1=E1+(I*R1)
V2=E2-(I*R2)


#Result:
print "(a)The current the 3 Ohms resistor is %.2f A." %(I_3)
print "   The current the 6 Ohms resistor is %.2f A." %(I_6)
print "   The current the 4.5 Ohms resistor is %.2f A." %(I)
print "(b)The potential difference across cell 1 is %.2f V and across cell 2 is %.2f V." %(V1,V2)
print "   Note that the pd of cell E1 is greater than its EMF.This is so because the cell is working as a load."

(a)The current the 3 Ohms resistor is 0.33 A.
   The current the 6 Ohms resistor is 0.17 A.
   The current the 4.5 Ohms resistor is 0.50 A.
(b)The potential difference across cell 1 is 4.25 V and across cell 2 is 7.50 V.
   Note that the pd of cell E1 is greater than its EMF.This is so because the cell is working as a load.

Example 3.32,Page number: 88

In [1]:
#Question:
"""Finding the potential of point A."""

#Variable Declaration:
E1=8.0                #EMF of first source(in Volts)
E2=6.0                #EMF of second source(in Volts)
E3=4.0                #EMF of third source(in Volts)
R1=5.0                #Resistance of resistor 1(in Ohms) 
R2=9.0                #Resistance of resistor 2(in Ohms)


#Calculations:
V_A=E3+E2-E1


#Result:
print "The potential of point A is %.2f V." %(V_A)

The potential of point A is 2.00 V.

Example 3.33,Page number: 88

In [5]:
#Question:
"""Finding the energy stored in a capacitor."""

#Variable Declaration:
C=4e-06               #Capacitance of the capacitor(in Farad)


#Calculations:
""" Applying KCL to the node at the top of the capacitor, current through 5 Ohms resistance is 1+2=3 A. 
    Applying KCL to the node at the bottom of the capacitor, current through 2 Ohms resistance is 2-1=1 A."""
V=(3*5)+(3*1)+(1*2)
W=0.5*C*V*V


#Result:
print "The energy stored in the capacitor is %e J." %(W)

The energy stored in the capacitor is 8.000000e-04 J.

Example 3.34,Page number: 88

In [29]:
#Question:
"""Finding the potential differnce between two points."""

#Variable Declaration:
E1=3.0                #EMF of cell 1(in Volts)
E2=2.0                #EMF of cell 2in Volts)
E3=1.0                #EMF of cell 3in Volts)
r1=1.0                #Internal resistance of cell 1(in Ohms)
r2=1.0                #Internal resistance of cell 2(in Ohms)
r3=1.0                #Internal resistance of cell 3(in Ohms) 
R=1.0                 #Resistance of resistor R(in Ohms)


#Calculations:
"""Assigning loop currents as I1 and I2 and writing KVL equations for the two loops,

   E1-(I1*r1)-((I1-I2)*r2)-E2=0; which is (-2*I1)+I2+1=0; 
   
   E2-(I2*r3)-((I2-I1)*r2)-E3=0; which is (-2*I2)+I1+1=0; """

eq1=-1
eq2=-1
I1=((2*eq1)+eq2)/(-3.0)
I2=((2*eq2)+eq1)/(-3.0)
Vab=E2+((I1-I2)*r2)+(0*R)

V_CB=E2
V_B=0.0
V_C=V_CB+V_B
I3=(V_C-E1)/r1
I4=(V_C-E3)/r3
I6=V_C/R
I5=I3+I4+I6


#Result:
print "(a)The potential difference between points A and B is %.2f V." %(Vab) 
print "   The current through r1 is %.2f A,the current through r2=%.2f A." %(I1,(I1-I2)) 
print "   The current through r3=%.2f A and the current through R is 0(no closed loop)." %(I2)
print "(b)The current through E1 is %.2f A." %(I3) 
print "   The current through E2 is %.2f A." %(I5)
print "   The current through E3 is %.2f A." %(I4)
print "   The current through R is %.2f A." %(I6)

(a)The potential difference between points A and B is 2.00 V.
   The current through r1 is 1.00 A,the current through r2=0.00 A.
   The current through r3=1.00 A and the current through R is 0(no closed loop).
(b)The current through E1 is -1.00 A.
   The current through E2 is 2.00 A.
   The current through E3 is 1.00 A.
   The current through R is 2.00 A.

Example 3.35,Page number: 89

In [18]:
#Question:
"""Finding the potential diiference between two points."""

#Calculations:
pd_9=0.0
pd_13=0.0
pd_10=6.0*10.0
pd_5=5.0*8.0
pd_18=15.0
Vab=pd_5-pd_10-pd_18

#Result:
print "The voltage drop Vab is %.2f V." %(Vab)
print "The resistors 4 Ohms,11 Ohms,9 Ohms,18 Ohms and 13 Ohms have no effect on the voltage drop Vab."

The voltage drop Vab is -35.00 V.
The resistors 4 Ohms,11 Ohms,9 Ohms,18 Ohms and 13 Ohms have no effect on the voltage drop Vab.

Example 3.36,Page number: 90

In [33]:
#Question:
"""Finding the voltage of dependent voltage source."""

from sympy import *;

#Calculations:
"""Apply KVL for the loops 1 and 2."""
print "The current I1 is expressed in Amperes and the voltage V1 is expressed in Volts."""
V1,I1 = symbols('V1 I1')
solve([V1+(2*I1),24-(16*I1)-(4*V1)], [V1,I1]) 

The current I1 is expressed in Amperes and the voltage V1 is expressed in Volts.
Out[33]:
{I1: 3, V1: -6}

Example 3.37,Page number: 90

In [37]:
#Question:
"""Finding the voltage V."""

from sympy import *;

#Calculations:
"""Apply KCL at the upper node and KVL equation for the left loop."""
print "The currents I1 and I2 are expressed in Amperes and the voltage V is expressed in Volts."""
I1,I2,V = symbols('I1 I2 V')
solve([I1-I2+(10*I2),36-(20e03*I1),V+(10*I2*(5e03))], [I1,I2,V]) 

The currents I1 and I2 are expressed in Amperes and the voltage V is expressed in Volts.
Out[37]:
{I1: 0.00180000000000000, I2: -0.000200000000000000, V: 10.0000000000000}

Example 3.38,Page number: 90

In [16]:
#Question:
"""Finding the loop currents."""

from sympy import *;

#Calculations:
"""Apply KVL for the loops 1 and 2."""
print "The loop currents i1 and i2 are expressed in Amperes."
i1,i2 = symbols('i1 i2')
solve([(3*i2)-((i2-i1)*1)-(i2*2)-3,4-(i1*1)-((i1-i2)*1)-(3*i2)], [i1,i2]) 

The loop currents i1 and i2 are expressed in Amperes.
Out[16]:
{i1: 3, i2: -1}

Example 3.39,Page number: 91

In [40]:
#Question:
"""Finding the nodal voltages."""

from sympy import *;

#Calculations:
"""Applying KCL at top node and solving for nodal voltages."""
print "All the voltages are expressed in Volts."""
print "The current I is in Amperes."
V1,V2,I = symbols('V1 V2 I')
solve([9-I-(V1/10)+(3*I),9-(4*V1/5)-(V1/10),V2-V1-108],[V1,V2,I])

All the voltages are expressed in Volts.
The current I is in Amperes.
Out[40]:
{I: -4, V1: 10, V2: 118}

Example 3.40,Page number: 91

In [15]:
#Question:
"""Finding the current I by nodal analysis."""

from sympy import *;

#Calculations:
"""Apply KCL at the nodes 1 and 2."""
print "The nodal voltages V1 and V2 are expressed in Volts."
print "The current I is expressed in Amperes."
V1,V2,I = symbols('V1 V2 I')
solve([(2*V1)-(V2)-12,(-2*V1)+(4*V2)-6,I-((V1-V2)/3)], [V1,V2,I]) 

The nodal voltages V1 and V2 are expressed in Volts.
The current I is expressed in Amperes.
Out[15]:
{I: 1, V1: 9, V2: 6}

Example 3.41,Page number: 91

In [14]:
#Question:
"""Finding the current I in the circuit."""

from sympy import *;

#Calculations:
"""Apply KCL at the nodes 1 and 2."""
print "The nodal voltages V1 and V2 are expressed in Volts."
print "The current I is expressed in Amperes."
V1,V2,I = symbols('V1 V2 I')
solve([(1*V1)-(0.5*V2)-6,(-0.5*V1)+(0.625*V2)-3,I-((V1-V2)/2)], [V1,V2,I]) 

The nodal voltages V1 and V2 are expressed in Volts.
The current I is expressed in Amperes.
Out[14]:
{I: -1.00000000000000, V1: 14.0000000000000, V2: 16.0000000000000}

Example 3.42,Page number: 92

In [12]:
#Question:
"""Finding the current through resistor."""

from sympy import *;

#Calculations:
"""Apply KCL at the nodes 1,2. and 3"""
print "The nodal voltages V1,V2 and V3 are expressed in Volts."
print "The current I flowing through the 2 Ohms resistor is expressed in Amperes." 
V1,V2,V3,I = symbols('V1 V2 V3 I')
solve([(7*V1)-(3*V2)-(4*V3)+11,(-3*V1)+(6*V2)-(2*V3)-3,(-4*V1)-(2*V2)+(11*V3)-25,I-(2*(V2-V3))], [V1,V2,V3,I]) 

The nodal voltages V1,V2 and V3 are expressed in Volts.
The current I flowing through the 2 Ohms resistor is expressed in Amperes.
Out[12]:
{I: -2, V1: 1, V2: 2, V3: 3}

Example 3.43,Page number: 92

In [11]:
#Question:
"""Finding the nodal voltages in the circuit."""

from sympy import *;

#Calculations:
Ieq=13.0*5.0 #Equivalent current source
"""Apply KCL at the nodes 1 and 2."""
print "The nodal voltages V1 and V2 are expressed in Volts."
V1,V2 = symbols('V1 V2')
solve([(9*V1)-(5*V2)-10,(-5*V1)+(11*V2)-52], [V1,V2]) 

The nodal voltages V1 and V2 are expressed in Volts.
Out[11]:
{V1: 5, V2: 7}