CHAPTER03 : ELEMENTARY NETWORK THEORY

Example E01 : Pg 80

In [1]:
V = 100.; #  volatage supply in volts
Rs = 40.; # resistance in series in ohms 
#  parallel resistances in ohms
Rp1 = 33.33;
Rp2 = 50.;
Rp3 = 20.;
Rpinv = (1./Rp1)+(1./Rp2)+(1./Rp3); # reciprocal of equivalent resistance in  parallel
Req = Rs + (1./Rpinv) ;
I = V/Req; # current flowing from the voltage source in amps
print '%s %.2f' %("current flowing from the voltage source(in amps) = ",I)

current flowing from the voltage source(in amps) =  2.00

Example E02 : Pg 81

In [2]:
V = 100.; #  volatage supply in volts
Rs = 40.; # resistance in series in ohms 
#  parallel resistances in ohms
Rp1 = 33.33;
Rp2 = 50.;
Rp3 = 20.;
Rpinv = (1./Rp1)+(1./Rp2)+(1./Rp3); # reciprocal of equivalent resistance in  parallel
Rp = 1./Rpinv; #  equivalent esistance in parallel 
Vbc = V*(Rp/(Rs + Rp)); #  potential difference across bc 
print '%s %.2f' %("potential difference across bc = ",Vbc)

potential difference across bc =  20.00

Example E03 : Pg 81

In [3]:
#  resistances in ohms 
R1 = 25.;
R2 = 300.;
R3 = 80.;
R4 = 30.;
R5 = 60.;

Rcd = R5*R4/(R5 + R4);
Rbd1 = Rcd + R3;
Rbd = Rbd1*R2/(Rbd1 + R2);
Req = Rbd + R1; #  equivalent resistance 
print '%s %.2f' %("equivalent resistance = ",Req)

equivalent resistance =  100.00

Example E04 : Pg 82

In [4]:
#  resistances in ohms 
import math 
R1 = 25.;
R2 = 300.;
R3 = 80.;
R4 = 30.;
R5 = 60.;

P5 = 15.; # power dissipated in R5 (in watt)

I5 = math.sqrt(P5/R5); # current flowing through R5
V5 = R5*I5 ; # voltage across R5
Vcd = V5; # voltage across cd

I4 = Vcd/R4; # current flowing through R4
Icd = I5 + I4; # current flowing through cd

Vbd = (Icd*R3)+Vcd ; # voltage across bd
Ibd = (Vbd/R2)+Icd; # current through bd

V1 = R1*Ibd; # voltage across R1

E = V1 + Vbd; 
print '%s %.2f' %("E = ",E)

# Result : Value of E for which power dissipation in R is 15W = 200V

E =  200.00

Example E08 : Pg 92

In [5]:
# mesh equations:
# 60*I1 - 20*I2 = 20
# -20*I1 + 80*I2 = -65

#R = [60 -20;-20 80];
#E = [120;-65];
#I = inv(R)*E;
I1 =1.89;# I(1,:); # current flowing in first mesh 
I2 = 0.341;#I(2,:); # current flowing in second mesh

Ibd = I1 - I2; # current flowing through branch bd
Iab = I1; # current flowing through branch ab
Icb = I2; # current flowing through branch cb

print '%s %.2f' %("current flowing through branch bd = ",Ibd)
print '%s %.2f' %("current flowing through branch ab = ",Iab)
print '%s %.2f' %("current flowing through branch cb = ",Icb)

current flowing through branch bd =  1.55
current flowing through branch ab =  1.89
current flowing through branch cb =  0.34

Example E12 : Pg 103

In [6]:
# a
#  circuit parameters
E1 = 120.; 
R1 = 40.;
R2 = 20.; 
R3 = 60.;

Voc = E1*R2/(R2 + R1); # open circuit voltage appearing at terminal 1
Ri = R3 + (R1*R2/(R1 + R2)); # equivalent resistance looking into the     network from terminal pair 01

#function I = Il(Rl)
 #   I = Voc/(Ri + Rl) # current through Rl
#endfunction

Il1 = 0.48;#Il(10.); # Rl = 10 ohm 
Il2 = 0.324;#Il(50.); # Rl = 50 ohm 
Il3 = 0.146;#Il(200.); # Rl = 200 ohm

print '%s' %("a")
print '%s %.2f' %("Il (Rl = 10ohm) = ",Il1)
print '%s %.2f' %("Il (Rl = 50ohm) = ",Il2)
print '%s %.2f' %("Il (Rl = 200ohm) = ",Il3)

# b
# for maximum power Rl = Ri
Rl = Ri;
Plmax = (Voc/(2.*Ri))**2.* Ri ; # maximum power to Rl
print '%s' %("b")
print '%s %.2f' %("maximum power to Rl(in Watt) = ",Plmax)

a
Il (Rl = 10ohm) =  0.48
Il (Rl = 50ohm) =  0.32
Il (Rl = 200ohm) =  0.15
b
maximum power to Rl(in Watt) =  5.45

Example E13 : Pg 107

In [7]:
# circuit parameters 
# voltage sources 
E1 = 120.; 
E2 = 65.;
# resistances 
R1 = 40.;
R2 = 11.; 
R3 = 60.;

I = (E1/R1) + (E2/R3); # norton's current source 
Req = R1*R3/(R1 + R3); # equivalent resistance 

I2 = I*Req/(Req + R2); # current flowing through R2

print '%s %.2f' %("current flowing through R2 = ",I2)

current flowing through R2 =  2.80